[Sundstrom_Ted.]_Mathematical_Reasoning_Writing - Copy.pdf

Mathematical Reasoning
Writing and Proof
Version 2.0
June 9, 2014
Ted Sundstrom
Grand Valley State University

Ted Sundstrom
Department of Mathematics
Grand Valley State University
Allendale, MI 49401
mathreasoning@gmail.com
Mathematical Reasoning: Writing and Proof
Previous versions of this book were published by Pearson Education, Inc.
Changes Made in Version 2.0
There are no changes in content between Version 1.1 of this book and Version 2.0.
The only change is that Appendix C, Answers and Hints for Selected Exercises,
now contains solutions and hints for more exercises. Those exercises with an an-
swer or a hint in the appendix are preceded by a star .?
/.
License
This work is licensed under the Creative Commons Attribution-NonCommercial-
ShareAlike 3.0 Unported License. The graphic
that appears throughout the text shows that the work is licensed with the Creative
Commons, that the work may be used for free by any party so long as attribution
is given to the author(s), that the work and its derivatives are used in the spirit of
“share and share alike,” and that no party other than the author(s) may sell this
work or any of its derivatives for profit. Full details may be found by visiting
http://creativecommons.org/licenses/by-nc-sa/3.0/
or sending a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain
View, California, 94041, USA.

Contents
Note to Students vi
Preface viii
1 Introduction to Writing Proofs in Mathematics 1
1.1 Statements and Conditional Statements . . . . . . . . . . . . . . . 1
1.2 Constructing Direct Proofs . . . . . . . . . . . . . . . . . . . . . 15
1.3 Chapter 1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . 31
2 Logical Reasoning 33
2.1 Statements and Logical Operators . . . . . . . . . . . . . . . . . 33
2.2 Logically Equivalent Statements . . . . . . . . . . . . . . . . . . 43
2.3 Open Sentences and Sets . . . . . . . . . . . . . . . . . . . . . . 52
2.4 Quantifiers and Negations . . . . . . . . . . . . . . . . . . . . . . 63
2.5 Chapter 2 Summary . . . . . . . . . . . . . . . . . . . . . . . . . 80
3 Constructing and Writing Proofs in Mathematics 82
3.1 Direct Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
3.2 More Methods of Proof . . . . . . . . . . . . . . . . . . . . . . . 102
3.3 Proof by Contradiction . . . . . . . . . . . . . . . . . . . . . . . 116
3.4 Using Cases in Proofs . . . . . . . . . . . . . . . . . . . . . . . . 131
3.5 The Division Algorithm and Congruence . . . . . . . . . . . . . . 141
iii

iv Contents
3.6 Review of Proof Methods . . . . . . . . . . . . . . . . . . . . . . 158
3.7 Chapter 3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . 166
4 Mathematical Induction 169
4.1 The Principle of Mathematical Induction . . . . . . . . . . . . . . 169
4.2 Other Forms of Mathematical Induction . . . . . . . . . . . . . . 188
4.3 Induction and Recursion . . . . . . . . . . . . . . . . . . . . . . 200
4.4 Chapter 4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . 213
5 Set Theory 215
5.1 Sets and Operations on Sets . . . . . . . . . . . . . . . . . . . . . 215
5.2 Proving Set Relationships . . . . . . . . . . . . . . . . . . . . . . 230
5.3 Properties of Set Operations . . . . . . . . . . . . . . . . . . . . 244
5.4 Cartesian Products . . . . . . . . . . . . . . . . . . . . . . . . . 254
5.5 Indexed Families of Sets . . . . . . . . . . . . . . . . . . . . . . 264
5.6 Chapter 5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . 277
6 Functions 281
6.1 Introduction to Functions . . . . . . . . . . . . . . . . . . . . . . 281
6.2 More about Functions . . . . . . . . . . . . . . . . . . . . . . . . 294
6.3 Injections, Surjections, and Bijections . . . . . . . . . . . . . . . 307
6.4 Composition of Functions . . . . . . . . . . . . . . . . . . . . . . 323
6.5 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 334
6.6 Functions Acting on Sets . . . . . . . . . . . . . . . . . . . . . . 349
6.7 Chapter 6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . 359
7 Equivalence Relations 362
7.1 Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362
7.2 Equivalence Relations . . . . . . . . . . . . . . . . . . . . . . . . 375
7.3 Equivalence Classes . . . . . . . . . . . . . . . . . . . . . . . . . 387

Contents v
7.4 Modular Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . 400
7.5 Chapter 7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . 411
8 Topics in Number Theory 414
8.1 The Greatest Common Divisor . . . . . . . . . . . . . . . . . . . 414
8.2 Prime Numbers and Prime Factorizations . . . . . . . . . . . . . 426
8.3 Linear Diophantine Equations . . . . . . . . . . . . . . . . . . . 439
8.4 Chapter 8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . 449
9 Finite and Infinite Sets 452
9.1 Finite Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452
9.2 Countable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 462
9.3 Uncountable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . 476
9.4 Chapter 9 Summary . . . . . . . . . . . . . . . . . . . . . . . . . 490
A Guidelines for Writing Mathematical Proofs 492
B Answers for the Progress Checks 497
C Answers and Hints for Selected Exercises 536
D List of Symbols 582
Index 585

Note to Students
This book may be different than other mathematics textbooks you have used since
one of the main goals of this book is to help you to develop the ability to construct
and write mathematical proofs. So this book is not just about mathematical content
but is also about the process of doing mathematics. Along the way, you will also
learn some important mathematical topics that will help you in your future study
of mathematics.
This book is designed not to be just casually read but rather to be engaged. It
may seem like a cliché (because it is in almost every mathematics book now) but
there is truth in the statement that mathematicsis not a spectator sport. To learn and
understand mathematics, you must engage in the process of doing mathematics. So
you must actively read and study the book, which means to have a pencil and paper
with you and be willing to follow along and fill in missing details. This type of
engagement is not easy and is often frustrating, but if you do so, you will learn a
great deal about mathematics and more importantly, about doing mathematics.
Recognizing that actively studying a mathematics book is often not easy, sev-
eral features of the textbook have been designed to help you become more engaged
as you study the material. Some of the features are:
Preview Activities. With the exception of Sections 1.1 and 3.6, each section
has exactly two preview activities. Some Preview Activities will review prior
mathematical work that is necessary for the new section. This prior work
may contain material from previous mathematical courses or it may contain
material covered earlier in this text. Other preview activities will introduce
new concepts and definitions that will be used when that section is discussed
in class. It is very important that you work on these preview activities before
starting the rest of the section. Please note that answers to these preview
activities are not included in the text. This book is designed to be used for
a course and it is left up to the discretion of each individual instructor as to
how to distribute the answers to the preview activities.
vi

Note to Students vii
Progress Checks. Several Progress Checks are included in each section.
These are either short exercises or short activities designed to help you de-
termine if you are understandingthe material as you are studyingthe material
in the section. As such, it is important to work through these progress checks
to test your understanding, and if necessary, study the material again before
proceeding further. So it is important to attempt these progress checks before
checking the answers, which are are provided in Appendix B.
Chapter Summaries. To assist you with studying the material in the text,
there is a summary at the end of each chapter. The summaries usually list
the important definitions introduced in the chapter and the important results
proven in the chapter. If appropriate, the summary also describes the impor-
tant proof techniques discussed in the chapter.
Answers for Selected Exercises. Answers or hints for several exercises are
included in an Appendix C. Those exercises with an answer or a hint in the
appendix are preceded by a star .?/. The main change in Version 2.0 of this
textbook from the previous versions is the addition of more exercises with
answers or hints in the appendix.
Although not part of the textbook, there are now 107 online ideos with about
14 hours of content that span the first seven chapters of this book. These videos
are freely available online at Grand Valley’s Department of Mathematics YouTube
channel on this playlist:
http://www.youtube.com/playlist?list=PL2419488168AE7001
These online videos were created and developed by Dr. Robert Talbert of Grand
Valley State University.
There is also a web site for the textbook at
https://sites.google.com/site/mathematicalreasoning3ed/
You may find some things there that could be of help. For example, there currently
is a link to study guides for the sections of this textbook. Good luck with your
study of mathematics and please make use of the online videos and the resources
available in the textbook and at the website for the textbook. If there are things that
you think would be good additions to the book or the web site, please feel free to
send me a message at mathreasoning@gmail.com.

Preface
Mathematical Reasoning: Writing and Proof is designed to be a text for the first
course in the college mathematics curriculum that introduces students to the pro-
cesses of constructing and writing proofs and focuses on the formal development
of mathematics. The primary goals of the text are to help students:
Develop logical thinking skills and to develop the ability to think more ab-
stractly in a proof oriented setting.
Develop the ability to construct and write mathematical proofs using stan-
dard methods of mathematical proof including direct proofs, proof by con-
tradiction, mathematical induction, case analysis, and counterexamples.
Develop the ability to read and understand written mathematical proofs.
Develop talents for creative thinking and problem solving.
Improve their quality of communication in mathematics. This includes im-
proving writing techniques, reading comprehension, and oral communica-
tion in mathematics.
Better understand the nature of mathematics and its language.
Another important goal of this text is to provide students with material that will be
needed for their further study of mathematics.
This type of course has now become a standard part of the mathematics major at
many colleges and universities. It is often referred to as a “transition course” from
the calculus sequence to the upper-level courses in the major. The transition is from
the problem-solving orientation of calculus to the more abstract and theoretical
upper-level courses. This is needed today because many students complete their
study of calculus without seeing a formal proof or having constructed a proof of
their own. This is in contrast to many upper-level mathematics courses, where
viii

Preface ix
the emphasis is on the formal development of abstract mathematical ideas, and
the expectations are that students will be able to read and understand proofs and be
able to construct and write coherent, understandable mathematical proofs. Students
should be able to use this text with a background of one semester of calculus.
Important Features of the Book
Following are some of the important features of this text that will help with the
transition from calculus to upper-level mathematics courses.
1. Emphasis on Writing in Mathematics
Issues dealing with writing mathematical exposition are addressed through-
out the book. Guidelines for writing mathematical proofs are incorporated
into the book. These guidelines are introduced as needed and begin in Sec-
tion 1.2. Appendix A contains a summary of all the guidelines for writing
mathematical proofs that are introduced throughout the text. In addition, ev-
ery attempt has been made to ensure that every completed proof presented in
this text is written according to these guidelines. This provides students with
examples of well-written proofs.
One of the motivating factors for writing this book was to develop a textbook
for the course “Communicating in Mathematics” at Grand Valley State Uni-
versity. This course is part of the university’s Supplemental Writing Skills
Program, and there was no text that dealt with writing issues in mathematics
that was suitable for this course. This is why some of the writing guidelines
in the text deal with the use of L
A
TEXor a word processor that is capable of
producing the appropriate mathematical symbols and equations. However,
the writing guidelines can easily be implemented for courses where students
do not have access to this type of word processing.
2. Instruction in the Process of Constructing Proofs
One of the primary goals of this book is to develop students’ abilities to
construct mathematical proofs. Another goal is to develop their abilities to
write the proof in a coherent manner that conveys an understanding of the
proof to the reader. These are two distinct skills.
Instruction on how to write proofs begins in Section 1.2 and is developed fur-
ther in Chapter 3. In addition, Chapter 4 is devoted to developing students’
abilities to construct proofs using mathematical induction.

x Preface
Students are introduced to a method to organize their thought processes when
attempting to construct a proof that uses a so-called know-show table. (See
Section 1.2 and Section 3.1.) Students use this table to work backward from
what it is they are trying to prove while at the same time working forward
from the assumptions of the problem. The know-show tables are used quite
extensively in Chapters 1 and 3. However, the explicit use of know-show ta-
bles is gradually reduced and these tables are rarely used in the later chapters.
One reason for this is that these tables may work well when there appears
to be only one way of proving a certain result. As the proofs become more
complicated or other methods of proof (such as proofs using cases) are used,
these know-show tables become less useful.
So the know-show tables are not to be considered an absolute necessity in
using the text. However, they are useful for students beginning to learn how
to construct and write proofs. They provide a convenient way for students to
organize their work. More importantly, they introduce students to a way of
thinking about a problem. Instead of immediately trying to write a complete
proof, the know-show table forces students to stop, think, and ask questions
such as
Just exactly what is it that I am trying to prove?
How can I prove this?
What methods do I have that may allow me to prove this?
What are the assumptions?
How can I use these assumptions to prove the result?
Being able to ask these questions is a big step in constructing a proof. The
next task is to answer the questions and to use those answers to construct a
proof.
3. Emphasis on Active Learning
One of the underlying premises of this text is that the best way to learn and
understand mathematics is to be actively involved in the learning process.
However, it is unlikely that students will learn all the mathematics in a given
course on their own. Students actively involved in learning mathematics
need appropriate materials that will provide guidance and support in their
learning of mathematics. There are several ways this text promotes activite
learning.

Preface xi
With the exception of Sections 1.1 and 3.6, each section has exactly
two preview activities. These Preview Activities should be completed
by the students prior to the classroom discussion of the section. The
purpose of the Preview Activities is to prepare students to participate
in the classroom discussion of the section. Some Preview Activities
will review prior mathematical work that is necessary for the new sec-
tion. This prior work may contain material from previous mathemati-
cal courses or it may contain material covered earlier in this text. Other
preview activities will introduce new concepts and definitions that will
be used when that section is discussed in class.
Several Progress Checks are included in each section. These are either
short exercises or short activities designed to help the students deter-
mine if they are understanding the material as it is presented. Some
progress checks are also intended to prepare the student for the next
topic in the section. Answers to the Progress Checks are provided in
Appendix B.
Explorations and activities are included at the end of the exercises of
each section. These activities can be done individually or in a collab-
orative learning setting, where students work in groups to brainstorm,
make conjectures, test each others’ ideas, reach consensus, and, it is
hoped, develop sound mathematical arguments to support their work.
These activities can also be assigned as homework in addition to the
other exercises at the end of each section.
4. Other Important Features of the Book
Several sections of the text include exercises called Evaluationof Proofs.
(The first such exercise appears in Section 3.1.) For these exercises,
there is a proposed proof of a proposition. However, the proposition
may be true or may be false. If a propositionis false, the proposed proof
is, of course, incorrect, and the student is asked to find the error in the
proof and then provide a counterexample showing that the proposition
is false. However, if the proposition is true, the proof may be incorrect
or not well written. In keeping with the emphasis on writing, students
are then asked to correct the proof and/or provide a well-written proof
according to the guidelines established in the book.
To assist students with studying the material in the text, there is a sum-
mary at the end of each chapter. The summaries usually list the im-
portant definitions introduced in the chapter and the important results

xii Preface
proven in the chapter. If appropriate, the summary also describes the
important proof techniques discussed in the chapter.
Answers or hints for several exercises are included in an appendix. This
was done in response to suggestions from many students at Grand Val-
ley and some students from other institutionswho were using the book.
In addition, those exercises with an answer or a hint in the appendix are
preceded by a star .?/.
Content and Organization
Mathematical content is needed as a vehicle for learning how to construct and write
proofs. The mathematical content for this text is drawn primarily from elementary
number theory, including congruence arithmetic; elementary set theory; functions,
including injections, surjections, and the inverse of a function; relations and equiv-
alence relations; further topics in number theory such as greatest common divisors
and prime factorizations; and cardinality of sets, including countable and uncount-
able sets. This material was chosen because it can be used to illustrate a broad
range of proof techniques and it is needed as a prerequisite for many upper-level
mathematics courses.
The chapters in the text can roughly be divided into the following classes:
Constructing and Writing Proofs: Chapters 1, 3, and 4
Logic: Chapter 2
Mathematical Content: Chapters 5, 6, 7, 8, and 9
The first chapter sets the stage for the rest of the book. It introduces students to
the use of conditional statements in mathematics, begins instruction in the process
of constructing a direct proof of a conditional statement, and introduces many of
the writing guidelines that will be used throughout the rest of the book. This is not
meant to be a thorough introduction to methods of proof. Before this is done, it is
necessary to introduce the students to the parts of logic that are needed to aid in the
construction of proofs. This is done in Chapter 2.
Students need to learn some logic and gain experience in the traditional lan-
guage and proof methods used in mathematics. Since this is a text that deals with
constructing and writing mathematical proofs, the logic that is presented in Chap-
ter 2 is intended to aid in the construction of proofs. The goals are to provide

Preface xiii
students with a thorough understanding of conditional statements, quantifiers, and
logical equivalencies. Emphasis is placed on writing correct and useful negations
of statements, especially those involving quantifiers. The logical equivalencies that
are presented provide the logical basis for some of the standard proof techniques,
such as proof by contrapositive, proof by contradiction, and proof using cases.
The standard methods for mathematical proofs are discussed in detail in Chap-
ter 3. The mathematical content that is introduced to illustrate these proof methods
includes some elementary number theory, including congruence arithmetic. These
concepts are used consistently throughout the text as a way to demonstrate ideas in
direct proof, proof by contrapositive, proof by contradiction, proof using cases, and
proofs using mathematical induction. This gives students a strong introduction to
important mathematical ideas while providing the instructor a consistent reference
point and an example of how mathematical notation can greatly simplify a concept.
The three sections of Chapter 4 are devoted to proofs using mathematical in-
duction. Again, the emphasis is not only on understanding mathematical induction
but also on developing the ability to construct and write proofs that use mathemat-
ical induction.
The last five chapters are considered “mathematical content” chapters. Con-
cepts of set theory are introduced in Chapter 5, and the methods of proof studied
in Chapter 3 are used to prove results about sets and operations on sets. The idea
of an “element-chasing proof” is also introduced in Section 5.2.
Chapter 6 provides a thorough study of functions. Functions are studied be-
fore relations in order to begin with the more specific notion with which students
have some familiarity and move toward the more general notion of a relation. The
concept of a function is reviewed but with attention paid to being precise with
terminology and is then extended to the general definition of a function. Various
proof techniques are employed in the study of injections, surjections, composition
of functions, inverses of functions, and functions acting on sets.
Chapter 7 introduces the concepts of relations and equivalence relations. Sec-
tion 7.4 is included to provide a link between the concept of an equivalence relation
and the number theory that has been discussed throughout the text.
Chapter 8 continues the study of number theory. The highlights include prob-
lems dealing with greatest common divisors, prime numbers, the Fundamental
Theorem of Arithmetic, and linear Diophantine equations.
Finally, Chapter 9 deals with further topics in set theory, focusing on cardinal-
ity, finite sets, countable sets, and uncountable sets.

xiv Preface
Designing a Course
Most instructors who use this text will design a course specifically suited to their
needs and the needs of their institution. However, a standard one-semester course
in constructing and writing proofs could cover the first six chapters of the text
and at least one of Chapter 7, Chapter 8, or Chapter 9. A class consisting of well-
prepared and motivated students could cover two of the last three chapters. If either
of these options is a bit too ambitious, Sections 4.3, 5.5, 6.6, 7.4, and 8.3 can be
considered optional sections. These are interesting sections that contain important
material, but the content of these sections is not essential to study the material in
the rest of the book.
Supplementary Materials for the Instructor
The instructor’s manual for this text includes suggestions on how to use the text,
how to incorporate writing into the course, and how to use the preview activities
and activities. The manual also includes solutions for all of the preview activities,
activities, and exercises. In addition, for each section, there is a description of
the purpose of each preview activity and how it is used in the corresponding sec-
tion, and there are suggestions about how to use each activity in that section. The
intention is to make it as easy as possible for the instructor to use the text in an
active learning environment. These activities can also be used in a more traditional
lecture-discussion course. In that case, some of the activities would be discussed
in class or assigned as homework.
PDF files are also available to instructors who use the text to assist them in
posting solutions to a course web page or distributing printed solutions to students.
For each section, there is a file containing the solutions of the preview activities,
and for each activity in the text, there is a file containing the solutions for that
activity. Instructors can contact the author through his e-mail address (mathreason-
ing@gmail.com) for access to the files.

Chapter 1
Introduction to
Writing Proofs in Mathematics
1.1 Statements and Conditional Statements
Much of our work in mathematics deals with statements. In mathematics, a state-
ment is a declarative sentence that is either true or false but not both. A statement
is sometimes called a proposition. The key is that there must be no ambiguity. To
be a statement, a sentence must be true or false, and it cannot be both. So a sen-
tence such as “The sky is beautiful” is not a statement since whether the sentence
is true or not is a matter of opinion. A question such as “Is it raining?” is not a
statement because it is a question and is not declaring or asserting that something
is true.
Some sentences that are mathematical in nature often are not statements be-
cause we may not know precisely what a variable represents. For example, the
equation 2x C 5 D 10 is not a statement since we do not know what x represents.
If we substitute a specific value for x (such as x D 3), then the resulting equation,
23C5 D 10 is a statement (which is a false statement). Following are some more
examples:
There exists a real number x such that 2x C 5 D 10.
This is a statement because either such a real number exists or such a real
number does not exist. In this case, this is a true statement since such a real
number does exist, namely x D 2:5.
1

2 Chapter 1. Introduction to Writing Proofs in Mathematics
For each real number x, 2x C 5 D 2

x C
5
2

.
This is a statement since either the sentence 2x C 5 D 2

x C
5
2

is true
when any real number is substituted for x (in which case, the statement is
true) or there is at least one real number that can be substituted for x and
produce a false statement (in which case, the statement is false). In this case,
the given statement is true.
Solve the equation x2
7x C 10 D 0.
This is not a statement since it is a directive. It does not assert that something
is true.
.a C b/2
D a2
C b2
is not a statement since it is not known what a and b
represent. However, the sentence, “There exist real numbers a and b such
that .aCb/2
D a2
Cb2
” is a statement. In fact, this is a true statement since
there are such integers. For example, if a D 1 and b D 0, then .a C b/2 D
a2 C b2.
Compare the statement in the previous item to the statement, “For all real
numbers a and b, .a C b/2
D a2
C b2
.” This is a false statement since there
are values for a and b for which .a C b/2 ¤ a2 C b2. For example, if a D 2
and b D 3, then .a C b/2
D 52
D 25 and a2
C b2
D 22
C 32
D 13.
Progress Check 1.1 (Statements)
Which of the following sentences are statements? Do not worry about determining
whether a statement is true or false; just determine whether each sentence is a
statement or not.
1. 3 C 4 D 8.
2. 2 7 C 8 D 22.
3. .x 1/ D
p
x C 11.
4. 2x C 5y D 7.
5. There are integers x and y such that 2x C 5y D 7:
6. There are integers x and y such that 23x C 37y D 52:
7. Given a line L and a point P not on that line, there is a unique line through
P that does not intersect L.
8. .a C b/3
D a3 C 3a2b C 3ab2 C b3:
9. .a C b/3
D a3
C 3a2
b C 3ab2
C b3
for all real numbers a and b.

1.1. Statements and Conditional Statements 3
10. The derivative of f.x/ D sin x is f 0.x/ D cos x.
11. Does the equation 3x2
5x 7 D 0 have two real number solutions?
12. If ABC is a right triangle with right angle at vertex B, and if D is the
midpoint of the hypotenuse, then the line segment connecting vertex B to D
is half the length of the hypotenuse.
13. There do not exist three integers x, y, and z such that x3 C y3 D z3:
How Do We Decide If a Statement Is True or False?
In mathematics, we often establish that a statement is true by writing a mathemat-
ical proof. To establish that a statement is false, we often find a so-called coun-
terexample. (These ideas will be explored later in this chapter.) So mathematicians
must be able to discover and construct proofs. In addition, once the discovery has
been made, the mathematician must be able to communicate this discovery to oth-
ers who speak the language of mathematics. We will be dealing with these ideas
throughout the text.
For now, we want to focus on what happens before we start a proof. One thing
that mathematicians often do is to make a conjecture beforehand as to whether
the statement is true or false. This is often done through exploration. The role of
exploration in mathematics is often difficult because the goal is not to find a specific
answer but simply to investigate. Following are some techniques of exploration that
might be helpful.
Techniques of Exploration
Guesswork and conjectures. Formulate and write down questions and con-
jectures. When we make a guess in mathematics, we usually call it a conjec-
ture.
Examples. Constructing appropriate examples is extremely important.
Exploration often requires looking at lots of examples. In this way, we can
gather information that provides evidence that a statement is true, or we
might find an example that shows the statement is false. This type of ex-
ample is called a counterexample.
For example, if someone makes the conjecture that sin.2x/ D 2 sin.x/, for
all real numbers x, we can test this conjecture by substituting specific values

for x. One way to do this is to choose values of x for which sin.x/ is known.
Using x D

4
, we see that
sin

2

4

D sin

2

D 1; and
2 sin

4

D 2
p
2
2
!
D
p
2:
Since 1 ¤
p
2, these calculations show that this conjecture is false. How-
ever, if we do not find a counterexample for a conjecture, we usually cannot
claim the conjecture is true. The best we can say is that our examples indi-
cate the conjecture is true. As an example, consider the conjecture that
If x and y are odd integers, then x C y is an even integer.
We can do lots of calculations, such as 3C7 D 10 and 5C11 D 16, and find
that every time we add two odd integers, the sum is an even integer. However,
it is not possible to test every pair of odd integers, and so we can only say
that the conjecture appears to be true. (We will prove that this statement is
true in the next section.)
Use of prior knowledge. This also is very important. We cannot start from
square one every time we explore a statement. We must make use of our ac-
quired mathematical knowledge. For the conjecture that sin .2x/ D 2 sin.x/,
for all real numbers x, we might recall that there are trigonometric identities
called “double angle identities.” We may even remember the correct identity
for sin.2x/, but if we do not, we can always look it up. We should recall (or
find) that
for all real numbers x; sin.2x/ D 2 sin.x/cos.x/:
We could use this identity to argue that the conjecture “for all real numbers
x, sin.2x/ D 2 sin.x/” is false, but if we do, it is still a good idea to give a
specific counterexample as we did before.
Cooperation and brainstorming. Working together is often more fruitful
than working alone. When we work with someone else, we can compare
notes and articulate our ideas. Thinking out loud is often a useful brain-
storming method that helps generate new ideas.

Progress Check 1.2 (Explorations)
Use the techniques of exploration to investigate each of the following statements.
Can you make a conjecture as to whether the statement is true or false? Can you
determine whether it is true or false?
1. .a C b/2
D a2 C b2, for all real numbers a and b.
2. There are integers x and y such that 2x C 5y D 41.
3. If x is an even integer, then x2
is an even integer.
4. If x and y are odd integers, then x y is an odd integer.
Conditional Statements
One of the most frequently used types of statements in mathematics is the so-called
conditional statement. Given statements P and Q, a statement of the form “If P
then Q” is called a conditional statement. It seems reasonable that the truth value
(true or false) of the conditional statement “If P then Q” depends on the truth val-
ues of P and Q. The statement “If P then Q” means that Q must be true whenever
P is true. The statement P is called the hypothesis of the conditional statement,
and the statement Q is called the conclusion of the conditional statement. Since
conditional statements are probably the most important type of statement in math-
ematics, we give a more formal definition.
Definition. A conditional statement is a statement that can be written in
the form “If P then Q,” where P and Q are sentences. For this conditional
statement, P is called the hypothesis and Q is called the conclusion.
Intuitively, “If P then Q” means that Q must be true whenever P is true.
Because conditional statements are used so often, a symbolic shorthand notation is
used to represent the conditional statement “If P then Q.” We will use the notation
P ! Q to represent “If P then Q.” When P and Q are statements, it seems
reasonable that the truth value (true or false) of the conditional statement P ! Q
depends on the truth values of P and Q. There are four cases to consider:
P is true and Q is true.
P is true and Q is false.
P is false and Q is true.
P is false and Q is false.

The conditional statement P ! Q means that Q is true whenever P is true.
It says nothing about the truth value of Q when P is false. Using this as a guide,
we define the conditional statement P ! Q to be false only when P is true and
Q is false, that is, only when the hypothesis is true and the conclusion is false. In
all other cases, P ! Q is true. This is summarized in Table 1.1, which is called
a truth table for the conditional statement P ! Q. (In Table 1.1, T stands for
“true” and F stands for “false.”)
P Q P ! Q
T T T
T F F
F T T
F F T
Table 1.1: Truth Table for P ! Q
The important thing to remember is that the conditional statement P ! Q
has its own truth value. It is either true or false (and not both). Its truth value
depends on the truth values for P and Q, but some find it a bit puzzling that the
conditional statement is considered to be true when the hypothesis P is false. We
will provide a justification for this through the use of an example.
Example 1.3 Suppose that I say
“If it is not raining, then Daisy is riding her bike.”
We can represent this conditional statement as P ! Q where P is the statement,
“It is not raining” and Q is the statement, “Daisy is riding her bike.”
Although it is not a perfect analogy, think of the statement P ! Q as being
false to mean that I lied and think of the statement P ! Q as being true to mean
that I did not lie. We will now check the truth value of P ! Q based on the truth
values of P and Q.
1. Suppose that both P and Q are true. That is, it is not raining and Daisy is
riding her bike. In this case, it seems reasonable to say that I told the truth
and that P ! Q is true.
2. Suppose that P is true and Q is false or that it is not raining and Daisy is not
riding her bike. It would appear that by making the statement, “If it is not

raining, then Daisy is riding her bike,” that I have not told the truth. So in
this case, the statement P ! Q is false.
3. Now suppose that P is false and Q is true or that it is raining and Daisy
is riding her bike. Did I make a false statement by stating that if it is not
raining, then Daisy is riding her bike? The key is that I did not make any
statement about what would happen if it was raining, and so I did not tell
a lie. So we consider the conditional statement, “If it is not raining, then
Daisy is riding her bike,” to be true in the case where it is raining and Daisy
is riding her bike.
4. Finally, suppose that both P and Q are false. That is, it is raining and Daisy
is not riding her bike. As in the previous situation, since my statement was
P ! Q, I made no claim about what would happen if it was raining, and so
I did not tell a lie. So the statement P ! Q cannot be false in this case and
so we consider it to be true.
Progress Check 1.4 (Explorations with Conditional Statements)
1. Consider the following sentence:
If x is a positive real number, then x2
C 8x is a positive real number.
Although the hypothesis and conclusion of this conditional sentence are not
statements, the conditional sentence itself can be considered to be a state-
ment as long as we know what possible numbers may be used for the vari-
able x. From the context of this sentence, it seems that we can substitute any
positive real number for x. We can also substitute 0 for x or a negative real
number for x provided that we are willing to work with a false hypothesis
in the conditional statement. (In Chapter 2, we will learn how to be more
careful and precise with these types of conditional statements.)
(a) Notice that if x D 3, then x2
C 8x D 15, which is negative. Does
this mean that the given conditional statement is false?
(b) Notice that if x D 4, then x2 C 8x D 48, which is positive. Does this
mean that the given conditional statement is true?
(c) Do you think this conditional statement is true or false? Record the
results for at least five different examples where the hypothesis of this
conditional statement is true.

2. “If n is a positive integer, then .n2 nC41/ is a prime number.” (Remember
that a prime number is a positive integer greater than 1 whose only positive
factors are 1 and itself.)
To explore whether or not this statement is true, try using (and recording
your results) for n D 1, n D 2, n D 3, n D 4, n D 5, and n D 10. Then
record the results for at least four other values of n. Does this conditional
statement appear to be true?
Further Remarks about Conditional Statements
1. The conventions for the truth value of conditional statements may seem a
bit strange,especially the fact that the conditional statement is true when the
hypothesis of the conditional statement is false. The following example is
meant to show that this makes sense.
Suppose that Ed has exactly $52 in his wallet. The following four state-
ments will use the four possible truth combinations for the hypothesis and
conclusion of a conditional statement.
If Ed has exactly $52 in his wallet, then he has $20 in his wallet. This
is a true statement. Notice that both the hypothesis and the conclusion
are true.
If Ed has exactly $52 in his wallet, then he has $100 in his wallet. This
statement is false. Notice that the hypothesis is true and the conclusion
is false.
If Ed has $100 in his wallet, then he has at least $50 in his wallet. This
statement is true regardless of how much money he has in his wallet.
In this case, the hypothesis is false and the conclusion is true.
If Ed has $100 in his wallet, then he has at least $80 in his wallet. This
statement is true regardless of how much money he has in his wallet.
In this case, the hypothesis is false and the conclusion is false.
This is admittedly a contrived example but it does illustrate that the conven-
tions for the truth value of a conditional statement make sense. The message
is that in order to be complete in mathematics, we need to have conventions
about when a conditional statement is true and when it is false.
2. The fact that there is only one case when a conditionalstatement is false often
provides a method to show that a given conditional statement is false. In

Progress Check 1.4, you were asked if you thought the following conditional
statement was true or false.
If n is a positive integer, then n2 n C 41

is a prime number.
Perhaps for all of the values you tried for n, n2 n C 41

turned out to be
a prime number. However, if we try n D 41, we get
n2
n C 41 D 412
41 C 41
n2
n C 41 D 412
:
So in the case where n D 41, the hypothesis is true (41 is a positive integer)
and the conclusion is false 412
is not prime

. Therefore, 41 is a counterex-
ample for this conjecture and the conditional statement
“If n is a positive integer, then n2
n C 41

is a prime number”
is false. There are other counterexamples (such as n D 42, n D 45, and
n D 50), but only one counterexample is needed to prove that the statement
is false.
3. Although one example can be used to prove that a conditional statement is
false, in most cases, we cannot use examples to prove that a conditional
statement is true. For example, in Progress Check 1.4, we substituted val-
ues for x for the conditional statement “If x is a positive real number, then
x2 C 8x is a positive real number.” For every positive real number used
for x, we saw that x2
C 8x was positive. However, this does not prove the
conditional statement to be true because it is impossible to substitute every
positive real number for x. So, although we may believe this statement is
true, to be able to conclude it is true, we need to write a mathematical proof.
Methods of proof will be discussed in Section 1.2 and Chapter 3.
Progress Check 1.5 (Working with a Conditional Statement)
The following statement is a true statement, which is proven in many calculus texts.
If the function f is differentiable at a, then the function f is continuous at a.
Using only this true statement, is it possible to make a conclusion about the func-
tion in each of the following cases?

1. It is known that the function f , where f.x/ D sin x, is differentiable at 0.
2. It is known that the function f , where f.x/ D 3
p
x, is not differentiable at
0.
3. It is known that the function f , where f.x/ D jxj, is continuous at 0.
4. It is known that the function f , where f.x/ D
jxj
x
is not continuous at 0.
Closure Properties of Number Systems
The primary number system used in algebra and calculus is the real number sys-
tem. We usually use the symbol R to stand for the set of all real numbers. The real
numbers consist of the rational numbers and the irrational numbers. The rational
numbers are those real numbers that can be written as a quotient of two integers
(with a nonzero denominator), and the irrational numbers are those real numbers
that cannot be written as a quotient of two integers. That is, a rational number can
be written in the form of a fraction, and an irrational number cannot be written in
the form of a fraction. Some common irrational numbers are
p
2, , and e. We
usually use the symbol Q to represent the set of all rational numbers. (The letter
Q is used because rational numbers are quotients of integers.) There is no standard
symbol for the set of all irrational numbers.
Perhaps the most basic number system used in mathematics is the set of nat-
ural numbers. The natural numbers consist of the positive whole numbers such
as 1, 2, 3, 107, and 203. We will use the symbol N to stand for the set of natural
numbers. Another basic number system that we will be working with is the set of
integers. The integers consist of zero, the positive whole numbers, and the nega-
tives of the positive whole numbers. If n is an integer, we can write n D
n
1
. So
each integer is a rational number and hence also a real number.
We will use the letter Z to stand for the set of integers. (The letter Z is from the
German word, Zahlen, for numbers.) Three of the basic properties of the integers
are that the set Z is closed under addition, the set Z is closed under multiplica-
tion, and the set of integers is closed under subtraction. This means that
If x and y are integers, then x C y is an integer;
If x and y are integers, then x y is an integer; and

If x and y are integers, then x y is an integer.
Notice that these so-called closure properties are defined in terms of conditional
statements. This means that if we can find one instance where the hypothesisis true
and the conclusion is false, then the conditional statement is false.
Example 1.6 (Closure)
1. In order for the set of natural numbers to be closed under subtraction, the
following conditional statement would have to be true: If x and y are natural
numbers, then x y is a natural number. However, since 5 and 8 are nat-
ural numbers, 5 8 D 3, which is not a natural number, this conditional
statement is false. Therefore, the set of natural numbers is not closed under
subtraction.
2. We can use the rules for multiplying fractions and the closure rules for the
integers to show that the rational numbers are closed under multiplication. If
a
b
and
c
d
are rational numbers (so a, b, c, and d are integers and b and d are
not zero), then
a
b

c
d
D
ac
bd
:
Since the integers are closed under multiplication, we know that ac and bd
are integers and since b ¤ 0 and d ¤ 0, bd ¤ 0. Hence,
ac
bd
is a rational
number and this shows that the rational numbers are closed under multipli-
cation.
Progress Check 1.7 (Closure Properties)
Answer each of the following questions.
1. Is the set of rational numbers closed under addition? Explain.
2. Is the set of integers closed under division? Explain.
3. Is the set of rational numbers closed under subtraction? Explain.

Exercises for Section 1.1
? 1. Which of the following sentences are statements?
(a) 32 C 42 D 52.
(b) a2 C b2 D c2.
(c) There exists integers a, b, and c such that a2 D b2 C c2.
(d) If x2 D 4, then x D 2.
(e) For each real number x, if x2 D 4, then x D 2.
(f) For each real number t, sin2
t C cos2
t D 1.
(g) sinx sin

4

.
(h) If n is a prime number, then n2
has three positive factors.
(i) 1 C tan2
D sec2
.
(j) Every rectangle is a parallelogram.
(k) Every even natural number greater than or equal to 4 is the sum of two
prime numbers.
? 2. Identify the hypothesis and the conclusion for each of the following condi-
tional statements.
(a) If n is a prime number, then n2
has three positive factors.
(b) If a is an irrational number and b is an irrational number, then a b is
an irrational number.
(c) If p is a prime number, then p D 2 or p is an odd number.
(d) If p is a prime number and p ¤ 2, then p is an odd number.
(e) If p ¤ 2 and p is an even number, then p is not prime.
? 3. Determine whether each of the following conditional statements is true or
false.
(a) If 10 7, then 3 D 4.
(b) If 7 10, then 3 D 4.
(c) If 10 7, then 3 C 5 D 8.
(d) If 7 10, then 3 C 5 D 8.
?
4. Determine the conditions under which each of the following conditional sen-
tences will be a true statement.

(a) If a C 2 D 5, then 8 5. (b) If 5 8, then a C 2 D 5.
5. Let P be the statement “Student X passed every assignment in Calculus I,”
and let Q be the statement “Student X received a grade of C or better in
Calculus I.”
(a) What does it mean for P to be true? What does it mean for Q to be
true?
(b) Suppose that Student X passed every assignment in Calculus I and
received a grade of B , and that the instructor made the statement
P ! Q. Would you say that the instructor lied or told the truth?
(c) Suppose that Student X passed every assignment in Calculus I and
received a grade of C , and that the instructor made the statement
(d) Now suppose that Student X did not pass two assignments in Calculus
I and received a grade of D, and that the instructor made the statement
(e) How are Parts (5b), (5c), and (5d) related to the truth table for P ! Q?
6. Following is a statement of a theorem which can be proven using calculus or
precalculus mathematics. For this theorem, a, b, and c are real numbers.
Theorem If f is a quadratic function of the form
f .x/ D ax2 C bx C c and a 0, then the function f has a
maximum value when x D
b
2a
.
Using only this theorem, what can be concluded about the functions given
by the following formulas?
? (a) g .x/ D 8x2 C 5x 2
? (b) h .x/ D
1
3
x2 C 3x
? (c) k .x/ D 8x2 5x 7
(d) j .x/ D
71
99
x2 C 210
(e) f .x/ D 4x2 3x C 7
(f) F .x/ D x4 C x3 C 9
7. Following is a statement of a theorem which can be proven using the quadratic
formula. For this theorem, a, b, and c are real numbers.
Theorem If f is a quadratic function of the form
f .x/ D ax2 C bx C c and ac 0, then the function f has two
x-intercepts.
Using only this theorem, what can be concluded about the functions given
by the following formulas?

(a) g .x/ D 8x2 C 5x 2
(b) h .x/ D
1
3
x2 C 3x
(c) k .x/ D 8x2 5x 7
(d) j .x/ D
71
99
x2 C 210
(e) f .x/ D 4x2
3x C 7
(f) F .x/ D x4 C x3 C 9
8. Following is a statement of a theorem about certain cubic equations. For this
theorem, b represents a real number.
Theorem A. If f is a cubic function of the form f.x/ D x3
x C b and
b 1, then the function f has exactly one x-intercept.
Following is another theorem about x-intercepts of functions:
Theorem B. If f and g are functions with g.x/ D k f.x/, where k is a
nonzero real number, then f and g have exactly the same x-intercepts.
Using only these two theorems and some simple algebraic manipulations,
what can be concluded about the functions given by the following formulas?
(a) f.x/ D x3 x C 7
(b) g.x/ D x3
C x C 7
(c) h.x/ D x3
C x 5
(d) k.x/ D 2x3 C 2x C 3
(e) r.x/ D x4
x C 11
(f) F.x/ D 2x3
2x C 7
?
9. (a) Is the set of natural numbers closed under division?
(b) Is the set of rational numbers closed under division?
(c) Is the set of nonzero rational numbers closed under division?
(d) Is the set of positive rational numbers closed under division?
(e) Is the set of positive real numbers closed under subtraction?
(f) Is the set of negative rational numbers closed under division?
(g) Is the set of negative integers closed under addition?
Explorations and Activities
10. Exploring Propositions. In Progress Check 1.2, we used exploration to
show that certain statements were false and to make conjectures that certain
statements were true. We can also use exploration to formulate a conjecture
that we believe to be true. For example, if we calculate successive powers of
2, 21
; 22
; 23
; 24
; 25
; : : :

and examine the units digits of these numbers, we
could make the following conjectures (among others):

1.2. Constructing Direct Proofs 15
If n is a natural number, then the units digit of 2n must be 2, 4, 6, or 8.
The units digits of the successive powers of 2 repeat according to the
pattern “2, 4, 8, 6.”
(a) Is it possible to formulate a conjecture about the units digits of succes-
sive powers of 4 41
; 42
; 43
; 44
; 45
; : : :

? If so, formulate at least one
conjecture.
(b) Is it possible to formulate a conjecture about the units digit of numbers
of the form 7n 2n, where n is a natural number? If so, formulate a
conjecture in the form of a conditional statement in the form “If n is a
natural number, then : : : .”
(c) Let f .x/ D e2x
. Determine the first eight derivatives of this function.
What do you observe? Formulate a conjecture that appears to be true.
The conjecture should be written as a conditional statement in the form,
“If n is a natural number, then : : : .”
1.2 Constructing Direct Proofs
Preview Activity 1 (Definition of Even and Odd Integers)
Definitions play a very important role in mathematics. A direct proof of a proposi-
tion in mathematics is often a demonstration that the proposition follows logically
from certain definitions and previously proven propositions. A definition is an
agreement that a particular word or phrase will stand for some object, property, or
other concept that we expect to refer to often. In many elementary proofs, the an-
swer to the question, “How do we prove a certain proposition?”, is often answered
by means of a definition. For example, in Progress Check 1.2 on page 5, all of the
examples you tried should have indicated that the following conditional statement
is true:
If x and y are odd integers, then x y is an odd integer.
In order to construct a mathematical proof of this conditional statement, we need
a precise definition of what it means to say that an integer is an even integer and
what it means to say that an integer is an odd integer.
Definition. An integer a is an even integer provided that there exists an
integer n such that a D 2n. An integer a is an odd integer provided there
exists an integer n such that a D 2n C 1.

Using this definition, we can conclude that the integer 16 is an even integer since
16 D 2 8 and 8 is an integer. By answering the following questions, you should
obtain a better understanding of these definitions. These questions are not here just
to have questions in the textbook. Constructing and answering such questions is
a way in which many mathemticians will try to gain a better understanding of a
definition.
1. Use the definitions given above to
(a) Explain why 28, 42, 24, and 0 are even integers.
(b) Explain why 51, 11, 1, and 1 are odd integers.
It is important to realize that mathematical definitions are not made randomly. In
most cases, they are motivated by a mathematical concept that occurs frequently.
2. Are the definitions of even integers and odd integers consistent with your
previous ideas about even and odd integers?
Preview Activity 2 (Thinking about a Proof)
Consider the following proposition:
Proposition. If x and y are odd integers, then x y is an odd integer.
Think about how you might go about proving this proposition. A direct proof of
a conditional statement is a demonstration that the conclusion of the conditional
statement follows logically from the hypothesis of the conditional statement. Defi-
nitions and previously proven propositionsare used to justify each step in the proof.
To help get started in proving this proposition, answer the following questions:
1. The proposition is a conditional statement. What is the hypothesis of this
conditional statement? What is the conclusion of this conditional statement?
2. If x D 2 and y D 3, then x y D 6. Does this example prove that the
proposition is false? Explain.
3. If x D 5 and y D 3, then x y D 15. Does this example prove that the
proposition is true? Explain.
In order to prove this proposition, we need to prove that whenever both x and y are
odd integers, x y is an odd integer. Since we cannot explore all possible pairs of
integer values for x and y, we will use the definition of an odd integer to help us
construct a proof.

4. To start a proof of this proposition, we will assume that the hypothesis of the
conditional statement is true. So in this case, we assume that both x and y
are odd integers. We can then use the definition of an odd integer to conclude
that there exists an integer m such that x D 2m C 1. Now use the definition
of an odd integer to make a conclusion about the integer y.
Note: The definition of an odd integer says that a certain other integer exists.
This definition may be applied to both x and y. However, do not use the
same letter in both cases. To do so would imply that x D y and we have not
made that assumption. To be more specific, if x D 2mC1 and y D 2mC1,
then x D y.
5. We need to prove that if the hypothesis is true, then the conclusion is true.
So, in this case, we need to prove that x y is an odd integer. At this point,
we usually ask ourselves a so-called backward question. In this case, we
ask, “Under what conditions can we conclude that x y is an odd integer?”
Use the definition of an odd integer to answer this question, and be careful
to use a different letter for the new integer than was used in Part (4).
Properties of Number Systems
At the end of Section 1.1, we introduced notations for the standard number systems
we use in mathematics. We also discussed some closure properties of the standard
number systems. For this text, it is assumed that the reader is familiar with these
closure properties and the basic rules of algebra that apply to all real numbers. That
is, it is assumed the reader is familiar with the properties of the real numbers shown
in Table 1.2.
Constructing a Proof of a Conditional Statement
In order to prove that a conditional statement P ! Q is true, we only need to
prove that Q is true whenever P is true. This is because the conditional statement
is true whenever the hypothesis is false. So in a direct proof of P ! Q, we assume
that P is true, and using this assumption, we proceed through a logical sequence
of steps to arrive at the conclusion that Q is true.
Unfortunately, it is often not easy to discover how to start this logical sequence
of steps or how to get to the conclusion that Q is true. We will describe a method
of exploration that often can help in discovering the steps of a proof. This method

For all real numbers x, y, and z
Identity Properties x C 0 D x and x 1 D x
Inverse Properties x C . x/ D 0 and if x ¤ 0, then x
1
x
D 1.
Commutative
Properties
x C y D y C x and xy D yx
Associative
Properties
.x C y/ C z D x C .y C z/ and .xy/ z D x .yz/
Distributive
Properties
x .y C z/ D xy C xz and .y C z/ x D yx C zx
Table 1.2: Properties of the Real Numbers
will involve working forward from the hypothesis, P , and backward from the con-
clusion, Q. We will use a device called the “know-show table” to help organize
our thoughts and the steps of the proof. This will be illustrated with the proposition
from Preview Activity 2.
Proposition. If x and y are odd integers, then x y is an odd integer.
The first step is to identify the hypothesis, P , and the conclusion,Q, of the condi-
tional statement. In this case, we have the following:
P : x and y are odd integers. Q: x y is an odd integer.
We now treat P as what we know (we have assumed it to be true) and treat Q as
what we want to show (that is, the goal). So we organize this by using P as the first
step in the know portion of the table and Q as the last step in the show portion of
the table. We will put the know portion of the table at the top and the show portion
of the table at the bottom.
Step Know Reason
P x and y are odd integers. Hypothesis
P1
:
:
:
:
:
:
:
:
:
Q1
Q x y is an odd integer. ?
Step Show Reason

We have not yet filled in the reason for the last step because we do not yet know
how we will reach the goal. The idea now is to ask ourselves questions about what
we know and what we are trying to prove. We usually start with the conclusion
that we are trying to prove by asking a so-called backward question. The basic
form of the question is, “Under what conditions can we conclude that Q is true?”
How we ask the question is crucial since we must be able to answer it. We should
first try to ask and answer the question in an abstract manner and then apply it to
the particular form of statement Q.
In this case, we are trying to prove that some integer is an odd integer. So our
backward question could be, “How do we prove that an integer is odd?” At this
time, the only way we have of answering this question is to use the definition of an
odd integer. So our answer could be, “We need to prove that there exists an integer
q such that the integer equals 2q C 1.” We apply this answer to statement Q and
insert it as the next to last line in the know-show table.
Step Know Reason
P1
:
:
:
:
:
:
:
:
:
Q1 There exists an integer q such
that xy D 2q C 1.
Q x y is an odd integer. Definition of an odd integer
Step Show Reason
We now focus our effort on proving statement Q1 since we know that if we can
prove Q1, then we can conclude that Q is true. We ask a backward question
about Q1 such as, “How can we prove that there exists an integer q such that
x y D 2q C 1?” We may not have a ready answer for this question, and so we
look at the know portion of the table and try to connect the know portion to the
show portion. To do this, we work forward from step P , and this involves asking
a forward question. The basic form of this type of question is, “What can we
conclude from the fact that P is true?” In this case, we can use the definition of an
odd integer to conclude that there exist integers m and n such that x D 2mC1 and
y D 2n C 1. We will call this Step P1 in the know-show table. It is important to
notice that we were careful not to use the letter q to denote these integers. If we had
used q again, we would be claiming that the same integer that gives x y D 2q C 1
also gives x D 2q C 1. This is why we used m and n for the integers x and y since
there is no guarantee that x equals y. The basic rule of thumb is to use a different
symbol for each new object we introduce in a proof. So at this point, we have:

Step P1. We know that there exist integers m and n such that x D 2m C 1
and y D 2n C 1.
Step Q1. We need to prove that there exists an integer q such that
x y D 2q C 1.
We must always be looking for a way to link the “know part” to the “show part”.
There are conclusions we can make from P1, but as we proceed, we must always
keep in mind the form of statement in Q1. The next forward question is, “What
can we conclude about x y from what we know?” One way to answer this is
to use our prior knowledge of algebra. That is, we can first use substitution to
write x y D .2m C 1/ .2n C 1/. Although this equation does not prove that
x y is odd, we can use algebra to try to rewrite the right side of this equation
.2m C 1/ .2n C 1/ in the form of an odd integer so that we can arrive at step Q1.
We first expand the right side of the equation to obtain
x y D .2m C 1/.2n C 1/
D 4mn C 2m C 2n C 1
Now compare the right side of the last equation to the right side of the equation in
step Q1. Sometimes the difficult part at this point is the realization that q stands
for some integer and that we only have to show that x y equals two times some
integer plus one. Can we now make that conclusion? The answer is yes because
we can factor a 2 from the first three terms on the right side of the equation and
obtain
x y D 4mn C 2m C 2n C 1
D 2.2mn C m C n/ C 1
We can now complete the table showing the outline of the proof as follows:

Step Know Reason
P1 There exist integers m and n
such that x D 2m C 1 and
y D 2n C 1.
Definition of an odd integer.
P 2 xy D .2m C 1/ .2n C 1/ Substitution
P 3 xy D 4mn C 2m C 2n C 1 Algebra
P 4 xy D 2 .2mn C m C n/ C 1 Algebra
P 5 .2mn C m C n/ is an integer. Closure properties of the
integers
Q1 There exists an integer q such
that xy D 2q C 1.
Use q D .2mn C m C n/
Q x y is an odd integer. Definition of an odd integer
It is very important to realize that we have only constructed an outline of a
proof. Mathematical proofs are not written in table form. They are written in
narrative form using complete sentences and correct paragraph structure, and they
follow certain conventions used in writing mathematics. In addition, most proofs
are written only from the forward perspective. That is, although the use of the
backward process was essential in discovering the proof, when we write the proof
in narrative form, we use the forward process described in the preceding table. A
completed proof follows.
Theorem 1.8. If x and y are odd integers, then x y is an odd integer.
Proof. We assume that x and y are odd integers and will prove that x y is an odd
integer. Since x and y are odd, there exist integers m and n such that
x D 2m C 1 and y D 2n C 1:
Using algebra, we obtain
x y D .2m C 1/ .2n C 1/
D 4mn C 2m C 2n C 1
D 2 .2mn C m C n/ C 1:
Since m and n are integers and the integers are closed under addition and multipli-
cation, we conclude that .2mn C m C n/ is an integer. This means that x y has

been written in the form .2q C 1/ for some integer q, and hence, x y is an odd
integer. Consequently, it has been proven that if x and y are odd integers, then x y
is an odd integer.
Writing Guidelines for Mathematics Proofs
At the risk of oversimplification, doing mathematics can be considered to have two
distinct stages. The first stage is to convince yourself that you have solved the
problem or proved a conjecture. This stage is a creative one and is quite often how
mathematics is actually done. The second equally important stage is to convince
other people that you have solved the problem or proved the conjecture. This
second stage often has little in common with the first stage in the sense that it does
not really communicate the process by which you solved the problem or proved
the conjecture. However, it is an important part of the process of communicating
mathematical results to a wider audience.
A mathematical proof is a convincing argument (within the accepted stan-
dards of the mathematical community) that a certain mathematical statement is
necessarily true. A proof generally uses deductive reasoning and logic but also
contains some amount of ordinary language (such as English). A mathematical
proof that you write should convince an appropriate audience that the result you
are proving is in fact true. So we do not consider a proof complete until there is
a well-written proof. So it is important to introduce some writing guidelines. The
preceding proof was written according to the following basic guidelines for writing
proofs. More writing guidelines will be given in Chapter 3.
1. Begin with a carefully worded statement of the theorem or result to be
proven. This should be a simple declarative statement of the theorem or
result. Do not simply rewrite the problem as stated in the textbook or given
on a handout. Problems often begin with phrases such as “Show that” or
“Prove that.” This should be reworded as a simple declarative statement of
the theorem. Then skip a line and write “Proof” in italics or boldface font
(when using a word processor). Begin the proof on the same line. Make
sure that all paragraphs can be easily identified. Skipping a line between
paragraphs or indenting each paragraph can accomplish this.
As an example, an exercise in a text might read, “Prove that if x is an odd
integer, then x2
is an odd integer.” This could be started as follows:
Theorem. If x is an odd integer, then x2
is an odd integer.

Proof: We assume that x is an odd integer : : :
2. Begin the proof with a statement of your assumptions. Follow the state-
ment of your assumptions with a statement of what you will prove.
Theorem. If x is an odd integer, then x2 is an odd integer.
Proof. We assume that x is an odd integer and will prove that x2
is an odd
integer.
3. Use the pronoun “we.” If a pronoun is used in a proof, the usual convention
is to use “we” instead of “I.” The idea is to stress that you and the reader
are doing the mathematics together. It will help encourage the reader to
continue working through the mathematics. Notice that we started the proof
of Theorem 1.8 with “We assume that : : : .”
4. Use italics for variables when using a word processor. When using a
word processor to write mathematics, the word processor needs to be capa-
ble of producing the appropriate mathematical symbols and equations. The
mathematics that is written with a word processor should look like typeset
mathematics. This means that italics font is used for variables, boldface font
is used for vectors, and regular font is used for mathematical terms such as
the names of the trigonometric and logarithmic functions.
For example, we do not write sin (x) or sin (x). The proper way to typeset
this is sin.x/.
5. Display important equations and mathematical expressions. Equations
and manipulations are often an integral part of mathematical exposition. Do
not write equations, algebraic manipulations,or formulas in one column with
reasons given in another column. Important equations and manipulations
should be displayed. This means that they should be centered with blank
lines before and after the equation or manipulations, and if the left side of
the equations do not change, it should not be repeated. For example,
x y D .2m C 1/ .2n C 1/
D 4mn C 2m C 2n C 1
D 2 .2mn C m C n/ C 1:
Since m and n are integers, we conclude that : : : .
6. Tell the reader when the proof has been completed. Perhaps the best
way to do this is to simply write, “This completes the proof.” Although it

may seem repetitive, a good alternative is to finish a proof with a sentence
that states precisely what has been proven. In any case, it is usually good
practice to use some “end of proof symbol” such as .
Progress Check 1.9 (Proving Propositions)
Construct a know-show table for each of the following propositions and then write
a formal proof for one of the propositions.
1. If x is an even integer and y is an even integer, then x C y is an even integer.
2. If x is an even integer and y is an odd integer, then x C y is an odd integer.
3. If x is an odd integer and y is an odd integer, then x C y is an even integer.
Some Comments about Constructing Direct Proofs
1. When we constructed the know-show table prior to writing a proof for Theo-
rem 1.8, we had only one answer for the backward question and one answer
for the forward question. Often, there can be more than one answer for these
questions. For example, consider the following statement:
If x is an odd integer, then x2
is an odd integer.
The backward question for this could be, “How do I prove that an integer is
an odd integer?” One way to answer this is to use the definition of an odd
integer, but another way is to use the result of Theorem 1.8. That is, we can
prove an integer is odd by proving that it is a product of two odd integers.
The difficulty then is deciding which answer to use. Sometimes we can
tell by carefully watching the interplay between the forward process and the
backward process. Other times, we may have to work with more than one
possible answer.
2. Sometimes we can use previously proven results to answer a forward ques-
tion or a backward question. This was the case in the example given in
Comment (1), where Theorem 1.8 was used to answer a backward question.
3. Although we start with two separate processes (forward and backward), the
key to constructing a proof is to find a way to link these two processes. This
can be difficult. One way to proceed is to use the know portion of the table
to motivate answers to backward questions and to use the show portion of
the table to motivate answers to forward questions.

4. Answering a backward question can sometimes be tricky. If the goal is the
statement Q, we must construct the know-show table so that if we know that
Q1 is true, then we can conclude that Q is true. It is sometimes easy to
answer this in a way that if it is known that Q is true, then we can conclude
that Q1 is true. For example, suppose the goal is to prove
y2
D 4;
where y is a real number. A backward question could be, “How do we
prove the square of a real number equals four?” One possible answer is to
prove that the real number equals 2. Another way is to prove that the real
number equals 2. This is an appropriate backward question, and these are
appropriate answers.
However, if the goal is to prove
y D 2;
where y is a real number, we could ask, “How do we prove a real number
equals 2?” It is not appropriate to answer this question with “prove that the
square of the real number equals 4.” This is because if y2
D 4, then it is not
necessarily true that y D 2.
5. Finally, it is very important to realize that not every proof can be constructed
by the use of a simple know-show table. Proofs will get more complicated
than the ones that are in this section. The main point of this section is not
the know-show table itself, but the way of thinking about a proof that is in-
dicated by a know-show table. In most proofs, it is very important to specify
carefully what it is that is being assumed and what it is that we are trying
to prove. The process of asking the “backward questions” and the “forward
questions” is the important part of the know-show table. It is very impor-
tant to get into the “habit of mind” of working backward from what it is we
are trying to prove and working forward from what it is we are assuming.
Instead of immediately trying to write a complete proof, we need to stop,
think, and ask questions such as
Just exactly what is it that I am trying to prove?
How can I prove this?
What methods do I have that may allow me to prove this?
What are the assumptions?
How can I use these assumptions to prove the result?

Progress Check 1.10 (Exploring a Proposition)
Construct a table of values for 3m2
C 4m C 6

using at least six different integers
for m. Make one-half of the values for m even integers and the other half odd
integers. Is the following proposition true or false?
If m is an odd integer, then 3m2 C 4m C 6

is an odd integer.
Justify your conclusion. This means that if the proposition is true, then you should
write a proof of the proposition. If the proposition is false, you need to provide an
example of an odd integer for which 3m2
C 4m C 6

is an even integer.
Progress Check 1.11 (Constructing and Writing a Proof)
The Pythagorean Theorem for right triangles states that if a and b are the lengths
of the legs of a right triangle and c is the length of the hypotenuse, then a2
C b2
D
c2. For example, if a D 5 and b D 12 are the lengths of the two sides of a right
triangle and if c is the length of the hypotenuse, then the c2
D 52
C 122
and so
c2 D 169. Since c is a length and must be positive, we conclude that c D 13.
Construct and provide a well-written proof for the following proposition.
Proposition. If m is a real number and m, m C 1, and m C 2 are the lengths of the
three sides of a right triangle, then m D 3.
Although this proposition uses different mathematical concepts than the one used
in this section, the process of constructing a proof for this proposition is the same
forward-backward method that was used to construct a proof for Theorem 1.8.
However, the backward question, “How do we prove that m D 3?” is simple
but may be difficult to answer. The basic idea is to develop an equation from the
forward process and show that m D 3 is a solution of that equation.
1. Construct a know-show table for each of the following statements and then
write a formal proof for one of the statements.
?
(a) If m is an even integer, then m C 1 is an odd integer.
(b) If m is an odd integer, then m C 1 is an even integer.

(a) If x is an even integer and y is an even integer, then x C y is an even
integer.
(b) If x is an even integer and y is an odd integer, then x C y is an odd
integer.
? (c) If x is an odd integer and y is an odd integer, then x C y is an even
integer.
? (a) If m is an even integer and n is an integer, then m n is an even integer.
?
(b) If n is an even integer, then n2
is an even integer.
(c) If n is an odd integer, then n2 is an odd integer.
4. Construct a know-show table and write a complete proof for each of the
following statements:
? (a) If m is an even integer, then 5m C 7 is an odd integer.
(b) If m is an odd integer, then 5m C 7 is an even integer.
(c) If m and n are odd integers, then mn C 7 is an even integer.
5. Construct a know-show table and write a complete proof for each of the
following statements:
(a) If m is an even integer, then 3m2
C 2m C 3 is an odd integer.
? (b) If m is an odd integer, then 3m2 C 7m C 12 is an even integer.
6. In this section, it was noted that there is often more than one way to answer a
backward question. For example, if the backward question is, “How can we
prove that two real numbers are equal?”, one possible answer is to prove that
their difference equals 0. Another possible answer is to prove that the first is
less than or equal to the second and that the second is less than or equal to
the first.
?
(a) Give at least one more answer to the backward question, “How can we
prove that two real numbers are equal?”

(b) List as many answers as you can for the backward question, “How can
we prove that a real number is equal to zero?”
(c) List as many answers as you can for the backward question, “How can
we prove that two lines are parallel?”
? (d) List as many answers as you can for the backward question, “How can
we prove that a triangle is isosceles?”
7. Are the following statements true or false? Justify your conclusions.
(a) If a, b and c are integers, then ab C ac is an even integer.
(b) If b and c are odd integers and a is an integer, then ab C ac is an even
integer.
8. Is the following statement true or false? Justify your conclusion.
If a and b are nonnegative real numbers and a C b D 0, then a D 0.
Either give a counterexample to show that it is false or outline a proof by
completing a know-show table.
9. An integer a is said to be a type 0 integer if there exists an integer n such
that a D 3n. An integer a is said to be a type 1 integer if there exists an
integer n such that a D 3n C 1. An integer a is said to be a type 2 integer
if there exists an integer m such that a D 3m C 2.
? (a) Give examples of at least four different integers that are type 1 integers.
(b) Give examples of at least four different integers that are type 2 integers.
? (c) By multiplying pairs of integers from the list in Exercise (9a), does it
appear that the following statement is true or false?
If a and b are both type 1 integers, then a b is a type 1 integer.
10. Use the definitions in Exercise (9) to help write a proof for each of the fol-
lowing statements:
? (a) If a and b are both type 1 integers, then a C b is a type 2 integer.
(b) If a and b are both type 2 integers, then a C b is a type 1 integer.
(c) If a is a type 1 integer and b is a type 2 integer, then a b is a type 2
integer.
(d) If a and b are both type 2 integers, then a b is type 1 integer.

11. Let a, b, and c be real numbers with a ¤ 0. The solutions of the quadratic
equation ax2
C bx C c D 0 are given by the quadratic formula, which
states that the solutions are x1 and x2, where
x1 D
b C
p
b2 4ac
2a
and x2 D
b
p
b2 4ac
2a
:
(a) Prove that the sum of the two solutions of the quadratic equation
ax2
C bx C c D 0 is equal to
b
a
.
(b) Prove that the product of the two solutions of the quadratic equation
ax2
C bx C c D 0 is equal to
c
a
.
12. (a) See Exercise (11) for the quadratic formula, which gives the solutions
to a quadratic equation. Let a, b, and c be real numbers with a ¤ 0.
The discriminant of the quadratic equation ax2
CbxCc D 0 is defined
to be b2 4ac. Explain how to use this discriminant to determine if
the quadratic equation has two real number solutions, one real number
solution, or no real number solutions.
(b) Prove that if a, b, and c are real numbers with a 0 and c 0, then
one solutions of the quadratic equation ax2 C bx C c D 0 is a positive
real number.
(c) Prove that if a, b, and c are real numbers with a ¤ 0, b 0, and
b 2
p
ac, then the quadratic equation ax2 C bx C c D 0 has no real
number solutions.
13. Pythagorean Triples. Three natural numbers a, b, and c with a b c
are said to form a Pythagorean triple provided that a2
C b2
D c2
. For
example, 3, 4, and 5 form a Pythagorean triple since 32 C 42 D 52. The
study of Pythagorean triples began with the development of the Pythagorean
Theorem for right triangles, which states that if a and b are the lengths
of the legs of a right triangle and c is the length of the hypotenuse, then
a2 C b2 D c2. For example, if the lengths of the legs of a right triangle are
4 and 7 units, then c2
D 42
C 72
D 63, and the length of the hypotenuse
must be
p
63 units (since the length must be a positive real number). Notice
that 4, 7, and
p
63 are not a Pythagorean triple since
p
63 is not a natural
number.

(a) Verify that each of the following triples of natural numbers form a
Pythagorean triple.
3, 4, and 5
6, 8, and 10
8, 15, and 17
10, 24, and 26
12, 35, and 37
14, 48, and 50
(b) Does there exist a Pythagorean triple of the form m, mC 7, and mC 8,
where m is a natural number? If the answer is yes, determine all such
Pythagorean triples. If the answer is no, prove that no such Pythagorean
triple exists.
(c) Does there exist a Pythagorean triple of the form m, mC11, and mC12,
where m is a natural number? If the answer is yes, determine all such
Pythagorean triples. If the answer is no, prove that no such Pythagorean
triple exists.
14. More Work with Pythagorean Triples. In Exercise (13), we verified that
each of the following triples of natural numbers are Pythagorean triples:
3, 4, and 5
6, 8, and 10
8, 15, and 17
10, 24, and 26
12, 35, and 37
14, 48, and 50
(a) Focus on the least even natural number in each of these Pythagorean
triples. Let n be this even number and find m so that n D 2m. Now try
to write formulas for the other two numbers in the Pythagorean triple
in terms of m. For example, for 3, 4, and 5, n D 4 and m D 2, and for
8, 15, and 17, n D 8 and m D 4. Once you think you have formulas,
test your results with m D 10. That is, check to see that you have a
Pythagorean triple whose smallest even number is 20.
(b) Write a proposition and then write a proof of the proposition. The
proposition should be in the form: If m is a natural number and m 2,
then ......

1.3. Chapter 1 Summary 31
1.3 Chapter 1 Summary
Important Definitions
Statement, page 1
Conditional statement, page 5
Even integer, page 15
Odd integer, page 15
Pythagorean triple, page 29
Important Number Systems and Their Properties
The natural numbers, N; the integers, Z; the rational numbers, Q; and the
real numbers, R. See page 10
Closure Properties of the Number Systems
Number System Closed Under
Natural Numbers, N addition and multiplication
Integers, Z addition, subtraction, and multiplication
Rational Numbers, Q
addition, subtraction, multiplication, and
division by nonzero rational numbers
Real Numbers, R
addition, subtraction, multiplication, and
division by nonzero real numbers
Inverse, commutative, associative, and distributiveproperties of the real num-
bers. See page 18.
Important Theorems and Results
Exercise (1), Section 1.2
If m is an even integer, then m C 1 is an odd integer.
If m is an odd integer, then m C 1 is an even integer.
If x is an even integer and y is an even integer, then x C y is an even integer.
If x is an even integer and y is an odd integer, then x C y is an odd integer.
If x is an odd integer and y is an odd integer, then x C y is an even integer.
Exercise (3), Section 1.2. If x is an even integer and y is an integer, then
x y is an even integer.

Theorem 1.8. If x is an odd integer and y is an odd integer, then x y is an
odd integer.
The Pythagorean Theorem, page 26. If a and b are the lengths of the legs
of a right triangle and c is the length of the hypotenuse, then a2
C b2
D c2
.

Chapter 2
Logical Reasoning
2.1 Statements and Logical Operators
Preview Activity 1 (Compound Statements)
Mathematicians often develop ways to construct new mathematical objects from
existing mathematical objects. It is possible to form new statements from existing
statements by connecting the statements with words such as “and” and “or” or by
negating the statement. A logical operator (or connective) on mathematical state-
ments is a word or combination of words that combines one or more mathematical
statements to make a new mathematical statement. A compound statement is a
statement that contains one or more operators. Because some operators are used so
frequently in logic and mathematics, we give them names and use special symbols
to represent them.
The conjunction of the statements P and Q is the statement “P and Q”
and its denoted by P ^ Q . The statement P ^ Q is true only when both P
and Q are true.
The disjunction of the statements P and Q is the statement “P or Q” and
its denoted by P _ Q . The statement P _ Q is true only when at least one
of P or Q is true.
The negation (of a statement) of the statement P is the statement “not P ”
and is denoted by :P . The negation of P is true only when P is false, and
:P is false only when P is true.
The implicaton or conditional is the statement “If P then Q” and is de-
noted by P ! Q . The statement P ! Q is often read as “P implies Q,
33

34 Chapter 2. Logical Reasoning
and we have seen in Section 1.1 that P ! Q is false only when P is true
and Q is false.
Some comments about the disjunction.
It is important to understand the use of the operator “or.” In mathematics, we use
the “inclusive or” unless stated otherwise. This means that P _ Q is true when
both P and Q are true and also when only one of them is true. That is, P _ Q is
true when at least one of P or Q is true, or P _ Q is false only when both P and
Q are false.
A different use of the word “or” is the “exclusive or.” For the exclusive or, the
resulting statement is false when both statements are true. That is, “P exclusive or
Q” is true only when exactly one of P or Q is true. In everyday life, we often use
the exclusive or. When someone says, “At the intersection, turn left or go straight,”
this person is using the exclusive or.
Some comments about the negation. Although the statement, :P , can be read
as “It is not the case that P ,” there are often betters ways to say or write this in
English. For example, we would usually say (or write):
The negation of the statement, “391 is prime” is “391 is not prime.”
The negation of the statement, “12 9” is “12 9.”
1. For the statements
P : 15 is odd Q: 15 is prime
write each of the following statements as English sentences and determine
whether they are true or false.
(a) P ^ Q. (b) P _ Q. (c) P ^ :Q. (d) :P _ :Q.
2. For the statements
P : 15 is odd R: 15 17
write each of the following statements in symbolic form using the operators
^, _, and :.

2.1. Statements and Logical Operators 35
(a) 15 17.
(b) 15 is odd or 15 17.
(c) 15 is even or 15 17.
(d) 15 is odd and 15 17.
Preview Activity 2 (Truth Values of Statements)
We will use the following two statements for all of this Preview Activity:
P is the statement “It is raining.”
Q is the statement “Daisy is playing golf.”
In each of the following four parts, a truth value will be assigned to statements P
and Q. For example, in Question (1), we will assume that each statement is true. In
Question (2), we will assume that P is true and Q is false. In each part, determine
the truth value of each of the following statements:
(a) (P ^ Q) It is raining and Daisy is playing golf.
(b) (P _ Q) It is raining or Daisy is playing golf.
(c) (P ! Q) If it is raining, then Daisy is playing golf.
(d) (:P ) It is not raining.
Which of the four statements [(a) through (d)] are true and which are false in each
of the following four situations?
1. When P is true (it is raining) and Q is true (Daisy is playing golf).
2. When P is true (it is raining) and Q is false (Daisy is not playing golf).
3. When P is false (it is not raining) and Q is true (Daisy is playing golf).
4. When P is false (it is not raining) and Q is false (Daisy is not playing golf).
In the preview activities for this section, we learned about compound state-
ments and their truth values. This information can be summarized with truth tables
as is shown below.

P :P
T F
F T
P Q P ^ Q
T T T
T F F
F T F
F F F
P Q P _ Q
T T T
T F T
F T T
F F F
P Q P ! Q
T T T
T F F
F T T
F F T
Rather than memorizing the truth tables, for many people it is easier to remem-
ber the rules summarized in Table 2.1.
Operator Symbolic Form Summary of Truth Values
Conjunction P ^ Q True only when both P and Q are true
Disjunction P _ Q False only when both P and Q are
false
Negation :P Opposite truth value of P
Conditional P ! Q False only when P is true and Q is
false
Table 2.1: Truth Values for Common Connectives
Other Forms of Conditional Statements
Conditional statements are extremely important in mathematics because almost all
mathematical theorems are (or can be) stated in the form of a conditional statement
in the following form:
If “certain conditions are met,” then “something happens.”
It is imperative that all students studying mathematics thoroughly understand the
meaning of a conditional statement and the truth table for a conditional statement.

We also need to be aware that in the English language, there are other ways for
expressing the conditional statement P ! Q other than “If P , then Q.” Following
are some common ways to express the conditional statement P ! Q in the English
language:
If P , then Q.
P implies Q.
P only if Q.
Q if P .
Whenever P is true, Q is true.
Q is true whenever P is true.
Q is necessary for P . (This means that if P is true, then Q is necessarily
true.)
P is sufficient for Q. (This means that if you want Q to be true, it is suffi-
cient to show that P is true.)
In all of these cases, P is the hypothesis of the conditional statement and Q is
the conclusion of the conditional statement.
Progress Check 2.1 (The “Only If” Statement)
Recall that a quadrilateral is a four-sided polygon. Let S represent the following
true conditional statement:
If a quadrilateral is a square, then it is a rectangle.
Write this conditional statement in English using
1. the word “whenever”
2. the phrase “only if”
3. the phrase “is necessary for”
4. the phrase “is sufficient for”
Constructing Truth Tables
Truth tables for compound statements can be constructed by using the truth tables
for the basic connectives. To illustrate this, we will construct a truth table for
.P ^ :Q/ ! R. The first step is to determine the number of rows needed.
For a truth table with two different simple statements, four rows are needed
since there are four different combinations of truth values for the two state-
ments. We should be consistent with how we set up the rows. The way we

will do it in this text is to label the rows for the first statement with (T, T, F,
F) and the rows for the second statement with (T, F, T, F). All truth tables in
the text have this scheme.
For a truth table with three different simple statements, eight rows are needed
since there are eight different combinations of truth values for the three state-
ments. Our standard scheme for this type of truth table is shown in Table 2.2.
The next step is to determine the columns to be used. One way to do this is to
work backward from the form of the given statement. For .P ^ :Q/ ! R, the
last step is to deal with the conditional operator .!/. To do this, we need to know
the truth values of .P ^ :Q/ and R. To determine the truth values for .P ^ :Q/,
we need to apply the rules for the conjunction operator .^/ and we need to know
the truth values for P and :Q.
Table 2.2 is a completed truth table for .P ^ :Q/ ! R with the step numbers
indicated at the bottom of each column. The step numbers correspond to the order
in which the columns were completed.
P Q R :Q P ^ :Q .P ^ :Q/ ! R
T T T F F T
T T F F F T
T F T T T T
T F F T T F
F T T F F T
F T F F F T
F F T T F T
F F F T F T
1 1 1 2 3 4
Table 2.2: Truth Table for .P ^ :Q/ ! R
When completing the column for P ^ :Q, remember that the only time the
conjunction is true is when both P and :Q are true.
When completing the column for .P ^ :Q/ ! R, remember that the only
time the conditional statement is false is when the hypothesis .P ^ :Q/ is
true and the conclusion, R, is false.
The last column entered is the truth table for the statement .P ^ :Q/ ! R using
the set up in the first three columns.

Progress Check 2.2 (Constructing Truth Tables)
Construct a truth table for each of the following statements:
1. P ^ :Q
2. : .P ^ Q/
3. :P ^ :Q
4. :P _ :Q
Do any of these statements have the same truth table?
The Biconditional Statement
Some mathematical results are stated in the form “P if and only if Q” or “P is
necessary and sufficient for Q.” An example would be, “A triangle is equilateral
if and only if its three interior angles are congruent.” The symbolic form for the
biconditional statement “P if and only if Q” is P $ Q. In order to determine
a truth table for a biconditional statement, it is instructive to look carefully at the
form of the phrase “P if and only if Q.” The word “and” suggests that this state-
ment is a conjunction. Actually it is a conjunction of the statements “P if Q” and
“P only if Q.” The symbolic form of this conjunction is Œ.Q ! P / ^ .P ! Q/.
Progress Check 2.3 (The Truth Table for the Biconditional Statement)
Complete a truth table for Œ.Q ! P / ^ .P ! Q/. Use the following columns:
P , Q, Q ! P , P ! Q, and Œ.Q ! P / ^ .P ! Q/. The last column of this
table will be the truth table for P $ Q.
Other Forms of the Biconditional Statement
As with the conditional statement, there are some common ways to express the
biconditional statement, P $ Q, in the English language. For example,
P if and only if Q.
P implies Q and Q implies P .
P is necessary and sufficient for
Q.

Tautologies and Contradictions
Definition. A tautology is a compound statement S that is true for all pos-
sible combinations of truth values of the component statements that are part
of S. A contradiction is a compound statement that is false for all possible
combinations of truth values of the component statements that are part of S.
That is, a tautology is necessarily true in all circumstances, and a contradiction
is necessarily false in all circumstances.
Progress Check 2.4 (Tautologies and Contradictions) For statements P and Q:
1. Use a truth table to show that .P _ :P / is a tautology.
2. Use a truth table to show that .P ^ :P / is a contradiction.
3. Use a truth table to determine if P ! .P _Q/ is a tautology, a contradiction,
or neither.
?
1. Suppose that Daisy says, “If it does not rain, then I will play golf.” Later in
the day you come to know that it did rain but Daisy still played golf. Was
Daisy’s statement true or false? Support your conclusion.
? 2. Suppose that P and Q are statements for which P ! Q is true and for
which :Q is true. What conclusion (if any) can be made about the truth
value of each of the following statements?
(a) P (b) P ^ Q (c) P _ Q
3. Suppose that P and Q are statements for which P ! Q is false. What
conclusion (if any) can be made about the truth value of each of the following
statements?
(a) :P ! Q (b) Q ! P (c) P _ Q
4. Suppose that P and Q are statements for which Q is false and :P ! Q is
true (and it is not known if R is true or false). What conclusion (if any) can
be made about the truth value of each of the following statements?

(a) :Q ! P
(b) P
? (c) P ^ R
(d) R ! :P
?
5. Construct a truth table for each of the following statements:
(a) P ! Q
(b) Q ! P
(c) :P ! :Q
(d) :Q ! :P
6. Construct a truth table for each of the following statements:
(a) P _ :Q
(b) : .P _ Q/
(c) :P _ :Q
(d) :P ^ :Q
?
7. Construct truth tables for P ^ .Q _ R/ and .P ^ Q/ _ .P ^ R/. What do
you observe?
8. Suppose each of the following statements is true.
Laura is in the seventh grade.
Laura got an A on the mathematics test or Sarah got an A on the math-
ematics test.
If Sarah got an A on the mathematics test, then Laura is not in the
seventh grade.
If possible, determine the truth value of each of the following statements.
Carefully explain your reasoning.
(a) Laura got an A on the mathematics test.
(b) Sarah got an A on the mathematics test.
(c) Either Laura or Sarah did not get an A on the mathematics test.
9. Let P stand for “the integer x is even,” and let Q stand for “x2
is even.”
Express the conditional statement P ! Q in English using
(a) The “if then” form of the conditional statement
(b) The word “implies”

? (c) The “only if” form of the conditional statement
?
(d) The phrase “is necessary for”
(e) The phrase “is sufficient for”
10. Repeat Exercise (9) for the conditional statement Q ! P .
? 11. For statements P and Q, use truth tables to determine if each of the follow-
ing statements is a tautology, a contradiction, or neither.
(a) :Q _ .P ! Q/.
(b) Q ^ .P ^ :Q/.
(c) .Q ^ P / ^ .P ! :Q/.
(d) :Q ! .P ^ :P /.
12. For statements P , Q, and R:
(a) Show that Œ.P ! Q/ ^ P ! Q is a tautology. Note: In symbolic
logic, this is an important logical argment form called modus ponens.
(b) Show that Œ.P ! Q/ ^ .Q ! R/ ! .P ! R/ is a tautology. Note:
In symbolic logic, this is an important logical argment form called syl-
logism.
13. Working with Conditional Statements. Complete the following table:
English Form Hypothesis Conclusion Symbolic Form
If P , then Q. P Q P ! Q
Q only if P . Q P Q ! P
P is necessary for Q.
P is sufficient for Q.
Q is necessary for P .
P implies Q.
P only if Q.
P if Q.
If Q then P .
If :Q, then :P .
If P , then Q ^ R.
If P _ Q, then R.

2.2. Logically Equivalent Statements 43
14. Working with Truth Values of Statements. Suppose that P and Q are true
statements, that U and V are false statements, and that W is a statement and
it is not known if W is true or false.
Which of the following statements are true, which are false, and for which
statements is it not possible to determine if it is true or false? Justify your
conclusions.
(a) .P _ Q/ _ .U ^ W /
(b) P ^ .Q ! W /
(c) P ^ .W ! Q/
(d) W ! .P ^ U /
(e) W ! .P ^ :U /
(f) .:P _ :U / ^ .Q _ :V /
(g) .P ^ :V / ^ .U _ W /
(h) .P _ :Q/ ! .U ^ W /
(i) .P _ W / ! .U ^ W /
(j) .U ^ :V / ! .P ^ W /
2.2 Logically Equivalent Statements
Preview Activity 1 (Logically Equivalent Statements)
In Exercises (5) and (6) from Section 2.1, we observed situations where two differ-
ent statements have the same truth tables. Basically, this means these statements
are equivalent, and we make the following definition:
Definition. Two expressions are logically equivalent provided that they have
the same truth value for all possible combinations of truth values for all vari-
ables appearing in the two expressions. In this case, we write X Y and say
that X and Y are logically equivalent.
1. Complete truth tables for : .P ^ Q/ and :P _ :Q.
2. Are the expressions : .P ^ Q/ and :P _ :Q logically equivalent?
3. Suppose that the statement “I will play golf and I will mow the lawn” is false.
Then its negation is true. Write the negation of this statement in the form of
a disjunction. Does this make sense?
Sometimes we actually use logical reasoning in our everyday living! Perhaps
you can imagine a parent making the following two statements.

Statement 1 If you do not clean your room, then you cannot
watch TV.
Statement 2 You clean your room or you cannot watch TV.
4. Let P be “you do not clean your room,” and let Q be “you cannot watch TV.”
Use these to translate Statement 1 and Statement 2 into symbolic forms.
5. Construct a truth table for each of the expressions you determined in Part (4).
Are the expressions logically equivalent?
6. Assume that Statement 1 and Statement 2 are false. In this case, what is
the truth value of P and what is the truth value of Q? Now, write a true
statement in symbolic form that is a conjunction and involves P and Q.
7. Write a truth table for the (conjunction) statement in Part (6) and compare it
to a truth table for : .P ! Q/. What do you observe?
Preview Activity 2 (Converse and Contrapositive)
We now define two important conditional statements that are associated with a
given conditional statement.
Definition. If P and Q are statements, then
The converse of the conditional statement P ! Q is the conditional
statement Q ! P .
The contrapositive of the conditional statement P ! Q is the condi-
tional statement :Q ! :P .
1. For the following, the variable x represents a real number. Label each of the
following statements as true or false.
(a) If x D 3, then x2
D 9.
(b) If x2
D 9, then x D 3.
(c) If x2
¤ 9, then x ¤ 3.
(d) If x ¤ 3, then x2
¤ 9.
2. Which statement in the list of conditional statements in Part (1) is the con-
verse of Statement (1a)? Which is the contrapositive of Statement (1a)?
3. Complete appropriate truth tables to show that
P ! Q is logically equivalent to its contrapositive :Q ! :P .

P ! Q is not logically equivalent to its converse Q ! P .
In Preview Activity 1, we introduced the concept of logically equivalent ex-
pressions and the notation X Y to indicate that statements X and Y are logi-
cally equivalent. The following theorem gives two important logical equivalencies.
They are sometimes referred to as De Morgan’s Laws.
Theorem 2.5 (De Morgan’s Laws)
For statements P and Q,
The statement : .P ^ Q/ is logically equivalent to :P _ :Q. This can be
written as : .P ^ Q/ :P _ :Q.
The statement : .P _ Q/ is logically equivalent to :P ^ :Q. This can be
written as : .P _ Q/ :P ^ :Q.
The first equivalency in Theorem 2.5 was established in Preview Activity 1.
Table 2.3 establishes the second equivalency.
P Q P _ Q : .P _ Q/ :P :Q :P ^ :Q
T T T F F F F
T F T F F T F
F T T F T F F
F F F T T T T
Table 2.3: Truth Table for One of De Morgan’s Laws
It is possible to develop and state several different logical equivalencies at this
time. However, we will restrict ourselves to what are considered to be some of the
most important ones. Since many mathematical statements are written in the form
of conditional statements, logical equivalencies related to conditional statements
are quite important.
Logical Equivalencies Related to Conditional Statements
The first two logical equivalencies in the following theorem were established in
Preview Activity 1, and the third logical equivalency was established in Preview
Activity 2.

Theorem 2.6. For statements P and Q,
1. The conditional statement P ! Q is logically equivalent to :P _ Q.
2. The statement : .P ! Q/ is logically equivalent to P ^ :Q.
3. The conditional statement P ! Q is logically equivalent to its contraposi-
tive :Q ! :P .
The Negation of a Conditional Statement
The logical equivalency : .P ! Q/ P ^ :Q is interesting because it shows
us that the negation of a conditional statement is not another conditional state-
ment. The negation of a conditional statement can be written in the form of a
conjunction. So what does it mean to say that the conditional statement
If you do not clean your room, then you cannot watch TV,
is false? To answer this, we can use the logical equivalency : .P ! Q/ P ^
:Q. The idea is that if P ! Q is false, then its negation must be true. So the
negation of this can be written as
You do not clean your room and you can watch TV.
For another example, consider the following conditional statement:
If 5 3, then . 5/2
. 3/2
.
This conditional statement is false since its hypothesis is true and its conclusion
is false. Consequently, its negation must be true. Its negation is not a conditional
statement. The negation can be written in the form of a conjunction by using the
logical equivalency : .P ! Q/ P ^ :Q. So, the negation can be written as
follows:
5 3 and :

. 5/2
. 3/2

.
However, the second part of this conjunction can be written in a simpler manner by
noting that “not less than” means the same thing as “greater than or equal to.” So
we use this to write the negation of the original conditional statement as follows:

5 3 and . 5/2
. 3/2
.
This conjunction is true since each of the individual statements in the conjunction
is true.
Another Method of Establishing Logical Equivalencies
We have seen that it often possible to use a truth table to establish a logical equiva-
lency. However, it is also possible to prove a logical equivalency using a sequence
of previously established logical equivalencies. For example,
P ! Q is logically equivalent to :P _ Q. So
: .P ! Q/ is logically equivalent to : .:P _ Q/.
Hence, by one of De Morgan’s Laws (Theorem 2.5), : .P ! Q/ is logically
equivalent to : .:P/ ^ :Q.
This means that : .P ! Q/ is logically equivalent to P ^ :Q.
The last step used the fact that : .:P / is logically equivalent to P .
When proving theorems in mathematics, it is often important to be able to
decide if two expressions are logically equivalent. Sometimes when we are at-
tempting to prove a theorem, we may be unsuccessful in developing a proof for the
original statement of the theorem. However, in some cases, it is possible to prove
an equivalent statement. Knowing that the statements are equivalent tells us that if
we prove one, then we have also proven the other. In fact, once we know the truth
value of a statement, then we know the truth value of any other logically equivalent
statement. This is illustrated in Progress Check 2.7.
Progress Check 2.7 (Working with a Logical Equivalency)
In Section 2.1, we constructed a truth table for .P ^ :Q/ ! R. See page 38.
1. Although it is possible to use truth tables to show that P ! .Q _ R/ is
logically equivalent to .P ^ :Q/ ! R, we instead use previously proven
logical equivalencies to prove this logical equivalency. In this case, it may
be easier to start working with .P ^ :Q/ ! R. Start with
.P ^ :Q/ ! R : .P ^ :Q/ _ R;
which is justified by the logical equivalency established in Part (5) of Preview
Activity 1. Continue by using one of De Morgan’s Laws on : .P ^ :Q/.

2. Let a and b be integers. Suppose we are trying to prove the following:
If 3 is a factor of a b, then 3 is a factor of a or 3 is a factor of b.
Explain why we will have proven this statement if we prove the following:
If 3 is a factor of a b and 3 is not a factor of a, then 3 is a factor of b.
As we will see, it is often difficult to construct a direct proof for a condi-
tional statement of the form P ! .Q _ R/. The logical equivalency in Progress
Check 2.7 gives us another way to attempt to prove a statement of the form
P ! .Q _ R/. The advantage of the equivalent form, .P ^ :Q/ ! R, is that
we have an additional assumption, :Q, in the hypothesis. This gives us more
information with which to work.
Theorem 2.8 states some of the most frequently used logical equivalencies used
when writing mathematical proofs.
Theorem 2.8 (Important Logical Equivalencies)
For statements P , Q, and R,
De Morgan’s Laws : .P ^ Q/ :P _ :Q
: .P _ Q/ :P ^ :Q
Conditional Statements P ! Q :Q ! :P (contrapositive)
P ! Q :P _ Q
: .P ! Q/ P ^ :Q
Biconditional Statement .P $ Q/ .P ! Q/ ^ .Q ! P /
Double Negation : .:P / P
Distributive Laws P _ .Q ^ R/ .P _ Q/ ^ .P _ R/
P ^ .Q _ R/ .P ^ Q/ _ .P ^ R/
Conditionals with P ! .Q _ R/ .P ^ :Q/ ! R
Disjunctions .P _ Q/ ! R .P ! R/ ^ .Q ! R/
We have already established many of these equivalencies. Others will be es-
tablished in the exercises.
?
1. Write the converse and contrapositive of each of the following conditional
statements.

(a) If a D 5, then a2 D 25.
(b) If it is not raining, then Laura is playing golf.
(c) If a ¤ b, then a4
¤ b4
.
(d) If a is an odd integer, then 3a is an odd integer.
? 2. Write each of the conditional statements in Exercise (1) as a logically equiva-
lent disjunction, and write the negation of each of the conditional statements
in Exercise (1) as a conjunction.
3. Write a useful negation of each of the following statements. Do not leave a
negation as a prefix of a statement. For example, we would write the negation
of “I will play golf and I will mow the lawn” as “I will not play golf or I will
not mow the lawn.”
?
(a) We will win the first game and we will win the second game.
? (b) They will lose the first game or they will lose the second game.
? (c) If you mow the lawn, then I will pay you $20.
?
(d) If we do not win the first game, then we will not play a second game.
?
(e) I will wash the car or I will mow the lawn.
(f) If you graduate from college, then you will get a job or you will go to
graduate school.
(g) If I play tennis, then I will wash the car or I will do the dishes.
(h) If you clean your room or do the dishes, then you can go to see a movie.
(i) It is warm outside and if it does not rain, then I will play golf.
4. Use truth tables to establish each of the following logical equivalencies deal-
ing with biconditional statements:
(a) .P $ Q/ .P ! Q/ ^ .Q ! P /
(b) .P $ Q/ .Q $ P /
(c) .P $ Q/ .:P $ :Q/
5. Use truth tables to prove each of the distributive laws from Theorem 2.8.
(a) P _ .Q ^ R/ .P _ Q/ ^ .P _ R/
(b) P ^ .Q _ R/ .P ^ Q/ _ .P ^ R/

6. Use truth tables to prove the followinglogical equivalency from Theorem 2.8:
Œ.P _ Q/ ! R .P ! R/ ^ .Q ! R/ :
? 7. Use previously proven logical equivalencies to prove each of the following
logical equivalencies about conditionals with conjunctions:
(a) Œ.P ^ Q/ ! R .P ! R/ _ .Q ! R/
(b) ŒP ! .Q ^ R/ .P ! Q/ ^ .P ! R/
8. If P and Q are statements, is the statement .P _ Q/ ^ : .P ^ Q/ logically
equivalent to the statement .P ^ :Q/_.Q ^ :P /? Justify your conclusion.
9. Use previously proven logical equivalencies to prove each of the following
logical equivalencies:
(a) Œ:P ! .Q ^ :Q/ P
(b) .P $ Q/ .:P _ Q/ ^ .:Q _ P /
(c) : .P $ Q/ .P ^ :Q/ _ .Q ^ :P /
(d) .P ! Q/ ! R .P ^ :Q/ _ R
(e) .P ! Q/ ! R .:P ! R/ ^ .Q ! R/
(f) Œ.P ^ Q/ ! .R _ S/ Œ.:R ^ :S/ ! .:P _ :Q/
(g) Œ.P ^ Q/ ! .R _ S/ Œ.P ^ Q ^ :R/ ! S
(h) Œ.P ^ Q/ ! .R _ S/ .:P _ :Q _ R _ S/
(i) : Œ.P ^ Q/ ! .R _ S/ .P ^ Q ^ :R ^ :S/
?
10. Let a be a real number and let f be a real-valued function defined on an
interval containing x D a. Consider the following conditional statement:
If f is differentiable at x D a, then f is continuous at x D a.
Which of the following statements have the same meaning as this conditional
statement and which ones are negations of this conditional statement?
Note: This is not asking which statements are true and which are false. It is
asking which statements are logically equivalent to the given statement. It
might be helpful to let P represent the hypothesis of the given statement, Q
represent the conclusion, and then determine a symbolic representation for
each statement. Instead of using truth tables, try to use already established
logical equivalencies to justify your conclusions.

(a) If f is continuous at x D a, then f is differentiable at x D a.
(b) If f is not differentiable at x D a, then f is not continuous at x D a.
(c) If f is not continuous at x D a, then f is not differentiable at x D a.
(d) f is not differentiable at x D a or f is continuous at x D a.
(e) f is not continuous at x D a or f is differentiable at x D a.
(f) f is differentiable at x D a and f is not continuous at x D a.
11. Let a; b, and c be integers. Consider the following conditional statement:
If a divides bc, then a divides b or a divides c.
Which of the following statements have the same meaning as this conditional
statement and which ones are negations of this conditional statement?
The note for Exercise (10) also applies to this exercise.
(a) If a divides b or a divides c, then a divides bc.
(b) If a does not divide b or a does not divide c, then a does not divide bc.
(c) a divides bc, a does not divide b, and a does not divide c.
? (d) If a does not divide b and a does not divide c, then a does not divide
bc.
(e) a does not divide bc or a divides b or a divides c.
(f) If a divides bc and a does not divide c, then a divides b.
(g) If a divides bc or a does not divide b, then a divides c.
12. Let x be a real number. Consider the following conditional statement:
If x3 x D 2x2 C 6, then x D 2 or x D 3.
Which of the following statements have the same meaning as this condi-
tional statement and which ones are negations of this conditional statement?
Explain each conclusion. (See the note in the instructions for Exercise (10).)
(a) If x ¤ 2 and x ¤ 3, then x3
x ¤ 2x2
C 6.
(b) If x D 2 or x D 3, then x3
x D 2x2
C 6.
(c) If x ¤ 2 or x ¤ 3, then x3
x ¤ 2x2
C 6.
(d) If x3 x D 2x2 C 6 and x ¤ 2, then x D 3.
(e) If x3 x D 2x2 C 6 or x ¤ 2, then x D 3.
(f) x3 x D 2x2 C 6, x ¤ 2, and x ¤ 3.
(g) x3 x ¤ 2x2 C 6 or x D 2 or x D 3.

13. Working with a Logical Equivalency. Suppose we are trying to prove the
following for integers x and y:
If x y is even, then x is even or y is even.
We notice that we can write this statement in the following symbolic form:
P ! .Q _ R/ ;
where P is “x y is even,” Q is “x is even,” and R is “y is even.”
(a) Write the symbolic form of the contrapositive of P ! .Q _ R/. Then
use one of De Morgan’s Laws (Theorem 2.5) to rewrite the hypothesis
of this conditional statement.
(b) Use the result from Part (13a) to explain why the given statement is
logically equivalent to the following statement:
If x is odd and y is odd, then x y is odd.
The two statements in this activity are logically equivalent. We now have the
choice of proving either of these statements. If we prove one, we prove the
other, or if we show one is false, the other is also false. The second statement
is Theorem 1.8, which was proven in Section 1.2.
2.3 Open Sentences and Sets
Preview Activity 1 (Sets and Set Notation)
The theory of sets is fundamental to mathematics in the sense that many areas
of mathematics use set theory and its language and notation. This language and
notation must be understood if we are to communicate effectively in mathematics.
At this point, we will give a very brief introduction to some of the terminology
used in set theory.
A set is a well-defined collection of objects that can be thought of as a single
entity itself. For example, we can think of the set of integers that are greater than
4. Even though we cannot write down all the integers that are in this set, it is still
a perfectly well-defined set. This means that if we are given a specific integer, we
can tell whether or not it is in the set of all even integers.

2.3. Open Sentences and Sets 53
The most basic way of specifying the elements of a set is to list the elements
of that set. This works well when the set contains only a small number of objects.
The usual practice is to list these elements between braces. For example, if the set
C consists of the integer solutions of the equation x2
D 9, we would write
C D f 3; 3g :
For larger sets, it is sometimes inconvenient to list all of the elements of the
set. In this case, we often list several of them and then write a series of three dots
(: : :) to indicate that the pattern continues. For example,
D D f1; 3; 5; 7; : : :49g
is the set of all odd natural numbers from 1 to 49, inclusive.
For some sets, it is not possible to list all of the elements of a set; we then
list several of the elements in the set and again use a series of three dots (: : :) to
indicate that the pattern continues. For example, if F is the set of all even natural
numbers, we could write
F D f2; 4; 6; : : :g :
We can also use the three dots before listing specific elements to indicate the
pattern prior to those elements. For example, if E is the set of all even integers, we
could write
E D f: : : 6; 4; 2; 0; 2; 4; 6; : : :g :
Listing the elements of a set inside braces is called the roster method of specifying
the elements of the set. We will learn other ways of specifying the elements of a
set later in this section.
1. Use the roster method to specify the elements of each of the following sets:
(a) The set of real numbers that are solutions of the equation x2
5x D 0.
(b) The set of natural numbers that are less than or equal to 10.
(c) The set of integers that are greater than 2.
2. Each of the following sets is defined using the roster method. For each set,
determine four elements of the set other than the ones listed using the roster
method.

A D f1; 4; 7; 10; : : :g
B D f2; 4; 8; 16; : : :g
C D f: : : ; 8; 6; 4; 2; 0g
D D f: : : ; 9; 6; 3; 0; 3; 6; 9; : : :g
Preview Activity 2 (Variables)
Not all mathematical sentences are statements. For example, an equation such as
x2
5 D 0
is not a statement. In this sentence, the symbol x is a variable. It represents a
number that may be chosen from some specified set of numbers. The sentence
(equation) becomes true or false when a specific number is substituted for x.
1. (a) Does the equation x2
25 D 0 become a true statement if 5 is sub-
stituted for x?
(b) Does the equation x2 25 D 0 become a true statement if
p
5 is
substituted for x?
Definition. A variable is a symbol representing an unspecified object that can
be chosen from a given set U . The set U is called the universal set for the
variable. It is the set of specified objects from which objects may be chosen
to substitutefor the variable. A constant is a specific member of the universal
set.
Some sets that we will use frequently are the usual number systems. Recall
that we use the symbol R to stand for the set of all real numbers, the symbol Q to
stand for the set of all rational numbers, the symbol Z to stand for the set of all
integers, and the symbol N to stand for the set of all natural numbers.
A variable is a symbol representing an unspecified object that can be chosen
from some specified set of objects. This specified set of objects is agreed to in
advance and is frequently called the universal set.
2. What real numbers will make the sentence “y2 2y 15 D 0” a true
statement when substituted for y?
3. What natural numbers will make the sentence “y2 2y 15 D 0” a true
statement when substituted for y?
4. What real numbers will make the sentence “
p
x is a real number” a true
statement when substituted for x?

5. What real numbers will make the sentence “sin2
x C cos2 x D 1” a true
statement when substituted for x?
6. What natural numbers will make the sentence “
p
n is a natural number” a
true statement when substituted for n?
7. What real numbers will make the sentence
Z y
0
t2
dt 9
a true statement when substituted for y?
Some Set Notation
In Preview Activity 1, we indicated that a set is a well-defined collection of objects
that can be thought of as an entity itself.
If A is a set and y is one of the objects in the set A, we write y 2 A and read
this as “y is an element of A” or “y is a member of A.” For example, if B is
the set of all integers greater than 4, then we could write 5 2 B and 10 2 B.
If an object z is not an element in the set A, we write z … A and read this as
“z is not an element of A.” For example, if B is the set of all integers greater
than 4, then we could write 2 … B and 4 … B.
When working with a mathematical object, such as set, we need to define when
two of these objects are equal. We are also often interested in whether or not one
set is contained in another set.
Definition. Two sets, A and B, are equal when they have precisely the same
elements. In this case, we write A D B . When the sets A and B are not
equal, we write A ¤ B.
The set A is a subset of a set B provided that each element of A is an element
of B. In this case, we write A B and also say that A is contained in B.
When A is not a subset of B, we write A 6 B.
Using these definitions, we see that for any set A, A D A and since it is true that
for each x 2 U , if x 2 A, then x 2 A, we also see that A A. That is, any set is
equal to itself and any set is a subset of itself. For some specific examples, we see
that:

f1; 3; 5g D f3; 5; 1g
f4; 8; 12g D f4; 4; 8; 12; 12g
f5; 10g D f5; 10; 5g
f5; 10g ¤ f5; 10; 15g but f5; 10g f5; 10; 15g and f5; 10; 15g 6 f5; 10g.
In each of the first three examples, the two sets have exactly the same elements
even though the elements may be repeated or written in a different order.
Progress Check 2.9 (Set Notation)
1. Let A D f 4; 2; 0; 2; 4; 6; 8; : : :g. Use correct set notation to indicate
which of the following integers are in the set A and which are not in the
set A. For example, we could write 6 2 A and 5 … A.
10 22 13 3 0 12
2. Use correct set notation (using D or ) to indicate which of the following
sets are equal and which are subsets of one of the other sets.
A D f3; 6; 9g B D f6; 9; 3; 6g
C D f3; 6; 9; : : :g D D f3; 6; 7; 9g
E D f9; 12; 15; : : :g F D f9; 7; 6; 2g
Variables and Open Sentences
As we have seen in the Preview Activities, not all mathematical sentences are state-
ments. This is often true if the sentence contains a variable. The following termi-
nology is useful in working with sentences and statements.
Definition. An open sentence is a sentence P.x1; x2; : : :; xn/ involving vari-
ables x1; x2; : : :; xn with the property that when specific values from the uni-
versal set are assigned to x1; x2; : : :; xn, then the resulting sentence is either
true or false. That is, the resulting sentence is a statement. An open sentence
is also called a predicate or a propositional function.
Notation: One reason an open sentence is sometimes called a propositional func-
tion is the fact that we use function notation P.x1; x2; : : :; xn/ for an open sentence

in n variables. When there is only one variable, such as x, we write P.x/, which is
read “P of x.” In this notation, x represents an arbitrary element of the universal
set, and P.x/ represents a sentence. When we substitute a specific element of the
universal set for x, the resulting sentence becomes a statement. This is illustrated
in the next example.
Example 2.10 (Open Sentences)
If the universal set is R, then the sentence “x2 3x 10 D 0” is an open sentence
involving the one variable x.
If we substitute x D 2, we obtain the false statement “22
3 2 10 D 0.”
If we substitute x D 5, we obtain the true statement “52 3 5 10 D 0.”
In this example, we can let P.x/ be the predicate “x2 3x 10 D 0 ” and then
say that P.2/ is false and P.5/ is true.
Using similar notation, we can let Q.x; y/ be the predicate “x C2y D 7.” This
predicate involves two variables. Then,
Q.1; 1/ is false since “1 C 2 1 D 7” is false; and
Q.3; 2/ is true since “3 C 2 2 D 7” is true.
Progress Check 2.11 (Working with Open Sentences)
1. Assume the universal set for all variables is Z and let P.x/ be the predicate
“x2 4.”
(a) Find two values of x for which P.x/ is false.
(b) Find two values of x for which P.x/ is true.
(c) Use the roster method to specify the set of all x for which P.x/ is true.
2. Assume the universal set for all variables is Z, and let R.x; y; z/ be the
predicate “x2 C y2 D z2.”
(a) Find two different examples for which R.x; y; z/ is false.
(b) Find two different examples for which R.x; y; z/ is true.
Without using the term, Example 2.10 and Progress Check 2.11 (and Preview
Activity 2) dealt with a concept called the truth set of a predicate.

Definition. The truth set of an open sentence with one variable is the
collection of objects in the universal set that can be substituted for the variable
to make the predicate a true statement.
One part of elementary mathematics consists of learning how to solve equations.
In more formal terms, the process of solving an equation is a way to determine the
truth set for the equation, which is an open sentence. In this case, we often call the
truth set the solution set. Following are three examples of truth sets.
If the universal set is R, then the truth set of the equation 3x 8 D 10 is the
set f6g.
If the universal set is R, then the truth set of the equation “x2
3x 10 D 0”
is f 2; 5g.
If the universal set is N, then the truth set of the open sentence “
p
n 2 N” is
f1; 4; 9; 16; : : :g.
Set Builder Notation
Sometimes it is not possible to list all the elements of a set. For example, if the
universal set is R, we cannot list all the elements of the truth set of “x2 4.” In
this case, it is sometimes convenient to use the so-called set builder notation in
which the set is defined by stating a rule that all elements of the set must satisfy. If
P.x/ is a predicate in the variable x, then the notation
fx 2 U j P.x/g
stands for the set of all elements x in the universal set U for which P.x/ is true.
If it is clear what set is being used for the universal set, this notation is sometimes
shortened to fx j P.x/g. This is usually read as “the set of all x such that P.x/.”
The vertical bar stands for the phrase “such that.” Some writers will use a colon (:)
instead of the vertical bar.
For a non-mathematical example, P could be the property that a college student
is a mathematics major. Then fx j P.x/g denotes the set of all college students who
are mathematics majors. This could be written as
fx j x is a college student who is a mathematics majorg :

Example 2.12 (Truth Sets)
Assume the universal set is R and P.x/ is “x2
4.” We can describe the truth set
of P.x/ as the set of all real numbers whose square is less than 4. We can also use
set builder notation to write the truth set of P.x/ as
˚
x 2 R j x2
4 :
However, if we solve the inequality x2 4, we obtain 2 x 2. So we could
also write the truth set as
fx 2 R j 2 x 2g :
We could read this as the set of all real numbers that are greater than 2 and less
than 2. We can also write
˚
x 2 R j x2
4 D fx 2 R j 2 x 2g :
Progress Check 2.13 (Working with Truth Sets)
Let P.x/ be the predicate “x2
9.”
1. If the universal set is R, describe the truth set of P.x/ using English and
write the truth set of P.x/ using set builder notation.
2. If the universal set is Z, then what is the truth set of P.x/? Describe this set
using English and then use the roster method to specify all the elements of
this truth set.
3. Are the truth sets in Parts (1) and (2) equal? Explain.
So far, our standard form for set builder notation has been fx 2 U j P.x/g. It
is sometimes possible to modify this form and put the predicate first. For example,
the set
A D f3n C 1 j n 2 Ng
describes the set of all natural numbers of the form 3nC1 for some natural number.
By substituting 1, 2, 3, 4, and so on, for n, we can use the roster method to write
A D f3n C 1 j n 2 Ng D f4; 7; 10; 13; : : :g :
We can sometimes “reverse this process” by starting with a set specified by the
roster method and then writing the same set using set builder notation.

Example 2.14 (Set Builder Notation)
Let B D f: : : 11; 7; 3; 1; 5; 9; 13; : : :g. The key to writing this set using set
builder notation is to recognize the pattern involved. We see that once we have an
integer in B, we can obtain another integer in B by adding 4. This suggests that
the predicate we will use will involve multiplying by 4.
Since it is usually easier to work with positive numbers, we notice that 1 2 B
and 5 2 B. Notice that
1 D 4 0 C 1 and 5 D 4 1 C 1.
This suggests that we might try f4n C 1 j n 2 Zg. In fact, by trying other inte-
gers for n, we can see that
B D f: : : 11; 7; 3; 1; 5; 9; 13; :: :g D f4n C 1 j n 2 Zg :
Progress Check 2.15 (Set Builder Notation)
Each of the following sets is defined using the roster method.
A D f1; 5; 9; 13; : : :g
B D f: : : 8; 6; 4; 2; 0g
C D

p
2;
p
2
3
;
p
2
5
; : : :

D D f1; 3; 9; 27; : : :g
1. Determine four elements of each set other than the ones listed using the roster
method.
2. Use set builder notation to describe each set.
The Empty Set
When a set contains no elements, we say that the set is the empty set. For example,
the set of all rational numbers that are solutions of the equation x2 D 2 is the
empty set since this equation has no solutions that are rational numbers.
In mathematics, the empty set is usually designated by the symbol ;. We usu-
ally read the symbol ; as “the empty set” or “the null set.” (The symbol ; is
actually the last letter in the Danish-Norwegian alphabet.)

When the Truth Set Is the Universal Set
The truth set of a predicate can be the universal set. For example, if the universal
set is the set of real numbers R, then the truth set of the predicate “x C 0 D x” is
R.
Notice that the sentence “x C 0 D x” has not been quantified and a particular
element of the universal set has not been substitutedfor the variable x. Even though
the truth set for this sentence is the universal set, we will adopt the convention
that unless the quantifier is stated explicitly, we will consider the sentence to be a
predicate or open sentence. So, with this convention, if the universal set is R, then
x C 0 D x is a predicate;
For each real number x, .x C 0 D x/ is a statement.
?
1. Use the roster method to specify the elements in each of the following sets
and then write a sentence in English describing the set.
(a)
˚
x 2 R j 2x2
C 3x 2 D 0
(b)
˚
x 2 Z j 2x2
C 3x 2 D 0
(c)
˚
x 2 Z j x2
25
(d)
˚
x 2 N j x2
25
(e) fy 2 Q j jy 2j D 2:5g
(f) fy 2 Z j jy 2j 2:5g
?
2. Each of the following sets is defined using the roster method.
A D f1; 4; 9; 16; 25; : : :g
B D
˚
: : : ; 4
; 3
; 2
; ; 0
C D f3; 9; 15; 21; 27 : : :g
D D f0; 4; 8; : : :; 96; 100g
(a) Determine four elements of each set other than the ones listed using the
roster method.
(b) Use set builder notation to describe each set.
? 3. Let A D
n
x 2 R j x .x C 2/2
x 3
2

D 0
o
. Which of the following sets
are equal to the set A and which are subsets of A?

(a) f 2; 0; 3g
(b)

3
2
; 2; 0

(c)

2; 2; 0;
3
2

(d)

2;
3
2

4. Use the roster method to specify the truth set for each of the following open
sentences. The universal set for each open sentence is the set of integers Z.
? (a) n C 7 D 4.
?
(b) n2
D 64.
(c)
p
n 2 N and n is less than 50.
(d) n is an odd integer that is greater than 2 and less than 14.
(e) n is an even integer that is greater than 10.
5. Use set builder notation to specify the following sets:
? (a) The set of all integers greater than or equal to 5.
(b) The set of all even integers.
? (c) The set of all positive rational numbers.
(d) The set of all real numbers greater than 1 and less than 7.
? (e) The set of all real numbers whose square is greater than 10.
6. For each of the following sets, use English to describe the set and when
appropriate, use the roster method to specify all of the elements of the set.
(a) fx 2 Rj 3 x 5g
(b) fx 2 Zj 3 x 5g
(c)
˚
x 2 Rj x2
D 16
(d)
˚
x 2 Rj x2
C 16 D 0
(e) fx 2 Zj x is odd g
(f) fx 2 Rj 3x 4 17g
7. Closure Explorations. In Section 1.1, we studied some of the closure prop-
erties of the standard number systems. (See page 10.) We can extend this
idea to other sets of numbers. So we say that:
A set A of numbers is closed under addition provided that whenever
x and y are are in the set A, x C y is in the set A.

2.4. Quantifiers and Negations 63
A set A of numbers is closed under multiplicationprovided that when-
ever x and y are are in the set A, x y is in the set A.
A set A of numbers is closed under subtraction provided that when-
ever x and y are are in the set A, x y is in the set A.
For each of the following sets, make a conjecture about whether or not it is
closed under addition and whether or not it is closed under multiplication. In
some cases, you may be able to find a counterexample that will prove the set
is not closed under one of these operations.
(a) The set of all odd natural num-
bers
(b) The set of all even integers
(c) A D f1; 4; 7; 10; 13; : : :g
(d) B D f: : : ; 6; 3; 0; 3; 6; 9; : : :g
(e) C D f3n C 1j n 2 Zg
(f) D D

1
2n
ˇ
ˇ
ˇ
ˇ n 2 N

2.4 Quantifiers and Negations
Preview Activity 1 (An Introduction to Quantifiers)
We have seen that one way to create a statement from an open sentence is to substi-
tute a specific element from the universal set for each variable in the open sentence.
Another way is to make some claim about the truth set of the open sentence. This
is often done by using a quantifier. For example, if the universal set is R, then the
following sentence is a statement.
For each real number x, x2 0.
The phrase “For each real number x” is said to quantify the variablethat follows it
in the sense that the sentence is claiming that something is true for all real numbers.
So this sentence is a statement (which happens to be false).
Definition. The phrase “for every” (or its equivalents) is called a universal
quantifier. The phrase “there exists” (or its equivalents) is called an existen-
tial quantifier. The symbol 8 is used to denote a universal quantifier, and the
symbol 9 is used to denote an existential quantifier.
Using this notation, the statement “For each real number x, x2 0” could be
written in symbolic form as: .8x 2 R/ x2
0

. The following is an example of
a statement involving an existential quantifier.

There exists an integer x such that 3x 2 D 0.
This could be written in symbolic form as
.9x 2 Z/ .3x 2 D 0/ :
This statement is false because there are no integers that are solutions of the linear
equation 3x 2 D 0. Table 2.4 summarizes the facts about the two types of
quantifiers.
A statement
involving
Often has the form The statement is true
provided that
A universal
quantifier:
.8x; P.x//
“For every x, P.x/,”
where P.x/ is a
predicate.
Every value of x in
the universal set makes
P.x/ true.
An existential
quantifier:
.9x; P.x//
“There exists an x such
that P.x/,” where P.x/
is a predicate.
There is at least one
value of x in the
universal set that makes
P.x/ true.
Table 2.4: Properties of Quantifiers
In effect, the table indicates that the universally quantified statement is true pro-
vided that the truth set of the predicate equals the universal set, and the existentially
quantified statement is true provided that the truth set of the predicate contains at
least one element.
Each of the following sentences is a statement or an open sentence. Assume that
the universal set for each variable in these sentences is the set of all real numbers.
If a sentence is an open sentence (predicate), determine its truth set. If a sentence
is a statement, determine whether it is true or false.
1. .8a 2 R/ .a C 0 D a/.
2. 3x 5 D 9.
3.
p
x 2 R.
4. sin.2x/ D 2.sinx/.cos x/.
5. .8x 2 R/ .sin.2x/ D 2.sinx/.cos x//.

6. .9x 2 R/ x2 C 1 D 0

.
7. .8x 2 R/ x3
x2

.
8. x2
C 1 D 0.
9. If x2 1, then x 1.
10. .8x 2 R/ If x2 1; then x 1

.
Preview Activity 2 (Attempting to Negate Quantified Statements)
1. Consider the following statement written in symbolic form:
.8x 2 Z/ .x is a multiple of 2/.
(a) Write this statement as an English sentence.
(b) Is the statement true or false? Why?
(c) How would you write the negation of this statement as an English sen-
tence?
(d) If possible, write your negation of this statement from part (2) symbol-
ically (using a quantifier).
2. Consider the following statement written in symbolic form:
.9x 2 Z/ x3 0

.
(a) Write this statement as an English sentence.
(b) Is the statement true or false? Why?
(c) How would you write the negation of this statement as an English sen-
tence?
(d) If possible, write your negation of this statement from part (2) symbol-
ically (using a quantifier).
We introduced the concepts of open sentences and quantifiers in Section 2.3.
Review the definitions given on pages 54, 58, and 63.
Forms of Quantified Statements in English
There are many ways to write statements involving quantifiers in English. In some
cases, the quantifiers are not apparent, and this often happens with conditional

statements. The following examples illustrate these points. Each example contains
a quantified statement written in symbolic form followed by several ways to write
the statement in English.
1. .8x 2 R/ x2 0

.
For each real number x, x2 0.
The square of every real number is greater than 0.
The square of a real number is greater than 0.
If x 2 R, then x2
0.
In the second to the last example, the quantifier is not stated explicitly. Care
must be taken when reading this because it really does say the same thing as
the previous examples. The last example illustrates the fact that conditional
statements often contain a “hidden” universal quantifier.
If the universal set is R, then the truth set of the open sentence x2
0 is the
set of all nonzero real numbers. That is, the truth set is
fx 2 R j x ¤ 0g :
So the preceding statements are false. For the conditional statement, the
example using x D 0 produces a true hypothesis and a false conclusion.
This is a counterexample that shows that the statement with a universal
quantifier is false.
2. .9x 2 R/ x2
D 5

.
There exists a real number x such that x2
D 5.
x2 D 5 for some real number x.
There is a real number whose square equals 5.
The second example is usually not used since it is not considered good writ-
ing practice to start a sentence with a mathematical symbol.
If the universal set is R, then the truth set of the predicate “x2
D 5” is
n p
5;
p
5
o
. So these are all true statements.

Negations of Quantified Statements
In Preview Activity 1, we wrote negations of some quantified statements. This
is a very important mathematical activity. As we will see in future sections, it
is sometimes just as important to be able to describe when some object does not
satisfy a certain property as it is to describe when the object satisfies the property.
Our next task is to learn how to write negations of quantified statements in a useful
English form.
We first look at the negation of a statement involving a universal quantifier. The
general form for such a statement can be written as .8x 2 U / .P.x//, where P.x/
is an open sentence and U is the universal set for the variable x. When we write
: .8x 2 U / ŒP .x/ ;
we are asserting that the statement .8x 2 U / ŒP.x/ is false. This is equivalent to
saying that the truth set of the open sentence P.x/ is not the universal set. That is,
there exists an element x in the universal set U such that P.x/ is false. This in turn
means that there exists an element x in U such that :P.x/ is true, which is equiv-
alent to saying that .9x 2 U / Œ:P.x/ is true. This explains why the following
result is true:
: .8x 2 U / ŒP.x/ .9x 2 U / Œ:P.x/ :
Similarly, when we write
: .9x 2 U / ŒP.x/ ;
we are asserting that the statement .9x 2 U / ŒP.x/ is false. This is equivalent to
saying that the truth set of the open sentence P.x/ is the empty set. That is, there
is no element x in the universal set U such that P.x/ is true. This in turn means
that for each element x in U , :P.x/ is true, and this is equivalent to saying that
.8x 2 U / Œ:P.x/ is true. This explains why the following result is true:
: .9x 2 U / ŒP.x/ .8x 2 U / Œ:P.x/ :
We summarize these results in the following theorem.
Theorem 2.16. For any open sentence P.x/,
: .8x 2 U / ŒP.x/ .9x 2 U / Œ:P.x/ , and
: .9x 2 U / ŒP.x/ .8x 2 U / Œ:P.x/ :

Example 2.17 (Negations of Quantified Statements)
Consider the following statement: .8x 2 R/ x3
x2

.
We can write this statement as an English sentence in several ways. Following
are two different ways to do so.
For each real number x, x3
x2
.
If x is a real number, then x3 is greater than or equal to x2.
The second statement shows that in a conditional statement, there is often a hidden
universal quantifier. This statement is false since there are real numbers x for which
x3
is not greater than or equal to x2
. For example, we could use x D 1 or x D 1
2.
This means that the negation must be true. We can form the negation as follows:
: .8x 2 R/ x3
x2

.9x 2 R/ : x3
x2

:
In most cases, we want to write this negation in a way that does not use the nega-
tion symbol. In this case, we can now write the open sentence : x3
x2

as
x3
x2

. (That is, the negation of “is greater than or equal to” is “is less than.”)
So we obtain the following:
: .8x 2 R/ x3
x2

.9x 2 R/ x3
x2

:
The statement .9x 2 R/ x3 x2

could be written in English as follows:
There exists a real number x such that x3
x2
.
There exists an x such that x is a real number and x3
x2
.
Progress Check 2.18 (Negating Quantified Statements)
For each of the following statements
Write the statement in the form of an English sentence that does not use the
symbols for quantifiers.
Write the negation of the statement in a symbolic form that does not use the
negation symbol.
Write the negation of the statement in the form of an English sentence that
does not use the symbols for quantifiers.

1. .8a 2 R/ .a C 0 D a/.
2. .8x 2 R/ Œsin.2x/ D 2.sin x/.cos x/.
3. .8x 2 R/ tan2
x C 1 D sec2
x

.
4. .9x 2 Q/ x2 3x 7 D 0

.
5. .9x 2 R/ x2 C 1 D 0

.
Counterexamples and Negations of Conditional Statements
The real number x D 1 in the previous example was used to show that the state-
ment .8x 2 R/ x3
x2

is false. This is called a counterexample to the state-
ment. In general, a counterexample to a statement of the form .8x/ ŒP.x/ is an
object a in the universal set U for which P.a/ is false. It is an example that proves
that .8x/ ŒP.x/ is a false statement, and hence its negation, .9x/ Œ:P.x/, is a
true statement.
In the preceding example, we also wrote the universally quantified statement as
a conditional statement. The number x D 1 is a counterexample for the statement
If x is a real number, then x3
is greater than or equal to x2
.
So the number 1 is an example that makes the hypothesis of the conditional state-
ment true and the conclusion false. Remember that a conditional statement often
contains a “hidden” universal quantifier. Also, recall that in Section 2.2 we saw
that the negation of the conditional statement “If P then Q” is the statement “P
and not Q.” Symbolically, this can be written as follows:
: .P ! Q/ P ^ :Q:
So when we specifically include the universal quantifier, the symbolic form of the
negation of a conditional statement is
: .8x 2 U / ŒP.x/ ! Q.x/ .9x 2 U / : ŒP.x/ ! Q.x/
.9x 2 U / ŒP.x/ ^ :Q.x/ :
That is,
: .8x 2 U / ŒP.x/ ! Q.x/ .9x 2 U / ŒP.x/ ^ :Q.x/ :

Progress Check 2.19 (Using Counterexamples)
Use counterexamples to explain why each of the following statements is false.
1. For each integer n, n2
C n C 1

is a prime number.
2. For each real number x, if x is positive, then 2x2 x.
Quantifiers in Definitions
Definitions of terms in mathematics often involve quantifiers. These definitions are
often given in a form that does not use the symbols for quantifiers. Not only is it
important to know a definition, it is also important to be able to write a negation of
the definition. This will be illustrated with the definition of what it means to say
that a natural number is a perfect square.
Definition. A natural number n is a perfect square provided that there exists
a natural number k such that n D k2
.
This definition can be written in symbolic form using appropriate quantifiers
as follows:
A natural number n is a perfect square provided .9k 2 N/ n D k2

.
We frequently use the following steps to gain a better understanding of a defi-
nition.
1. Examples of natural numbers that are perfect squares are 1, 4, 9, and 81 since
1 D 12, 4 D 22, 9 D 32, and 81 D 92.
2. Examples of natural numbers that are not perfect squares are 2, 5, 10, and
50.
3. This definition gives two “conditions.” One is that the natural number n is a
perfect square and the other is that there exists a natural number k such that
n D k2
. The definition states that these mean the same thing. So when we
say that a natural number n is not a perfect square, we need to negate the
condition that there exists a natural number k such that n D k2
. We can use
the symbolic form to do this.

: .9k 2 N/ n D k2

.8k 2 N/ n ¤ k2

Notice that instead of writing : n D k2

, we used the equivalent form of
n ¤ k2

. This will be easier to translate into an English sentence. So we
can write,
A natural number n is not a perfect square provided that for every nat-
ural number k, n ¤ k2
.
The preceding method illustrates a good method for trying to understand a new
definition. Most textbooks will simply define a concept and leave it to the reader
to do the preceding steps. Frequently, it is not sufficient just to read a definition
and expect to understand the new term. We must provide examples that satisfy the
definition, as well as examples that do not satisfy the definition, and we must be
able to write a coherent negation of the definition
Progress Check 2.20 (Multiples of Three)
Definition. An integer n is a multiple of 3 provided that there exists an integer
k such that n D 3k.
1. Write this definition in symbolic form using quantifiers by completing the
following:
An integer n is a multiple of 3 provided that ....
2. Give several examples of integers (including negative integers) that are mul-
tiples of 3.
3. Give several examples of integers (including negative integers) that are not
multiples of 3.
4. Use the symbolic form of the definition of a multiple of 3 to complete the
following sentence: “An integer n is not a multiple of 3 provided that ....”
5. Without using the symbols for quantifiers, complete the following sentence:
“An integer n is not a multiple of 3 provide that ....”

Statements with More than One Quantifier
When a predicate contains more than one variable, each variable must be quantified
to create a statement. For example, assume the universal set is the set of integers,
Z, and let P .x; y/ be the predicate, “x C y D 0.” We can create a statement from
this predicate in several ways.
1. .8x 2 Z/ .8y 2 Z/ .x C y D 0/.
We could read this as, “For all integers x and y, x C y D 0.” This is a
false statement since it is possible to find two integers whose sum is not zero
.2 C 3 ¤ 0/.
2. .8x 2 Z/ .9y 2 Z/ .x C y D 0/.
We could read this as, “For every integer x, there exists an integer y such
that x C y D 0.” This is a true statement.
3. .9x 2 Z/ .8y 2 Z/ .x C y D 0/.
We could read this as, “There exists an integer x such that for each integer
y, x C y D 0.” This is a false statement since there is no integer whose sum
with each integer is zero.
4. .9x 2 Z/ .9y 2 Z/ .x C y D 0/.
We could read this as, “There exist integers x and y such that
x C y D 0.” This is a true statement. For example, 2 C . 2/ D 0.
When we negate a statement with more than one quantifier, we consider each quan-
tifier in turn and apply the appropriate part of Theorem 2.16. As an example, we
will negate Statement (3) from the preceding list. The statement is
.9x 2 Z/ .8y 2 Z/ .x C y D 0/ :
We first treat this as a statement in the following form: .9x 2 Z/ .P.x// where
P.x/ is the predicate .8y 2 Z/ .x C y D 0/. Using Theorem 2.16, we have
: .9x 2 Z/ .P.x// .8x 2 Z/ .:P.x// :
Using Theorem 2.16 again, we obtain the following:
:P.x/ : .8y 2 Z/ .x C y D 0/
.9y 2 Z/ : .x C y D 0/
.9y 2 Z/ .x C y ¤ 0/ :

Combining these two results, we obtain
: .9x 2 Z/ .8y 2 Z/ .x C y D 0/ .8x 2 Z/ .9y 2 Z/ .x C y ¤ 0/ :
The results are summarized in the following table.
Symbolic Form English Form
Statement .9x 2 Z/ .8y 2 Z/ .x C y D 0/
There exists an integer
x such that for each
integer y, x C y D 0.
Negation .8x 2 Z/ .9y 2 Z/ .x C y ¤ 0/
For each integer x, there
exists an integer y such
that x C y ¤ 0.
Since the given statement is false, its negation is true.
We can construct a similar table for each of the four statements. The next table
shows Statement (2), which is true, and its negation, which is false.
Symbolic Form English Form
Statement .8x 2 Z/ .9y 2 Z/ .x C y D 0/
For every integer x,
there exists an integer y
such that x C y D 0.
Negation .9x 2 Z/ .8y 2 Z/ .x C y ¤ 0/
There exists an integer
x such that for every
integer y, x C y ¤ 0.
Progress Check 2.21 (Negating a Statement with Two Quantifiers)
Write the negation of the statement
.8x 2 Z/ .8y 2 Z/ .x C y D 0/
in symbolic form and as a sentence written in English.
Writing Guideline
Try to use English and minimize the use of cumbersome notation. Do not use the
special symbols for quantifiers 8 (for all), 9 (there exists), Ö (such that), or )

(therefore) in formal mathematical writing. It is often easier to write and usually
easier to read, if the English words are used instead of the symbols. For example,
why make the reader interpret
.8x 2 R/ .9y 2 R/ .x C y D 0/
when it is possible to write
For each real number x, there exists a real number y such that x C y D 0,
or, more succinctly (if appropriate),
Every real number has an additive inverse.
? 1. For each of the following, write the statement as an English sentence and
then explain why the statement is false.
(a) .9x 2 Q/ x2
3x 7 D 0

.
(b) .9x 2 R/ x2
C 1 D 0

.
(c) .9m 2 N/ m2
1

.
2. For each of the following, use a counterexample to show that the statement
is false. Then write the negation of the statement in English, without using
symbols for quantifiers.
?
(a) .8m 2 Z/ m2
is even

.
? (b) .8x 2 R/ x2 0

.
(c) For each real number x,
p
x 2 R.
(d) .8m 2 Z/
m
3
2 Z

.
(e) .8a 2 Z/
p
a2 D a

.
?
(f) .8x 2 R/ tan2
x C 1 D sec2
x

.
3. For each of the following statements
Write the statement as an English sentence that does not use the sym-
bols for quantifiers.

Write the negation of the statement in symbolicform in which the nega-
tion symbol is not used.
Write a useful negation of the statement in an English sentence that
does not use the symbols for quantifiers.
?
(a) .9x 2 Q/

x
p
2

.
(b) .8x 2 Q/ x2
2 ¤ 0

.
?
(c) .8x 2 Z/ .x is even or x is odd/.
(d) .9x 2 Q/
p
2 x
p
3

. Note: The sentence “
p
2 x
p
3” is
actually a conjuction. It means
p
2 x and x
p
3.
? (e) .8x 2 Z/ If x2 is odd, then x is odd

.
(f) .8n 2 N/ [If n is a perfect square, then .2n 1/ is not a prime num-
ber].
(g) .8n 2 N/ n2
n C 41 is a prime number

.
?
(h) .9x 2 R/ .cos.2x/ D 2.cos x//.
4. Write each of the following statements as an English sentence that does not
use the symbols for quantifiers.
?
(a) .9m 2 Z/ .9n 2 Z/ .m n/
(b) .9m 2 Z/ .8n 2 Z/ .m n/
(c) .8m 2 Z/ .9n 2 Z/ .m n/
(d) .8m 2 Z/ .8n 2 Z/ .m n/
? (e) .9n 2 Z/ .8m 2 Z/ m2 n

(f) .8n 2 Z/ .9m 2 Z/ m2 n

?
5. Write the negation of each statement in Exercise (4) in symbolic form and
as an English sentence that does not use the symbols for quantifiers.
?
6. Assume that the universal set is Z. Consider the following sentence:
.9t 2 Z/ .t x D 20/ :
(a) Explain why this sentence is an open sentence and not a statement.
(b) If 5 is substituted for x, is the resulting sentence a statement? If it is a
statement, is the statement true or false?
(c) If 8 is substituted for x, is the resulting sentence a statement? If it is a
(d) If 2 is substituted for x, is the resulting sentence a statement? If it is
a statement, is the statement true or false?

(e) What is the truth set of the open sentence .9t 2 Z/ .t x D 20/?
7. Assume that the universal set is R. Consider the following sentence:
.9t 2 R/ .t x D 20/ :
(a) Explain why this sentence is an open sentence and not a statement.
(b) If 5 is substituted for x, is the resulting sentence a statement? If it is a
(c) If is substituted for x, is the resulting sentence a statement? If it is a
(d) If 0 is substituted for x, is the resulting sentence a statement? If it is a
(e) What is the truth set of the open sentence .9t 2 R/ .t x D 20/?
8. Let Z be the set of all nonzero integers.
(a) Use a counterexample to explain why the following statement is false:
For each x 2 Z
, there exists a y 2 Z
such that xy D 1.
(b) Write the statement in part (a) in symbolic form using appropriate sym-
bols for quantifiers.
(c) Write the negation of the statement in part (b) in symbolic form using
appropriate symbols for quantifiers.
(d) Write the negation from part (c) in English without usings the symbols
for quantifiers.
9. An integer m is said to have the divides propertyprovided that for all integers
a and b, if m divides ab, then m divides a or m divides b.
(a) Using the symbols for quantifiers, write what it means to say that the
integer m has the divides property.
(b) Using the symbols for quantifiers, write what it means to say that the
integer m does not have the divides property.
(c) Write an English sentence stating what it means to say that the integer
m does not have the divides property.
10. In calculus, we define a function f with domain R to be strictly increasing
provided that for all real numbers x and y, f .x/ f .y/ whenever x y.
Complete each of the following sentences using the appropriate symbols for
quantifiers:

? (a) A function f with domain R is strictly increasing provided that : : : :
(b) A function f with domain R is not strictly increasing provided
that : : ::
Complete the following sentence in English without using symbols for quan-
tifiers:
(c) A function f with domain R is not strictly increasing provided
that : : ::
11. In calculus, we define a function f to be continuous at a real number a
provided that for every 0, there exists a ı 0 such that if jx aj ı,
then jf .x/ f .a/j .
Note: The symbol is the lowercase Greek letter epsilon, and the symbol ı
is the lowercase Greek letter delta.
Complete each of the following sentences using the appropriate symbols for
quantifiers:
(a) A function f is continuous at the real number a provided that : : : :
(b) A function f is not continuous at the real number a provided that : : : :
Complete the following sentence in English without using symbols for quan-
tifiers:
(c) A function f is not continuous at the real number a provided that : : : :
12. The following exercises contain definitions or results from more advanced
mathematics courses. Even though we may not understand all of the terms
involved, it is still possible to recognize the structure of the given statements
and write a meaningful negation of that statement.
(a) In abstract algebra, an operation on a set A is called a commutative
operation provided that for all x; y 2 A, x y D y x. Carefully
explain what it means to say that an operation on a set A is not a
commutative operation.
(b) In abstract algebra, a ring consists of a nonempty set R and two opera-
tions called addition and multiplication. A nonzero element a in a ring
R is called a zero divisor provided that there exists a nonzero element
b in R such that ab D 0. Carefully explain what it means to say that a
nonzero element a in a ring R is not a zero divisor.

(c) A set M of real numbers is called a neighborhood of a real number a
provided that there exists a positive real number such that the open
interval .a ; a C / is contained in M. Carefully explain what it
means to say that a set M is not a neighborhood of a real number a.
(d) In advanced calculus, a sequence of real numbers fx1; x2; : : :; xk; : : :g
is called a Cauchy sequence provided that for each positive real num-
ber , there exists a natural number N such that for all m; n 2 N, if
m N and n N , then jxn xmj . Carefully explain what it
means to say that the sequence of real numbers fx1; x2; : : :; xk; : : :g is
not a Cauchy sequence.
13. Prime Numbers. The following definition of a prime number is very im-
portant in many areas of mathematics. We will use this definition at various
places in the text. It is introduced now as an example of how to work with a
definition in mathematics.
Definition. A natural number p is a prime number provided that it is greater
than 1 and the only natural numbers that are factors of p are 1 and p. A natural
number other than 1 that is not a prime number is a composite number. The
number 1 is neither prime nor composite.
Using the definition of a prime number, we see that 2, 3, 5, and 7 are prime
numbers. Also, 4 is a composite number since 4 D 2 2; 10 is a composite
number since 10 D 2 5; and 60 is a composite number since 60 D 4 15.
(a) Give examples of four natural numbers other than 2, 3, 5, and 7 that are
prime numbers.
(b) Explain why a natural number p that is greater than 1 is a prime number
provided that
For all d 2 N, if d is a factor of p, then d D 1 or d D p.
(c) Give examples of four natural numbers that are composite numbers and
explain why they are composite numbers.
(d) Write a useful description of what it means to say that a natural number
is a composite number (other than saying that it is not prime).

14. Upper Bounds for Subsets of R. Let A be a subset of the real numbers.
A number b is called an upper bound for the set A provided that for each
element x in A, x b.
(a) Write this definition in symbolic form by completing the following:
Let A be a subset of the real numbers. A number b is called an upper
bound for the set A provided that : : : :
(b) Give examples of three different upper bounds for the set
A D fx 2 R j 1 x 3g.
(c) Does the set B D fx 2 R j x 0g have an upper bound? Explain.
(d) Give examples of three different real numbers that are not upper bounds
for the set A D fx 2 R j 1 x 3g.
(e) Complete the following in symbolic form: “Let A be a subset of R. A
number b is not an upper bound for the set A provided that : : : :”
(f) Without using the symbols for quantifiers, complete the following sen-
tence: “Let A be a subset of R. A number b is not an upper bound for
the set A provided that : : : :”
(g) Are your examples in Part (14d) consistentwith your work in Part (14f)?
Explain.
15. Least Upper Bound for a Subset of R. In Exercise 14, we introduced the
definition of an upper bound for a subset of the real numbers. Assume that
we know this definition and that we know what it means to say that a number
is not an upper bound for a subset of the real numbers.
Let A be a subset of R. A real number ˛ is the least upper bound for A
provided that ˛ is an upper bound for A, and if ˇ is an upper bound for A,
then ˛ ˇ.
Note: The symbol ˛ is the lowercase Greek letter alpha, and the symbol ˇ
is the lowercase Greek letter beta.
If we define P .x/ to be “x is an upper bound for A,” then we can write the
definition for least upper bound as follows:
A real number ˛ is the least upper bound for A provided that
P .˛/ ^ Œ.8ˇ 2 R/ .P .ˇ/ ! .˛ ˇ//.
(a) Why is a universal quantifier used for the real number ˇ?
(b) Complete the following sentence in symbolic form: “A real number ˛
is not the least upper bound for A provided that : : : :”

(c) Complete the following sentence as an English sentence: “A real num-
ber ˛ is not the least upper bound for A provided that : : : :”
Logically equivalent statements,
page 43
Converse of a conditional state-
ment, page 44
Contrapositive of a conditional
statement, page 44
Equal sets, page 55
Variable, page 54
Universal set for a variable,
page 54
Constant, page 54
Predicate, page 54
Open sentence, page 54
Truth set of a predicate, page 58
Universal quantifier, page 63
Existential quantifier, page 63
Empty set, page 60
Counterexample, pages 66 and 69
Perfect square, page 70
Prime number, page 78
Composite number, page 78
Important Theorems and Results
Theorem 2.8. Important Logical Equivalencies. For statements P , Q, and R,

De Morgan’s Laws : .P ^ Q/ :P _ :Q
: .P _ Q/ :P ^ :Q
Conditional Statements P ! Q :Q ! :P (contrapositive)
P ! Q :P _ Q
: .P ! Q/ P ^ :Q
Biconditional Statement .P $ Q/ .P ! Q/ ^ .Q ! P /
Double Negation : .:P / P
Distributive Laws P _ .Q ^ R/ .P _ Q/ ^ .P _ R/
P ^ .Q _ R/ .P ^ Q/ _ .P ^ R/
Conditionals with P ! .Q _ R/ .P ^ :Q/ ! R
Disjunctions .P _ Q/ ! R .P ! R/ ^ .Q ! R/
Theorem 2.16. Negations of Quantified Statements. For any predicate P .x/,
: .8x/ ŒP.x/ .9x/ Œ:P.x/ , and
: .9x/ ŒP.x/ .8x/ Œ:P.x/ :
Important Set Theory Notation
Notation Description Page
y 2 A y is an element of the set A. 55
z … A z is not an element of the set A. 55
f g The roster method 53
fx 2 U j P.x/g Set builder notation 58

Chapter 3
Constructing and Writing Proofs
in Mathematics
3.1 Direct Proofs
Preview Activity 1 (Definition of Divides, Divisor, Multiple)
In Section 1.2, we studied the concepts of even integers and odd integers. The
definition of an even integer was a formalization of our concept of an even integer
as being one this is “divisible by 2,” or a “multiple of 2.” We could also say that if
“2 divides an integer,” then that integer is an even integer. We will now extend this
idea to integers other than 2. Following is a formal definition of what it means to
say that a nonzero integer m divides an integer n.
Definition. A nonzero integer m divides an integer n provided that there is
an integer q such that n D m q. We also say that m is a divisor of n, m is
a factor of n, and n is a multiple of m. The integer 0 is not a divisor of any
integer. If a and b are integers and a ¤ 0, we frequently use the notation a j b
as a shorthand for “a divides b.”
A Note about Notation: Be careful with the notation a j b. This does not represent
the rational number
a
b
. The notation a j b represents a relationship between the
integers a and b and is simply a shorthand for “a divides b.”
A Note about Definitions: Technically, a definition in mathematics should almost
always be written using “if and only if.” It is not clear why, but the convention in
82

3.1. Direct Proofs 83
mathematics is to replace the phrase “if and only if” with “if” or an equivalent. Per-
haps this is a bit of laziness or the “if and only if” phrase can be a bit cumbersome.
In this text, we will often use the phrase “provided that” instead.
The definition for “divides” can be written in symbolic form using appropriate
quantifiers as follows: A nonzero integer m divides an integer n provided that
.9q 2 Z/ .n D m q/.
1. Use the definition of divides to explain why 4 divides 32 and to explain why
8 divides 96.
2. Give several examples of two integers where the first integer does not divide
the second integer.
3. According to the definition of “divides,” does the integer 10 divide the inte-
ger 0? That is, is 10 a divisor of 0? Explain.
4. Use the definition of “divides” to complete the following sentence in sym-
bolic form: “The nonzero integer m does not divide the integer n means that
....”
5. Use the definition of “divides” to complete the following sentence without
using the symbols for quantifiers: “The nonzero integer m does not divide
the integer n : : : :”
6. Give three different examples of three integers where the first integer divides
the second integer and the second integer divides the third integer.
As we have seen in Section 1.2, a definition is frequently used when constructing
and writing mathematical proofs. Consider the following conjecture:
Conjecture: Let a, b, and c be integers with a ¤ 0 and b ¤ 0. If a divides b and
b divides c, then a divides c.
7. Explain why the examples you generated in part (6) provide evidence that
this conjecture is true.
In Section 1.2, we also learned how to use a know-show table to help organize our
thoughts when trying to construct a proof of a statement. If necessary, review the
appropriate material in Section 1.2.
8. State precisely what we would assume if we were trying to write a proof of
the preceding conjecture.

84 Chapter 3. Constructing and Writing Proofs in Mathematics
9. Use the definition of “divides” to make some conclusions based on your
assumptions in part (8).
10. State precisely what we would be trying to prove if we were trying to write
a proof of the conjecture.
11. Use the definition of divides to write an answer to the question, “How can
we prove what we stated in part (10)?
Preview Activity 2 (Calendars and Clocks)
This preview activity is intended to help with understanding the concept of congru-
ence, which will be studied at the end of this section.
1. Suppose that it is currently Tuesday.
(a) What day will it be 3 days from now?
(b) What day will it be 10 days from now?
(c) What day will it be 17 days from now? What day will it be 24 days
from now?
(d) Find several other natural numbers x such that it will be Friday x days
from now.
(e) Create a list (in increasing order) of the numbers 3; 10; 17; 24, and the
numbers you generated in Part (1d). Pick any two numbers from this
list and subtract one from the other. Repeat this several times.
(f) What do the numbers you obtained in Part (1e) have in common?
2. Suppose that we are using a twelve-hour clock with no distinction between
a.m. and p.m. Also, suppose that the current time is 5:00.
(a) What time will it be 4 hours from now?
(b) What time will it be 16 hours from now? What time will it be 28 hours
from now?
(c) Find several other natural numbers x such that it will be 9:00 x hours
from now.
(d) Create a list (in increasing order) of the numbers 4; 16; 28, and the
numbers you generated in Part (2c). Pick any two numbers from this
list and subtract one from the other. Repeat this several times.
(e) What do the numbers you obtained in Part (2d) have in common?

3. This is a continuation of Part (1). Suppose that it is currently Tuesday.
(a) What day was it 4 days ago?
(b) What day was it 11 days ago? What day was it 18 days ago?
(c) Find several other natural numbers x such that it was Friday x days
ago.
(d) Create a list (in increasing order) consisting of the numbers
18; 11; 4, the opposites of the numbers you generated in Part (3c)
and the positive numbers in the list from Part (1e). Pick any two num-
bers from this list and subtract one from the other. Repeat this several
times.
(e) What do the numbers you obtained in Part (3d) have in common?
Some Mathematical Terminology
In Section 1.2, we introduced the idea of a direct proof. Since then, we have used
some common terminology in mathematics without much explanation. Before we
proceed further, we will discuss some frequently used mathematical terms.
A proof in mathematics is a convincing argument that some mathematical
statement is true. A proof should contain enough mathematical detail to be con-
vincing to the person(s) to whom the proof is addressed. In essence, a proof is
an argument that communicates a mathematical truth to another person (who has
the appropriate mathematical background). A proof must use correct, logical rea-
soning and be based on previously established results. These previous results can
be axioms, definitions, or previously proven theorems. These terms are discussed
below.
Surprising to some is the fact that in mathematics, there are always undefined
terms. This is because if we tried to define everything, we would end up go-
ing in circles. Simply put, we must start somewhere. For example, in Euclidean
geometry, the terms “point,” “line,” and “contains” are undefined terms. In this
text, we are using our number systems such as the natural numbers and integers
as undefined terms. We often assume that these undefined objects satisfy certain
properties. These assumed relationships are accepted as true without proof and
are called axioms (or postulates). An axiom is a mathematical statement that is
accepted without proof. Euclidean geometry starts with undefined terms and a set
of postulates and axioms. For example, the following statement is an axiom of
Euclidean geometry:

Given any two distinct points, there is exactly one line that contains these two
points.
The closure properties of the number systems discussed in Section 1.1 and the
properties of the number systems in Table 1.2 on page 18 are being used as
axioms in this text.
A definition is simply an agreement as to the meaning of a particular term.
For example, in this text, we have defined the terms “even integer” and “odd inte-
ger.” Definitions are not made at random, but rather, a definition is usually made
because a certain property is observed to occur frequently. As a result, it becomes
convenient to give this property its own special name. Definitions that have been
made can be used in developing mathematical proofs. In fact, most proofs require
the use of some definitions.
In dealing with mathematical statements, we frequently use the terms “con-
jecture,” “theorem,” “proposition,” “lemma,” and “corollary.” A conjecture is a
statement that we believe is plausible. That is, we think it is true, but we have not
yet developed a proof that it is true. A theorem is a mathematical statement for
which we have a proof. A term that is often considered to be synonymous with
“theorem” is proposition.
Often the proof of a theorem can be quite long. In this case, it is often easier
to communicate the proof in smaller “pieces.” These supporting pieces are often
called lemmas. A lemma is a true mathematical statement that was proven mainly
to help in the proof of some theorem. Once a given theorem has been proven, it
is often the case that other propositions follow immediately from the fact that the
theorem is true. These are called corollaries of the theorem. The term corollary is
used to refer to a theorem that is easily proven once some other theorem has been
proven.
Constructing Mathematical Proofs
To create a proof of a theorem, we must use correct logical reasoning and mathe-
matical statements that we already accept as true. These statements include axioms,
definitions, theorems, lemmas, and corollaries.
In Section 1.2, we introduced the use of a know-show table to help us organize
our work when we are attempting to prove a statement. We also introduced some
guidelines for writing mathematical proofs once we have created the proof. These
guidelines should be reviewed before proceeding.

Please remember that when we start the process of writing a proof, we are
essentially “reporting the news.” That is, we have already discovered the proof,
and now we need to report it. This reporting often does not describe the process of
discovering the news (the investigative portion of the process).
Quite often, the first step is to develop a conjecture. This is often done after
working within certain objects for some time. This is what we did in Preview Ac-
tivity 1 when we used examples to provide evidence that the following conjecture
is true:
Conjecture: Let a, b, and c be integers with a ¤ 0 and b ¤ 0. If a divides b and
b divides c, then a divides c.
Before we try to prove a conjecture, we should make sure we have explored
some examples. This simply means to construct some specific examples where the
integers a, b, and c satisfy the hypothesis of the conjecture in order to see if they
also satisfy the conclusion. We did this for this conjecture in Preview Activity 1.
We will now start a know-show table for this conjecture.
Step Know Reason
P a; b; c 2 Z, a ¤ 0, b ¤ 0, a j b
and b j c
Hypothesis
P1
:
:
:
:
:
:
:
:
:
Q1
Q a j c
Step Show Reason
The backward question we ask is, “How can we prove that a divides c?” One
answer is to use the definition and show that there exists an integer q such that
c D a q. This could be step Q1 in the know-show table.
We now have to prove that a certain integer q exists, so we ask the question,
“How do we prove that this integer exists?” When we are at such a stage in the
backward process of a proof, we usually turn to what is known in order to prove
that the object exists or to find or construct the object we are trying to prove exists.
We often say that we try to “construct” the object or at least prove it exists from
the known information. So at this point, we go to the forward part of the proof to
try to prove that there exists an integer q such that c D a q.
The forward question we ask is, “What can we conclude from the facts that
a j b and b j c?” Again, using the definition, we know that there exist integers s

and t such that b D a s and c D b t. This could be step P1 in the know-show
table.
The key now is to determine how to get from P1 to Q1. That is, can we use the
conclusions that the integers s and t exist in order to prove that the integer q (from
the backward process) exists. Using the equation b D a s, we can substitute a s
for b in the second equation, c D b t. This gives
c D b t
D .a s/ t
D a.s t/:
The last step used the associative property of multiplication. (See Table 1.2 on
page 18.) This shows that c is equal to a times some integer. (This is because s t
is an integer by the closure property for integers.) So although we did not use the
letter q, we have arrived at step Q1. The completed know-show table follows.
Step Know Reason
P a; b; c 2 Z, a ¤ 0, b ¤ 0, a j b
and b j c
Hypothesis
P1 .9s 2 Z/ .b D a s/
.9t 2 Z/ .c D b t/
Definition of “divides”
P 2 c D .a s/ t Substitution for b
P 3 c D a .s t/ Associative property of
multiplication
Q1 .9q 2 Z/ .c D a q/ Step P 3 and the closure
properties of the integers
Q a j c Definition of “divides”
Notice the similarities between what we did for this proof and many of the proofs
about even and odd integers we constructed in Section 1.2. When we try to prove
that a certain object exists, we often use what is called the construction method
for a proof. The appearance of an existential quantifier in the show (or backward)
portion of the proof is usually the indicator to go to what is known in order to prove
the object exists.
We can now report the news by writing a formal proof.
Theorem 3.1. Let a, b, and c be integers with a ¤ 0 and b ¤ 0. If a divides b
and b divides c, then a divides c.

Proof. We assume that a, b, and c are integers with a ¤ 0 and b ¤ 0. We further
assume that a divides b and that b divides c. We will prove that a divides c.
Since a divides b and b divides c, there exist integers s and t such that
b D a s; and (1)
c D b t: (2)
We can now substitute the expression for b from equation (1) into equation (2).
This gives
c D .a s/ t:
Using the associate property for multiplication, we can rearrange the right side of
the last equation to obtain
c D a .s t/:
Because both s and t are integers, and since the integers are closed under multi-
plication, we know that s t 2 Z. Therefore, the previous equation proves that a
divides c. Consequently, we have proven that whenever a, b, and c are integers
with a ¤ 0 and b ¤ 0 such that a divides b and b divides c, then a divides c.
Writing Guidelines for Equation Numbers
We wrote the proof for Theorem 3.1 according to the guidelines introduced in
Section 1.2, but a new element that appeared in this proof was the use of equation
numbers. Following are some guidelines that can be used for equation numbers.
If it is necessary to refer to an equation later in a proof, that equation should
be centered and displayed. It should then be given a number. The number for the
equation should be written in parentheses on the same line as the equation at the
right-hand margin as in shown in the following example.
Since x is an odd integer, there exists an integer n such that
x D 2n C 1: (1)
Later in the proof, there may be a line such as
Then, using the result in equation (1), we obtain ....

Notice that we did not number every equation in Theorem 3.1. We should only
number those equations we will be referring to later in the proof, and we should
only number equations when it is necessary. For example, instead of numbering
an equation, it is often better to use a phrase such as, “the previous equation
proves that ...” or “we can rearrange the terms on the right side of the previous
equation.” Also, note that the word “equation” is not capitalized when we are
referring to an equation by number. Although it may be appropriate to use a
capital “E,” the usual convention in mathematics is not to capitalize.
Progress Check 3.2 (A Property of Divisors)
1. Give at least four different examples of integers a, b, and c with a ¤ 0 such
that a divides b and a divides c.
2. For each example in Part (1), calculate the sum b C c. Does the integer a
divide the sum b C c?
3. Construct a know-show table for the following proposition: For all integers
a, b, and c with a ¤ 0, if a divides b and a divides c, then a divides .b Cc/.
Using Counterexamples
In Section 1.2 and so far in this section, our focus has been on proving statements
that involve universal quantifiers. However, another important skill for mathemati-
cians is to be able to recognize when a statement is false and then to be able to prove
that it is false. For example, suppose we want to know if the following proposition
is true or false.
For each integer n, if 5 divides n2
1

, then 5 divides .n 1/.
Suppose we start trying to prove this proposition. In the backward process, we
would say that in order to prove that 5 divides .n 1/, we can show that there
exists an integer k such that
Q1 W n 1 D 5k or n D 5k C 1:
For the forward process, we could say that since 5 divides n2 1

, we know that
there exists an integer m such that
P1 W n2
1 D 5m or n2
D 5m C 1:

The problem is that there is no straightforward way to use P1 to prove Q1. At
this point, it would be a good idea to try some examples for n and try to find
situations in which the hypothesis of the proposition is true. (In fact, this should
have been done before we started trying to prove the proposition.) The following
table summarizes the results of some of these explorations with values for n.
n n2
1 Does 5 divide n2
1

n 1 Does 5 divide .n 1/
1 0 yes 0 yes
2 3 no 1 no
3 8 no 2 no
4 15 yes 3 no
We can stop exploring examples now since the last row in the table provides an
example where the hypothesis is true and the conclusion is false. Recall from
Section 2.4 (see page 69) that a counterexample for a statement of the form
.8x 2 U / .P.x// is an element a in the universal set for which P.a/ is false. So
we have actually proved that the negation of the proposition is true.
When using a counterexample to prove a statement is false, we do not use the
term “proof” since we reserve a proof for proving a proposition is true. We could
summarize our work as follows:
Conjecture. For each integer n, if 5 divides n2
1

, then 5 divides
.n 1/.
The integer n D 4 is a counterexample that proves this conjecture is false.
Notice that when n D 4, n2 1 D 15 and 5 divides 15. Hence, the
hypothesis of the conjecture is true in this case. In addition, n 1 D 3
and 5 does not divide 3 and so the conclusion of the conjecture is false in
this case. Since this is an example where the hypothesis is true and the
conclusion is false, the conjecture is false.
As a general rule of thumb, anytime we are trying to decide if a proposition is true
or false, it is a good idea to try some examples first. The examples that are chosen
should be ones in which the hypothesis of the proposition is true. If one of these
examples makes the conclusion false, then we have found a counterexample and
we know the proposition is false. If all of the examples produce a true conclusion,
then we have evidence that the proposition is true and can try to write a proof.
Progress Check 3.3 (Using a Counterexample)
Use a counterexample to prove the following statement is false.
For all integers a and b, if 5 divides a or 5 divides b, then 5 divides .5aCb/.

Congruence
What mathematicians call congruence is a concept used to describe cycles in the
world of the integers. For example, the day of the week is a cyclic phenomenon in
that the day of the week repeats every seven days. The time of the day is a cyclic
phenomenon because it repeats every 12 hours if we use a 12-hour clock or every
24 hours if we use a 24-hour clock. We explored these two cyclic phenomena in
Preview Activity 2.
Similar to what we saw in Preview Activity 2, if it is currently Monday, then it
will be Wednesday 2 days from now, 9 days from now, 16 days from now, 23 days
from now, and so on. In addition, it was Wednesday 5 days ago, 12 days ago, 19
days ago, and so on. Using negative numbers for time in the past, we generate the
following list of numbers:
: : :; 19; 12; 5; 2; 9; 16; 23; : : ::
Notice that if we subtract any number in the list above from any other number
in that list, we will obtain a multiple of 7. For example,
16 2 D 14 D 7 2
. 5/ .9/ D 14 D 7 . 2/
16 . 12/ D 28 D 7 4:
Using the concept of congruence, we would say that all the numbers in this
list are congruent modulo 7, but we first have to define when two numbers are
congruent modulo some natural number n.
Definition. Let n 2 N. If a and b are integers, then we say that a is congruent
to b modulo n provided that n divides a b. A standard notation for this is
a b .mod n/. This is read as “a is congruent to b modulo n” or “a is
congruent to b mod n.”
Notice that we can use the definition of divides to say that n divides .a b/ if and
only if there exists an integer k such that a b D nk. So we can write
a b .mod n/ means .9k 2 Z/ .a b D nk/ ; or
a b .mod n/ means .9k 2 Z/ .a D b C nk/ :

This means that in order to find integers that are congruent to b modulo n,
we only need to add multiples of n to b. For example, to find integers that are
congruent to 2 modulo 5, we add multiples of 5 to 2. This gives the following list:
: : : 13; 8; 3; 2; 7; 12; 17; : : ::
We can also write this using set notation and say that
fa 2 Zj a 2 .mod 5/g D f: : : 13; 8; 3; 2; 7; 12; 17; : : :g:
Progress Check 3.4 (Congruence Modulo 8)
1. Determine at least eight different integers that are congruent to 5 modulo 8.
2. Use set builder notation and the roster method to specify the set of all integers
that are congruent to 5 modulo 8.
3. Choose two integers that are congruent to 5 modulo 8 and add them. Then
repeat this for at least five other pairs of integers that are congruent to 5
modulo 8.
4. Explain why all of the sums that were obtained in Part (3) are congruent to
2 modulo 8.
We will study the concept of congruence modulo n in much more detail later
in the text. For now, we will work with the definition of congruence modulo n in
the context of proofs. For example, all of the examples used in Progress Check 3.4
should provide evidence that the following proposition is true.
Proposition 3.5. For all integers a and b, if a 5 .mod 8/ and b 5 .mod 8/,
then .a C b/ 2 .mod 8/.
Progress Check 3.6 (Proving Proposition 3.5)
We will use “backward questions” and “forward questions” to help construct a
proof for Proposition 3.5. So, we might ask, “How do we prove that .a C b/
2 .mod 8/?” One way to answer this is to use the definition of congruence and
state that .a C b/ 2 .mod 8/ provided that 8 divides .a C b 2/.
1. Use the definition of divides to determine a way to prove that 8 divides .a C
b 2/.

We now turn to what we know and ask, “What can we conclude from the as-
sumptions that a 5 .mod 8/ and b 5 .mod 8/.” We can again use the
definition of congruence and conclude that 8 divides .a 5/ and 8 divides .b 5/.
2. Use the definition of divides to make conclusions based on the facts that 8
divides .a 5/ and 8 divides .b 5/.
3. Solve an equation from part (2) for a and for b.
4. Use the results from part (3) to prove that 8 divides .a C b 2/.
5. Write a proof for Proposition 3.5.
Additional Writing Guidelines
We will now be writing many proofs, and it is important to make sure we write
according to accepted guidelines so that our proofs may be understood by oth-
ers. Some writing guidelines were introduced in Chapter 1. The first four writing
guidelines given below can be considered general guidelines, and the last three can
be considered as technical guidelines specific to writing in mathematics.
1. Know your audience. Every writer should have a clear idea of the intended
audience for a piece of writing. In that way, the writer can give the right
amount of information at the proper level of sophistication to communicate
effectively. This is especially true for mathematical writing. For example, if
a mathematician is writing a solution to a textbook problem for a solutions
manual for instructors, the writing would be brief with many details omitted.
However, if the writing was for a students’ solution manual, more details
would be included.
2. Use complete sentences and proper paragraph structure. Good grammar
is an important part of any writing. Therefore, conform to the accepted rules
of grammar. Pay careful attention to the structure of sentences. Write proofs
using complete sentences but avoid run-on sentences. Also, do not forget
punctuation, and always use a spell checker when using a word processor.
3. Keep it simple. It is often difficult to understand a mathematical argument
no matter how well it is written. Do not let your writing help make it more
difficult for the reader. Use simple, declarative sentences and short para-
graphs, each with a simple point.

4. Write a first draft of your proof and then revise it. Remember that a proof
is written so that readers are able to read and understand the reasoning in the
proof. Be clear and concise. Include details but do not ramble. Do not be
satisfied with the first draft of a proof. Read it over and refine it. Just like
any worthwhile activity, learning to write mathematics well takes practice
and hard work. This can be frustrating. Everyone can be sure that there will
be some proofs that are difficult to construct, but remember that proofs are a
very important part of mathematics. So work hard and have fun.
5. Do not use for multiplication or ˆ for exponents. Leave this type of
notation for writing computer code. The use of this notation makes it difficult
for humans to read. In addition, avoid using = for division when using a
complex fraction.
For example, it is very difficult to read x3 3x2 C 1=2

=.2x=3 7/; the
fraction
x3
3x2
C
1
2
2x
3
7
is much easier to read.
6. Do not use a mathematical symbol at the beginning of a sentence. For
example, we should not write, “Let n be an integer. n is an odd integer
provided that ....” Many people find this hard to read and often have to re-
read it to understand it. It would be better to write, “An integer n is an odd
integer provided that ....”
7. Use English and minimize the use of cumbersome notation. Do not use
the special symbols for quantifiers 8 (for all), 9 (there exists), Ö (such that),
or ) (therefore) in formal mathematical writing. It is often easier to write,
and usually easier to read, if the English words are used instead of the sym-
bols. For example, why make the reader interpret
.8x 2 R/ .9y 2 R/ .x C y D 0/
or, more succinctly (if appropriate),

?
1. Prove each of the following statements:
(a) For all integers a, b, and c with a ¤ 0, if a j b and a j c, then
a j .b c/.
(b) For each n 2 Z, if n is an odd integer, then n3 is an odd integer.
(c) For each integer a, if 4 divides .a 1/, then 4 divides a2
1

.
2. For each of the following, use a counterexample to prove the statement is
false.
? (a) For each odd natural number n, if n 3, then 3 divides n2 1

.
(b) For each natural number n, .3 2n
C 2 3n
C 1/ is a prime number.
(c) For all real numbers x and y,
p
x2 C y2 2xy.
?
(d) For each integer a, if 4 divides a2
1

, then 4 divides .a 1/.
3. Determine if each of the following statements is true or false. If a statement
is true, then write a formal proof of that statement, and if it is false, then
provide a counterexample that shows it is false.
(a) For all integers a, b, and c with a ¤ 0, if a j b, then a j .bc/.
?
(b) For all integers a and b with a ¤ 0, if 6 j .ab/, then 6 j a or 6 j b.
(c) For all integers a, b, and c with a ¤ 0, if a divides .b 1/ and a
divides .c 1/, then a divides .bc 1/.
? (d) For each integer n, if 7 divides n2 4

, then 7 divides .n 2/.
? (e) For every integer n, 4n2 C 7n C 6 is an odd integer.
?
(f) For every odd integer n, 4n2
C 7n C 6 is an odd integer.
? (g) For all integers a, b, and d with d ¤ 0, if d divides both a b and
a C b, then d divides a.
(h) For all integers a, b, and c with a ¤ 0, if a j .bc/, then a j b or a j c.
? 4. (a) If x and y are integers and xy D 1, explain why x D 1 or x D 1.
(b) Is the following proposition true or false?
For all nonzero integers a and b, if a j b and b j a, then a D ˙b.
?
5. Prove the following proposition:

Let a be an integer. If there exists an integer n such that
a j .4n C 3/ and a j .2n C 1/, then a D 1 or a D 1.
Hint: Use the fact that the only divisors of 1 are 1 and 1.
6. Determine if each of the following statements is true or false. If a statement
is true, then write a formal proof of that statement, and if it is false, then
provide a counterexample that shows it is false.
(a) For each integer a, if there exists an integer n such that a divides .8nC
7/ and a divides .4n C 1/, then a divides 5.
(b) For each integer a, if there exists an integer n such that a divides .9nC
5/ and a divides .6n C 1/, then a divides 7.
(c) For each integer n, if n is odd, then 8 divides n4 C 4n2 C 11

.
(d) For each integer n, if n is odd, then 8 divides n4 C n2 C 2n

.
7. Let a be an integer and let n 2 N.
(a) Prove that if a 0 .mod n/, then n j a.
(b) Prove that if n j a, then a 0 .mod n/.
?
8. Let a and b be integers. Prove that if a 2 .mod 3/ and b 2 .mod 3/,
then
(a) a C b 1 .mod 3/; (b) a b 1 .mod 3/.
9. Let a and b be integers. Prove that if a 7 .mod 8/ and b 3 .mod 8/,
then:
(a) a C b 2 .mod 8/; (b) a b 5 .mod 8/.
10. Determine if each of the following propositions is true or false. Justify each
conclusion.
(a) For all integers a and b, if ab 0 .mod 6/, then a 0 .mod 6/ or
b .mod 6/.
(b) For each integer a, if a 2 .mod 8/, then a2 4 .mod 8/.
(c) For each integer a, if a2
4 .mod 8/, then a 2 .mod 8/.

11. Let n be a natural number. Prove each of the following:
?
(a) For every integer a, a a .mod n/.
This is called the reflexive property of congruence modulo n.
? (b) For all integers a and b, if a b .mod n/, then b a .mod n/.
This is called the symmetric property of congruence modulo n.
(c) For all integers a, b, and c, if a b .mod n/ and b c .mod n/,
then a c .mod n/.
This is called the transitive property of congruence modulo n.
?
12. Let n be a natural number and let a, b, c, and d be integers. Prove each of
the following.
(a) If a b .mod n/ and c d .mod n/, then
.a C c/ .b C d/ .mod n/.
(b) If a b .mod n/ and c d .mod n/, then ac bd .mod n/.
13. (a) Let a, b, and c be real numbers with a ¤ 0. Explain how to use
a part of the quadratic formula (called the discriminant) to determine
if the quadratic equation ax2
C bx C c D 0 has two real number
solutions, one real number solution, or no real number solutions. (See
Exercise (11) in Section 1.2 for a statement of the quadratic formula.)
(b) Prove that if a, b, and c are real numbers for which a 0 and c 0,
then one solution of the quadratic equation ax2 C bx C c D 0 is a pos-
itive real number.
(c) Prove that if a, b, and c are real numbers, if a ¤ 0, b 0 and
b
2

p
ac, then the quadratic equation ax2 C bx C c D 0 has no
real number solution.
14. Let h and k be real numbers and let r be a positive number. The equation for
a circle whose center is at the point .h; k/ and whose radius is r is
.x h/2
C .y k/2
D r2
:
We also know that if a and b are real numbers, then
The point .a; b/ is inside the circle if .a h/2
C .b k/2
r2
.
The point .a; b/ is on the circle if .a h/2
C .b k/2
D r2
.

The point .a; b/ is outside the circle if .a h/2
C .b k/2
r2.
Prove that all points on or inside the circle whose equation is .x 1/2
C
.y 2/2
D 4 are inside the circle whose equation is x2 C y2 D 26.
15. Let r be a positive real number. The equation for a circle of radius r whose
center is the origin is x2
C y2
D r2
.
(a) Use implicit differentiation to determine
dy
dx
.
(b) Let .a; b/ be a point on the circle with a ¤ 0 and b ¤ 0. Determine
the slope of the line tangent to the circle at the point .a; b/.
(c) Prove that the radius of the circle to the point .a; b/ is perpendicular to
the line tangent to the circle at the point .a; b/. Hint: Two lines (neither
of which is horizontal) are perpendicular if and only if the products of
their slopes is equal to 1.
16. Determine if each of the following statements is true or false. Provide a
counterexample for statements that are false and provide a complete proof
for those that are true.
(a) For all real numbers x and y,
p
xy
x C y
2
.
(b) For all real numbers x and y, xy

x C y
2
2
.
(c) For all nonnegative real numbers x and y,
p
xy
x C y
2
.
17. Use one of the true inequalities in Exercise (16) to prove the following propo-
sition.
For each real number a, the value of x that gives the maximum value
of y D x .a x/ is x D
a
2
.
18. (a) State the Pythagorean Theorem for right triangles.
The diagrams in Figure 3.1 will be used for the problems in this exer-
cise.
(b) In the diagram on the left, x is the length of a side of the equilateral
triangle and h is the length of an altitude of the equilateral triangle.
The labeling in the diagram shows the fact that the altitude intersects
the base of the equilateral triangle at the midpoint of the base. Use the

a
c
b
E
F
D
B
C
A
x x
x/2
x/2
h
Figure 3.1: Diagrams for Exercise (18)
Pythagorean Theorem to prove that the area of this equilateral triangle
is
p
3
4
x2.
(c) In the diagram on the right, 4ABC is a right triangle. In addition,
there has been an equilateral triangle constructed on each side of this
right triangle. Prove that the area of the equilateral triangle on the
hypotenuse is equal to the sum of the areas of the equilateral triangles
constructed on the other two sides of the right triangle.
19. Evaluation of proofs
This type of exercise will appear frequently in the book. In each case, there
is a proposed proof of a proposition. However, the proposition may be true
or may be false.
If a proposition is false, the proposed proof is, of course, incorrect. In
this situation, you are to find the error in the proof and then provide a
counterexample showing that the proposition is false.
If a propositionis true, the proposed proof may still be incorrect. In this
case, you are to determine why the proof is incorrect and then write a
correct proof using the writing guidelines that have been presented in
this book.
If a proposition is true and the proof is correct, you are to decide if the
proof is well written or not. If it is well written, then you simply must
indicate that this is an excellent proof and needs no revision. On the
other hand, if the proof is not well written, then you must then revise

the proof so by writing it according to the guidelines presented in this
text.
(a) Proposition. If m is an even integer, then .5m C 4/ is an even integer.
Proof. We see that 5m C 4 D 10n C 4 D 2 .5n C 2/. Therefore,
.5m C 4/ is an even integer.
(b) Proposition. For all real numbers x and y, if x ¤ y, x 0, and
y 0, then
x
y
C
y
x
2.
Proof. Since x and y are positive real numbers, xy is positive and we
can multiply both sides of the inequality by xy to obtain

x
y
C
y
x

xy 2 xy
x2
C y2
2xy:
By combining all terms on the left side of the inequality, we see that
x2 2xy C y2 0 and then by factoring the left side, we obtain
.x y/2
0. Since x ¤ y, .x y/ ¤ 0 and so .x y/2
0. This
proves that if x ¤ y, x 0, and y 0, then
x
y
C
y
x
2.
(c) Proposition. For all integers a, b, and c, if a j .bc/, then a j b or a j c.
Proof. We assume that a, b, and c are integers and that a divides bc.
So, there exists an integer k such that bc D ka. We now factor k as
k D mn, where m and n are integers. We then see that
bc D mna:
This means that b D ma or c D na and hence, a j b or a j c.
(d) Proposition. For all positive integers a, b, and c,

ab
c
D a.bc /
.
This proposition is false as is shown by the following counterexample:
If we let a D 2, b D 3, and c D 2, then

ab
c
D a.bc /
23
2
D 2.32
/
82
D 29
64 ¤ 512

20. Congruence Modulo 6.
(a) Find several integers that are congruent to 5 modulo 6 and then square
each of these integers.
(b) For each integer m from Part (20a), determine an integer k so that
0 k 6 and m2 k .mod 6/. What do you observe?
(c) Based on the work in Part (20b), complete the following conjecture:
For each integer m, if m 5 .mod 6/, then ....
(d) Complete a know-show table for the conjecture in Part (20c) or write a
proof of the conjecture.
21. Pythagorean Triples. Three natural numbers a, b, and c with a b c are
called a Pythagorean triple provided that a2
C b2
D c2
. See Exercise (13)
on page 29 in Section 1.2. Three natural numbers are called consecutive
natural numbers if they can be written in the form m, m C 1, and m C 2,
where m is a natural number.
(a) Determine all Pythagorean triples consisting of three consecutive nat-
ural numbers. (State a theorem and prove it.)
(b) Determine all Pythagorean triples that can be written in the form m,
m C 7, and m C 8, where m is a natural number. State a theorem and
prove it.
3.2 More Methods of Proof
Preview Activity 1 (Using the Contrapositive)
The following statement was proven in Exercise (3c) on page 27 in Section 1.2.
If n is an odd integer, then n2
is an odd integer.
Now consider the following proposition:
For each integer n, if n2 is an odd integer, then n is an odd integer.
1. After examining several examples, decide whether you think this proposition
is true or false.

3.2. More Methods of Proof 103
2. Try completing the following know-show table for a direct proof of this
proposition. The question is, “Can we perform algebraic manipulations to
get from the ‘know’ portion of the table to the ‘show’ portion of the table?”
Be careful with this! Remember that we are working with integers and we
want to make sure that we can end up with an integer q as stated in Step Q1.
Step Know Reason
P n2 is an odd integer. Hypothesis
P1 .9k 2 Z/ n2
D 2k C 1

Definition of “odd integer”
:
:
:
:
:
:
:
:
:
Q1 .9q 2 Z/ .n D 2q C 1/
Q n is an odd integer. Definition of “odd integer”
Step Show Reason
Recall that the contrapositive of the conditional statement P ! Q is the condi-
tional statement :Q ! :P . We have seen in Section 2.2 that the contrapositive
of a conditional statement is logically equivalent to the conditional statement. (It
might be a good idea to review Preview Activity 2 from Section 2.2 on page 44.)
Consider the following proposition once again:
For each integer n, if n2 is an odd integer, then n is an odd integer.
3. Write the contrapositive of this conditional statement. Remember that “not
odd” means “even.”
4. Complete a know-show table for the contrapositive statement from Part (3).
5. By completing the proof in Part (4), have you proven the given proposition?
That is, have you proven that if n2 is an odd integer, then n is an odd integer?
Explain.
Preview Activity 2 (A Biconditional Statement)
1. In Exercise (4a) from Section 2.2, we constructed a truth table to prove that
the biconditional statement, P $ Q, is logically equivalent to .P ! Q/ ^
.Q ! P /. Complete this exercise if you have not already done so.
2. Suppose that we want to prove a biconditionalstatement of the form P $ Q.
Explain a method for completing this proof based on the logical equivalency
in part (1).

3. Let n be an integer. Assume that we have completed the proofs of the fol-
lowing two statements:
If n is an odd integer, then n2 is an odd integer.
If n2 is an odd integer, then n is an odd integer.
(See Exercise (3c) from Section 1.2 and Preview Activity 1.) Have we com-
pleted the proof of the following proposition?
For each integer n, n is an odd integer if and only if n2
is an odd integer.
Explain.
Review of Direct Proofs
In Sections 1.2 and 3.1, we studied direct proofs of mathematical statements. Most
of the statements we prove in mathematics are conditional statements that can be
written in the form P ! Q. A direct proof of a statement of the form P ! Q
is based on the definition that a conditional statement can only be false when the
hypothesis, P , is true and the conclusion, Q, is false. Thus, if the conclusion is
true whenever the hypothesis is true, then the conditional statement must be true.
So, in a direct proof,
We start by assuming that P is true.
From this assumption, we logically deduce that Q is true.
We have used the so-called forward and backward method to discover how to log-
ically deduce Q from the assumption that P is true.
Proof Using the Contrapositive
As we saw in Preview Activity 1, it is sometimes difficult to construct a direct proof
of a conditional statement. This is one reason we studied logical equivalencies in
Section 2.2. Knowing that two expressions are logically equivalent tells us that if
we prove one, then we have also proven the other. In fact, once we know the truth
value of a statement, then we know the truth value of any other statement that is
logically equivalent to it.
One of the most useful logical equivalencies in this regard is that a conditional
statement P ! Q is logically equivalent to its contrapositive, :Q ! :P . This

means that if we prove the contrapositive of the conditional statement, then we
have proven the conditional statement. The following are some important points to
remember.
A conditional statement is logically equivalent to its contrapositive.
Use a direct proof to prove that :Q ! :P is true.
Caution: One difficulty with this type of proof is in the formation of correct
negations. (We need to be very careful doing this.)
We might consider using a proof by contrapositive when the statements P
and Q are stated as negations.
Writing Guidelines
One of the basic rules of writing mathematical proofs is to keep the reader in-
formed. So when we prove a result using the contrapositive, we indicate this within
the first few lines of the proof. For example,
We will prove this theorem by proving its contrapositive.
We will prove the contrapositive of this statement.
In addition, make sure the reader knows the status of every assertion that you
make. That is, make sure you state whether an assertion is an assumption of the
theorem, a previously proven result, a well-known result, or something from the
reader’s mathematical background. Following is a completed proof of a statement
from Preview Activity 1.
Theorem 3.7. For each integer n, if n2 is an even integer, then n is an even integer.
Proof. We will prove this result by proving the contrapositive of the statement,
which is
For each integer n, if n is an odd integer, then n2 is an odd integer.
However, in Theorem 1.8 on page 21, we have already proven that if x and y are
odd integers, then x y is an odd integer. So using x D y D n, we can conclude
that if n is an odd integer, then n n, or n2
, is an odd integer. We have thus proved
the contrapositive of the theorem, and consequently, we have proved that if n2 is
an even integer, then n is an even integer.

Using Other Logical Equivalencies
As was noted in Section 2.2, there are several different logical equivalencies. For-
tunately, there are only a small number that we often use when trying to write
proofs, and many of these are listed in Theorem 2.8 at the end of Section 2.2.
We will illustrate the use of one of these logical equivalencies with the following
proposition:
For all real numbers a and b, if a ¤ 0 and b ¤ 0, then ab ¤ 0.
First, notice that the hypothesis and the conclusion of the conditional statement are
stated in the form of negations. This suggests that we consider the contrapositive.
Care must be taken when we negate the hypothesis since it is a conjunction. We
use one of De Morgan’s Laws as follows:
: .a ¤ 0 ^ b ¤ 0/ .a D 0/ _ .b D 0/ :
Progress Check 3.8 (Using Another Logical Equivalency)
1. In English, write the contrapositive of, “For all real numbers a and b, if
a ¤ 0 and b ¤ 0, then ab ¤ 0.”
The contrapositive is a conditional statement in the form X ! .Y _ Z/. The
difficulty is that there is not much we can do with the hypothesis .ab D 0/ since
we know nothing else about the real numbers a and b. However, if we knew that a
was not equal to zero, then we could multiply both sides of the equation ab D 0 by
1
a
. This suggests that we consider using the following logical equivalency based
on a result in Theorem 2.8 on page 48:
X ! .Y _ Z/ .X ^ :Y / ! Z:
2. In English, use this logical equivalency, to write a statement that is logically
equivalent to the contrapositive from Part (1).
The logical equivalency in Part (2) makes sense because if we are trying to
prove Y _ Z, we only need to prove that at least one of Y or Z is true. So the idea
is to prove that if Y is false, then Z must be true.
3. Use the ideas presented in the progress check to complete the proof of the
following proposition.

Proposition 3.9. For all real numbers a and b, if a ¤ 0 and b ¤ 0, then
ab ¤ 0.
Proof. We will prove the contrapositive of this proposition, which is
For all real numbers a and b, if ab D 0, then a D 0 or b D 0.
This contrapositive, however, is logically equivalent to the following:
For all real numbers a and b, if ab D 0 and a ¤ 0, then b D 0.
To prove this, we let a and b be real numbers and assume that ab D 0 and a ¤ 0.
We can then multiply both sides of the equation ab D 0 by
1
a
. This gives
Now complete the proof.
:
:
:
Therefore, b D 0. This completes the proof of a statement that is logically
equivalent to the contrapositive, and hence, we have proven the proposition.
Proofs of Biconditional Statements
In Preview Activity 2, we used the following logical equivalency:
.P $ Q/ .P ! Q/ ^ .Q ! P / :
This logical equivalency suggests one method for proving a biconditionalstatement
written in the form “P if and only if Q.” This method is to construct separate
proofs of the two conditional statements P ! Q and Q ! P . For example, since
we have now proven each of the following:
For each integer n, if n is an even integer, then n2 is an even integer. (Exer-
cise (3c) on page 27 in Section 1.2)
For each integer n, if n2
is an even integer, then n is an even integer. (Theo-
rem 3.7)
we can state the following theorem.

Theorem 3.10. For each integer n, n is an even integer if and only if n2 is an even
integer.
Writing Guidelines
When proving a biconditional statement using the logical equivalency
.P $ Q/ .P ! Q/ ^ .Q ! P /, we actually need to prove two conditional
statements. The proof of each conditional statement can be considered as one of
two parts of the proof of the biconditional statement. Make sure that the start and
end of each of these parts is indicated clearly. This is illustrated in the proof of the
following proposition.
Proposition 3.11. Let x 2 R. The real number x equals 2 if and only if
x3
2x2
C x D 2.
Proof. We will prove this biconditional statement by proving the following two
conditional statements:
For each real number x, if x equals 2 , then x3 2x2 C x D 2.
For each real number x, if x3
2x2
C x D 2, then x equals 2.
For the first part, we assume x D 2 and prove that x3 2x2 C x D 2. We can
do this by substituting x D 2 into the expression x3
2x2
C x. This gives
x3
2x2
C x D 23
2 22

C 2
D 8 8 C 2
D 2:
This completes the first part of the proof.
For the second part, we assume that x3 2x2 C x D 2 and from this assumption,
we will prove that x D 2. We will do this by solving this equation for x. To do so,
we first rewrite the equation x3 2x2 C x D 2 by subtracting 2 from both sides:
x3
2x2
C x 2 D 0:
We can now factor the left side of this equation by factoring an x2
from the first
two terms and then factoring .x 2/ from the resulting two terms. This is shown

below.
x3
2x2
C x 2 D 0
x2
.x 2/ C .x 2/ D 0
.x 2/ x2
C 1

D 0
Now, in the real numbers, if a product of two factors is equal to zero, then one of
the factors must be zero. So this last equation implies that
x 2 D 0 or x2
C 1 D 0:
The equation x2
C 1 D 0 has no real number solution. So since x is a real
number, the only possibility is that x 2 D 0. From this we can conclude that x
must be equal to 2.
Since we have now proven both conditional statements, we have proven that
x D 2 if and only if x3 2x2 C x D 2.
Constructive Proofs
We all know how to solve an equation such as 3x C 8 D 23, where x is a real
number. To do so, we first add 8 to both sides of the equation and then divide
both sides of the resulting equation by 3. Doing so, we obtain the following result:
If x is a real number and 3x C 8 D 23, then x D 5.
Notice that the process of solving the equation actually does not prove that
x D 5 is a solution of the equation 3x C 8 D 23. This process really shows that if
there is a solution, then that solution must be x D 5. To show that this is a solution,
we use the process of substituting 5 for x in the left side of the equation as follows:
If x D 5, then
3x C 8 D 3 .5/ C 8 D 15 C 8 D 23:
This proves that x D 5 is a solution of the equation 3x C 8 D 23. Hence, we
have proven that x D 5 is the only real number solution of 3x C 8 D 23.
We can use this same process to show that any linear equation has a real number
solution. An equation of the form
ax C b D c;

where a, b, and c are real numbers with a ¤ 0, is called a linear equation in one
variable.
Proposition 3.12. If a, b, and c are real numbers with a ¤ 0, then the linear
equation ax C b D c has exactly one real number solution, which is x D
c b
a
.
Proof. Assume that a, b, and c are real numbers with a ¤ 0. We can solve the
linear equation ax C b D c by adding b to both sides of the equation and then
dividing both sides of the resulting equation by a .since a ¤ 0/, to obtain
x D
c b
a
:
This shows that if there is a solution, then it must be x D
c b
a
. We also see that
if x D
c b
a
, then
ax C b D a

c b
a

C b
D .c b/ C b
D c:
Therefore, the linear equation ax C b D c has exactly one real number solution
and the solution is x D
c b
a
.
The proof given for Proposition 3.12 is called a constructive proof. This is a
technique that is often used to prove a so-called existence theorem. The objective
of an existence theorem is to prove that a certain mathematical object exists. That
is, the goal is usually to prove a statement of the form
There exists an x such that P.x/.
For a constructive proof of such a proposition, we actually name, describe, or ex-
plain how to construct some object in the universe that makes P.x/ true. This is
what we did in Proposition 3.12 since in the proof, we actually proved that
c b
a
is a solution of the equation ax C b D c. In fact, we proved that this is the only
solution of this equation.

Nonconstructive Proofs
Another type of proof that is often used to prove an existence theorem is the so-
called nonconstructive proof. For this type of proof, we make an argument that an
object in the universal set that makes P.x/ true must exist but we never construct
or name the object that makes P.x/ true. The advantage of a constructive proof
over a nonconstructiveproof is that the constructive proof will yield a procedure or
algorithm for obtaining the desired object.
The proof of the Intermediate Value Theorem from calculus is an example of
a nonconstructiveproof. The Intermediate Value Theorem can be stated as follows:
If f is a continuous function on the closed interval Œa; b and if q is any real
number strictly between f.a/ and f.b/, then there exists a number c in the
interval .a; b/ such that f.c/ D q.
The Intermediate Value Theorem can be used to prove that a solution to some
equations must exist. This is shown in the next example.
Example 3.13 (Using the Intermediate Value Theorem)
Let x represent a real number. We will use the Intermediate Value Theorem to
prove that the equation x3 x C 1 D 0 has a real number solution.
To investigatesolutions of the equation x3 xC1 D 0, we will use the function
f.x/ D x3
x C 1:
Notice that f. 2/ D 5 and that f.0/ D 1. Since f. 2/ 0 and f.0/ 0, the
Intermediate Value Theorem tells us that there is a real number c between 2 and
0 such that f.c/ D 0. This means that there exists a real number c between 2
and 0 such that
c3
c C 1 D 0;
and hence c is a real number solution of the equation x3
x C 1 D 0. This proves
that the equation x3 x C 1 D 0 has at least one real number solution.
Notice that this proof does not tell us how to find the exact value of c. It does,
however, suggest a method for approximating the value of c. This can be done by
finding smaller and smaller intervals Œa; b such that f.a/ and f.b/ have opposite
signs.

?
1. Let n be an integer. Prove each of the following:
(a) If n is even, then n3
is even.
(b) If n3
is even, then n is even.
(c) The integer n is even if and only if n3 is an even integer.
(d) The integer n is odd if and only if n3 is an odd integer.
2. In Section 3.1, we defined congruence modulo n where n is a natural number.
If a and b are integers, we will use the notation a 6 b .mod n/ to mean that
a is not congruent to b modulo n.
?
(a) Write the contrapositive of the following conditional statement:
For all integers a and b, if a 6 0 .mod 6/ and b 6 0 .mod 6/,
then ab 6 0 .mod 6/.
(b) Is this statement true or false? Explain.
?
3. (a) Write the contrapositive of the following statement:
For all positive real numbers a and b, if
p
ab ¤
a C b
2
, then a ¤ b.
(b) Is this statement true or false? Prove the statement if it is true or provide
a counterexample if it is false.
?
(a) For each a 2 Z, if a 2 .mod 5/, then a2
4 .mod 5/.
(b) For each a 2 Z, if a2
4 .mod 5/, then a 2 .mod 5/.
(c) For each a 2 Z, a 2 .mod 5/ if and only if a2 4 .mod 5/.
5. Is the following proposition true or false?
For all integers a and b, if ab is even, then a is even or b is even.
Justify your conclusion by writing a proof if the proposition is true or by
providing a counterexample if it is false.
?
6. Consider the following proposition: For each integer a, a 3 .mod 7/ if
and only if a2 C 5a

3 .mod 7/.
(a) Write the proposition as the conjunction of two conditional statements.

(b) Determine if the two conditional statements in Part (a) are true or false.
If a conditional statement is true, write a proof, and if it is false, provide
a counterexample.
(c) Is the given proposition true or false? Explain.
7. Consider the following proposition: For each integer a, a 2 .mod 8/ if
and only if a2
C 4a

4 .mod 8/.
(a) Write the proposition as the conjunction of two conditional statements.
(b) Determine if the two conditional statements in Part (a) are true or false.
If a conditional statement is true, write a proof, and if it is false, provide
a counterexample.
(c) Is the given proposition true or false? Explain.
8. For a right triangle, suppose that the hypotenuse has length c feet and the
lengths of the sides are a feet and b feet.
(a) What is a formula for the area of this right triangle? What is an isosce-
les triangle?
(b) State the Pythagorean Theorem for right triangles.
?
(c) Prove that the right triangle described above is an isosceles triangle if
and only if the area of the right triangle is
1
4
c2.
?
9. A real number x is defined to be a rational number provided
there exist integers m and n with n ¤ 0 such that x D
m
n
:
A real number that is not a rational number is called an irrational number.
It is known that if x is a positive rational number, then there exist positive
integers m and n with n ¤ 0 such that x D
m
n
.
Is the following proposition true or false? Explain.
For each positive real number x, if x is irrational, then
p
x is irrational.
? 10. Is the following proposition true or false? Justify your conclusion.
For each integer n, n is even if and only if 4 divides n2.
11. Prove that for each integer a, if a2
1 is even, then 4 divides a2
1.

12. Prove that for all integers a and m, if a and m are the lengths of the sides of
a right triangle and m C 1 is the length of the hypotenuse, then a is an odd
integer.
If p; q 2 Q with p q, then there exists an x 2 Q with p x q.
14. Are the following propositions true or false? Justify your conclusion.
(a) There exist integers x and y such that 4x C 6y D 2.
(b) There exist integers x and y such that 6x C 15y D 2.
(c) There exist integers x and y such that 6x C 15y D 9.
? 15. Prove that there exists a real number x such that x3 4x2 D 7.
16. Let y1; y2; y3; y4 be real numbers. The mean, y, of these four numbers is
defined to be the sum of the four numbers divided by 4. That is,
y D
y1 C y2 C y3 C y4
4
:
Prove that there exists a yi with 1 i 4 such that yi y.
Hint: One way is to let ymax be the largest of y1; y2; y3; y4.
17. Let a and b be natural numbers such that a2
D b3
. Prove each of the
propositions in Parts (6a) through (6d). (The results of Exercise (1) and
Theorem 3.10 may be helpful.)
(a) If a is even, then 4 divides a.
? (b) If 4 divides a, then 4 divides b.
(c) If 4 divides b, then 8 divides a.
(d) If a is even, then 8 divides a.
(e) Give an example of natural numbers a and b such that a is even and
a2 D b3, but b is not divisible by 8.
?
Let a and b be integers with a ¤ 0. If a does not divide b, then the
equation ax3 C bx C .b C a/ D 0 does not have a solution that is a
natural number.
Hint: It may be necessary to factor a sum of cubes. Recall that
u3
C v3
D .u C v/ u2
uv C v2

:

19. Evaluation of Proofs
See the instructions for Exercise (19) on page 100 from Section 3.1.
(a) Proposition. If m is an odd integer, then .m C 6/ is an odd integer.
Proof. For m C 6 to be an odd integer, there must exist an integer n
such that
m C 6 D 2n C 1:
By subtracting 6 from both sides of this equation, we obtain
m D 2n 6 C 1
D 2 .n 3/ C 1:
By the closure properties of the integers, .n 3/ is an integer, and
hence, the last equation implies that m is an odd integer. This proves
that if m is an odd integer, then m C 6 is an odd integer.
(b) Proposition. For all integers m and n, if mn is an even integer, then m
is even or n is even.
Proof. For either m or n to be even, there exists an integer k such that
m D 2k or n D 2k. So if we multiply m and n, the product will contain
a factor of 2 and, hence, mn will be even.
20. Using a Logical Equivalency. Consider the following proposition:
Proposition. For all integers a and b, if 3 does not divide a and 3 does
not divide b, then 3 does not divide the product a b.
(a) Notice that the hypothesis of the proposition is stated as a conjunc-
tion of two negations (“3 does not divide a and 3 does not divide b”).
Also, the conclusion is stated as the negation of a sentence (“3 does
not divide the product a b.”). This often indicates that we should con-
sider using a proof of the contrapositive. If we use the symbolic form
.:Q ^ :R/ ! :P as a model for this proposition, what is P , what
is Q, and what is R?
(b) Write a symbolic form for the contrapositive of .:Q ^ :R/ ! :P .
(c) Write the contrapositive of the proposition as a conditional statement
in English.

We do not yet have all the tools needed to prove the proposition or its contra-
positive. However, later in the text, we will learn that the following proposi-
tion is true.
Proposition X. Let a be an integer. If 3 does not divide a, then there exist
integers x and y such that 3x C ay D 1.
(d) i. Find integers x and y guaranteed by Proposition X when a D 5.
ii. Find integers x and y guaranteed by Proposition X when a D 2.
iii. Find integers x and y guaranteed by Proposition X when a D 2.
(e) Assume that Proposition X is true and use it to help construct a proof
of the contrapositive of the given proposition. In doing so, you will
most likely have to use the logical equivalency P ! .Q _ R/
.P ^ :Q/ ! R.
3.3 Proof by Contradiction
Preview Activity 1 (Proof by Contradiction)
On page 40 in Section 2.1, we defined a tautology to be a compound state-
ment S that is true for all possible combinations of truth values of the component
statements that are part of S. We also defined contradiction to be a compound
statement that is false for all possible combinations of truth values of the compo-
nent statements that are part of S.
That is, a tautology is necessarily true in all circumstances, and a contradiction
is necessarily false in all circumstances.
1. Use truth tables to explain why .P _ :P / is a tautology and .P ^ :P / is a
contradiction.
Another method of proof that is frequently used in mathematics is a proof by
contradiction. This method is based on the fact that a statement X can only be true
or false (and not both). The idea is to prove that the statement X is true by showing
that it cannot be false. This is done by assuming that X is false and proving that this
leads to a contradiction. (The contradiction often has the form .R ^ :R/, where R
is some statement.) When this happens, we can conclude that the assumption that
the statement X is false is incorrect and hence X cannot be false. Since it cannot
be false, then X must be true.

3.3. Proof by Contradiction 117
A logical basis for the contradiction method of proof is the tautology
Œ:X ! C ! X;
where X is a statement and C is a contradiction. The following truth table estab-
lishes this tautology.
X C :X :X ! C .:X ! C / ! X
T F F T T
F F T F T
This tautology shows that if :X leads to a contradiction, then X must be true. The
previous truth table also shows that the statement :X ! C is logically equiva-
lent to X. This means that if we have proved that :X leads to a contradiction,
then we have proved statement X. So if we want to prove a statement X using a
proof by contradiction, we assume that :X is true and show that this leads to a
contradiction.
When we try to prove the conditional statement, “If P then Q” using a proof
by contradiction, we must assume that P ! Q is false and show that this leads to
a contradiction.
2. Use a truth table to show that : .P ! Q/ is logically equivalent to P ^:Q.
The preceding logical equivalency shows that when we assume that P ! Q is
false, we are assuming that P is true and Q is false. If we can prove that this leads
to a contradiction, then we have shown that : .P ! Q/ is false and hence that
P ! Q is true.
3. Give a counterexample to show that the following statement is false.
For each real number x,
1
x.1 x/
4.
4. When a statement is false, it is sometimes possible to add an assumption
that will yield a true statement. This is usually done by using a conditional
statement. So instead of working with the statement in (3), we will work with
a related statement that is obtained by adding an assumption (or assumptions)
to the hypothesis.
For each real number x, if 0 x 1, then
1
x.1 x/
4.

To begin a proof by contradiction for this statement, we need to assume the
negation of the statement. To do this, we need to negate the entire state-
ment, including the quantifier. Recall that the negation of a statement with
a universal quantifier is a statement that contains an existential quantifier.
(See Theorem 2.16 on page 67). With this in mind, carefully write down
all assumptions made at the beginning of a proof by contradiction for this
statement.
Preview Activity 2 (Constructing a Proof by Contradiction)
Proposition. For all real numbers x and y, if x ¤ y, x 0; and y 0, then
x
y
C
y
x
2:
To start a proof by contradiction, we assume that this statement is false; that is, we
assume the negation is true. Because this is a statement with a universal quantifier,
we assume that there exist real numbers x and y such that x ¤ y, x 0; y 0
and that
x
y
C
y
x
2: (Notice that the negation of the conditional sentence is a
conjunction.)
For this proof by contradiction, we will only work with the know column of a
know-show table. This is because we do not have a specific goal. The goal is to
obtain some contradiction, but we do not know ahead of time what that contradic-
tion will be. Using our assumptions, we can perform algebraic operations on the
inequality
x
y
C
y
x
2 (2)
until we obtain a contradiction.
1. Try the following algebraic operations on the inequality in (2). First, multi-
ply both sides of the inequality by xy, which is a positive real number since
x 0 and y 0. Then, subtract 2xy from both sides of this inequality and
finally, factor the left side of the resulting inequality.
2. Explain why the last inequality you obtained leads to a contradiction.
By obtaining a contradiction, we have proved that the proposition cannot be false,
and hence, must be true.

Writing Guidelines: Keep the Reader Informed
A very important piece of information about a proof is the method of proof to be
used. So when we are going to prove a result using the contrapositive or a proof by
contradiction, we indicate this at the start of the proof.
We will prove this result by proving the contrapositive of the statement.
We will prove this statement using a proof by contradiction.
We will use a proof by contradiction.
We have discussed the logic behind a proof by contradiction in the preview activ-
ities for this section. The basic idea for a proof by contradiction of a proposition
is to assume the proposition is false and show that this leads to a contradiction.
We can then conclude that the proposition cannot be false, and hence, must be
true. When we assume a proposition is false, we are, in effect, assuming that its
negation is true. This is one reason why it is so important to be able to write nega-
tions of propositions quickly and correctly. We will illustrate the process with the
proposition discussed in Preview Activity 1.
Proposition 3.14. For each real number x, if 0 x 1, then
1
x.1 x/
4.
Proof. We will use a proof by contradiction. So we assume that the proposition is
false, or that there exists a real number x such that 0 x 1 and
1
x.1 x/
4: (1)
We note that since 0 x 1, we can conclude that x 0 and that .1 x/ 0.
Hence, x.1 x/ 0 and if we multiply both sides of inequality (1) by x.1 x/,
we obtain
1 4x.1 x/:
We can now use algebra to rewrite the last inequality as follows:
1 4x 4x2
4x2
4x C 1 0
.2x 1/2
0
However, .2x 1/ is a real number and the last inequality says that a real number
squared is less than zero. This is a contradiction since the square of any real number

must be greater than or equal to zero. Hence, the proposition cannot be false, and
we have proved that for each real number x, if 0 x 1, then
1
x.1 x/
4.
Progress Check 3.15 (Starting a Proof by Contradiction)
One of the most important parts of a proof by contradiction is the very first part,
which is to state the assumptions that will be used in the proof by contradiction.
This usually involves writing a clear negation of the proposition to be proven. Re-
view De Morgan’s Laws and the negation of a conditional statement in Section 2.2.
(See Theorem 2.8 on page 48.) Also, review Theorem 2.16 (on page 67) and then
write a negation of each of the following statements. (Remember that a real number
is “not irrational” means that the real number is rational.)
1. For each real number x, if x is irrational, then 3
p
x is irrational.
2. For each real number x,

x C
p
2

is irrational or

x C
p
2

is irrational.
3. For all integers a and b, if 5 divides ab, then 5 divides a or 5 divides b.
4. For all real numbers a and b, if a 0 and b 0, then
2
a
C
2
b
¤
4
a C b
.
Important Note
A proof by contradiction is often used to prove a conditional statement
P ! Q when a direct proof has not been found and it is relatively easy to form the
negation of the proposition. The advantage of a proof by contradiction is that we
have an additional assumption with which to work (since we assume not only P but
also :Q). The disadvantage is that there is no well-defined goal to work toward.
The goal is simply to obtain some contradiction. There usually is no way of telling
beforehand what that contradiction will be, so we have to stay alert for a possible
absurdity. Thus, when we set up a know-show table for a proof by contradiction,
we really only work with the know portion of the table.
Progress Check 3.16 (Exploration and a Proof by Contradiction)
For each integer n, if n 2 .mod 4/, then n 6 3 .mod 6/.

1. Determine at least five different integers that are congruent to 2 modulo 4,
and determine at least five different integers that are congruent to 3 modulo
6. Are there any integers that are in both of these lists?
2. For this proposition, why does it seem reasonable to try a proof by contra-
diction?
3. For this proposition,state clearly the assumptions that need to be made at the
beginning of a proof by contradiction, and then use a proof by contradiction
to prove this proposition.
Proving that Something Does Not Exist
In mathematics, we sometimes need to prove that something does not exist or that
something is not possible. Instead of trying to construct a direct proof, it is some-
times easier to use a proof by contradiction so that we can assume that the some-
thing exists. For example, suppose we want to prove the following proposition:
Proposition 3.17. For all integers x and y, if x and y are odd integers, then there
does not exist an integer z such that x2 C y2 D z2.
Notice that the conclusion involves trying to prove that an integer with a certain
property does not exist. If we use a proof by contradiction, we can assume that such
an integer z exists. This gives us more with which to work.
Progress Check 3.18 Complete the following proof of Proposition 3.17:
Proof. We will use a proof by contradiction. So we assume that there exist integers
x and y such that x and y are odd and there exists an integer z such that x2
Cy2
D
z2. Since x and y are odd, there exist integers m and n such that x D 2m C 1 and
y D 2n C 1.
1. Use the assumptions that x and y are odd to prove that x2
C y2
is even and
hence, z2 is even. (See Theorem 3.7 on page 105.)
We can now conclude that z is even. (See Theorem 3.7 on page 105.) So there
exists an integer k such that z D 2k. If we substitute for x, y, and z in the
equation x2 C y2 D z2, we obtain
.2m C 1/2
C .2n C 1/2
D .2k/2
:

2. Use the previous equation to obtain a contradiction. Hint: One way is to use
algebra to obtain an equation where the left side is an odd integer and the
right side is an even integer.
Rational and Irrational Numbers
One of the most important ways to classify real numbers is as a rational number
or an irrational number. Following is the definition of rational (and irrational)
numbers given in Exercise (9) from Section 3.2.
Definition. A real number x is defined to be a rational number provided that
there exist integers m and n with n ¤ 0 such that x D
m
n
. A real number that
is not a rational number is called an irrational number.
This may seem like a strange distinctionbecause most people are quite familiar
with the rational numbers (fractions) but the irrational numbers seem a bit unusual.
However, there are many irrational numbers such as
p
2,
p
3, 3
p
2, , and the
number e. We are discussing these matters now because we will soon prove that
p
2 is irrational in Theorem 3.20.
We use the symbol Q to stand for the set of rational numbers. There is no
standard symbol for the set of irrational numbers. Perhaps one reason for this is
because of the closure properties of the rational numbers. We introduced closure
properties in Section 1.1, and the rational numbers Q are closed under addition,
subtraction, multiplication, and division by nonzero rational numbers. This means
that if x; y 2 Q, then
x C y, x y, and xy are in Q; and
If y ¤ 0, then
x
y
is in Q.
The basic reasons for these facts are that if we add, subtract, multiply, or divide
two fractions, the result is a fraction. One reason we do not have a symbol for
the irrational numbers is that the irrational numbers are not closed under these
operations. For example, we will prove that
p
2 is irrational in Theorem 3.20. We
then see that
p
2
p
2 D 2 and
p
2
p
2
D 1;

which shows that the product of irrational numbers can be rational and the quotient
of irrational numbers can be rational.
It is also important to realize that every integer is a rational number since any
integer can be written as a fraction. For example, we can write 3 D
3
1
. In general,
if n 2 Z, then n D
n
1
, and hence, n 2 Q.
Because the rational numbers are closed under the standard operations and the
definition of an irrational number simply says that the number is not rational, we
often use a proof by contradiction to prove that a number is irrational. This is
illustrated in the next proposition.
Proposition 3.19. . For all real numbers x and y, if x is rational and x ¤ 0 and
y is irrational, then x y is irrational.
Proof. We will use a proof by contradiction. So we assume that there exist real
numbers x and y such that x is rational, y is irrational, and x y is rational. Since
x ¤ 0, we can divide by x, and since the rational numbers are closed under di-
vision by nonzero rational numbers, we know that
1
x
2 Q. We now know that
x y and
1
x
are rational numbers and since the rational numbers are closed under
multiplication, we conclude that
1
x
.xy/ 2 Q:
However,
1
x
.xy/ D y and hence, y must be a rational number. Since a real num-
ber cannot be both rational and irrational, this is a contradiction to the assumption
that y is irrational. We have therefore proved that for all real numbers x and y, if
x is rational and x ¤ 0 and y is irrational, then x y is irrational.
The Square Root of 2 Is an Irrational Number
The proof that the square root of 2 is an irrational number is one of the classic
proofs in mathematics, and every mathematics student should know this proof.
This is why we will be doing some preliminary work with rational numbers and
integers before completing the proof. The theorem we will be proving can be stated
as follows:

Theorem 3.20. If r is a real number such that r2 D 2, then r is an irrational
number.
This is stated in the form of a conditional statement, but it basically means that
p
2 is irrational (and that
p
2 is irrational). That is,
p
2 cannot be written as a
quotient of integers with the denominator not equal to zero.
In order to complete this proof, we need to be able to work with some basic
facts that follow about rational numbers and even integers.
1. Each integer m is a rational number since m can be written as m D
m
1
.
2. Notice that
2
3
D
4
6
, since
4
6
D
2 2
3 2
D
2
2

2
3
D
2
3
We can also show that
15
12
D
5
4
,
10
8
D
5
4
, and
30
16
D
15
8
Item (2) was included to illustrate the fact that a rational number can be written
as a fraction in “lowest terms” with a positive denominator. This means that any
rational number can be written as a quotient
m
n
, where m and n are integers, n 0,
and m and n have no common factor greater than 1.
3. If n is an integer and n2
is even, what can be conclude about n. Refer to
Theorem 3.7 on page 105.
In a proof by contradiction of a conditional statement P ! Q, we assume the
negation of this statement or P ^ :Q. So in a proof by contradiction of Theo-
rem 3.20, we will assume that r is a real number, r2
D 2, and r is not irrational
(that is, r is rational).
Theorem 3.20. If r is a real number such that r2 D 2, then r is an irrational
number.
Proof. We will use a proof by contradiction. So we assume that the statement of
the theorem is false. That is, we assume that

r is a real number, r2 D 2, and r is a rational number.
Since r is a rational number, there exist integers m and n with n 0 such that
r D
m
n
and m and n have no common factor greater than 1. We will obtain a contradiction
by showingthat m and n must both be even. Squaring both sides of the last equation
and using the fact that r2
D 2, we obtain
2 D
m2
n2
m2
D 2n2
: (1)
Equation (1) implies that m2 is even, and hence, by Theorem 3.7, m must be an
even integer. This means that there exists an integer p such that m D 2p. We can
now substitute this into equation (1), which gives
.2p/2
D 2n2
4p2
D 2n2
: (2)
We can divide both sides of equation (2) by 2 to obtain n2 D 2p2. Consequently,
n2
is even and we can once again use Theorem 3.7 to conclude that m is an even
integer.
We have now established that both m and n are even. This means that 2 is a
common factor of m and n, which contradicts the assumption that m and n have no
common factor greater than 1. Consequently, the statement of the theorem cannot
be false, and we have proved that if r is a real number such that r2
D 2, then r is
an irrational number.
1. This exercise is intended to provide another rationale as to why a proof by
contradiction works.
Suppose that we are trying to prove that a statement P is true. Instead of
proving this statement, assume that we prove that the conditional statement
“If :P , then C ” is true, where C is some contradiction. Recall that a con-
tradiction is a statement that is always false.

? (a) In symbols, write a statement that is a disjunction and that is logically
equivalent to :P ! C .
(b) Since we have proven that :P ! C is true, then the disjunction in
Exercise (1a) must also be true. Use this to explain why the statement
P must be true.
(c) Now explain why P must be true if we prove that the negation of P
implies a contradiction.
2. Are the following statements true or false? Justify each conclusion.
? (a) For all integers a and b, if a is even and b is odd, then 4 does not divide
a2
C b2

.
? (b) For all integers a and b, if a is even and b is odd, then 6 does not divide
a2
C b2

.
(c) For all integers a and b, if a is even and b is odd, then 4 does not divide
a2
C 2b2

.
? (d) For all integers a and b, if a is odd and b is odd, then 4 divides
a2
C 3b2

.
?
3. Consider the following statement:
For each positive real number r, if r2 D 18, then r is irrational.
(a) If you were setting up a proof by contradiction for this statement, what
would you assume? Carefully write down all conditions that you would
assume.
(b) Complete a proof by contradiction for this statement.
4. Prove that the cube root of 2 is an irrational number. That is, prove that if r
is a real number such that r3
D 2, then r is an irrational number.
? 5. Prove the following propositions:
(a) For all real numbers x and y, if x is rational and y is irrational, then
x C y is irrational.
(b) For all nonzero real numbers x and y, if x is rational and y is irrational,
then
x
y
is irrational.
6. Are the following statements true or false? Justify each conclusion.
?
(a) For each positive real number x, if x is irrational, then x2
is irrational.

? (b) For each positive real number x, if x is irrational, then
p
x is irrational.
(c) For every pair of real numbers x and y, if x C y is irrational, then x is
irrational and y is irrational.
(d) For every pair of real numbers x and y, if x C y is irrational, then x is
irrational or y is irrational.
7. (a) Give an example that shows that the sum of two irrational numbers can
be a rational number.
(b) Now explain why the following proof that
p
2 C
p
5

is an irrational
number is not a valid proof: Since
p
2 and
p
5 are both irrational num-
bers, their sum is an irrational number. Therefore,
p
2 C
p
5

is an
irrational number.
Note: You may even assume that we have proven that
p
5 is an irra-
tional number. (We have not proven this.)
(c) Is the real number
p
2C
p
5 a rational number or an irrational number?
Justify your conclusion.
8. (a) Prove that for each real number x,

x C
p
2

is irrational or

x C
p
2

is irrational.
(b) Generalize the propositionin Part (a) for any irrational number (instead
of just
p
2) and then prove the new proposition.
9. Is the following statement true or false?
For all positive real numbers x and y,
p
x C y
p
x C
p
y.
10. Is the following proposition true or false? Justify your conclusion.
For each real number x, x .1 x/
1
4
.
?
11. (a) Is the base 2 logarithm of 32, log2.32/, a rational number or an irra-
tional number? Justify your conclusion.
(b) Is the base 2 logarithm of 3, log2.3/, a rational number or an irrational
number? Justify your conclusion.
? 12. In Exercise (15) in Section 3.2, we proved that there exists a real number
solution to the equation x3
4x2
D 7. Prove that there is no integer x such
that x3 4x2 D 7.

13. Prove each of the following propositions:
?
(a) For each real number , if 0

2
, then Œsin./ C cos./ 1.
(b) For all real numbers a and b, if a ¤ 0 and b ¤ 0, then
p
a2 C b2 ¤
a C b.
(c) If n is an integer greater than 2, then for all integers m, n does not
divide m or n C m ¤ nm.
(d) For all real numbers a and b, if a 0 and b 0, then
2
a
C
2
b
¤
4
a C b
:
? 14. Prove that there do not exist three consecutive natural numbers such that the
cube of the largest is equal to the sum of the cubes of the other two.
15. Three natural numbers a, b, and c with a b c are called a Pythagorean
triple provided that a2 C b2 D c2. For example, the numbers 3, 4, and 5
form a Pythagorean triple, and the numbers 5, 12, and 13 form a Pythagorean
triple.
(a) Verify that if a D 20, b D 21, and c D 29, then a2
C b2
D c2
, and
hence, 20, 21, and 29 form a Pythagorean triple.
(b) Determine two other Pythagorean triples. That is, find integers a; b,
and c such that a2 C b2 D c2.
(c) Is the following proposition true or false? Justify your conclusion.
For all integers a, b, and c, if a2 Cb2 D c2, then a is even or b is even.
16. Consider the following proposition: There are no integers a and b such that
b2 D 4a C 2.
(a) Rewrite this statement in an equivalent form using a universal quantifier
by completing the following:
For all integers a and b, : : : :
(b) Prove the statement in Part (a).
17. Is the following statement true or false? Justify your conclusion.
For each integer n that is greater than 1, if a is the smallest positive
factor of n that is greater than 1, then a is prime.
See Exercise (13) in Section 2.4 (page 78) for the definition of a prime num-
ber and the definition of a composite number.

18. A magic square is a square array of natural numbers whose rows, columns,
and diagonals all sum to the same number. For example, the following is a
3 by 3 magic square since the sum of 3 numbers in each row is equal to 15,
the sum of the 3 numbers in each column is equal to 15, and the sum of the
3 numbers in each diagonal is equal to 15.
8 3 4
1 5 9
6 7 2
Prove that the following 4 by 4 square cannot be completed to form a magic
square.
1 2
3 4 5
6 7 8
9 10
Hint: Assign each of the six blank cells in the square a name. One possibility
is to use a, b, c, d, e, and f .
19. Using only the digits 1 through 9 one time each, is it possibleto construct a 3
by 3 magic square with the digit 3 in the center square? That is, is it possible
to construct a magic square of the form
a b c
d 3 e
f g h
where a; b; c; d; e; f; g; h are all distinct digits, none of which is equal to 3?
Either construct such a magic square or prove that it is not possible.
(a) Proposition. For each real number x, if x is irrational and m is an
integer, then mx is irrational.
Proof. We assume that x is a real number and is irrational. This means
that for all integers a and b with b ¤ 0, x ¤
a
b
. Hence, we may
conclude that mx ¤
ma
b
and, therefore, mx is irrational.

(b) Proposition. For all real numbers x and y, if x is irrational and y is
rational, then x C y is irrational.
Proof. We will use a proof by contradiction. So we assume that the
proposition is false, which means that there exist real numbers x and y
where x … Q, y 2 Q, and x C y 2 Q. Since the rational numbers are
closed under subtraction and x C y and y are rational, we see that
.x C y/ y 2 Q:
However, .x C y/ y D x, and hence we can conclude that x 2 Q.
This is a contradiction to the assumption that x … Q. Therefore, the
proposition is not false, and we have proven that for all real numbers x
and y, if x is irrational and y is rational, then x C y is irrational.
(c) Proposition. For each real number x, x.1 x/
1
4
.
Proof. A proof by contradiction will be used. So we assume the propo-
sition is false. This means that there exists a real number x such that
x.1 x/
1
4
. If we multiply both sides of this inequality by 4, we
obtain 4x.1 x/ 1. However, if we let x D 3, we then see that
4x.1 x/ 1
4 3.1 3/ 1
12 1
The last inequality is clearly a contradiction and so we have proved the
proposition.
21. A Proof by Contradiction. Consider the following proposition:
Proposition. Let a, b, and c be integers. If 3 divides a, 3 divides b, and
c 1 .mod 3/, then the equation
ax C by D c
has no solution in which both x and y are integers.
Complete the following proof of this proposition:

3.4. Using Cases in Proofs 131
Proof. A proof by contradiction will be used. So we assume that the state-
ment is false. That is, we assume that there exist integers a, b, and c such
that 3 divides both a and b, that c 1 .mod 3/, and that the equation
ax C by D c
has a solution in which both x and y are integers. So there exist integers m
and n such that
am C bn D c:
Hint: Now use the facts that 3 divides a, 3 divides b, and c 1 .mod 3/.
22. Exploring a Quadratic Equation. Consider the following proposition:
Proposition. For all integers m and n, if n is odd, then the equation
x2
C 2mx C 2n D 0
has no integer solution for x.
(a) What are the solutions of the equation when m D 1 and n D 1? That
is, what are the solutions of the equation x2
C 2x 2 D 0?
(b) What are the solutions of the equation when m D 2 and n D 3? That
is, what are the solutions of the equation x2
C 4x C 2 D 0?
(c) Solve the resulting quadratic equation for at least two more examples
using values of m and n that satisfy the hypothesis of the proposition.
(d) For this proposition, why does it seem reasonable to try a proof by
contradiction?
(e) For this proposition, state clearly the assumptions that need to be made
at the beginning of a proof by contradiction.
(f) Use a proof by contradiction to prove this proposition.
3.4 Using Cases in Proofs
Preview Activity 1 (Using a Logical Equivalency)
1. Complete a truth table to show that .P _ Q/ ! R is logically equivalent to
.P ! R/ ^ .Q ! R/.

2. Suppose that you are trying to prove a statement that is written in the form
.P _ Q/ ! R. Explain why you can complete this proof by writing sepa-
rate and independent proofs of P ! R and Q ! R.
3. Now consider the following proposition:
Proposition. For all integers x and y, if xy is odd, then x is odd and y is
odd.
Write the contrapositive of this proposition.
4. Now prove that if x is an even integer, then xy is an even integer. Also,
prove that if y is an even integer, then xy is an even integer.
5. Use the results proved in part (4) and the explanation in part (2) to explain
why we have proved the contrapositive of the proposition in part (3).
Preview Activity 2 (Using Cases in a Proof)
The work in Preview Activity 1 was meant to introduce the idea of using cases
in a proof. The method of using cases is often used when the hypothesis of the
proposition is a disjunction. This is justified by the logical equivalency
Œ.P _ Q/ ! R Œ.P ! R/ ^ .Q ! R/ :
See Theorem 2.8 on page 48 and Exercise (6) on page 50.
In some other situations when we are trying to prove a proposition or a theorem
about an element x in some set U , we often run into the problem that there does
not seem to be enough information about x to proceed. For example, consider the
following proposition:
Proposition 1. If n is an integer, then n2
C n

is an even integer.
If we were trying to write a direct proof of this proposition,the only thing we could
assume is that n is an integer. This is not much help. In a situation such as this,
we will sometimes use cases to provide additional assumptions for the forward
process of the proof. Cases are usually based on some common properties that the
element x may or may not possess. The cases must be chosen so that they exhaust
all possibilities for the object x in the hypothesis of the original proposition. For
Proposition 1, we know that an integer must be even or it must be odd. We can thus
use the following two cases for the integer n:
The integer n is an even integer;

The integer n is an odd integer.
1. Complete the proof for the following proposition:
Proposition 2: If n is an even integer, then n2 C n is an even integer.
Proof. Let n be an even integer. Then there exists an integer m such that
n D 2m. Substituting this into the expression n2
C n yields ....
2. Construct a proof for the following proposition:
Proposition 3: If n is an odd integer, then n2
C n is an even integer.
3. Explain why the proofs of Proposition 2 and Proposition 3 can be used to
construct a proof of Proposition 1.
Some Common Situations to Use Cases
When using cases in a proof, the main rule is that the cases must be chosen so
that they exhaust all possibilities for an object x in the hypothesis of the original
proposition. Following are some common uses of cases in proofs.
When the hypothesis is, Case 1: n is an even integer.
“n is an integer.” Case 2: n is an odd integer.
When the hypothesis is, Case 1: m and n are even.
“m and n are integers.” Case 2: m is even and n is odd.
Case 3: m is odd and n is even.
Case 4: m and n are both odd.
When the hypothesis is, Case 1: x is rational.
“x is a real number.” Case 2: x is irrational.
When the hypothesis is, Case 1: x D 0. OR Case 1: x 0.
“x is a real number.” Case 2: x ¤ 0. Case 2: x D 0.
Case 3: x 0.

When the hypothesis is, Case 1: a D b. OR Case 1: a b.
“a and b are real Case 2: a ¤ b. Case 2: a D b.
numbers.” Case 3: a b.
Writing Guidelines for a Proof Using Cases
When writing a proof that uses cases, we use all the other writing guidelines. In
addition, we make sure that it is clear where each case begins. This can be done by
using a new paragraph with a label such as “Case 1,” or it can be done by starting
a paragraph with a phrase such as, “In the case where ....”
Progress Check 3.21 (Using Cases: n Is Even or n Is Odd)
Complete the proof of the following proposition:
Proposition. For each integer n, n2
5n C 7 is an odd integer.
Proof. Let n be an integer. We will prove that n2 5n C 7 is an odd integer by
examining the case where n is even and the case where n is odd.
Case 1. The integer n is even. In this case, there exists an integer m such that
n D 2m. Therefore, ....
As another example of using cases, consider a situation where we know that a and
b are real numbers and ab D 0. If we want to make a conclusion about b, the
temptation might be to divide both sides of the equation by a. However, we can
only do this if a ¤ 0. So, we consider two cases: one when a D 0 and the other
when a ¤ 0.
Proposition 3.22. For all real numbers a and b, if ab D 0, then a D 0 or b D 0.
Proof. We let a and b be real numbers and assume that ab D 0. We will prove
that a D 0 or b D 0 by considering two cases: (1) a D 0, and (2) a ¤ 0.
In the case where a D 0, the conclusion of the proposition is true and so there
is nothing to prove.
In the case where a ¤ 0, we can multiply both sides of the equation ab D 0

by
1
a
and obtain
1
a
ab D
1
a
0
b D 0:
So in both cases, a D 0 or b D 0, and this proves that for all real numbers a and b,
if ab D 0, then a D 0 or b D 0.
Absolute Value
Most students by now have studied the concept of the absolute value of a real
number. We use the notation jxj to stand for the absolute value of the real number
x. One way to think of the absolute value of x is as the “distance” between x and
0 on the number line. For example,
j5j D 5 and j 7j D 7:
Although this notion of absolute value is convenient for determining the absolute
value of a specific number, if we want to prove properties about absolute value, we
need a more careful and precise definition.
Definition. For x 2 R, we define jxj, called the absolute value of x, by
jxj D
(
x; if x 0;
x if x 0.
Let’s first see if this definition is consistent with our intuitive notion of absolute
value by looking at two specific examples.
Since 5 0, we see that j5j D 5, which should be no surprise.
Since 7 0, we see that j 7j D . 7/ D 7.
Notice that the definition of the absolute value of x is given in two parts, one for
when x 0 and the other for when x 0. This means that when attempting to
prove something about absolute value, we often uses cases. This will be illustrated
in Theorem 3.23.

Theorem 3.23. Let a be a positive real number. For each real number x,
1. jxj D a if and only if x D a or x D a.
2. j xj D jxj.
Proof. The proof of Part (2) is part of Exercise (10). We will prove Part (1).
We let a be a positive real number and let x 2 R. We will first prove that if
jxj D a, then x D a or x D a. So we assume that jxj D a. In the case where
x 0, we see that jxj D x, and since jxj D a, we can conclude that x D a.
In the case where x 0, we see that jxj D x. Since jxj D a, we can
conclude that x D a and hence that x D a. These two cases prove that if
jxj D a, then x D a or x D a.
We will now prove that if x D a or x D a, then jxj D a. We start by
assuming that x D a or x D a. Since the hypothesis of this conditional statement
is a disjunction, we use two cases. When x D a, we see that
jxj D jaj D a since a 0:
When x D a, we conclude that
jxj D j aj D . a/ since a 0;
and hence, jxj D a. This proves that if x D a or x D a, then jxj D a. Because
we have proven both conditional statements, we have proven that jxj D a if and
only if x D a or x D a.
Progress Check 3.24 (Equations Involving Absolute Values)
1. What is j4:3j and what is j j?
2. Use the properties of absolute value in Proposition 3.23 to help solve the
following equations for t, where t is a real number.
(a) jtj D 12.
(b) jt C 3j D 5.
(c) jt 4j D
1
5
.
(d) j3t 4j D 8.
Although solving equations involving absolute values may not seem to have
anything to do with writing proofs, the point of Progress Check 3.24 is to empha-
size the importance of using cases when dealing with absolutevalue. The following
theorem provides some important properties of absolute value.

Theorem 3.25. Let a be a positive real number. For all real numbers x and y,
1. jxj a if and only if a x a.
2. jxyj D jxj jyj.
3. jx C yj jxj C jyj. This is known as the Triangle Inequality.
Proof. We will prove Part (1). The proof of Part (2) is included in Exercise (10),
and the proof of Part (3) is Exercise (14). For Part (1), we will prove the bicondi-
tional proposition by proving the two associated conditional propositions.
So we let a be a positive real number and let x 2 R and first assume that
jxj a. We will use two cases: either x 0 or x 0.
In the case where x 0, we know that jxj D x and so the inequality jxj a
implies that x a. However, we also know that a 0 and that x 0.
Therefore, we conclude that a x and, hence, a x a.
When x 0, we see that jxj D x. Therefore, the inequality jxj a
implies that x a, which in turn implies that a x. In this case, we also
know that x a since x is negative and a is positive. Hence, a x a
So in both cases, we have proven that a x a and this proves that if
jxj a, then a x a. We now assume that a x a.
If x 0, then jxj D x and hence, jxj a.
If x 0, then jxj D x and so x D jxj. Thus, a jxj. By
multiplyingboth sides of the last inequality by 1, we conclude that jxj a.
These two cases prove that if a x a, then jxj a. Hence, we have
proven that jxj a if and only if a x a.
?
1. In Preview Activity 2, we proved that if n is an integer, then n2
C n is an
even integer. We define two integers to be consecutive integers if one of the
integers is one more than the other integer. This means that we can represent
consecutive integers as m and m C 1, where m is some integer.

Explain why the result proven in Preview Activity 2 can be used to prove
that the product of any two consecutive integers is divisible by 2.
? 2. Prove that if u is an odd integer, then the equation x2 C x u D 0 has no
solution that is an integer.
? 3. Prove that if n is an odd integer, then n D 4k C 1 for some integer k or
n D 4k C 3 for some integer k.
? 4. Prove the following proposition:
For each integer a, if a2 D a, then a D 0 or a D 1.
5. (a) Prove the following proposition:
For all integers a, b, and d with d ¤ 0, if d divides a or d divides
b, then d divides the product ab.
Hint: Notice that the hypothesis is a disjunction. So use two cases.
(b) Write the contrapositive of the proposition in Exercise (5a).
? (c) Write the converse of the proposition in Exercise (5a). Is the converse
true or false? Justify your conclusion.
6. Are the following propositions true or false? Justify all your conclusions. If
a biconditional statement is found to be false, you should clearly determine
if one of the conditional statements within it is true. In that case, you should
state an appropriate theorem for this conditional statement and prove it.
?
(a) For all integers m and n, m and n are consecutive integers if and only
if 4 divides m2 C n2 1

.
(b) For all integers m and n, 4 divides m2
n2

if and only if m and n
are both even or m and n are both odd.
7. Is the following proposition true or false? Justify your conclusion with a
counterexample or a proof.
For each integer n, if n is odd, then 8 j n2
1

.
? 8. Prove that there are no natural numbers a and n with n 2 and a2 C1 D 2n.
9. Are the following propositions true or false? Justify each conclusions with a
(a) For all integers a and b with a ¤ 0, the equation ax C b D 0 has a
rational number solution.

(b) For all integers a, b, and c, if a, b, and c are odd, then the equation
ax2
C bx C c D 0 has no solution that is a rational number.
Hint: Do not use the quadratic formula. Use a proof by contradiction
and recall that any rational number can be written in the form
p
q
, where
p and q are integers, q 0, and p and q have no common factor greater
than 1.
(c) For all integers a, b, c, and d, if a, b, c, and d are odd, then the
equation ax3
C bx2
C cx C d D 0 has no solution that is a rational
number.
10. ? (a) Prove Part (2) of Proposition 3.23.
For each x 2 R, j xj D jxj.
(b) Prove Part (2) of Theorem 3.25.
For all real numbers x and y, jxyj D jxj jyj.
11. Let a be a positive real number. In Part (1) of Theorem 3.25, we proved that
for each real number x, jxj a if and only if a x a. It is important
to realize that the sentence a x a is actually the conjunction of two
inequalities. That is, a x a means that a x and x a.
? (a) Complete the following statement: For each real number x, jxj a if
and only if ....
(b) Prove that for each real number x, jxj a if and only if a x a.
(c) Complete the following statement: For each real number x, jxj a if
and only if ....
12. Prove each of the following:
(a) For each nonzero real number x ,
ˇ
ˇx 1
ˇ
ˇ D
1
jxj
.
(b) For all real numbers x and y, jx yj jxj jyj.
Hint: An idea that is often used by mathematicians is to add 0 to an
expression “intelligently”. In this case, we know that . y/ C y D 0.
Start by adding this “version” of 0 inside the absolute value sign of jxj.
(c) For all real numbers x and y, jjxj jyjj jx yj.

(a) Proposition. For all nonzero integers a and b, if a C 2b ¤ 3 and
9a C 2b ¤ 1, then the equation ax3
C 2bx D 3 does not have a
solution that is a natural number.
Proof. We will prove the contrapositive, which is
For all nonzero integers a and b, if the equation ax3C2bx D 3 has
a solutionthat is a natural number, then aC2b D 3 or 9aC2b D 1.
So we let a and b be nonzero integers and assume that the natural
number n is a solution of the equation ax3 C 2bx D 3. So we have
an3
C 2bn D 3 or
n an2
C b

D 3:
So we can conclude that n D 3 and an2
C b D 1. Since we now have
the value of n, we can substitute it in the equation an3 C 2bn D 3 and
obtain 27a C 6b D 3. Dividing both sides of this equation by 3 shows
that 9a C 2b D 1. So there is no need for us to go any further, and this
concludes the proof of the contrapositive of the proposition.
(b) Proposition. For all nonzero integers a and b, if a C 2b ¤ 3 and
9a C 2b ¤ 1, then the equation ax3 C 2bx D 3 does not have a
solution that is a natural number.
Proof. We will use a proof by contradiction. Let us assume that there
exist nonzero integers a and b such that a C 2b D 3 and 9a C 2b D 1
and an3 C 2bn D 3, where n is a natural number. First, we will solve
one equation for 2b; doing this, we obtain
a C 2b D 3
2b D 3 a: (1)
We can now substitute for 2b in an3 C 2bn D 3. This gives
an3
C .3 a/n D 3
an3
C 3n an D 3
n an2
C 3 a

D 3: (2)
By the closure properties of the integers, an2
C 3 a

is an integer
and, hence, equation (2) implies that n divides 3. So n D 1 or n D 3.
When we substituten D 1 into the equation an3
C2bn D 3, we obtain
aC2b D 3. This is a contradiction since we are told in the proposition
that a C 2b ¤ 3. This proves that the negation of the proposition is
false and, hence, the proposition is true.

3.5. The Division Algorithm and Congruence 141
14. Proof of the Triangle Inequality.
(a) Verify that the triangle inequality is true for several different real num-
bers x and y. Be sure to have some examples where the real numbers
are negative.
(b) Explain why the following proposition is true: For each real number r,
jrj r jrj.
(c) Now let x and y be real numbers. Apply the result in Part (14b) to both
x and y. Then add the corresponding parts of the two inequalities to
obtain another inequality. Use this to prove that jx C yj jxj C jyj.
3.5 The Division Algorithm and Congruence
Preview Activity 1 (Quotients and Remainders)
1. Let a D 27 and b D 4. We will now determine several pairs of integers q
and r so that 27 D 4q C r. For example, if q D 2 and r D 19, we obtain
4 2 C 19 D 27. The following table is set up for various values of q. For
each q, determine the value of r so that 4q C r D 27.
q 1 2 3 4 5 6 7 8 9 10
r 19 5
4q C r 27 27 27 27 27 27 27 27 27 27
2. What is the smallest positive value for r that you obtained in your examples
from Part (1)?
Division is not considered an operation on the set of integers since the quotient
of two integers need not be an integer. However, we have all divided one integer
by another and obtained a quotient and a remainder. For example, if we divide
113 by 5, we obtain a quotient of 22 and a remainder of 3. We can write this as
113
5
D 22 C
3
5
. If we multiply both sides of this equation by 5 and then use the

distributive property to “clear the parentheses,” we obtain
5
113
5
D 5

22 C
3
5

113 D 5 22 C 3
This is the equation that we use when working in the integers since it involves only
multiplication and addition of integers.
3. What are the quotient and the remainder when we divide 27 by 4? How is
this related to your answer for Part (2)?
4. Repeat part (1) using a D 17 and b D 5. So the object is to find integers q
and r so that 17 D 5q C r. Do this by completing the following table.
q 7 6 5 4 3 2 1
r 18 7
5q C r 17 17 17 17 17 17 17
5. The convention we will follow is that the remainder will be the smallest
positive integer r for which 17 D 5q C r and the quotient will be the
corresponding value of q. Using this convention, what is the quotient and
what is the remainder when 17 is divided by 5?
Preview Activity 2 (Some Work with Congruence Modulo n)
1. Let n be a natural number and let a and b be integers.
(a) Write the definition of “a is congruent to b modulo n,” which is written
a b .mod n/.
(b) Use the definition of “divides” to complete the following:
When we write a b .mod n/, we may conclude that there exists
an integer k such that ....
We will now explore what happens when we multiply several pairs of integers
where the first one is congruent to 3 modulo 6 and the second is congruent to 5
modulo 6. We can use set builder notation and the roster method to specify the set
A of all integers that are congruent to 2 modulo 6 as follows:
A D fa 2 Z j a 3 .mod 6/g D f: : : 15; 9; 3; 3; 9; 15; 21; : : :g:

2. Use the roster method to specify the set B of all integers that are congruent
to 5 modulo 6.
B D fb 2 Z j b 5 .mod 6/g D :
Notice that 15 2 A and 11 2 B and that 15 C 11 D 26. Also notice that 26
2 .mod 6/ and that 2 is the smallest positive integer that is congruent to 26 (mod
6).
3. Now choose at least four other pairs of integers a and b where a 2 A and
b 2 B. For each pair, calculate .a C b/ and then determine the smallest
positive integer r for which .a C b/ r .mod 6/. Note: The integer r will
satisfy the inequalities 0 r 6.
4. Prove that for all integers a and b, if a 3 .mod 6/ and b 5 .mod 6/,
then .a C b/ 2 .mod 6/.
The Division Algorithm
Preview Activity 1 was an introduction to a mathematical result known as the Di-
vision Algorithm. One of the purposes of this preview activity was to illustrate that
we have already worked with this result, perhaps without knowing its name. For
example, when we divide 337 by 6, we often write
337
6
D 56 C
1
6
:
When we multiply both sides of this equation by 6, we get
337 D 6 56 C 1:
When we are working within the system of integers, the second equation is pre-
ferred over the first since the second one uses only integers and the operations of
addition and multiplication, and the integers are closed under addition and multi-
plication. Following is a complete statement of the Division Algorithm.
For all integers a and b with b 0, there exist unique integers q and r such
that
a D bq C r and 0 r b:

Some Comments about the Division Algorithm
1. The Division Algorithmcan be proven, but we have not yet studied the meth-
ods that are usually used to do so. In this text, we will treat the Division Al-
gorithm as an axiom of the integers. The work in Preview Activity 1 provides
some rationale that this is a reasonable axiom.
2. The statement of the Division Algorithm contains the new phrase, “there
exist unique integers q and r such that : : ::” This means that there is only
one pair of integers q and r that satisfy both the conditions a D bq C r and
0 r b. As we saw in Preview Activity 1, there are several different ways
to write the integer a in the form a D bq C r. However, there is only one
way to do this and satisfy the additional condition that 0 r b.
3. In light of the previous comment, when we speak of the quotient and the
remainder when we “divide an integer a by the positive integer b,” we will
always mean the quotient .q/ and the remainder .r/ guaranteed by the Divi-
sion Algorithm. So the remainder r is the least nonnegative integer such that
there exists an integer (quotient) q with a D bq C r.
4. If a 0, then we must be careful when writing the result of the Division
Algorithm. For example, in parts (4) and (5) of Preview Activity 1, with
a D 17 and b D 5, we obtained 17 D 5 . 4/ C 3, and so the quotient is
4 and the remainder is 3. Notice that this is different than the result from a
calculator, which would be
17
5
D 3:4. But this means
17
5
D

3 C
4
10

D 3
2
5
:
If we multiply both sides of this equation by 5, we obtain
17 D 5 . 3/ C . 2/ :
This is not the result guaranteed by the Division Algorithm since the value
of 2 does not satisfy the result of being greater than or equal to 0 and less
than 5.
5. One way to look at the DivisionAlgorithm is that the integer a is either going
to be a multiple of b, or it will lie between two multiples of b. Suppose that a
is not a multiple of b and that it lies between the multiples bq and b .q C 1/,
where q is some integer. This is shown on the number line in Figure 3.2.

bq a
r
b(q + 1)
Figure 3.2: Remainder for the Division Algorithm
If r represents the distance from b q to a, then
r D a b q; or
a D b q C r:
From the diagram, also notice that r is less than the distance between b q
and b .q C 1/. Algebraically, this distance is
b .q C 1/ b q D b q C b b q
D b:
Thus, in the case where a is not a multiple of b, we get 0 r b.
6. We have been implicitlyusing the fact that an integer cannot be both even and
odd. There are several ways to understand this fact, but one way is through
the Division Algorithm. When we classify an integer as even or odd, we are
doing so on the basis of the remainder (according to the Division Algorithm)
when the integer is “divided” by 2. If a 2 Z, then by the Division Algorithm
there exist unique integers q and r such that
a D 2q C r and 0 r 2:
This means that the remainder, r, can only be zero or one (and not both).
When r D 0, the integer is even, and when r D 1, the integer is odd.
Progress Check 3.26 (Using the Division Algorithm)
1. What are the possible remainders (according to the Division Algorithm)
when an integer is
(a) Divided by 4? (b) Divided by 9?
2. For each of the following, find the quotient and remainder (guaranteed by the
Division Algorithm) and then summarize the results by writing an equation
of the form a D bq C r, where 0 r b.

(a) When 17 is divided by 3.
(b) When 17 is divided by 3.
(c) When 73 is divided by 7.
(d) When 73 is divided by 7.
(e) When 436 is divided by 27.
(f) When 539 is divided by 110.
Using Cases Determined by the Division Algorithm
The Division Algorithm can sometimes be used to construct cases that can be used
to prove a statement that is true for all integers. We have done this when we divided
the integers into the even integers and the odd integers since even integers have a
remainder of 0 when divided by 2 and odd integers have a remainder o 1 when
divided by 2.
Sometimes it is more useful to divide the integer a by an integer other than 2.
For example, if a is divided by 3, there are three possible remainders: 0, 1, and
2. If a is divided by 4, there are four possible remainders: 0, 1, 2, and 3. The
remainders form the basis for the cases.
If the hypothesis of a proposition is that “n is an integer,” then we can use the
Division Algorithm to claim that there are unique integers q and r such that
n D 3q C r and 0 r 3:
We can then divide the proof into the following three cases: (1) r D 0; (2) r D 1;
and (3) r D 2. This is done in Proposition 3.27.
Proposition 3.27. If n is an integer, then 3 divides n3
n.
Proof. Let n be an integer. We will show that 3 divides n3
n by examining the
three cases for the remainder when n is divided by 3. By the Division Algorithm,
there exist unique integers q and r such that
n D 3q C r, and 0 r 3:
This means that we can consider the following three cases: (1) r D 0; (2) r D 1;
and (3) r D 2.
In the case where r D 0, we have n D 3q. By substituting this into the
expression n3
n, we get
n3
n D .3q/3
.3q/
D 27q3
3q
D 3 9q3
q

:

Since 9q3 q

is an integer, the last equation proves that 3 j n3 n

.
In the second case, r D 1 and n D 3q C 1. When we substitute this into
n3
n

, we obtain
n3
n D .3q C 1/3
.3q C 1/
D 27q3
27q2
C 27q C 1

.3q C 1/
D 27q3
C 27q2
C 6q
D 3 9q3
C 9q2
C 2q

:
Since 9q3 C 9q2 C 2q

is an integer, the last equation proves that 3 j n3 n

.
The last case is when r D 2. The details for this case are part of Exercise (1).
Once this case is completed, we will have proved that 3 divides n3
n in all three
cases. Hence, we may conclude that if n is an integer, then 3 divides n3 n.
Properties of Congruence
Most of the work we have done so far has involved using definitions to help prove
results. We will continue to prove some results but we will now prove some theo-
rems about congruence (Theorem 3.28 and Theorem 3.30) that we will then use to
help prove other results.
Let n 2 N. Recall that if a and b are integers, then we say that a is congruent
to b modulo n provided that n divides a b, and we write a b .mod n/. (See
Section 3.1.) We are now going to prove some properties of congruence that are
direct consequences of the definition. One of these properties was suggested by the
work in Preview Activity 2 and is Part (1) of the next theorem.
Theorem 3.28 (Properties of Congruence Modulo n). Let n be a natural number
and let a; b; c; and d be integers. If a b .mod n/ and c d .mod n/, then
1. .a C c/ .b C d/ .mod n/.
2. ac bd .mod n/.
3. For each m 2 N, am bm .mod n/.
Proof. We will prove Parts (2) and (3). The proof of Part (1) is Progress Check 3.29.
Let n be a natural number and let a; b; c; and d be integers. Assume that a
b .mod n/ and that c d .mod n/. This means that n divides a b and that

n divides c d. Hence, there exist integers k and q such that a b D nk and
c d D nq. We can then write a D b C nk and c D d C nq and obtain
ac D .b C nk/ .d C nq/
D bd C bnq C dnk C n2
kq
D bd C n .bq C dk C nkq/ :
By subtracting bd from both sides of the last equation, we see that
ac bd D n .bq C dk C nkq/ :
Since bq C dk C nkq is an integer, this proves that n j .ac bd/, and hence we
can conclude that ac bd .mod n/. This completes the proof of Part (2).
Part (2) basically means that if we have two congruences, we can multiply the
corresponding sides of these congruences to obtain another congruence. We have
assumed that a b .mod n/ and so we write this twice as follows:
a b .mod n/ ; and
a b .mod n/ :
If we now use the result in Part (2) and multiply the corresponding sides of these
two congruences, we obtain a2 b2 .mod n/. We can then use this congruence
and the congruence a b .mod n/ and the result in Part (2) to conclude that
a2
a b2
b .mod n/ ;
or that a3 b3 .mod n/. We can say that we can continue with this process to
prove Part (3), but this is not considered to be a formal proof of this result. To
construct a formal proof for this, we could use a proof by mathematical induction.
This will be studied in Chapter 4. See Exercise (13) in Section 4.1.
Progress Check 3.29 (Proving Part (1) of Theorem 3.28)
Prove part (1) of Theorem 3.28.
Exercise (11) in Section 3.1 gave three important properties of congruence modulo
n. Because of their importance, these properties are stated and proved in Theo-
rem 3.30. Please remember that textbook proofs are usually written in final form
of “reporting the news.” Before reading these proofs, it might be instructive to first
try to construct a know-show table for each proof.
Theorem 3.30 (Properties of Congruence Modulo n). Let n 2 N,
and let a, b, and c be integers.

1. For every integer a, a a .mod n/.
This is called the reflexive property of congruence modulo n.
2. If a b .mod n/, then b a .mod n/.
This is called the symmetric property of congruence modulo n.
3. If a b .mod n/ and b c .mod n/, then a c .mod n/.
This is called the transitive property of congruence modulo n.
Proof. We will prove the reflexive property and the transitive property. The proof
of the symmetric property is Exercise (3).
Let n 2 N, and let a 2 Z. We will show that a a .mod n/. Notice that
a a D 0 D n 0:
This proves that n divides .a a/ and hence, by the definition of congruence
modulo n, we have proven that a a .mod n/.
To prove the transitive property, we let n 2 N, and let a, b, and c be integers.
We assume that a b .mod n/ and that b c .mod n/. We will use the
definition of congruence modulo n to prove that a c .mod n/. Since a
b .mod n/ and b c .mod n/, we know that n j .a b/ and n j .b c/. Hence,
there exist integers k and q such that
a b D nk
b c D nq:
By adding the corresponding sides of these two equations, we obtain
.a b/ C .b c/ D nk C nq:
If we simplify the left side of the last equation and factor the right side, we get
a c D n .k C q/ :
By the closure property of the integers, .k C q/ 2 Z, and so this equation proves
that n j .a c/ and hence that a c .mod n/. This completes the proof of the
transitive property of congruence modulo n.

Using Cases Based on Congruence Modulo n
Notice that the set of all integers that are congruent to 2 modulo 7 is
fn 2 Z j n 2 .mod 7/g D f: : : ; 19; 12; 5; 2; 9; 16; 23; : ::g :
If we divideany integer in this set by 7 and write the result according to the Division
Algorithm, we will get a remainder of 2. For example,
2 D 7 0 C 2 5 D 7 . 1/ C 2
9 D 7 1 C 2 12 D 7 . 2/ C 2
16 D 7 2 C 2 19 D 7 . 3/ C 2
23 D 7 3 C 2:
Is this a coincidence or is this always true? Let’s look at the general case. For this,
let n be a natural number and let a 2 Z. By the Division Algorithm, there exist
unique integers q and r such that
a D nq C r and 0 r n:
By subtracting r from both sides of the equation a D nq C r, we obtain
a r D nq:
But this implies that n j .a r/ and hence that a r .mod n/. We have proven
the following result.
Theorem 3.31. Let n 2 N and let a 2 Z. If a D nq C r and 0 r n for some
integers q and r, then a r .mod n/.
This theorem says that an integer is congruent (mod n) to its remainder when it
is divided by n. Since this remainder is unique and since the only possible remain-
ders for division by n are 0; 1; 2; : ::; n 1, we can state the following result.
Corollary 3.32. If n 2 N, then each integer is congruent, modulo n, to precisely
one of the integers 0; 1; 2; : : :; n 1. That is, for each integer a, there exists a
unique integer r such that
a r .mod n/ and 0 r n:
Corollary 3.32 can be used to set up cases for an integer in a proof. If n 2 N
and a 2 Z, then we can consider n cases for a. The integer a could be congruent to

0; 1; 2; : : :; or n 1 modulo n. For example, if we assume that 5 does not divide an
integer a, then we know a is not congruent to 0 modulo 5, and hence, that a must
be congruent to 1, 2, 3, or 4 modulo 5. We can use these as 4 cases within a proof.
For example, suppose we wish to determine the values of a2
modulo 5 for integers
that are not congruent to 0 modulo 5. We begin by squaring some integers that are
not congruent to 0 modulo 5. We see that
12
D 1 and 1 1 .mod 5/ :
32
D 9 and 9 4 .mod 5/ :
62
D 36 and 36 1 .mod 5/ :
82
D 64 and 64 4 .mod 5/ :
92
D 81 and 81 1 .mod 5/ :
These explorations indicate that the following proposition is true and we will now
outline a method to prove it.
Proposition 3.33. For each integer a, if a 6 0 .mod 5/, then a2
1 .mod 5/ or
a2
4 .mod 5/.
Proof. We will prove this proposition using cases for a based on congruence mod-
ulo 5. In doing so, we will use the results in Theorem 3.28 and Theorem 3.30.
Because the hypothesis is a 6 0 .mod 5/, we can use four cases, which are: (1)
a 1 .mod 5/, (2) a 2 .mod 5/, (3) a 3 .mod 5/, and (4) a 4 .mod 5/.
Following are proofs for the first and fourth cases.
Case 1. .a 1 .mod 5//. In this case, we use Theorem 3.28 to conclude that
a2
12
.mod 5/ or a2
1 .mod 5/ :
This proves that if a 1 .mod 5/, then a2 1 .mod 5/.
a2
42
.mod 5/ or a2
16 .mod 5/ :
We also know that 16 1 .mod 5/. So we have a2 16 .mod 5/ and 16
1 .mod 5/, and we can now use the transitive property of congruence (Theo-
rem 3.30) to conclude that a2 1 .mod 5/. This proves that if a 4 .mod 5/,
then a2
1 .mod 5/.

Progress Check 3.34 (Using Properties of Congruence)
Complete a proof of Proposition 3.33 by completing proofs for the other two cases.
Note: It is possible to prove Proposition 3.33 using only the definition of congru-
ence instead of using the properties that we have proved about congruence. How-
ever, such a proof would involve a good deal of algebra. One of the advantages of
using the properties is that it avoids the use of complicated algebra in which it is
easy to make mistakes.
In the proof of Proposition 3.33, we used four cases. Sometimes it may seem
a bit overwhelming when confronted with a proof that requires several cases. For
example, if we want to prove something about some integers modulo 6, we may
have to use six cases. However, there are sometimes additional assumptions (or
conclusions) that can help reduce the number of cases that must be considered.
This will be illustrated in the next progress check.
Progress Check 3.35 (Using Cases Modulo 6)
Suppose we want to determine the possible values for a2
modulo 6 for odd integers
that are not multiples of 3. Before beginning to use congruence arithmetic (as in
the proof of Proposition 3.33) in each of the possible six cases, we can show that
some of the cases are not possibleunder these assumptions. (In some sense, we use
a short proof by contradiction for these cases.) So assume that a is an odd integer.
Then:
If a 0 .mod 6/, then there exists an integer k such that a D 6k. But then
a D 2.3k/ and hence, a is even. Since we assumed that a is odd, this case
is not possible.
If a 2 .mod 6/, then there exists an integer k such that a D 6k C 2. But
then a D 2.3k C 1/ and hence, a is even. Since we assumed that a is odd,
this case is not possible.
1. Prove that if a is an odd integer, then a cannot be congruent to 4 modulo 6.
2. Prove that if a is an integer and 3 does not divide a, then a cannot be con-
gruent to 3 modulo 6.
3. So if a is an odd integer that is not a multiple of 3, then a must be congruent
to 1 or 5 modulo 6. Use these two cases to prove the following proposition:
Proposition 3.36. For each integer a, if a is an odd integer that is not multiple of
3, then a2
1 .mod 6/.

1. Complete the details for the proof of Case 3 of Proposition 3.27.
?
2. (a) Use cases based on congruence modulo 3 and properties of congruence
to prove that for each integer n, n3 n .mod 3/.
(b) Explain why the result in Part (a) proves that for each integer n, 3 di-
vides n3
n

. Compare this to the proof of the same result in Propo-
sition 3.27.
? 3. Prove the symmetric property of congruence stated in Theorem 3.30.
?
4. Consider the following proposition: For each integer a, if 3 divides a2
, then
3 divides a.
(a) Write the contrapositive of this proposition.
(b) Prove the proposition by proving its contrapositive. Hint: Consider
using cases based on the Division Algorithm using the remainder for
“division by 3.” There will be two cases.
?
5. (a) Let n 2 N and let a 2 Z. Explain why n divides a if and only if
a 0 .mod n/.
(b) Let a 2 Z. Explain why if a 6 0 .mod 3/, then a 1 .mod 3/ or
a 2 .mod 3/.
(c) Is the following proposition true or false? Justify your conclusion.
For each a 2 Z, if a 6 0 .mod 3/, then a2 1 .mod 3/.
?
6. Prove the following proposition by proving its contrapositive. (Hint: Use
case analysis. There are several cases.)
For all integers a and b, if ab 0 .mod 3/, then a 0 .mod 3/ or
b 0 .mod 3/.
?
7. (a) Explain why the following proposition is equivalent to the proposition
in Exercise (6).
For all integers a and b, if 3 j ab , then 3 j a or 3 j b.
(b) Prove that for each integer a, if 3 divides a2, then 3 divides a.
8. ? (a) Prove that the real number
p
3 is an irrational number. That is, prove
that
If r is a positive real number such that r2
D 3, then r is irrational.

(b) Prove that the real number
p
12 is an irrational number.
? 9. Prove that for each natural number n,
p
3n C 2 is not a natural number.
10. Extending the idea in Exercise (1) of Section 3.4, we can represent three
consecutive integers as m, m C 1, and m C 2, where m is an integer.
(a) Explain why we can also represent three consecutive integers as k 1,
k, and k C 1, where k is an integer.
?
(b) Explain why Proposition 3.27 proves that the product of any three con-
secutive integers is divisible by 3.
? (c) Prove that the product of three consecutive integers is divisible by 6.
11. (a) Use the result in Proposition 3.33 to help prove that the integer m D
5; 344; 580; 232; 468; 953; 153 is not a perfect square. Recall that an
integer n is a perfect square provided that there exists an integer k such
that n D k2
. Hint: Use a proof by contradiction.
(b) Is the integer n D 782; 456; 231; 189; 002; 288; 438 a perfect square?
12. (a) Use the result in Proposition 3.33 to help prove that for each integer a,
if 5 divides a2
, then 5 divides a.
(b) Prove that the real number
p
13. (a) Prove that for each integer a, if a 6 0 .mod 7/, then a2
6 0 .mod 7/.
(b) Prove that for each integer a, if 7 divides a2
, then 7 divides a.
(c) Prove that the real number
p
14. ?
(a) If an integer has a remainder of 6 when it is divided by 7, is it possibleto
determine the remainder of the square of that integer when it is divided
by 7? If so, determine the remainder and prove that your answer is
correct.
(b) If an integer has a remainder of 11 when it is divided by 12, is it pos-
sible to determine the remainder of the square of that integer when it
is divided by 12? If so, determine the remainder and prove that your
answer is correct.
(c) Let n be a natural number greater than 2. If an integer has a remainder
of n 1 when it is divided by n, is it possibleto determine the remainder
of the square of that integer when it is divided by n? If so, determine
the remainder and prove that your answer is correct.

15. Let n be a natural number greater than 4 and let a be an integer that has a
remainder of n 2 when it is divided by n. Make whatever conclusions you
can about the remainder of a2 when it is divided by n. Justify all conclusions.
proof or a counterexample.
For each natural number n, if 3 does not divide n2 C 2

, then n is not
a prime number or n D 3.
17. (a) Is the following proposition true or false? Justify your conclusion with
a counterexample or a proof.
For each integer n, if n is odd, then n2 1 .mod 8/.
(b) Compare this proposition to the proposition in Exercise (7) from Sec-
tion 3.4. Are these two propositions equivalent? Explain.
(c) Is the following proposition true or false? Justify your conclusion with
a counterexample or a proof.
For each integer n, if n is odd and n is not a multiple of 3, then
n2
1 .mod 24/.
For all integers a and b, if 3 divides a2
C b2

, then 3 divides a and 3
divides b.
For each integer a, 3 divides a3 C 23a.
20. Are the following statements true or false? Either prove the statement is true
or provide a counterexample to show it is false.
(a) For all integers a and b, if a b 0 .mod 6/, then a 0 .mod 6/ or
b 0 .mod 6/.
(b) For all integers a and b, if a b 0 .mod 8/, then a 0 .mod 8/ or
b 0 .mod 8/.
(c) For all integers a and b, if a b 1 .mod 6/, then a 1 .mod 6/ or
b 1 .mod 6/.
(d) For all integers a and b, if ab 7 .mod 12/, then either a 1 .mod 12/
or a 7 .mod 12/.

21. (a) Determine several pairs of integers a and b such that a b .mod 5/.
For each such pair, calculate 4a C b, 3a C 2b, and 7a C 3b. Are each
of the resulting integers congruent to 0 modulo 5?
(b) Prove or disprove the following proposition:
Let m and n be integers such that .m C n/ 0 .mod 5/ and let
a; b; 2 Z. If a b .mod 5/, then .ma C nb/ 0 .mod 5/.
(a) Proposition. For all integers a and b, if .a C 2b/ 0 .mod 3/, then
.2a C b/ 0 .mod 3/.
Proof. We assume a; b 2 Z and .a C 2b/ 0 .mod 3/. This means
that 3 divides a C 2b and, hence, there exists an integer m such that
a C 2b D 3m. Hence, a D 3m 2b. For .2a C b/ 0 .mod 3/, there
exists an integer x such that 2a C b D 3x. Hence,
2.3m 2b/ C b D 3x
6m 3b D 3x
3.2m b/ D 3x
2m b D x:
Since .2m b/ is an integer, this proves that 3 divides .2a C b/ and
hence, .2a C b/ 0 .mod 3/.
(b) Proposition. For each integer m, 5 divides m5
m

.
Proof. Let m 2 Z. We will prove that 5 divides m5
m

by proving
that m5 m

0 .mod 5/. We will use cases.
For the first case, if m 0 .mod 5/, then m5 0 .mod 5/ and, hence,
m5
m

0 .mod 5/.
For the second case, if m 1 .mod 5/, then m5
1 .mod 5/ and,
hence, m5 m

.1 1/ .mod 5/, which means that m5 m

0 .mod 5/.
For the third case, if m 2 .mod 5/, then m5 32 .mod 5/ and,
hence, m5
m

.32 2/ .mod 5/, which means that m5
m

0 .mod 5/.

23. Using a Contradiction to Prove a Case Is Not Possible. Explore the state-
ments in Parts (a) and (b) by considering several examples where the hypoth-
esis is true.
(a) If an integer a is divisible by both 4 and 6, then it divisible by 24.
(b) If an integer a is divisible by both 2 and 3, then it divisible by 6.
(c) What can you conclude from the examples in Part (a)?
(d) What can you conclude from the examples in Part (b)?
The proof of the following proposition based on Part (b) uses cases. In this
proof, however, we use cases and a proof by contradiction to prove that a
certain integer cannot be odd. Hence, it must be even. Complete the proof
of the proposition.
Proposition. Let a 2 Z. If 2 divides a and 3 divides a, then 6 divides a.
Proof: Let a 2 Z and assume that 2 divides a and 3 divides a. We will prove
that 6 divides a. Since 3 divides a, there exists an integer n such that
a D 3n:
The integer n is either even or it is odd. We will show that it must be even by
obtaining a contradiction if it assumed to be odd. So, assume that n is odd.
(Now complete the proof.)
24. The Last Two Digits of a Large Integer.
Notice that 7; 381; 272 72 .mod 100/ since 7; 381; 272 72 D 7; 381; 200,
which is divisible by 100. In general, if we start with an integer whose deci-
mal representation has more than two digits and subtract the integer formed
by the last two digits, the result will be an integer whose last two digits are
00. This result will be divisible by 100. Hence, any integer with more than
2 digits is congruent modulo 100 to the integer formed by its last two digits.
(a) Start by squaring both sides of the congruence 34 81 .mod 100/ to
prove that 38
61 .mod 100/ and then prove that 316
21 .mod 100/.
What does this tell you about the last two digits in the decimal repre-
sentation of 316?
(b) Use the two congruences in Part (24a) and laws of exponents to deter-
mine r where 320 r .mod 100/ and r 2 Z with 0 r 100 . What
does this tell you about the last two digits in the decimal representation
of 320
?

(c) Determine the last two digits in the decimal representation of 3400.
(d) Determine the last two digits in the decimal representation of 4804.
Hint: One way is to determine the “mod 100 values” for 42
, 44
, 48
,
416, 432, 464, and so on. Then use these values and laws of exponents
to determine r, where 4804
r .mod 100/ and r 2 Z with 0 r
100.
(e) Determine the last two digits in the decimal representation of 33356.
(f) Determine the last two digits in the decimal representation of 7403
.
3.6 Review of Proof Methods
This section is different from others in the text. It is meant primarily as a review
of the proof methods studied in Chapter 3. So the first part of the section will be
a description of some of the main proof techniques introduced in Chapter 3. The
most important part of this section is the set of exercises since these exercises will
provide an opportunity to use the proof techniques that we have studied so far.
We will now give descriptions of three of the most common methods used to
prove a conditional statement.
Direct Proof of a Conditional Statement .P ! Q/
When is it indicated? This type of proof is often used when the hypothesis
and the conclusion are both stated in a “positive” manner. That is, no nega-
tions are evident in the hypothesis and conclusion. That is, no negations are
evident in the hypothesis and conclusion.
Description of the process. Assume that P is true and use this to conclude
that Q is true. That is, we use the forward-backward method and work
forward from P and backward from Q.
Why the process makes sense. We know that the conditional statement
P ! Q is automatically true when the hypothesis is false. Therefore, be-
cause our goal is to prove that P ! Q is true, there is nothing to do in the
case that P is false. Consequently, we may assume that P is true. Then, in
order for P ! Q to be true, the conclusion Q must also be true. (When P
is true, but Q is false, P ! Q is false.) Thus, we must use our assumption
that P is true to show that Q is also true.

3.6. Review of Proof Methods 159
Proof of a Conditional Statement .P ! Q/ Using the Contrapositive
When is it indicated? This type of proof is often used when both the hy-
pothesis and the conclusion are stated in the form of negations. This often
works well if the conclusion contains the operator “or”; that is, if the con-
clusion is in the form of a disjunction. In this case, the negation will be a
conjunction.
Description of the process. We prove the logically equivalent statement p
:Q ! :P . The forward-backward method is used to prove :Q ! :P .
That is, we work forward from :Q and backward from :P .
Why the process makes sense. When we prove :Q ! :P , we are also
proving P ! Q because these two statements are logically equivalent.
When we prove the contrapositive of P ! Q, we are doing a direct proof
of :Q ! :P . So we assume :Q because, when doing a direct proof, we
assume the hypothesis, and :Q is the hypothesis of the contrapositive. We
must show :P because it is the conclusion of the contrapositive.
Proof of .P ! Q/ Using a Proof by Contradiction
When is it indicated? This type of proof is often used when the conclusion
is stated in the form of a negation, but the hypothesis is not. This often works
well if the conclusion contains the operator “or”; that is, if the conclusion is
in the form of a disjunction. In this case, the negation will be a conjunction.
Description of the process. Assume P and :Q and work forward from
these two assumptions until a contradiction is obtained.
Why the process makes sense. The statement P ! Q is either true or
false. In a proof by contradiction, we show that it is true by eliminating the
only other possibility (that it is false). We show that P ! Q cannot be false
by assuming it is false and reaching a contradiction. Since we assume that
P ! Q is false, and the only way for a conditional statement to be false is
for its hypothesis to be true and its conclusion to be false, we assume that P
is true and that Q is false (or, equivalently, that :Q is true). When we reach
a contradiction, we know that our original assumption that P ! Q is false
is incorrect. Hence, P ! Q cannot be false, and so it must be true.

Other Methods of Proof
The methods of proof that were just described are three of the most common types
of proof. However, we have seen other methods of proof and these are described
below.
Proofs that Use a Logical Equivalency
As was indicated in Section 3.2, we can sometimes use of a logical equivalency to
help prove a statement. For example, in order to prove a statement of the form
P ! .Q _ R/ ; (1)
it is sometimes possible to use the logical equivalency
ŒP ! .Q _ R/ Œ.P ^ :Q/ ! R :
We would then prove the statement
.P ^ :Q/ ! R: (2)
Most often, this would use a direct proof for statement (2) but other methods could
also be used. Because of the logical equivalency, by proving statement (2), we have
also proven the statement (1).
Proofs that Use Cases
When we are trying to prove a proposition or a theorem, we often run into the
problem that there does not seem to be enough information to proceed. In this
situation, we will sometimes use cases to provide additional assumptions for the
forward process of the proof. When this is done, the original proposition is divided
into a number of separate cases that are proven independently of each other. The
cases must be chosen so that they exhaust all possibilities for the hypothesis of
the original proposition. This method of case analysis is justified by the logical
equivalency
.P _ Q/ ! R .P ! R/ ^ .Q ! R/ ;
which was established in Preview Activity 1 in Section 3.4.

Constructive Proof
This is a technique that is often used to prove a so-called existence theorem. The
objective of an existence theorem is to prove that a certain mathematical object
exists. That is, the goal is usually to prove a statement of the form
There exists an x such that P.x/.
For a constructive proof of such a proposition, we actually name, describe, or ex-
plain how to construct some object in the universe that makes P.x/ true.
Nonconstructive Proof
This is another type of proof that is often used to prove an existence theorem is
the so-called nonconstructive proof. For this type of proof, we make an argument
that an object in the universal set that makes P .x/ true must exist but we never
construct or name the object that makes P .x/ true.
1. Let h and k be real numbers and let r be a positive number. The equation for
a circle whose center is at the point .h; k/ and whose radius is r is
.x h/2
C .y k/2
D r2
:
We also know that if a and b are real numbers, then
The point .a; b/ is inside the circle if .a h/2
C .b k/2
r2.
The point .a; b/ is on the circle if .a h/2
C .b k/2
D r2.
The point .a; b/ is outside the circle if .a h/2
C .b k/2
r2
.
Prove that all points on or inside the circle whose equation is .x 1/2
C
.y 2/2
D 4 are inside the circle whose equation is x2
C y2
D 26.
2. Let r be a positive real number. The equation for a circle of radius r whose
center is the origin is x2 C y2 D r2.
(a) Use implicit differentiation to determine
dy
dx
.

(b) Let .a; b/ be a point on the circle with a ¤ 0 and b ¤ 0. Determine
the slope of the line tangent to the circle at the point .a; b/.
(c) Prove that the radius of the circle to the point .a; b/ is perpendicular to
the line tangent to the circle at the point .a; b/. Hint: Two lines (neither
of which is horizontal) are perpendicular if and only if the products of
their slopes is equal to 1.
(a) For each integer a, if 3 does not divide a, then 3 divides 2a2
C 1.
(b) For each integer a, if 3 divides 2a2
C 1, then 3 does not divide a.
(c) For each integer a, 3 does not divide a if and only if 3 divides 2a2
C 1.
4. Prove that for each real number x and each irrational number q, .x C q/ is
irrational or .x q/ is irrational.
5. Prove that there exist irrational numbers u and v such that uv
is a rational
number.
Hint: We have proved that
p
2 is irrational. For the real number q D
p
2
p
2
,
either q is rational or q is irrational. Use this disjunction to set up two cases.
6. Let a and b be natural numbers such that a2
D b3
. Prove each of the
propositions in Parts (6a) through (6d). (The results of Exercise (1) and
Theorem 3.10 from Section 3.2 may be helpful.)
(a) If a is even, then 4 divides a.
(b) If 4 divides a, then 4 divides b.
(c) If 4 divides b, then 8 divides a.
(d) If a is even, then 8 divides a.
(e) Give an example of natural numbers a and b such that a is even and
a2
D b3
, but b is not divisible by 8.
Let a and b be integers with a ¤ 0. If a does not divide b, then the
equation ax3 C bx C .b C a/ D 0 does not have a solution that is a
natural number.
Hint: It may be necessary to factor a sum of cubes. Recall that
u3
C v3
D .u C v/ u2
uv C v2

:

8. Recall that a Pythagorean triple consists of three natural numbers a, b, and
c such that a b c and a2
C b2
D c2
. Are the following propositions
true or false? Justify your conclusions.
(a) For all a; b; c 2 N such that a b c, if a, b, and c form a
Pythagorean triple, then 3 divides a or 3 divides b.
(b) For all a; b; c 2 N such that a b c, if a, b, and c form a
Pythagorean triple, then 5 divides a or 5 divides b or 5 divides c.
9. (a) Prove that there exists a Pythagorean triple a, b, and c, where a D 5
and b and c are consecutive natural numbers.
(b) Prove that there exists a Pythagorean triple a, b, and c, where a D 7
and b and c are consecutive natural numbers.
(c) Let m be an odd natural number that is greater than 1. Prove that there
exists a Pythagorean triple a, b, and c, where a D m and b and c are
consecutive natural numbers.
10. One of the most famous unsolved problems in mathematics is a conjecture
made by Christian Goldbach in a letter to Leonhard Euler in 1742. The
conjecture made in this letter is now known as Goldbach’s Conjecture. The
conjecture is as follows:
Every even integer greater than 2 can be expressed as the sum of two
(not necessarily distinct) prime numbers.
Currently, it is not known if this conjecture is true or false.
(a) Write 50, 142, and 150 as a sum of two prime numbers.
(b) Prove the following:
If Goldbach’s Conjecture is true, then every integer greater
than 5 can be written as a sum of three prime numbers.
(c) Prove the following:
If Goldbach’s Conjecture is true, then every odd integer greater
than 7 can be written as a sum of three odd prime numbers.
11. Two prime numbers that differ by 2 are called twin primes. For example,
3 and 5 are twin primes, 5 and 7 are twin primes, and 11 and 13 are twin
primes. Determine at least two other pairs of twin primes. Is the following
proposition true or false? Justify your conclusion.
For all natural numbers p and q if p and q are twin primes other than
3 and 5, then pq C 1 is a perfect square and 36 divides pq C 1.

(a) For all integers a and b, .a C b/2 a2 C b2

.mod 2/.
(b) For all integers a and b, .a C b/3 a3 C b3

.mod 3/.
(c) For all integers a and b, .a C b/4
a4
C b4

.mod 4/.
(d) For all integers a and b, .a C b/5
a5
C b5

.mod 5/.
If any of the statements above are false, write a new statement of the follow-
ing form that is true (and prove that it is true):
For all integers a and b, .a C b/n .an C something C bn/ .mod n/.
13. Let a, b, c, and d be real numbers with a ¤ 0 and let f .x/ D ax3 Cbx2 C
cx C d.
(a) Determine the derivative and second derivative of the cubic function
f .
(b) Prove that the cubic function f has at most two critical points and has
exactly one inflection point.
14. A Special Case of Fermat’s Last Theorem. We have already seen examples
of Pythagorean triples, which are natural numbers a, b, and c where a2
C
b2 D c2. For example, 3, 4, and 5 form a Pythagorean triple as do 5,
12, and 13. One of the famous mathematicians of the 17th century was
Pierre de Fermat (1601 – 1665). Fermat made an assertion that for each
natural number n with n 3, there are no integers a, b, and c for which
an C bn D cn. This assertion was discovered in a margin of one of Fermat’s
books after his death, but Fermat provided no proof. He did, however, state
that he had discovered truly remarkable proof but the margin did not contain
enough room for the proof.
This assertion became known as Fermat’s Last Theorem but it more prop-
erly should have been called Fermat’s Last Conjecture. Despite the efforts
of mathematicians, this “theorem” remained unproved until Andrew Wiles, a
British mathematician, first announced a proof in June of 1993. However, it
was soon recognized that this proof had a serious gap, but a widely accepted
version of the proof was published by Wiles in 1995. Wiles’ proof uses
many concepts and techniques that were unknown at the time of Fermat. We

cannot discuss the proof here, but we will explore and prove the following
proposition, which is a (very) special case of Fermat’s Last Theorem.
Proposition. There do not exist prime numbers a, b,and c such that
a3 C b3 D c3.
Although Fermat’s Last Theorem implies this proposition is true, we will use
a proof by contradiction to prove this proposition. For a proof by contradic-
tion, we assume that
there exist prime numbers a, b,and c such that a3
C b3
D c3
.
Since 2 is the only even prime number, we will use the following cases: (1)
a D b D 2; (2) a and b are both odd; and (3) one of a and b is odd and the
other one is 2.
(a) Show that the case where a D b D 2 leads to a contradiction and
hence, this case is not possible.
(b) Show that the case where a and b are both odd leads to a contradiction
and hence, this case is not possible.
(c) We now know that one of a or b must be equal to 2. So we assume that
b D 2 and that a is an odd prime. Substitute b D 2 into the equation
b3 D c3 a3 and then factor the expression c3 a3. Use this to obtain
a contradiction.
(d) Write a complete proof of the proposition.

Divides, divisor, page 82
Factor, multiple, page 82
Proof, page 85
Undefined term, page 85
Axiom, page 85
Definition, page 86
Conjecture, page 86
Theorem, page 86
Proposition, page 86
Lemma, page 86
Corollary, page 86
Congruence modulo n, page 92
Tautology, page 40
Contradiction, page 40
Absolute value, page 135
Important Theorems and Results about Even and Odd Integers
If m is an even integer, then m C 1 is an odd integer.
If m is an odd integer, then m C 1 is an even integer.
If x is an even integer and y is an even integer, then x C y is an even integer.
If x is an even integer and y is an odd integer, then x C y is an odd integer.
If x is an odd integer and y is an odd integer, then x C y is an even integer.
Exercise (3), Section 1.2. If x is an even integer and y is an integer, then
x y is an even integer.
Theorem 1.8. If x is an odd integer and y is an odd integer, then x y is an
odd integer.
Theorem 3.7. The integer n is an even integer if and only if n2 is an even
integer.
Preview Activity 2 in Section 3.2. The integer n is an odd integer if and
only if n2
is an odd integer.

Important Theorems and Results about Divisors
Theorem 3.1. For all integers a, b, and c with a ¤ 0, if a j b and b j c, then
a j c.
Exercise (3), Section 3.1. For all integers a, b, and c with a ¤ 0,
If a j b and a j c, then a j .b C c/.
If a j b and a j c, then a j .b c/.
Exercise (3a), Section 3.1. For all integers a, b, and c with a ¤ 0, if a j b,
then a j .bc/.
Exercise (4), Section 3.1. For all nonzero integers a and b, if a j b and
b j a, then a D ˙b.
Let a and b be integers with b 0. Then there exist unique integers q and r such
that
Important Theorems and Results about Congruence
Theorem 3.28. Let a; b; c 2 Z and let n 2 N. If a b .mod n/ and
c d .mod n/, then
.a C c/ .b C d/ .mod n/.
ac bd .mod n/.
For each m 2 N, am bm .mod n/.
Theorem 3.30. For all integers a, b, and c,
Reflexive Property. a a .mod n/.
Symmetric Property. If a b .mod n/, then b a .mod n/.
Transitive Property. If a b .mod n/ and b c .mod n/, then
a c .mod n/.
Theorem 3.31. Let a 2 Z and let n 2 N. If a D nq C r and 0 r n for
some integers q and r, then a r .mod n/.

Corollary 3.32. Each integer is congruent, modulo n, to precisely one of the
integers 0; 1; 2; : : :; n 1. That is, for each integer a, there exists a unique
integer r such that
a r .mod n/ and 0 r n:

Chapter 4
Mathematical Induction
4.1 The Principle of Mathematical Induction
Preview Activity 1 (Exploring Statements of the Form .8n 2 N/ .P.n//)
One of the most fundamental sets in mathematics is the set of natural numbers
N. In this section, we will learn a new proof technique, called mathematical in-
duction, that is often used to prove statements of the form .8n 2 N/ .P.n//. In
Section 4.2, we will learn how to extend this method to statements of the form
.8n 2 T / .P.n//, where T is a certain type of subset of the integers Z.
For each natural number n, let P.n/ be the following open sentence:
4 divides 5n
1

:
1. Does this open sentence become a true statement when n D 1? That is, is 1
in the truth set of P.n/?
2. Does this open sentence become a true statement when n D 2? That is, is 2
in the truth set of P.n/?
3. Choose at least four more natural numbers and determine whether the open
sentence is true or false for each of your choices.
All of the examples that were used should provide evidence that the following
proposition is true:
For each natural number n, 4 divides .5n
1/.
169

170 Chapter 4. Mathematical Induction
We should keep in mind that no matter how many examples we try, we cannot
prove this proposition with a list of examples because we can never check if 4
divides .5n 1/ for every natural number n. Mathematical induction will provide
a method for proving this proposition.
For another example, for each natural number n, we now let Q.n/ be the fol-
lowing open sentence:
12
C 22
C C n2
D
n.n C 1/.2n C 1/
6
: (1)
The expression on the left side of the previous equation is the sum of the squares
of the first n natural numbers. So when n D 1, the left side of equation (1) is 12.
When n D 2, the left side of equation (1) is 12
C 22
.
4. Does Q.n/ become a true statement when
n D 1? (Is 1 in the truth set of Q.n/?)
5. Choose at least four more natural numbers and determine whether the open
sentence is true or false for each of your choices. A table with the columns
n, 12 C 22 C C n2, and
n.n C 1/.2n C 1/
6
may help you organize your
work.
All of the examples we have explored, should indicate the following proposition is
true:
For each natural number n, 12
C 22
C C n2
D
n.n C 1/.2n C 1/
6
.
In this section, we will learn how to use mathematical induction to prove this state-
ment.
Preview Activity 2 (A Property of the Natural Numbers)
Intuitively, the natural numbers begin with the number 1, and then there is 2, then 3,
then 4, and so on. Does this process of “starting with 1” and “adding 1 repeatedly”
result in all the natural numbers? We will use the concept of an inductive set to
explore this idea in this activity.

4.1. The Principle of Mathematical Induction 171
Definition. A set T that is a subset of Z is an inductive set provided that for
each integer k, if k 2 T , then k C 1 2 T .
1. Carefully explain what it means to say that a subset T of the integers Z is
not an inductive set. This description should use an existential quantifier.
2. Use the definition of an inductive set to determine which of the following
sets are inductive sets and which are not. Do not worry about formal proofs,
but if a set is not inductive, be sure to provide a specific counterexample that
proves it is not inductive.
(a) A D f1; 2; 3; : : :; 20g
(b) The set of natural numbers, N
(c) B D fn 2 N j n 5g
(d) S D fn 2 Z j n 3g
(e) R D fn 2 Z j n 100g
(f) The set of integers, Z
(g) The set of odd natural numbers.
3. This part will explore one of the underlying mathematical ideas for a proof
by induction. Assume that T N and assume that 1 2 T and that T is an
inductive set. Use the definition of an inductive set to answer each of the
following:
(a) Is 2 2 T ? Explain.
(b) Is 3 2 T ? Explain.
(c) Is 4 2 T ? Explain.
(d) Is 100 2 T ? Explain.
(e) Do you think that T D N? Ex-
plain.
Inductive Sets
The two open sentences in Preview Activity 1 appeared to be true for all values of
n in the set of natural numbers, N. That is, the examples in this preview activity
provided evidence that the following two statements are true.
For each natural number n, 4 divides .5n
1/.
For each natural number n, 12
C 22
C C n2
D
n.n C 1/.2n C 1/
6
.
One way of proving statements of this form uses the concept of an inductive set
introduced in Preview Activity 2. The idea is to prove that if one natural number

makes the open sentence true, then the next one also makes the open sentence
true. This is how we handle the phrase “and so on” when dealing with the natural
numbers. In Preview Activity 2, we saw that the number systems N and Z and
other sets are inductive. What we are trying to do is somehow distinguish N from
the other inductive sets. The way to do this was suggested in Part (3) of Preview
Activity 2. Although we will not prove it, the following statement should seem
true.
Statement 1: For each subset T of N, if 1 2 T and T is inductive, then T D N.
Notice that the integers, Z, and the set S D fn 2 Z j n 3g both contain 1 and
both are inductive, but they both contain numbers other than natural numbers. For
example, the following statement is false:
Statement 2: For each subset T of Z, if 1 2 T and T is inductive, then T D Z.
The set S D fn 2 Z j n 3g D f 3; 2; 1; 0; 1; 2; 3; : : : g is a counterexample
that shows that this statement is false.
Progress Check 4.1 (Inductive Sets)
Suppose that T is an inductive subset of the integers. Which of the following
statements are true, which are false, and for which ones is it not possible to tell?
1. 1 2 T and 5 2 T.
2. If 1 2 T , then 5 2 T.
3. If 5 … T, then 2 … T.
4. For each integer k, if k 2 T, then
k C 7 2 T .
5. For each integer k, k … T or
k C 1 2 T .
6. There exists an integer k such that
k 2 T and k C 1 … T.
7. For each integer k, if k C 1 2 T,
then k 2 T.
8. For each integer k, if k C 1 … T,
then k … T.
The Principle of Mathematical Induction
Although we proved that Statement (2) is false, in this text, we will not prove
that Statement (1) is true. One reason for this is that we really do not have a
formal definition of the natural numbers. However, we should be convinced that

Statement (1) is true. We resolve this by making Statement (1) an axiom for the
natural numbers so that this becomes one of the defining characteristics of the
natural numbers.
The Principle of Mathematical Induction
If T is a subset of N such that
1. 1 2 T, and
2. For every k 2 N, if k 2 T, then .k C 1/ 2 T,
then T D N.
Using the Principle of Mathematical Induction
The primary use of the Principle of Mathematical Induction is to prove statements
of the form
.8n 2 N/ .P .n// ;
where P.n/ is some open sentence. Recall that a universally quantified statement
like the preceding one is true if and only if the truth set T of the open sentence
P.n/ is the set N. So our goal is to prove that T D N, which is the conclusion of
the Principle of Mathematical Induction. To verify the hypothesis of the Principle
of Mathematical Induction, we must
1. Prove that 1 2 T. That is, prove that P.1/ is true.
2. Prove that if k 2 T, then .k C 1/ 2 T. That is, prove that if P.k/ is true,
then P.k C 1/ is true.
The first step is called the basis step or the initial step, and the second step is
called the inductive step. This means that a proof by mathematical induction will
have the following form:

Procedure for a Proof by Mathematical Induction
To prove: .8n 2 N/ .P .n//
Basis step: Prove P.1/.
Inductive step: Prove that for each k 2 N,
if P.k/ is true, then P.k C 1/ is true.
We can then conclude that P.n/ is true for all n 2 N.
Note that in the inductive step, we want to prove that the conditional statement “for
each k 2 N, if P.k/ then P.k C 1/” is true. So we will start the inductive step by
assuming that P.k/ is true. This assumption is called the inductive assumption
or the inductive hypothesis.
The key to constructing a proof by induction is to discover how P.k C 1/ is
related to P.k/ for an arbitrary natural number k. For example, in Preview Activ-
ity 1, one of the open sentences P.n/ was
12
C 22
C C n2
D
n.n C 1/.2n C 1/
6
:
Sometimes it helps to look at some specific examples such as P.2/ and P.3/. The
idea is not just to do the computations, but to see how the statements are related.
This can sometimes be done by writing the details instead of immediately doing
computations.
P.2/ is 12
C 22
D
2 3 5
6
P.3/ is 12
C 22
C 32
D
3 4 7
6
In this case, the key is the left side of each equation. The left side of P.3/ is
obtained from the left side of P.2/ by adding one term, which is 32. This suggests
that we might be able to obtain the equation for P.3/ by adding 32
to both sides of
the equation in P.2/. Now for the general case, if k 2 N, we look at P.k C 1/ and
compare it to P.k/.
P.k/ is 12
C 22
C C k2
D
k.k C 1/.2k C 1/
6
P.k C 1/ is 12
C 22
C C .k C 1/2
D
.k C 1/ Œ.k C 1/ C 1 Œ2 .k C 1/ C 1
6

The key is to look at the left side of the equation for P.k C 1/ and realize what this
notation means. It means that we are adding the squares of the first .k C 1/ natural
numbers. This means that we can write
12
C 22
C C .k C 1/2
D 12
C 22
C C k2
C .k C 1/2
:
This shows us that the left side of the equation for P.k C 1/ can be obtained from
the left side of the equation for P.k/ by adding .k C 1/2
. This is the motivation
for proving the inductive step in the following proof.
Proposition 4.2. For each natural number n,
12
C 22
C C n2
D
n.n C 1/.2n C 1/
6
:
Proof. We will use a proof by mathematical induction. For each natural number
n, we let P.n/ be
12
C 22
C C n2
D
n.n C 1/.2n C 1/
6
:
We first prove that P.1/ is true. Notice that
1 .1 C 1/ .2 1 C 1/
6
D 1. This shows
that
12
D
1 .1 C 1/ .2 1 C 1/
6
;
which proves that P.1/ is true.
For the inductive step, we prove that for each k 2 N, if P.k/ is true, then P.k C1/
is true. So let k be a natural number and assume that P.k/ is true. That is, assume
that
12
C 22
C C k2
D
k.k C 1/.2k C 1/
6
: (1)
The goal now is to prove that P .k C 1/ is true. That is, it must be proved that
12
C 22
C C k2
C .k C 1/2
D
.k C 1/ Œ.k C 1/ C 1 Œ2.k C 1/ C 1
6
D
.k C 1/ .k C 2/ .2k C 3/
6
: (2)

To do this, we add .k C 1/2
to both sides of equation (1) and algebraically rewrite
the right side of the resulting equation. This gives
12
C 22
C C k2
C .k C 1/2
D
k.k C 1/.2k C 1/
6
C .k C 1/2
D
k.k C 1/.2k C 1/ C 6.k C 1/2
6
D
.k C 1/ Œk.2k C 1/ C 6.k C 1/
6
D
.k C 1/ 2k2
C 7k C 6

6
D
.k C 1/.k C 2/.2k C 3/
6
:
Comparing this result to equation (2), we see that if P.k/ is true, then P.k C 1/ is
true. Hence, the inductive step has been established, and by the Principle of Mathe-
matical Induction, we have proved that for each natural number n,
12
C 22
C C n2
D
n.n C 1/.2n C 1/
6
.
Writing Guideline
The proof of Proposition 4.2 shows a standard way to write an induction proof.
When writing a proof by mathematical induction, we should follow the guideline
that we always keep the reader informed. This means that at the beginning of the
proof, we should state that a proof by induction will be used. We should then
clearly define the open sentence P.n/ that will be used in the proof.
Summation Notation
The result in Proposition 4.2 could be written using summation notation as follows:
For each natural number n,
n
P
j D1
j 2
D
n.n C 1/.2n C 1/
6
.

In this case, we use j for the index for the summation, and the notation
n
P
j D1
j 2
tells
us to add all the values of j 2
for j from 1 to n, inclusive. That is,
n
X
j D1
j 2
D 12
C 22
C C n2
:
So in the proof of Proposition4.2, we would let P.n/ be
n
P
j D1
j 2
D
n.n C 1/.2n C 1/
6
,
and we would use the fact that for each natural number k,
kC1
X
j D1
j 2
D
0
@
k
X
j D1
j 2
1
A C .k C 1/2
:
Progress Check 4.3 (An Example of a Proof by Induction)
1. Calculate 1 C 2 C 3 C C n and
n.n C 1/
2
for several natural numbers n.
What do you observe?
2. Use mathematical induction to prove that 1 C 2 C 3 C C n D
n .n C 1/
2
.
To do this, let P.n/ be the open sentence, “1C2C3C Cn D
n .n C 1/
2
.”
For the basis step, notice that the equation 1 D
1 .1 C 1/
2
shows that P.1/ is
true. Now let k be a natural number and assume that P.k/ is true. That is,
assume that
1 C 2 C 3 C C k D
k .k C 1/
2
;
and complete the proof.
Some Comments about Mathematical Induction
1. The basis step is an essential part of a proof by induction. See Exercise (19)
for an example that shows that the basis step is needed in a proof by induc-
tion.
2. Exercise (20) provides an example that shows the inductive step is also an
essential part of a proof by mathematical induction.

3. It is important to remember that the inductive step in an induction proof is
a proof of a conditional statement. Although we did not explicitly use the
forward-backward process in the inductive step for Proposition 4.2, it was
implicitly used in the discussion prior to Proposition 4.2. The key question
was, “How does knowing the sum of the first k squares help us find the sum
of the first .k C 1/ squares?”
4. When proving the inductive step in a proof by induction, the key question is,
How does knowing P.k/ help us prove P.k C 1/?
In Proposition 4.2, we were able to see that the way to answer this question
was to add a certain expression to both sides of the equation given in P.k/.
Sometimes the relationshipbetween P.k/ and P.k C 1/ is not as easy to see.
For example, in Preview Activity 1, we explored the following proposition:
For each natural number n, 4 divides .5n 1/.
This means that the open sentence, P.n/, is “4 divides .5n
1/.” So in the
inductive step, we assume k 2 N and that 4 divides

5k 1

. This means
that there exists an integer m such that
5k
1 D 4m: (1)
In the backward process, the goal is to prove that 4 divides

5kC1
1

. This
can be accomplished if we can prove that there exists an integer s such that
5kC1
1 D 4s: (2)
We now need to see if there is anything in equation (1) that can be used in
equation (2). The key is to find something in the equation 5k 1 D 4m that
is related to something similar in the equation 5kC1
1 D 4s. In this case,
we notice that
5kC1
D 5 5k
:
So if we can solve 5k
1 D 4m for 5k
, we could make a substitution for 5k
.
This is done in the proof of the following proposition.
Proposition 4.4. For every natural number n, 4 divides .5n 1/.

Proof. (Proof by Mathematical Induction) For each natural number n, let P.n/ be
“4 divides .5n
1/.” We first prove that P .1/ is true. Notice that when n D 1,
.5n 1/ D 4. Since 4 divides 4, P .1/ is true.
For the inductive step, we prove that for all k 2 N, if P .k/ is true, then
P .k C 1/ is true. So let k be a natural number and assume that P.k/ is true. That
is, assume that
4 divides

5k
1

:
This means that there exists an integer m such that
5k
1 D 4m:
Thus,
5k
D 4m C 1: (1)
In order to prove that P.k C 1/ is true, we must show that 4 divides

5kC1
1

.
Since 5kC1
D 5 5k
, we can write
5kC1
1 D 5 5k
1: (2)
We now substitute the expression for 5k from equation (1) into equation (2). This
gives
5kC1
1 D 5 5k
1
D 5.4m C 1/ 1
D .20m C 5/ 1
D 20m C 4
D 4.5m C 1/ (3)
Since .5m C 1/ is an integer, equation (3) shows that 4 divides

5kC1
1

. There-
fore, if P.k/ is true, then P.k C 1/ is true and the inductive step has been estab-
lished. Thus, by the Principle of Mathematical Induction, for every natural number
n, 4 divides .5n
1/.
Proposition4.4 was stated in terms of “divides.” We can use congruence to state
a proposition that is equivalent to Proposition 4.4. The idea is that the sentence,
4 divides .5n
1/ means that 5n
1 .mod 4/. So the following proposition is
equivalent to Proposition 4.4.
Proposition 4.5. For every natural number n, 5n
1 .mod 4/.

Since we have proved Proposition4.4, we have in effect proved Proposition4.5.
However, we could have proved Proposition 4.5 first by using the results in Theo-
rem 3.28 on page 147. This will be done in the next progress check.
Progress Check 4.6 (Proof of Proposition 4.5) .
To prove Proposition 4.5, we let P.n/ be 5n
1 .mod 4/ and notice that P.1/ is
true since 5 1 .mod 4/. For the inductive step, let k be a natural number and
assume that P.k/ is true. That is, assume that 5k
1 .mod 4/.
1. What must be proved in order to prove that P.k C 1/ is true?
2. Since 5kC1
D 5 5k
, multiply both sides of the congruence 5k
1 .mod 4/
by 5. The results in Theorem 3.28 on page 147 justify this step.
3. Now complete the proof that for each k 2 N, if P.k/ is true, then P.k C 1/
is true and complete the induction proof of Proposition 4.5.
It might be nice to compare the proofs of Propositions 4.4 and 4.5 and decide which
one is easier to understand.
? 1. Which of the following sets are inductive sets? Explain.
(a) Z
(b) fx 2 N j x 4g
(c) fx 2 Z j x 10g
(d) f1; 2; 3; : : :; 500g
? 2. (a) Can a finite, nonempty set be inductive? Explain.
(b) Is the empty set inductive? Explain.
3. Use mathematical induction to prove each of the following:
? (a) For each natural number n, 2 C 5 C 8 C C .3n 1/ D
n .3n C 1/
2
.
(b) For each natural number n, 1 C 5 C 9 C C .4n 3/ D n .2n 1/.
(c) For each natural number n, 13 C 23 C 33 C C n3 D

n .n C 1/
2
2
.
4. Based on the results in Progress Check 4.3 and Exercise (3c), if n 2 N, is
there any conclusion that can be made about the relationship between the
sum 13
C 23
C 33
C C n3

and the sum .1 C 2 C 3 C C n/?

5. Instead of using induction, we can sometimes use previously proven results
about a summation to obtain results about a different summation.
(a) Use the result in Progress Check 4.3 to prove the following proposition:
For each natural number n, 3 C 6 C 9 C C 3n D
3n .n C 1/
2
.
(b) Subtract n from each side of the equation in Part (a). On the left side of
this equation, explain why this can be done by subtracting 1 from each
term in the summation.
(c) Algebraically simplify the right side of the equation in Part (b) to ob-
tain a formula for the sum 2 C 5 C 8 C .3n 1/. Compare this to
Exercise (3a).
? 6. (a) Calculate 1 C 3 C 5 C C .2n 1/ for several natural numbers n.
(b) Based on your work in Exercise (6a), if n 2 N, make a conjecture about
the value of the sum 1 C 3 C 5 C C .2n 1/ D
n
P
j D1
.2j 1/.
(c) Use mathematical induction to prove your conjecture in Exercise (6b).
?
7. In Section 3.1, we defined congruence modulo n for a natural number n, and
in Section 3.5, we used the Division Algorithm to prove that each integer
is congruent, modulo n, to precisely one of the integers 0; 1; 2; : : :; n 1
(Corollary 3.32).
(a) Find the value of r so that 4 r .mod 3/ and r 2 f0; 1; 2g.
(b) Find the value of r so that 42
r .mod 3/ and r 2 f0; 1; 2g.
(c) Find the value of r so that 43
r .mod 3/ and r 2 f0; 1; 2g.
(d) For two other values of n, find the value of r so that 4n
r .mod 3/
and r 2 f0; 1; 2g.
(e) If n 2 N, make a conjecture concerning the value of r where
4n
r .mod 3/ and r 2 f0; 1; 2g. This conjecture should be written
as a self-contained proposition including an appropriate quantifier.
(f) Use mathematical induction to prove your conjecture.
? (a) For each natural number n, 3 divides .4n 1/.
(b) For each natural number n, 6 divides n3 n

.

9. In Exercise (7), we proved that for each natural number n, 4n 1 .mod 3/.
Explain how this result is related to the proposition in Exercise (8a).
10. Use mathematical induction to prove that for each natural number n, 3 di-
vides n3 C 23n. Compare this proof to the proof from Exercise (19) in
Section 3.5.
11. (a) Calculate the value of 5n
2n
for n D 1, n D 2, n D 3, n D 4, n D 5,
and n D 6.
(b) Based on your work in Part (a), make a conjecture about the values of
5n 2n for each natural number n.
(c) Use mathematical induction to prove your conjecture in Part (b).
12. Let x and y be integers. Prove that for each natural number n, .x y/
divides .xn yn/. Explain why your conjecture in Exercise (11) is a special
case of this result.
?
13. Prove Part (3) of Theorem 3.28 from Section 3.4. Let n 2 N and let a and b
be integers. For each m 2 N, if a b .mod n/, then am bm .mod n/.
? 14. Use mathematical induction to prove that the sum of the cubes of any three
consecutive natural numbers is a multiple of 9.
15. Let a be a real number. We will explore the derivatives of the function
f .x/ D eax
. By using the chain rule, we see that
d
dx
eax

D aeax
:
Recall that the second derivative of a function is the derivative of the deriva-
tive function. Similarly, the third derivative is the derivative of the second
derivative.
(a) What is
d2
dx2
.eax
/, the second derivative of eax
?
(b) What is
d3
dx3
.eax
/, the third derivative of eax
?
(c) Let n be a natural number. Make a conjecture about the nth derivative
of the function f .x/ D eax
. That is, what is
dn
dxn
.eax
/? This con-
jecture should be written as a self-contained proposition including an
appropriate quantifier.
(d) Use mathematical induction to prove that your conjecture.

16. In calculus, it can be shown that
Z
sin2
x dx D
x
2
1
2
sinx cos x C c and
Z
cos2
x dx D
x
2
C
1
2
sinx cos x C c:
Using integration by parts, it is also possible to prove that for each natural
number n,
Z
sinn
x dx D
1
n
sinn 1
x cos x C
n 1
n
Z
sinn 2
x dx and
Z
cosn
x dx D
1
n
cosn 1
x sinx C
n 1
n
Z
cosn 2
x dx:
(a) Determine the values of
Z =2
0
sin2
x dx and
Z =2
0
sin4
x dx:
(b) Use mathematical induction to prove that for each natural number n,
Z =2
0
sin2n
x dx D
1 3 5 .2n 1/
2 4 6 .2n/

2
and
Z =2
0
sin2nC1
x dx D
2 4 6 .2n/
1 3 5 .2n C 1/
:
These are known as the Wallis sine formulas.
(c) Use mathematical induction to prove that
Z =2
0
cos2n
x dx D
1 3 5 .2n 1/
2 4 6 .2n/

2
and
Z =2
0
cos2nC1
x dx D
2 4 6 .2n/
1 3 5 .2n C 1/
:
These are known as the Wallis cosine formulas.
17. (a) Why is it not possible to use mathematical induction to prove a propo-
sition of the form
.8x 2 Q/ .P.x// ;
where P.x/ is some predicate?

(b) Why is it not possible to use mathematical induction to prove a propo-
sition of the form
For each real number x with x 1, P.x/,
where P.x/ is some predicate?
(a) For each natural number n, 1 C 4 C 7 C C .3n 2/ D
n .3n 1/
2
.
Proof. We will prove this proposition using mathematical induction.
So we let P.n/ be the open sentence
1 C 4 C 7 C C .3n 2/:
Using n D 1, we see that 3n 2 D 1 and hence, P .1/ is true.
We now assume that P.k/ is true. That is,
1 C 4 C 7 C C .3k 2/ D
k .3k 1/
2
:
We then see that
1 C 4 C 7 C C .3k 2/ C .3.k C 1/ 2/ D
.k C 1/ .3k C 2/
2
k .3k 1/
2
C .3k C 1/ D
.k C 1/ .3k C 2/
2
3k2
k

C .6k C 2/
2
D
3k2
C 5k C 2
2
3k2 C 5k C 2
2
D
3k2 C 5k C 2
2
:
We have thus proved that P.k C 1/ is true, and hence, we have proved
the proposition.
(b) For each natural number n, 1 C 4 C 7 C C .3n 2/ D
n .3n 1/
2
.
So we let
P.n/ D 1 C 4 C 7 C C .3n 2/:

Using n D 1, we see that P.1/ D 1 and hence, P.1/ is true.
We now assume that P.k/ is true. That is,
1 C 4 C 7 C C .3k 2/ D
k .3k 1/
2
:
We then see that
P.k C 1/ D 1 C 4 C 7 C C .3k 2/ C .3.k C 1/ 2/
D
k .3k 2/
2
C 3 .k C 1/ 2
D
3k2 k C 6k C 6 4
2
D
3k2
C 5k C 2
2
D
.k C 1/ .3k C 2/
2
:
We have thus proved that P.k C 1/ is true, and hence, we have proved
the proposition.
(c) All dogs are the same breed.
For each natural number n, we let P.n/ be
Any set of n dogs consists entirely of dogs of the same breed.
We will prove that for each natural number n, P.n/ is true, which will
prove that all dogs are the same breed. A set with only one dog consists
entirely of dogs of the same breed and, hence, P.1/ is true.
So we let k be a natural number and assume that P.k/ is true, that
is, that every set of k dogs consists of dogs of the same breed. Now
consider a set D of k C 1 dogs, where
D D fd1; d2; : : :; dk; dkC1g:
If we remove the dog d1 from the set D, we then have a set D1 of k
dogs, and using the assumption that P.k/ is true, these dogs must all
be of the same breed. Similarly, if we remove dkC1 from the set D, we
again have a set D2 of k dogs, and these dogs must all be of the same

breed. Since D D D1 [ D2, we have proved that all of the dogs in D
must be of the same breed.
This proves that if P.k/ is true, then P.k C 1/ is true and, hence, by
mathematical induction, we have proved that for each natural number
n, any set of n dogs consists entirely of dogs of the same breed.
19. The Importance of the Basis Step. Most of the work done in constructing a
proof by induction is usually in proving the inductive step. This was certainly
the case in Proposition 4.2. However, the basis step is an essential part of the
proof. Without it, the proof is incomplete. To see this, let P.n/ be
1 C 2 C C n D
n2
C n C 1
2
:
(a) Let k 2 N. Complete the following proof that if P.k/ is true, then
P.k C 1/ is true.
Let k 2 N. Assume that P.k/ is true. That is, assume that
1 C 2 C C k D
k2 C k C 1
2
: (1)
The goal is to prove that P.k C 1/ is true. That is, we need to prove
that
1 C 2 C C k C .k C 1/ D
.k C 1/2
C .k C 1/ C 1
2
: (2)
To do this, we add .k C 1/ to both sides of equation (1). This gives
1 C 2 C C k C .k C 1/ D
k2
C k C 1
2
C .k C 1/
D :
(b) Is P.1/ true? Is P.2/ true? What about P.3/ and P.4/? Explain how
this shows that the basis step is an essential part of a proof by induction.
20. Regions of a Circle. Place n equally spaced points on a circle and connect
each pair of points with the chord of the circle determined by that pair of
points. See Figure 4.1.

1 point, 1 region 2 points, 2 regions
3 points, 4 regions 4 points, ? regions
Figure 4.1: Regions of Circles
Count the number of distinct regions within each circle. For example, with
three points on the circle, there are four distinct regions. Organize your data
in a table with two columns: “Number of Points on the Circle” and “Number
of Distinct Regions in the Circle.”
(a) How many regions are there when there are four equally spaced points
on the circle?
(b) Based on the work so far, make a conjecture about how many distinct
regions would you get with five equally spaced points.
(c) Based on the work so far, make a conjecture about how many distinct
regions would you get with six equally spaced points.
(d) Figure 4.2 shows the figures associated with Parts (b) and (c). Count
the number of regions in each case. Are your conjectures correct or
incorrect?
(e) Explain why this activity shows that the inductive step is an essential
part of a proof by mathematical induction.

Figure 4.2: Regions of Circles
4.2 Other Forms of Mathematical Induction
Preview Activity 1 (Exploring a Proposition about Factorials)
Definition. If n is a natural number, we define n factorial, denoted by nŠ , to
be the product of the first n natural numbers. In addition, we define 0Š to be
equal to 1.
Using this definition, we see that
0Š D 1 3Š D 1 2 3 D 6
1Š D 1 4Š D 1 2 3 4 D 24
2Š D 1 2 D 2 5Š D 1 2 3 4 5 D 120:
In general, we write nŠ D 1 2 3 .n 1/ n or nŠ D n .n 1/ 2 1. Notice
that for any natural number n, nŠ D n .n 1/Š.
1. Compute the values of 2n and nŠ for each natural number n with 1 n 7.
Now let P.n/ be the open sentence, “nŠ 2n
.”
2. Which of the statements P.1/ through P.7/ are true?
3. Based on the evidence so far, does the following proposition appear to be
true or false? For each natural number n with n 4, nŠ 2n.
Let k be a natural number with k 4. Suppose that we want to prove that if P.k/
is true, then P.k C 1/ is true. (This could be the inductive step in an induction

4.2. Other Forms of Mathematical Induction 189
proof.) To do this, we would be assuming that kŠ 2k and would need to prove
that .k C 1/Š 2kC1
. Notice that if we multiply both sides of the inequality
kŠ 2k by .k C 1/, we obtain
.k C 1/ kŠ .k C 1/2k
: (1)
4. In the inequality in (1), explain why .k C 1/ kŠ D .k C 1/Š.
5. Now look at the right side of the inequality in (1). Since we are assuming
that k 4, we can conclude that .k C 1/ 2. Use this to help explain why
.k C 1/2k 2kC1.
6. Now use the inequality in (1) and the work in steps (4) and (5) to explain
why .k C 1/Š 2kC1
.
Preview Activity 2 (Prime Factors of a Natural Number)
Recall that a natural number p is a prime number provided that it is greater than
1 and the only natural numbers that divide p are 1 and p. A natural number other
than 1 that is not a prime number is a composite number. The number 1 is neither
prime nor composite.
1. Give examples of four natural numbers that are prime and four natural num-
bers that are composite.
2. Write each of the natural numbers 20, 40, 50, and 150 as a product of prime
numbers.
3. Do you think that any composite number can be written as a product of prime
numbers?
4. Write a useful description of what it means to say that a natural number is a
composite number (other than saying that it is not prime).
5. Based on your work in Part (2), do you think it would be possible to use
induction to prove that any composite number can be written as a product of
prime numbers?
The Domino Theory
Mathematical induction is frequently used to prove statements of the form
.8n 2 N/ .P.n// ; (1)

where P.n/ is an open sentence. This means that we are proving that every state-
ment in the following infinite list is true.
P.1/; P.2/; P.3/; : : : (2)
The inductive step in a proof by induction is to prove that if one statement in this
infinite list of statements is true, then the next statement in the list must be true.
Now imagine that each statement in (2) is a domino in a chain of dominoes. When
we prove the inductive step, we are proving that if one domino is knocked over,
then it will knock over the next one in the chain. Even if the dominoes are set up
so that when one falls, the next one will fall, no dominoes will fall unless we start
by knocking one over. This is why we need the basis step in an induction proof.
The basis step guarantees that we knock over the first domino. The inductive step,
then, guarantees that all dominoes after the first one will also fall.
Now think about what would happen if instead of knocking over the first domino,
we knock over the sixth domino. If we also prove the inductive step, then we would
know that every domino after the sixth domino would also fall. This is the idea of
the Extended Principle of Mathematical Induction. It is not necessary for the basis
step to be the proof that P.1/ is true. We can make the basis step be the proof that
P.M/ is true, where M is some natural number. The Extended Principle of Math-
ematical Induction can be generalized somewhat by allowing M to be any integer.
We are still only concerned with those integers that are greater than or equal to M.
The Extended Principle of Mathematical Induction
Let M be an integer. If T is a subset of Z such that
1. M 2 T, and
2. For every k 2 Z with k M, if k 2 T, then .k C 1/ 2 T,
then T contains all integers greater than or equal to M. That is,
fn 2 Z j n Mg T.
Using the Extended Principle of Mathematical Induction
The primary use of the Principle of Mathematical Induction is to prove statements
of the form
.8n 2 Z; with n M / .P.n//;
where M is an integer and P.n/ is some open sentence. (In most induction proofs,
we will use a value of M that is greater than or equal to zero.) So our goal is to

prove that the truth set T of the predicate P.n/ contains all integers greater than or
equal to M. So to verify the hypothesis of the Extended Principle of Mathematical
Induction, we must
1. Prove that M 2 T. That is, prove that P.M/ is true.
2. Prove that for every k 2 Z with k M, if k 2 T, then .k C 1/ 2 T. That is,
prove that if P.k/ is true, then P.k C 1/ is true.
As before, the first step is called the basis step or the initial step, and the sec-
ond step is called the inductive step. This means that a proof using the Extended
Principle of Mathematical Induction will have the following form:
Let M be an integer. To prove: .8n 2 Z with n M/ .P.n//
Basis step: Prove P.M/.
Inductive step: Prove that for every k 2 Z with k M,
We can then conclude that P.n/ is true for all n 2 Z with n M.
This is basically the same procedure as the one for using the Principle of Math-
ematical Induction. The only difference is that the basis step uses an integer M
other than 1. For this reason, when we write a proof that uses the Extended Princi-
ple of Mathematical Induction, we often simply say we are going to use a proof by
mathematical induction. We will use the work from Preview Activity 1 to illustrate
such a proof.
Proposition 4.7. For each natural number n with n 4, nŠ 2n
.
Proof. We will use a proof by mathematical induction. For this proof, we let
P.n/ be “nŠ 2n.”
We first prove that P.4/ is true. Using n D 4, we see that 4Š D 24 and 24 D 16.
This means that 4Š 24
and, hence, P .4/ is true.
For the inductive step, we prove that for all k 2 N with k 4, if P.k/ is true,
then P.k C 1/ is true. So let k be a natural number greater than or equal to 4, and
assume that P.k/ is true. That is, assume that
kŠ 2k
: (1)

The goal is to prove that P.k C 1/ is true or that .k C 1/Š 2kC1. Multiplying
both sides of inequality (1) by k C 1 gives
.k C 1/ kŠ .k C 1/ 2k
; or
.k C 1/Š .k C 1/ 2k
: (2)
Now, k 4. Thus, k C 1 2, and hence .k C 1/ 2k
2 2k
. This means that
.k C 1/ 2k
2kC1
: (3)
Inequalities (2) and (3) show that
.k C 1/Š 2kC1
;
and this proves that if P.k/ is true, then P.k C 1/ is true. Thus, the inductive step
has been established, and so by the Extended Principle of Mathematical Induction,
nŠ 2n for each natural number n with n 4.
Progress Check 4.8 (Formulating Conjectures)
Formulate a conjecture (with an appropriate quantifier) that can be used as an an-
swer to each of the following questions.
1. For which natural numbers n is 3n
greater than 1 C 2n
?
2. For which natural numbers n is 2n greater than .n C 1/2
?
3. For which natural numbers n is

1 C
1
n
n
less than n?
The Second Principle of Mathematical Induction
Let P.n/ be
n is a prime number or n is a product of prime numbers.
(This is related to the work in Preview Activity 2.)
Suppose we would like to use induction to prove that P.n/ is true for all natural
numbers greater than 1. We have seen that the idea of the inductive step in a proof

by induction is to prove that if one statement in an infinite list of statements is true,
then the next statement must also be true. The problem here is that when we factor
a composite number, we do not get to the previous case. For example, if assume
that P.39/ is true and we want to prove that P.40/ is true, we could factor 40 as
40 D 2 20. However, the assumption that P.39/ is true does not help us prove
that P.40/ is true.
This work is intended to show the need for another principle of induction. In
the inductive step of a proof by induction, we assume one statement is true and
prove the next one is true. The idea of this new principle is to assume that all of the
previous statements are true and use this assumption to prove the next statement is
true. This is stated formally in terms of subsets of natural numbers in the Second
Principle of Mathematical Induction. Rather than stating this principle in two ver-
sions, we will state the extended version of the Second Principle. In many cases,
we will use M D 1 or M D 0.
The Second Principle of Mathematical Induction
1. M 2 T, and
2. For every k 2 Z with k M, if fM; M C 1; : : :; kg T, then
.k C 1/ 2 T,
then T contains all integers greater than or equal to M. That is,
fn 2 Z j n Mg T.
Using the Second Principle of Mathematical Induction
The primary use of mathematical induction is to prove statements of the form
.8n 2 Z; with n M / .P.n// ;
where M is an integer and P.n/ is some predicate. So our goal is to prove that the
truth set T of the predicate P.n/ contains all integers greater than or equal to M.
To use the Second Principle of Mathematical Induction, we must
1. Prove that M 2 T. That is, prove that P.M/ is true.
2. Prove that for every k 2 N, if k M and fM; M C 1; : : :; kg T, then
.k C 1/ 2 T. That is, prove that if P.M/; P.M C 1/; : : :; P.k/ are true,
then P.k C 1/ is true.

As before, the first step is called the basis step or the initial step, and the
second step is called the inductive step. This means that a proof using the Second
Principle of Mathematical Induction will have the following form:
Let M be an integer. To prove: .8n 2 Z with n M / .P.n//
Inductive step: Let k 2 Z with k M. Prove that if
P.M/; P.M C 1/; : : :; P.k/ are true, then
P.k C 1/ is true.
We will use this procedure to prove the propositionsuggestedin Preview Activity2.
Theorem 4.9. Each natural number greater than 1 is either a prime number or is
a product of prime numbers.
Proof. We will use the Second Principle of Mathematical Induction. We let P.n/
be
n is either a prime number or n is a product of prime numbers.
For the basis step, P.2/ is true since 2 is a prime number.
To prove the inductive step, we let k be a natural number with k 2. We assume
that P.2/; P.3/; : : :; P.k/ are true. That is, we assume that each of the natural
numbers 2; 3; : : :; k is a prime number or a product of prime numbers. The goal
is to prove that P.k C 1/ is true or that .k C 1/ is a prime number or a product of
prime numbers.
Case 1: If .k C 1/ is a prime number, then P.k C 1/ is true.
Case 2: If .k C 1/ is not a prime number, then .k C 1/ can be factored into a
product of natural numbers with each one being less than .k C 1/. That is, there
exist natural numbers a and b with
k C 1 D a b; and 1 a k and 1 b k:

Using the inductive assumption, this means that P.a/ and P.b/ are both true.
Consequently, a and b are prime numbers or are products of prime numbers. Since
k C 1 D a b, we conclude that .k C 1/ is a product of prime numbers. That is,
we conclude that P .k C 1/ is true. This proves the inductive step.
Hence, by the Second Principle of Mathematical Induction, we conclude that
P.n/ is true for all n 2 N with n 2, and this means that each natural number
greater than 1 is either a prime number or is a product of prime numbers.
We will conclude this section with a progress check that is really more of an
activity. We do this rather than including the activity at the end of the exercises
since this activity illustrates a use of the Second Principle of Mathematical Induc-
tion in which it is convenient to have the basis step consist of the proof of more
than one statement.
Progress Check 4.10 (Using the Second Principle of Induction)
Consider the following question:
For which natural numbers n do there exist nonnegative integers x and y
such that n D 3x C 5y?
To help answer this question, we will let Z
D fx 2 Z j x 0g, and let P .n/ be
There exist x; y 2 Z
such that n D 3x C 5y.
Notice that P.1/ is false since if both x and y are zero, then 3x C 5y D 0 and
if either x 0 or y 0, then 3x C 5y 3. Also notice that P.6/ is true since
6 D 3 2 C 5 0 and P.8/ is true since 8 D 3 1 C 5 1.
1. Explain why P.2/, P.4/, and P.7/ are false and why P.3/ and P.5/ are
true.
2. Explain why P.9/, P.10/, P.11/, and P.12/ are true.
We could continue trying to determine other values of n for which P.n/ is true.
However, let us see if we can use the work in part (2) to determine if P.13/ is true.
Notice that 13 D 3 C 10 and we know that P.10/ is true. We should be able to use
this to prove that P.13/ is true. This is formalized in the next part.
3. Let k 2 N with k 10. Prove that if P.8/, P.9/, : : :, P.k/ are true, then
P.k C 1/ is true. Hint: k C 1 D 3 C .k 2/.

4. Prove the following proposition using mathematical induction. Use the Sec-
ond Principle of Induction and have the basis step be a proof that P.8/, P.9/,
and P.10/ are true. (The inductive step is part (3).)
Proposition 4.11. For each n 2 N with n 8, there exist nonnegative
integers x and y such that n D 3x C 5y.
? (a) For each natural number n with n 2, 3n 1 C 2n.
(b) For each natural number n with n 6, 2n .n C 1/2
.
(c) For each natural number n with n 3,

1 C
1
n
n
n.
? 2. For which natural numbers n is n2 2n? Justify your conclusion.
3. For which natural numbers n is nŠ 3n? Justify your conclusion.
4. (a) Verify that

1
1
4

D
3
4
and that

1
1
4

1
1
9

D
4
6
.
(b) Verify that

1
1
4

1
1
9

1
1
16

D
5
8
and that

1
1
4

1
1
9

1
1
16

1
1
25

D
6
10
.
(c) For n 2 N with n 2, make a conjecture about a formula for the
product

1
1
4

1
1
9

1
1
16

1
1
n2

.
(d) Based on your work in Parts (4a) and (4b), state a proposition and then
use the Extended Principle of Mathematical Induction to prove your
proposition.
?
For each nonnegative integer n, 8n j .4n/Š:
6. Let y D ln x.

(a) Determine
dy
dx
,
d2y
dx2
,
d3y
dx3
, and
d4y
dx4
.
(b) Let n be a natural number. Formulate a conjecture for a formula for
dn
y
dxn
. Then use mathematical induction to prove your conjecture.
7. For which natural numbers n do there exist nonnegative integers x and y
such that n D 4x C 5y? Justify your conclusion.
?
8. Can each natural number greater than or equal to 4 be written as the sum
of at least two natural numbers, each of which is a 2 or a 3? Justify your
conclusion. For example, 7 D 2C2C3, and 17 D 2C2C2C2C3C3C3.
9. Can each natural number greater than or equal to 6 be written as the sum
of at least two natural numbers, each of which is a 2 or a 5? Justify your
conclusion. For example, 6 D 2C2C2, 9 D 2C2C5, and 17 D 2C5C5C5.
10. Use mathematical induction to prove the following proposition:
Let x be a real number with x 0. Then for each natural number n
with n 2, .1 C x/n
1 C nx.
Explain where the assumption that x 0 was used in the proof.
11. Prove that for each odd natural number n with n 3,

1 C
1
2

1
1
3

1 C
1
4

1 C
. 1/n
n

D 1:
?
12. Prove that for each natural number n,
any set with n elements has
n .n 1/
2
two-element subsets.
13. Prove or disprove each of the following propositions:
(a) For each n 2 N,
1
1 2
C
1
2 3
C C
1
n.n C 1/
D
n
n C 1
.
(b) For each natural number n with n 3,
1
3 4
C
1
4 5
C C
1
n.n C 1/
D
n 2
3n C 3
:
(c) For each n 2 N, 12C23C34C Cn.nC1/ D
n.n C 1/.n C 2/
3
.


n3
3
C
n2
2
C
7n
6

is a natural number.

n5
5
C
n4
2
C
n3
3
n
30

is an integer.
? 16. (a) Prove that if n 2 N, then there exists an odd natural number m and a
nonnegative integer k such that n D 2k
m.
(b) For each n 2 N, prove that there is only one way to write n in the form
described in Part (a). To do this, assume that n D 2k
m and n D 2q
p
where m and p are odd natural numbers and k and q are nonnegative
integers. Then prove that k D q and m D p.
(a) For each natural number n with n 2, 2n 1 C n.
Proof. We let k be a natural number and assume that 2k
1 C k.
Multiplying both sides of this inequality by 2, we see that 2kC1
2 C 2k. However, 2 C 2k 2 C k and, hence,
2kC1
1 C .k C 1/:
By mathematical induction, we conclude that 2n 1 C n.
(b) Each natural number greater than or equal to 6 can be written as the
sum of natural numbers, each of which is a 2 or a 5.
Proof. We will use a proof by induction. For each natural number n,
we let P.n/ be, “There exist nonnegative integers x and y such that
n D 2x C 5y.” Since
6 D 3 2 C 0 5 7 D 2 C 5
8 D 4 2 C 0 5 9 D 2 2 C 1 5
we see that P.6/; P.7/; P.8/, and P.9/ are true.
We now suppose that for some natural number k with k 10 that
P.6/; P.7/; : : :P.k/ are true. Now
k C 1 D .k 4/ C 5:

Since k 10, we see that k 4 6 and, hence, P.k 4/ is true. So
k 4 D 2x C 5y and, hence,
k C 1 D .2x C 5y/ C 5
D 2x C 5 .y C 1/:
This proves that P.kC1/ is true, and hence, by the Second Principle of
Mathematical Induction, we have proved that for each natural number
n with n 6, there exist nonnegative integers x and y such that n D
2x C 5y.
18. The Sum of the Angles of a Convex Quadrilateral. There is a famous
theorem in Euclidean geometry that states that the sum of the interior angles
of a triangle is 180ı.
(a) Use the theorem about triangles to determine the sum of the angles of
a convex quadrilateral. Hint: Draw a convex quadrilateral and draw a
diagonal.
(b) Use the result in Part (1) to determine the sum of the angles of a convex
pentagon.
(c) Use the result in Part (2) to determine the sum of the angles of a convex
hexagon.
(d) Let n be a natural number with n 3. Make a conjecture about the sum
of the angles of a convex polygon with n sides and use mathematical
induction to prove your conjecture.
19. De Moivre’s Theorem. One of the most interesting results in trigonometry
is De Moivre’s Theorem, which relates the complex number i to the trigono-
metric functions. Recall that the number i is the complex number whose
square is 1, that is, i2
D 1. One version of the theorem can be stated as
follows:
If x is a real number, then for each nonnegative integer n,
Œcos x C i.sinx/n
D cos.nx/ C i.sin.nx//:
This theorem is named after Abraham de Moivre (1667 – 1754), a French
mathematician.

(a) The proof of De Moivre’s Theorem requires the use of the trigonomet-
ric identities for the sine and cosine of the sum of two angles. Use the
Internet or a book to find identities for sin.˛ C ˇ/ and cos.˛ C ˇ/.
(b) To get a sense of how things work, expand Œcos x C i.sinx/2
and write
the result in the form a C bi. Then use the identities from part (1) to
prove that Œcos x C i.sinx/2
D cos.2x/ C i.sin.2x//.
(c) Use mathematical induction to prove De Moivre’s Theorem.
4.3 Induction and Recursion
Preview Activity 1 (Recursively Defined Sequences)
In a proof by mathematical induction, we “start with a first step” and then prove
that we can always go from one step to the next step. We can use this same idea to
define a sequence as well. We can think of a sequence as an infinite list of numbers
that are indexed by the natural numbers (or some infinite subset of N [ f0g). We
often write a sequence in the following form:
a1; a2; : : :; an; : : ::
The number an is called the nth
term of the sequence. One way to define a se-
quence is to give a specific formula for the nth term of the sequence such as
an D
1
n
.
Another way to define a sequence is to give a specific definition of the first term
(or the first few terms) and then state, in general terms, how to determine anC1 in
terms of n and the first n terms a1; a2; : : :; an. This process is known as definition
by recursion and is also called a recursive definition. The specific definition of
the first term is called the initial condition, and the general definition of anC1 in
terms of n and the first n terms a1; a2; : : :; an is called the recurrence relation.
(When more than one term is defined explicitly, we say that these are the initial
conditions.) For example, we can define a sequence recursively as follows:
b1 D 16, and for each n 2 N, bnC1 D
1
2
bn.

4.3. Induction and Recursion 201
Using n D 1 and then n D 2, we then see that
b2 D
1
2
b1 b3 D
1
2
b2
D
1
2
16 D
1
2
8
D 8 D 4
1. Calculate b4 through b10. What seems to be happening to the values of bn
as n gets larger?
2. Define a sequence recursively as follows:
T1 D 16, and for each n 2 N, TnC1 D 16 C
1
2
Tn.
Then T2 D 16 C
1
2
T1 D 16 C 8 D 24. Calculate T3 through T10. What
seems to be happening to the values of Tn as n gets larger?
The sequences in Parts (1) and (2) can be generalized as follows: Let a and r be
real numbers. Define two sequences recursively as follows:
a1 D a, and for each n 2 N, anC1 D r an.
S1 D a, and for each n 2 N, SnC1 D a C r Sn.
3. Determine formulas (in terms of a and r) for a2 through a6. What do you
think an is equal to (in terms of a, r, and n)?
4. Determine formulas (in terms of a and r) for S2 through S6. What do you
think an is equal to (in terms of a, r, and n)?
In Preview Activity 1 in Section 4.2, for each natural number n, we defined nŠ,
read n factorial, as the product of the first n natural numbers. We also defined 0Š
to be equal to 1. Now recursively define a sequence of numbers a0; a1; a2; : : : as
follows:
a0 D 1, and
for each nonnegative integer n, anC1 D .n C 1/ an.
Using n D 0, we see that this implies that a1 D 1 a0 D 1 1 D 1. Then using
n D 1, we see that
a2 D 2a1 D 2 1 D 2:

5. Calculate a3; a4; a5, and a6.
6. Do you think that it is possible to calculate a20 and a100? Explain.
7. Do you think it is possibleto calculate an for any natural number n? Explain.
8. Compare the values of a0; a1; a2; a3; a4; a5, and a6 with those of
0Š; 1Š; 2Š; 3Š; 4Š; 5Š, and 6Š. What do you observe? We will use mathematical
induction to prove a result about this sequence in Exercise (1).
Preview Activity 2 (The Fibonacci Numbers)
The Fibonacci numbers are a sequence of natural numbers f1; f2; f3; : : :; fn; : : :
defined recursively as follows:
f1 D 1 and f2 D 1, and
For each natural number n, fnC2 D fnC1 C fn.
In words, the recursion formula states that for any natural number n with n 3,
the nth
Fibonacci number is the sum of the two previous Fibonacci numbers. So
we see that
f3 D f2 C f1 D 1 C 1 D 2;
f4 D f3 C f2 D 2 C 1 D 3; and
f5 D f4 C f3 D 3 C 2 D 5:
1. Calculate f6 through f20.
2. Which of the Fibonacci numbers f1 through f20 are even? Which are mul-
tiples of 3?
3. For n D 2, n D 3, n D 4, and n D 5, how is the sum of the first .n 1/
Fibonacci numbers related to the .n C 1/st
Fibonacci number?
4. Record any other observations about the values of the Fibonacci numbers
or any patterns that you observe in the sequence of Fibonacci numbers. If
necessary, compute more Fibonacci numbers.

The Fibonacci Numbers
The Fibonacci numbers form a famous sequence in mathematics that was inves-
tigated by Leonardo of Pisa (1170 – 1250), who is better known as Fibonacci.
Fibonacci introduced this sequence to the Western world as a solution of the fol-
lowing problem:
Suppose that a pair of adult rabbits (one male, one female) produces a pair
of rabbits (one male, one female) each month. Also, suppose that newborn
rabbits become adults in two months and produce another pair of rabbits.
Starting with one adult pair of rabbits, how many pairs of rabbits will be
produced each month for one year?
Since we start with one adult pair, there will be one pair produced the first
month, and since there is still only one adult pair, one pair will also be produced in
the second month (since the new pair produced in the first month is not yet mature).
In the third month, two pairs will be produced, one by the original pair and one by
the pair which was produced in the first month. In the fourth month, three pairs
will be produced, and in the fifth month, five pairs will be produced.
The basic rule is that in a given month after the first two months, the number of
adult pairs is the number of adult pairs one month ago plus the number of pairs born
two months ago. This is summarized in Table 4.1, where the number of pairs pro-
duced is equal to the number of adult pairs, and the number of adult pairs follows
the Fibonacci sequence of numbers that we developed in Preview Activity 2.
Months 1 2 3 4 5 6 7 8 9 10
Adult Pairs 1 1 2 3 5 8 13 21 34 55
Newborn Pairs 1 1 2 3 5 8 13 21 34 55
Month-Old Pairs 0 1 1 2 3 5 8 13 21 34
Table 4.1: Fibonacci Numbers
Historically, it is interesting to note that Indian mathematicians were studying
these types of numerical sequences well before Fibonacci. In particular, about fifty
years before Fibonacci introduced his sequence, Acharya Hemachandra (1089 –
1173) considered the followingproblem, which is from the biography of Hemachan-
dra in the MacTutor History of Mathematics Archive at http://www-groups.dcs.st-
and.ac.uk/ history/Biographies/Hemachandra.html.

Suppose we assume that lines are composed of syllables which are either
short or long. Suppose also that each long syllable takes twice as long to
articulate as a short syllable. A line of length n contains n units where each
short syllable is one unit and each long syllable is two units. Clearly a line
of length n units takes the same time to articulate regardless of how it is
composed. Hemchandra asks: How many different combinations of short
and long syllables are possible in a line of length n?
This is an important problem in the Sanskrit language since Sanskrit meters are
based on duration rather than on accent as in the English Language. The answer
to this question generates a sequence similar to the Fibonacci sequence. Suppose
that hn is the number of patterns of syllables of length n. We then see that h1 D 1
and h2 D 2. Now let n be a natural number and consider pattern of length n C 2.
This pattern either ends in a short syllable or a long syllable. If it ends in a short
syllable and this syllable is removed, then there is a pattern of length n C 1, and
there are hnC1 such patterns. Similarly, if it ends in a long syllable and this syllable
is removed, then there is a pattern of length n, and there are hn such patterns. From
this, we conclude that
hnC2 D hnC1 C hn:
This actually generates the sequence 1, 2, 3, 5, 8, 13, 21, ldots. For more informa-
tion about Hemachandra, see the article Math for Poets and Drummers by Rachel
Wells Hall in the February 2008 issue of Math Horizons.
We will continue to use the Fibonacci sequence in this book. This sequence
may not seem all that important or interesting. However, it turns out that this se-
quence occurs in nature frequently and has applications in computer science. There
is even a scholarly journal, The Fibonacci Quarterly, devoted to the Fibonacci
numbers.
The sequence of Fibonacci numbers is one of the most studied sequences in
mathematics, due mainly to the many beautiful patterns it contains. Perhaps one
observation you made in Preview Activity 2 is that every third Fibonacci number
is even. This can be written as a proposition as follows:
For each natural number n; f3n is an even natural number:
As with many propositions associated with definitions by recursion, we can prove
this using mathematical induction. The first step is to define the appropriate open
sentence. For this, we can let P.n/ be, “f3n is an even natural number.”
Notice that P.1/ is true since f3 D 2. We now need to prove the inductive
step. To do this, we need to prove that for each k 2 N,

That is, we need to prove that for each k 2 N, if f3k is even, then f3.kC1/ is even.
So let’s analyze this conditional statement using a know-show table.
Step Know Reason
P f3k is even. Inductive hypothesis
P1 .9m 2 N/ .f3k D 2m/ Definition of “even integer”
:
:
:
:
:
:
:
:
:
Q1 .9q 2 N/ f3.kC1/ D 2q

Q f3.kC1/ is even. Definition of ”even integer”
Step Show Reason
The key question now is, “Is there any relation between f3.kC1/ and fk?” We
can use the recursion formula that defines the Fibonacci sequence to find such a
relation.
The recurrence relation for the Fibonacci sequence states that a Fibonacci num-
ber (except for the first two) is equal to the sum of the two previous Fibonacci num-
bers. If we write 3 .k C 1/ D 3k C 3, then we get f3.kC1/ D f3kC3. For f3kC3,
the two previous Fibonacci numbers are f3kC2 and f3kC1. This means that
f3kC3 D f3kC2 C f3kC1:
Using this and continuing to use the Fibonacci relation, we obtain the following:
f3.kC1/ D f3kC3
D f3kC2 C f3kC1
D .f3kC1 C f3k/ C f3kC1:
The preceding equation states that f3.kC1/ D 2f3kC1 C f3k. This equation
can be used to complete the proof of the induction step.
Progress Check 4.12 (Every Third Fibonacci Number Is Even)
Complete the proof of Proposition 4.13.
Proposition 4.13. For each natural number n, the Fibonacci number f3n is an
even natural number.
Hint: We have already defined the predicate P .n/ to be used in an induction proof
and have proved the basis step. Use the information in and after the preceding
know-show table to help prove that if f3k is even, then f3.kC1/ is even.

Geometric Sequences and Geometric Series
Let a; r 2 R. The following sequence was introduced in Preview Activity 1.
Initial condition: a1 D a.
Recurrence relation: For each n 2 N, anC1 D r an.
This is a recursive definition for a geometric sequence with initial term a and
(common) ratio r. The basic idea is that the next term in the sequence is obtained
by multiplying the previous term by the ratio r. The work in Preview Activity 1
suggests that the following proposition is true.
Theorem 4.14. Let a; r 2 R. If a geometric sequence is defined by a1 D a and
for each n 2 N, anC1 D r an, then for each n 2 N, an D a rn 1
.
The proof of this proposition is Exercise (6).
Another sequence that was introduced in Preview Activity 1 is related to geo-
metric series and is defined as follows:
Initial condition: S1 D a.
Recurrence relation: For each n 2 N, SnC1 D a C r Sn.
For each n 2 N, the term Sn is a (finite) geometric series with initial term a and
(common) ratio r. The work in Preview Activity 1 suggests that the following
proposition is true.
Theorem 4.15. Let a; r 2 R. If the sequence S1; S2; : : :; Sn; : : : is defined by
S1 D a and for each n 2 N, SnC1 D a C r Sn, then for each n 2 N, Sn D
a C a r C a r2
C C a rn 1
. That is, the geometric series Sn is the sum of
the first n terms of the corresponding geometric sequence.
The proof of Proposition 4.15 is Exercise (7). The recursive definition of a
geometric series and Proposition 4.15 give two different ways to look at geometric
series. Proposition 4.15 represents a geometric series as the sum of the first n
terms of the corresponding geometric sequence. Another way to determine the
sum a geometric series is given in Theorem 4.16, which gives a formula for the
sum of a geometric series that does not use a summation.
Theorem 4.16. Let a; r 2 R and r ¤ 1. If the sequence S1; S2; : : :; Sn; : : : is
defined by S1 D a and for each n 2 N, SnC1 D a C r Sn, then for each n 2 N,
Sn D a

1 rn
1 r

.
The proof of Proposition 4.16 is Exercise (8).

? 1. For the sequence a0;a1; a2; : : :; an; : : : , assume that a0 D 1 and that for
each n 2 N [ f0g, anC1 D .n C 1/ an. Use mathematical induction to prove
that for each n 2 N [ f0g, an D nŠ.
2. Assume that f1; f2; : : :; fn; : : : are the Fibonacci numbers. Prove each of
the following:
?
(a) For each n 2 N, f4n is a multiple of 3.
(b) For each n 2 N, f5n is a multiple of 5.
? (c) For each n 2 N with n 2, f1 C f2 C C fn 1 D fnC1 1.
(d) For each n 2 N, f1 C f3 C C f2n 1 D f2n.
(e) For each n 2 N, f2 C f4 C C f2n D f2nC1 1.
?
(f) For each n 2 N, f 2
1 C f 2
2 C C f 2
n D fnfnC1.
(g) For each n 2 N such that n 6 0 .mod 3/, fn is an odd integer.
3. Use the result in Part (f) of Exercise (2) to prove that
f 2
1 C f 2
2 C C f 2
n C f 2
nC1
f 2
1 C f 2
2 C C f 2
n
D 1 C
fnC1
fn
:
4. The quadratic formula can be used to show that ˛ D
1 C
p
5
2
and ˇ D
1
p
5
2
are the two real number solutions of the quadratic equation x2
x 1 D 0.
Notice that this implies that
˛2
D ˛ C 1; and
ˇ2
D ˇ C 1:
It may be surprising to find out that these two irrational numbers are closely
related to the Fibonacci numbers.
(a) Verify that f1 D
˛1 ˇ1
˛ ˇ
and that f2 D
˛2 ˇ2
˛ ˇ
.
(b) (This part is optional,but it may help with the inductionproof in part (c).)
Work with the relation f3 D f2 Cf1 and substitutethe expressions for
f1 and f2 from part (a). Rewrite the expression as a single fraction and
then in the numerator use ˛2 C ˛ D ˛.˛ C 1/ and a similar equation
involving ˇ. Now prove that f3 D
˛3
ˇ3
˛ ˇ
.

(c) Use induction to prove that for each natural number n, if ˛ D
1 C
p
5
2
and ˇ D
1
p
5
2
, then fn D
˛n
ˇn
˛ ˇ
. Note: This formula for the nth
Fibonacci number is known as Binet’s formula, named after the French
mathematician Jacques Binet (1786 – 1856).
5. Is the following conjecture true or false? Justify your conclusion.
Conjecture. Let f1; f2; : : :; fm; : : : be the sequence of the Fibonacci num-
bers. For each natural number n, the numbers fnfnC3, 2fnC1fnC2, and
f 2
nC1 C f 2
nC2

form a Pythagorean triple.
?
6. Prove Proposition 4.14. Let a; r 2 R. If a geometric sequence is defined
by a1 D a and for each n 2 N, anC1 D r an, then for each n 2 N,
an D a rn 1
.
7. Prove Proposition 4.15. Let a; r 2 R. If the sequence S1; S2; : : :; Sn; : : : is
defined by S1 D a and for each n 2 N, SnC1 D a C r Sn, then for each
n 2 N, Sn D a C a r C a r2
C C a rn 1
. That is, the geometric series
Sn is the sum of the first n terms of the corresponding geometric sequence.
? 8. Prove Proposition4.16. Let a; r 2 R and r ¤ 1. If the sequence S1; S2; : : :; Sn; : : :
is defined by S1 D a and for each n 2 N, SnC1 D a C r Sn, then for each
n 2 N, Sn D a

1 rn
1 r

.
9. For the sequence a1; a2; : : :; an; : : : , assume that a1 D 2 and that for each
n 2 N, anC1 D an C 5.
? (a) Calculate a2 through a6.
? (b) Make a conjecture for a formula for an for each n 2 N.
(c) Prove that your conjecture in Exercise (9b) is correct.
10. The sequence in Exercise (9) is an example of an arithmetic sequence. An
arithmetic sequence is defined recursively as follows:
Let c and d be real numbers. Define the sequence a1; a2; : : :; an; : : : by
a1 D c and for each n 2 N, anC1 D an C d.
(a) Determine formulas for a3 through a8.
(b) Make a conjecture for a formula for an for each n 2 N.

11. For the sequence a1; a2; : : :; an; : : : , assume that a1 D 1, a2 D 5, and that
for each n 2 N, anC1 D an C 2an 1. Prove that for each natural number n,
an D 2n C . 1/n.
? 12. For the sequence a1; a2; : : :; an; : : : , assume that a1 D 1 and that for each
n 2 N, anC1 D
p
5 C an.
(a) Calculate, or approximate, a2 through a6.
(b) Prove that for each n 2 N, an 3.
for each n 2 N, anC2 D 3anC1 2an.
?
(a) Calculate a3 through a6.
? (b) Make a conjecture for a formula for an for each n 2 N.
for each n 2 N, anC2 D
1
2

anC1 C
2
an

.
?
(a) Calculate a3 through a6.
(b) Prove that for each n 2 N, 1 an 2.
15. For the sequence a1; a2; : : :; an; : : :, assume that a1 D 1, a2 D 1, a3 D 1,
and for that each natural number n,
anC3 D anC2 C anC1 C an:
(a) Compute a4, a5, a6, and a7.
(b) Prove that for each natural number n with n 1, an 2n 2.
16. For the sequence a1; a2; : : :; an; : : :, assume that a1 D 1, and that for each
natural number n,
anC1 D an C n nŠ:
(a) Compute nŠ for the first 10 natural numbers.
? (b) Compute an for the first 10 natural numbers.
(c) Make a conjecture about a formula for an in terms of n that does not
involve a summation or a recursion.

(d) Prove your conjecture in Part (c).
17. For the sequence a1; a2; : : :; an; : : : , assume that a1 D 1, a2 D 1, and for
each n 2 N, anC2 D anC1 C 3an. Determine which terms in this sequence
are divisible by 4 and prove that your answer is correct.
18. The Lucas numbers are a sequence of natural numbers L1; L2; L3; : : :; Ln; : : :;
which are defined recursively as follows:
L1 D 1 and L2 D 3, and
For each natural number n, LnC2 D LnC1 C Ln.
List the first 10 Lucas numbers and the first ten Fibonacci numbers and then
prove each of the following propositions. The Second Principle of Mathe-
matical Induction may be needed to prove some of these propositions.
?
(a) For each natural number n, Ln D 2fnC1 fn.
(b) For each n 2 N with n 2, 5fn D Ln 1 C LnC1.
(c) For each n 2 N with n 3, Ln D fnC2 fn 2.
19. There is a formula for the Lucas numbers similar to the formula for the Fi-
bonacci numbers in Exercise (4). Let ˛ D
1 C
p
5
2
and ˇ D
1
p
5
2
. Prove
that for each n 2 N, Ln D ˛n
C ˇn
.
20. Use the result in Exercise (19), previously proven results from Exercise (18),
or mathematical induction to prove each of the following results about Lucas
numbers and Fibonacci numbers.
(a) For each n 2 N, Ln D
f2n
fn
.
(b) For each n 2 N, fnC1 D
fn C Ln
2
.
(c) For each n 2 N, LnC1 D
Ln C 5fn
2
.
(d) For each n 2 N with n 2, Ln D fnC1 C fn 1.

(a) Let fn be the nth Fibonacci number, and let ˛ be the positive solution
of the equation x2
D x C1. So ˛ D
1 C
p
5
2
. For each natural number
n, fn ˛n 1
.
Proof. We will use a proof by mathematical induction. For each natu-
ral number n, we let P.n/ be, “fn ˛n 1
.”
We first note that P.1/ is true since f1 D 1 and ˛0 D 1. We also notice
that P.2/ is true since f2 D 1 and, hence, f2 ˛1
.
We now let k be a natural number with k 2 and assume that P.1/,
P.2/, ..., P.k/ are all true. We now need to prove that P .k C 1/ is
true or that fkC1 ˛k.
Since P.k 1/ and P.k/ are true, we know that fk 1 ˛k 2
and
fk ˛k 1. Therefore,
fkC1 D fk C f k 1
fkC1 ˛k 1
C ˛k 2
fkC1 ˛k 2
.˛ C 1/:
We now use the fact that ˛ C 1 D ˛2 and the preceding inequality to
obtain
fkC1 ˛k 2
˛2
fkC1 ˛k
:
This proves that if P.1/, P.2/, ..., P.k/ are true, then P.k C 1/ is
true. Hence, by the Second Principle of Mathematical Induction, we
conclude that or each natural number n, fn ˛n 1.
22. Compound Interest. Assume that R dollars is deposited in an account that
has an interest rate of i for each compounding period. A compounding pe-
riod is some specified time period such as a month or a year.
For each integer n with n 0, let Vn be the amount of money in an account
at the end of the nth compounding period. Then
V1 D R C i R
D R .1 C i/
V2 D V1 C i V1
D .1 C i/ V1
D .1 C i/2
R:

(a) Explain why V3 D V2CiV2. Then use the formula for V2 to determine
a formula for V3 in terms of i and R.
(b) Determine a recurrence relation for VnC1 in terms of i and Vn.
(c) Write the recurrence relation in Part (22b) so that it is in the form of a
recurrence relation for a geometric sequence. What is the initial term
of the geometric sequence and what is the common ratio?
(d) Use Proposition 4.14 to determine a formula for Vn in terms of I, R,
and n.
23. The Future Value of an Ordinary Annuity. For an ordinary annuity, R
dollars is deposited in an account at the end of each compounding period. It
is assumed that the interest rate, i, per compounding period for the account
remains constant.
Let St represent the amount in the account at the end of the tth compounding
period. St is frequently called the future value of the ordinary annuity.
So S1 D R. To determine the amount after two months, we first note that
the amount after one month will gain interest and grow to .1 C i/S1. In
addition, a new deposit of R dollars will be made at the end of the second
month. So
S2 D R C .1 C i/S1:
(a) For each n 2 N, use a similar argument to determine a recurrence
relation for SnC1 in terms of R, i, and Sn.
(b) By recognizing this as a recursion formula for a geometric series, use
Proposition 4.16 to determine a formula for Sn in terms of R, i, and
n that does not use a summation. Then show that this formula can be
written as
Sn D R

.1 C i/n
1
i

:
(c) What is the future value of an ordinary annuity in 20 years if $200 dol-
lars is deposited in an account at the end of each month where the in-
terest rate for the account is 6% per year compounded monthly? What
is the amount of interest that has accumulated in this account during
the 20 years?

Inductive set, page 171
Factorial, page 201
Recursive definition, page 200
Fibonacci numbers, page 202
Geometric sequence, page 206
Geometric series, page 206
The Various Forms of Mathematical Induction
1. The Principle of Mathematical Induction
If T is a subset of N such that
(a) 1 2 T, and
(b) For every k 2 N, if k 2 T, then .k C 1/ 2 T,
then T D N.
Procedure for a Proof by Mathematical Induction
To prove .8n 2 N/ .P.n//
Basis step: Prove P.1/.
Inductive step: Prove that for each k 2 N, if P.k/ is true, then
P.k C 1/ is true.
2. The Extended Principle of Mathematical Induction
(a) M 2 T , and
(b) For every k 2 Z with k M, if k 2 T , then .k C 1/ 2 T ,
then T contains all integers greater than or equal to M.
Let M be an integer. To prove .8n 2 Z with n M/ .P.n//
Inductive step: Prove that for every k 2 Z with k M, if P.k/
is true, then P.k C 1/ is true.

3. The Second Principle of Mathematical Induction
(a) M 2 T, and
(b) For every k 2 Z with k M, if fM; M C 1; : : :; kg T, then
.k C 1/ 2 T,
then T contains all integers greater than or equal to M.
Let M be an integer. To prove .8n 2 Z with n M / .P.n//
Inductive step: Let k 2 Z with k M. Prove that
if P.M/; P.M C 1/; : : :; P.k/ are true, then
P.k C 1/ is true.
Important Results
Theorem 4.9. Each natural number greater than 1 is either a prime number
or is a product of prime numbers.
Theorem 4.14. Let a; r 2 R. If a geometric sequence is defined by a1 D a
and for each n 2 N, anC1 D r an, then for each n 2 N, an D a rn 1.
Theorem 4.15. Let a; r 2 R. If the sequence S1; S2; : : :; Sn; : : : is defined
by S1 D a and for each n 2 N, SnC1 D a C r Sn, then for each n 2 N,
Sn D a C a r C a r2
C C a rn 1
. That is, the geometric series Sn is
the sum of the first n terms of the corresponding geometric sequence.
Theorem 4.16. Let a; r 2 R and r ¤ 1. If the sequence S1; S2; : : :; Sn; : : :
is defined by S1 D a and for each n 2 N, SnC1 D a C r Sn, then for each
n 2 N, Sn D a

1 rn
1 r

.

Chapter 5
Set Theory
5.1 Sets and Operations on Sets
Preview Activity 1 (Set Operations)
Before beginning this section, it would be a good idea to review sets and set nota-
tion, including the roster method and set builder notation, in Section 2.3.
In Section 2.1, we used logical operators (conjunction, disjunction, negation)
to form new statements from existing statements. In a similar manner, there are
several ways to create new sets from sets that have already been defined. In fact,
we will form these new sets using the logical operators of conjunction (and), dis-
junction (or), and negation (not). For example, if the universal set is the set of
natural numbers N and
A D f1; 2; 3; 4; 5; 6g and B D f1; 3; 5; 7; 9g;
The set consisting of all natural numbers that are in A and are in B is the set
f1; 3; 5g;
The set consisting of all natural numbers that are in A or are in B is the set
f1; 2; 3; 4; 5; 6; 7; 9g; and
The set consisting of all natural numbers that are in A and are not in B is the
set f2; 4; 6g.
These sets are examples of some of the most common set operations, which are
given in the following definitions.
215

216 Chapter 5. Set Theory
Definition. Let A and B be subsets of some universal set U. The intersection
of A and B, written A B and read “A intersect B,” is the set of all elements
that are in both A and B. That is,
A B D fx 2 U j x 2 A and x 2 Bg:
The union of A and B, written A [ B and read “A union B,” is the set of all
elements that are in A or in B. That is,
A [ B D fx 2 U j x 2 A or x 2 BgŠ:
Definition. Let A and B be subsets of some universal set U. The set differ-
ence of A and B, or relative complement of B with respect to A, written
A B and read “A minus B” or “the complement of B with respect to A,” is
the set of all elements in A that are not in B. That is,
A B D fx 2 U j x 2 A and x … Bg:
The complement of the set A, written Ac
and read “the complement of A,”
is the set of all elements of U that are not in A. That is,
Ac
D fx 2 U j x … Ag:
For the rest of this preview activity, the universal set is U D f0; 1; 2; 3; : : :; 10g,
and we will use the following subsets of U :
A D f0; 1; 2; 3; 9g and B D f2; 3; 4; 5; 6g.
So in this case, A B D fx 2 U j x 2 A and x 2 Bg D f2; 3g. Use the roster
method to specify each of the following subsets of U .
1. A [ B 2. Ac
3. Bc
We can now use these sets to form even more sets. For example,
A Bc
D f0; 1; 2; 3; 9g f0; 1; 7; 8; 9; 10g D f0; 1; 9g:
Use the roster method to specify each of the following subsets of U .

5.1. Sets and Operations on Sets 217
4. A [ Bc 5. Ac Bc 6. Ac [ Bc 7. .A B/c
Preview Activity 2 (Venn Diagrams for Two Sets)
In Preview Activity 1, we worked with verbal and symbolic definitions of set op-
erations. However, it is also helpful to have a visual representation of sets. Venn
diagrams are used to represent sets by circles (or some other closed geometric
shape) drawn inside a rectangle. The points inside the rectangle represent the uni-
versal set U , and the elements of a set are represented by the points inside the
circle that represents the set. For example, Figure 5.1 is a Venn diagram showing
two sets.
U
A B
3
2
1
4
Figure 5.1: Venn Diagram for Two Sets
In Figure 5.1, the elements of A are represented by the points inside the left circle,
and the elements of B are represented by the points inside the right circle. The four
distinct regions in the diagram are numbered for reference purposes only. (The
numbers do not represent elements in a set.) The following table describes the four
regions in the diagram.
Region Elements of U Set
1 In A and not in B A B
2 In A and in B A B
3 In B and not in A B A
4 Not in A and not in B Ac
Bc
We can use these regions to represent other sets. For example, the set A [ B is
represented by regions 1, 2, and 3 or the shaded region in Figure 5.2.

B
U
3
4
A
1 2
Figure 5.2: Venn Diagram for A [ B
Let A and B be subsets of a universal set U . For each of the following, draw a
Venn diagram for two sets and shade the region that represent the specified set. In
addition, describe the set using set builder notation.
1. Ac
2. Bc
3. Ac
[ B
4. Ac
[ Bc
5. .A B/c
6. .A [ B/ .A B/
Set Equality, Subsets, and Proper Subsets
In Section 2.3, we introduced some basic definitions used in set theory, what it
means to say that two sets are equal and what it means to say that one set is a
subset of another set. See the definitions on page 55. We need one more definition.
Definition. Let A and B be two sets contained in some universal set U . The
set A is a proper subset of B provided that A B and A ¤ B. When A is a
proper subset of B, we write A B.
One reason for the definition of proper subset is that each set is a subset of
itself. That is,
If A is a set, then A A.
However, sometimes we need to indicate that a set X is a subset of Y but X ¤ Y .
For example, if
X D f1; 2g and Y D f0; 1; 2; 3g;

then X Y . We know that X Y since each element of X is an element of Y ,
but X ¤ Y since 0 2 Y and 0 … X. (Also, 3 2 Y and 3 … X.) Notice that the
notations A B and A B are used in a manner similar to inequality notation
for numbers (a b and a b).
It is often very important to be able to describe precisely what it means to say
that one set is not a subset of the other. In the preceding example, Y is not a subset
of X since there exists an element of Y (namely, 0) that is not in X.
In general, the subset relation is described with the use of a universal quantifier
since A B means that for each element x of U , if x 2 A, then x 2 B. So when
we negate this, we use an existential quantifier as follows:
A B means .8x 2 U/ Œ.x 2 A/ ! .x 2 B/.
A 6 B means : .8x 2 U / Œ.x 2 A/ ! .x 2 B/
.9x 2 U / : Œ.x 2 A/ ! .x 2 B/
.9x 2 U / Œ.x 2 A/ ^ .x … B/.
So we see that A 6 B means that there exists an x in U such that x 2 A and
x … B.
Notice that if A D ;, then the conditionalstatement, “For each x 2 U , if x 2 ;,
then x 2 B” must be true since the hypothesis will always be false. Another way
to look at this is to consider the following statement:
; 6 B means that there exists an x 2 ; such that x … B.
However, this statement must be false since there does not exist an x in ;. Since
this is false, we must conclude that ; B. Although the facts that ; B and
B B may not seem very important, we will use these facts later, and hence we
summarize them in Theorem 5.1.
Theorem 5.1. For any set B, ; B and B B.
In Section 2.3, we also defined two sets to be equal when they have precisely
the same elements. For example,
˚
x 2 R j x2
D 4 D f 2; 2g:
If the two sets A and B are equal, then it must be true that every element of A is
an element of B, that is, A B, and it must be true that every element of B is

an element of A, that is, B A. Conversely, if A B and B A, then A and
B must have precisely the same elements. This gives us the following test for set
equality:
Theorem 5.2. Let A and B be subsets of some universal set U. Then A D B if
and only if A B and B A.
Progress Check 5.3 (Using Set Notation)
Let the universal set be U D f1; 2; 3; 4; 5; 6g, and let
A D f1; 2; 4g; B D f1; 2; 3; 5g; C D
˚
x 2 U j x2
2 :
In each of the following, fill in the blank with one or more of the symbols ;
; D; ¤; 2; or … so that the resulting statement is true. For each blank, include all
symbols that result in a true statement. If none of these symbols makes a true
statement, write nothing in the blank.
A B ; A
5 B f5g B
A C f1; 2g C
f1; 2g A f4; 2; 1g A
6 A B ;
More about Venn Diagrams
In Preview Activity 2, we learned how to use Venn diagrams as a visual represen-
tation for sets, set operations, and set relationships. In that preview activity, we
restricted ourselves to using two sets. We can, of course, include more than two
sets in a Venn diagram. Figure 5.3 shows a general Venn diagram for three sets
(including a shaded region that corresponds to A C ).
In this diagram, there are eight distinct regions, and each region has a unique
reference number. For example, the set A is represented by the combination of re-
gions 1, 2, 4, and 5, whereas the set C is represented by the combination of regions
4, 5, 6, and 7. This means that the set A C is represented by the combination of
regions 4 and 5. This is shown as the shaded region in Figure 5.3.
Finally, Venn diagrams can also be used to illustrate special relationships be-
tween sets. For example, if A B, then the circle representing A should be
completely contained in the circle for B. So if A B, and we know nothing about

A B
1
2
3
4
5
6
7
8
C
U
Figure 5.3: Venn Diagram for A C
U
B
A
C
Figure 5.4: Venn Diagram Showing A B
any relationship between the set C and the sets A and B, we could use the Venn
diagram shown in Figure 5.4.
Progress Check 5.4 (Using Venn Diagrams)
Let A, B, and C be subsets of a universal set U .
1. For each of the following, draw a Venn diagram for three sets and shade the
region(s) that represent the specified set.
(a) .A B/ C
(b) .A B/ [ C
(c) .Ac
[ B/
(d) Ac
.B [ C /

2. Draw the most general Venn diagram showing B .A [ C /.
3. Draw the most general Venn diagram showing A .Bc
[ C /.
The Power Set of a Set
The symbol 2 is used to describe a relationship between an element of the univer-
sal set and a subset of the universal set, and the symbol is used to describe a
relationship between two subsets of the universal set. For example, the number 5
is an integer, and so it is appropriate to write 5 2 Z. It is not appropriate, however,
to write 5 Z since 5 is not a set. It is important to distinguish between 5 and
f5g. The difference is that 5 is an integer and f5g is a set consisting of one ele-
ment. Consequently, it is appropriate to write f5g Z, but it is not appropriate to
write f5g 2 Z. The distinctionbetween these two symbols .5 and f5g/ is important
when we discuss what is called the power set of a given set.
Definition. If A is a subset of a universal set U , then the set whose members
are all the subsets of A is called the power set of A. We denote the power set
of A by P.A/ . Symbolically, we write
P.A/ D fX U j X Ag:
That is, X 2 P.A/ if and only if X A.
When dealing with the power set of A, we must always remember that ; A
and A A. For example, if A D fa; bg, then the subsets of A are
;; fag; fbg; fa; bg: (1)
We can write this as
P.A/ D f;; fag; fbg; fa; bgg:
Now let B D fa; b; cg. Notice that B D A [ fcg. We can determine the subsets
of B by starting with the subsets of A in (1). We can form the other subsets of B
by taking the union of each set in (1) with the set fcg. This gives us the following
subsets of B.
fcg; fa; cg; fb; cg; fa; b; cg: (2)
So the subsets of B are those sets in (1) combined with those sets in (2). That is,
the subsets of B are
;; fag; fbg; fa; bg ; fcg; fa; cg; fb; cg; fa; b; cg; (3)

which means that
P.B/ D f;; fag; fbg; fa; bg; fcg; fa; cg; fb; cg; fa; b; cgg:
Notice that we could write
fa; cg B or that fa; cg 2 P.B/:
Also, notice that A has two elements and A has four subsets, and B has three ele-
ments and B has eight subsets. Now, let n be a nonnegative integer. The following
result can be proved using mathematical induction. (See Exercise 17).)
Theorem 5.5. Let n be a nonnegative integer and let T be a subset of some uni-
versal set. If the set T has n elements, then the set T has 2n
subsets. That is, P.T /
has 2n elements.
The Cardinality of a Finite Set
In our discussion of the power set, we were concerned with the number of ele-
ments in a set. In fact, the number of elements in a finite set is a distinguishing
characteristic of the set, so we give it the following name.
Definition. The number of elements in a finite set A is called the cardinality
of A and is denoted by card .A/.
For example, card .;/ D 0; card .fa; bg/ D 2; card .P.fa; bg// D 4.
Theoretical Note: There is a mathematical way to distinguish between finite and
infinite sets, and there is a way to define the cardinality of an infinite set. We will
not concern ourselves with this at this time. More about the cardinality of finite
and infinite sets is discussed in Chapter 9.
Standard Number Systems
We can use set notation to specify and help describe our standard number systems.
The starting point is the set of natural numbers, for which we use the roster
method.
N D f1; 2; 3; 4; : : : g

The integers consist of the natural numbers, the negatives of the natural numbers,
and zero. If we let N D f: : :; 4; 3; 2; 1g, then we can use set union and
write
Z D N [ f0g [ N:
So we see that N Z, and in fact, N Z.
We need to use set builder notation for the set Q of all rational numbers,
which consists of quotients of integers.
Q D
n m
n
ˇ
ˇ
ˇ m; n 2 Z and n ¤ 0
o
Since any integer n can be written as n D
n
1
, we see that Z Q.
We do not yet have the tools to give a complete description of the real numbers.
We will simply say that the real numbers consist of the rational numbers and the
irrational numbers. In effect, the irrational numbers are the complement of the
set of rational numbers Q in R. So we can use the notation Qc
D fx 2 R j x … Qg
and write
R D Q [ Qc
and Q Qc
D ;:
A number system that we have not yet discussed is the set of complex numbers.
The complex numbers, C, consist of all numbers of the form aCbi, where a; b 2 R
and i D
p
1 (or i2 D 1). That is,
C D
n
a C bi
ˇ
ˇ
ˇ a; b 2 R and i D
p
1
o
:
We can add and multiply complex numbers as follows: If a; b; c; d 2 R, then
.a C bi/ C .c C di/ D .a C c/ C .b C d/ i; and
.a C bi/ .c C di/ D ac C adi C bci C bdi2
D .ac bd/ C .ad C bc/ i:
? 1. Assume the universal set is the set of real numbers. Let
A D f 3; 2; 2; 3g; B D
˚
x 2 Rj x2
D 4 or x2
D 9 ;
C D
˚
x 2 R j x2
C 2 D 0 ; D D fx 2 R j x 0g:
Respond to each of the following questions. In each case, explain your an-
swer.

(a) Is the set A equal to the set B?
(b) Is the set A a subset of the set B?
(c) Is the set C equal to the set D?
(d) Is the set C a subset of the set D?
(e) Is the set A a subset of the set D?
?
2. (a) Explain why the set fa; bg is equal to the set fb; ag.
(b) Explain why the set fa; b; b; a; cg is equal to the set fb; c; ag.
?
3. Assume that the universal set is the set of integers. Let
A D f 3; 2; 2; 3g; B D
˚
x 2 Z j x2
9 ;
C D fx 2 Zj x 3g; D D f1; 2; 3; 4g:
In each of the following, fill in the blank with one or more of the symbols ,
, 6, D , ¤, 2, or … so that the resulting statement is true. For each blank,
include all symbols that result in a true statement. If none of these symbols
makes a true statement, write nothing in the blank.
A B ; A
5 C f5g C
A C f1; 2g B
f1; 2g A f3; 2; 1g D
4 B D ;
card .A/ card .D/ card .A/ card .B/
A P.A/ A P.B/
?
4. Write all of the proper subset relations that are possible using the sets of
numbers N, Z, Q, and R.
?
5. For each statement, write a brief, clear explanation of why the statement is
true or why it is false.
(a) The set fa; bg is a subset of fa; c; d; eg.
(b) The set f 2; 0; 2g is equal to
˚
x 2 Z j x is even and x2
5 .
(c) The empty set ; is a subset of f1g.
(d) If A D fa; bg, then the set fag is a subset of P.A/.

6. Use the definitions of set intersection, set union, and set difference to write
useful negations of these definitions. That is, complete each of the following
sentences
?
(a) x … A B if and only if : : ::
(b) x … A [ B if and only if : : ::
(c) x … A B if and only if : : ::
?
7. Let U D f1; 2; 3; 4; 5; 6; 7; 8; 9; 10g, and let
A D f3; 4; 5; 6; 7g; B D f1; 5; 7; 9g;
C D f3; 6; 9g; D D f2; 4; 6; 8g:
Use the roster method to list all of the elements of each of the following sets.
(a) A B
(b) A [ B
(c) .A [ B/c
(d) Ac Bc
(e) .A [ B/ C
(f) A C
(g) B C
(h) .A C/ [ .B C/
(i) B D
(j) .B D/c
(k) A D
(l) B D
(m) .A D/ [ .B D/
(n) .A [ B/ D
8. Let U D N, and let
A D fx 2 N j x 7g; B D fx 2 N j x is oddg;
C D fx 2 N j x is a multiple of 3g; D D fx 2 N j x is eveng:
Use the roster method to list all of the elements of each of the following sets.
(a) A B
(b) A [ B
(c) .A [ B/c
(d) Ac Bc
(e) .A [ B/ C
(f) .A C/ [ .B C /
(g) B D
(h) .B D/c
(i) A D
(j) B D
(k) .A D/ [ .B D/
(l) .A [ B/ D

9. Let P , Q, R, and S be subsets of a universal set U . Assume that
.P Q/ .R S/.
(a) Complete the following sentence:
For each x 2 U , if x 2 .P Q/, then : : : :
? (b) Write a useful negation of the statement in Part (9a).
(c) Write the contrapositive of the statement in Part (9a).
10. Let U be the universal set. Consider the following statement:
For all A, B, and C that are subsets of U , if A B, then Bc Ac.
?
(a) Identify three conditional statements in the given statement.
(b) Write the contrapositive of this statement.
(c) Write the negation of this statement.
11. Let A, B, and C be subsets of some universal set U . Draw a Venn diagram
for each of the following situations.
(a) A C
(b) A B D ;
(c) A 6 B; B 6 A; C A; and C 6 B
(d) A B; C B; and A C D ;
12. Let A, B, and C be subsets of some universal set U . For each of the fol-
lowing, draw a general Venn diagram for the three sets and then shade the
indicated region.
(a) A B
(b) A C
(c) .A B/ [ .A C /
(d) B [ C
(e) A .B [ C /
(f) .A B/ C
13. We can extend the idea of consecutive integers (See Exercise (10) in Sec-
tion 3.5) to represent four consecutive integers as m, m C 1, m C 2, and
m C 3, where m is an integer. There are other ways to represent four con-
secutive integers. For example, if k 2 Z, then k 1, k, k C 1, and k C 2 are
four consecutive integers.
(a) Prove that for each n 2 Z, n is the sum of four consecutive integers if
and only if n 2 .mod 4/.

(b) Use set builder notation or the roster method to specify the set of inte-
gers that are the sum of four consecutive integers.
(c) Specify the set of all natural numbers that can be written as the sum of
four consecutive natural numbers.
(d) Prove that for each n 2 Z, n is the sum of eight consecutive integers if
and only if n 4 .mod 8/.
(e) Use set builder notation or the roster method to specify the set of inte-
gers that are the sum of eight consecutive integers.
(f) Specify the set of all natural numbers can be written as the sum of eight
consecutive natural numbers.
14. One of the properties of real numbers is the so-called Law of Trichotomy,
which states that if a; b 2 R, then exactly one of the following is true:
a b; a D b; a b.
Is the following proposition concerning sets true or false? Either provide a
proof that it is true or a counterexample showing it is false.
If A and B are subsets of some universal set, then exactly one of the follow-
ing is true:
A B; A D B; B A.
15. Intervals of Real Numbers. In previous mathematics courses, we have fre-
quently used subsets of the real numbers called intervals. There are some
common names and notations for intervals. These are given in the following
table, where it is assumed that a and b are real numbers and a b.
Interval
Notation Set Notation Name
.a; b/ D fx 2 R j a x bg Open interval from a to b
Œa; b D fx 2 R j a x bg Closed interval from a to b
Œa; b/ D fx 2 R j a x bg Half-open interval
.a; b D fx 2 R j a x bg Half-open interval
.a; C1/ D fx 2 Rj x ag Open ray
. 1; b/ D fx 2 Rj x bg Open ray
Œa; C1/ D fx 2 Rj x ag Closed ray
. 1; b D fx 2 Rj x bg Closed ray

(a) Is .a; b/ a proper subset of .a; b? Explain.
(b) Is Œa; b a subset of .a; C1/? Explain.
(c) Use interval notation to describe
i. the intersection of the interval Œ 3; 7 with the interval .5; 9;
ii. the union of the interval Œ 3; 7 with the interval .5; 9;
iii. the set difference Œ 3; 7 .5; 9.
(d) Write the set fx 2 R j jxj 0:01g using interval notation.
(e) Write the set fx 2 R j jxj 2g as the union of two intervals.
16. More Work with Intervals. For this exercise, use the interval notation de-
scribed in Exercise 15.
(a) Determine the intersection and union of Œ2; 5 and Œ 1; C1/.
(b) Determine the intersection and union of Œ2; 5 and Œ3:4; C1/.
(c) Determine the intersection and union of Œ2; 5 and Œ7; C1/.
Now let a, b, and c be real numbers with a b.
(d) Explain why the intersection of Œa; b and Œc; C1/ is either a closed
interval, a set with one element, or the empty set.
(e) Explain why the union of Œa; b and Œc; C1/ is either a closed ray or
the union of a closed interval and a closed ray.
17. Proof of Theorem 5.5. To help with the proof by induction of Theorem 5.5,
we first prove the following lemma. (The idea for the proof of this lemma
was illustratedwith the discussionof power set after the definition on page 222.)
Lemma 5.6. Let A and B be subsets of some universal set. If A D B [ fxg,
where x … B, then any subset of A is either a subset of B or a set of the form
C [ fxg, where C is a subset of B.
Proof. Let A and B be subsets of some universal set, and assume that A D
B [ fxg where x … B. Let Y be a subset of A. We need to show that Y is a
subset of B or that Y D C [ fxg, where C is some subset of B. There are
two cases to consider: (1) x is not an element of Y, and (2) x is an element
of Y.
Case 1: Assume that x … Y. Let y 2 Y . Then y 2 A and y ¤ x. Since
A D B [ fxg;

this means that y must be in B. Therefore, Y B.
Case 2: Assume that x 2 Y. In this case, let C D Y fxg. Then every
element of C is an element of B. Hence, we can conclude that C B and
that Y D C [ fxg.
Cases (1) and (2) show that if Y A, then Y B or Y D C [ fxg, where
C B.
To begin the induction proof of Theorem 5.5, for each nonnegative integer
n, we let P.n/ be, “If a finite set has exactly n elements, then that set has
exactly 2n
subsets.”
(a) Verify that P.0/ is true. (This is the basis step for the induction proof.)
(b) Verify that P.1/ and P.2/ are true.
(c) Now assume that k is a nonnegative integer and assume that P.k/is
true. That is, assume that if a set has k elements, then that set has 2k
subsets. (This is the inductive assumption for the induction proof.)
Let T be a subset of the universal set with card .T / D k C 1, and let
x 2 T . Then the set B D T fxg has k elements.
Now use the inductive assumption to determine how many subsets B
has. Then use Lemma 5.6 to prove that T has twice as many subsets
as B. This should help complete the inductive step for the induction
proof.
5.2 Proving Set Relationships
Preview Activity 1 (Working with Two Specific Sets)
Let S be the set of all integers that are multiples of 6, and let T be the set of all
even integers.
1. List at least four different positive elements of S and at least four different
negative elements of S. Are all of these integers even?
2. Use the roster method to specify the sets S and T . (See Section 2.3 for
a review of the roster method.) Does there appear to be any relationship
between these two sets? That is, does it appear that the sets are equal or that
one set is a subset of the other set?

5.2. Proving Set Relationships 231
3. Use set builder notation to specify the sets S and T . (See Section 2.3 for a
review of the set builder notation.)
4. Using appropriate definitions, describe what it means to say that an integer
x is a multiple of 6 and what it means to say that an integer y is even.
5. In order to prove that S is a subset of T , we need to prove that for each
integer x, if x 2 S, then x 2 T .
Complete the know-show table in Table 5.1 for the proposition that S is a
subset of T .
This table is in the form of a proof method called the choose-an-element
method. This method is frequently used when we encounter a universal
quantifier in a statement in the backward process. (In this case, this is Step
Q1.) The key is that we have to prove something about all elements in Z.
We can then add something to the forward process by choosing an arbitrary
element from the set S. (This is done in Step P1.) This does not mean that
we can choose a specific element of S. Rather, we must give the arbitrary
element a name and use only the properties it has by being a member of the
set S. In this case, the element is a multiple of 6.
Step Know Reason
P S is the set of all integers that
are multiples of 6. T is the set of
all even integers.
Hypothesis
P1 Let x 2 S. Choose an arbitrary element
of S.
P 2 .9m 2 Z/ .x D 6m/ Definition of “multiple”
:
:
:
:
:
:
:
:
:
Q2 x is an element of T . x is even
Q1 .8x 2 Z/ Œ.x 2 S/ ! .x 2 T / Step P1 and Step Q2
Q S T . Definition of “subset”
Step Show Reason
Table 5.1: Know-show table for Preview Activity 1

Preview Activity 2 (Working with Venn Diagrams)
1. Draw a Venn diagram for two sets, A and B, with the assumption that A is
a subset of B. On this Venn diagram, lightly shade the area corresponding
to Ac
. Then, determine the region on the Venn diagram that corresponds to
Bc. What appears to be the relationship between Ac and Bc? Explain.
2. Draw a general Venn diagram for two sets, A and B. First determine the
region that corresponds to the set A B and then, on the Venn diagram, shade
the region corresponding to A .A B/ and shade the region corresponding
to A B. What appears to be the relationship between these two sets?
Explain.
In this section, we will learn how to prove certain relationships about sets. Two
of the most basic types of relationships between sets are the equality relation and
the subset relation. So if we are asked a question of the form, “How are the sets A
and B related?”, we can answer the question if we can prove that the two sets are
equal or that one set is a subset of the other set. There are other ways to answer this,
but we will concentrate on these two for now. This is similar to asking a question
about how two real numbers are related. Two real numbers can be related by the
fact that they are equal or by the fact that one number is less than the other number.
The Choose-an-Element Method
The method of proof we will use in this section can be called the choose-an-
element method. This method was introduced in Preview Activity 1. This method
is frequently used when we encounter a universal quantifier in a statement in the
backward process. This statement often has the form
For each element with a given property, something happens.
Since most statements with a universal quantifier can be expressed in the form of a
conditional statement, this statement could have the following equivalent form:
If an element has a given property, then something happens.
We will illustrate this with the proposition from Preview Activity 1. This proposi-
tion can be stated as follows:

Let S be the set of all integers that are multiples of 6, and let T be the set of
all even integers. Then S is a subset of T.
In Preview Activity 1, we worked on a know-show table for this proposition. The
key was that in the backward process, we encountered the following statement:
Each element of S is an element of T or, more precisely, if x 2 S, then
x 2 T.
In this case, the “element” is an integer, the “given property” is that it is an ele-
ment of S, and the “something that happens” is that the element is also an element
of T . One way to approach this is to create a list of all elements with the given
property and verify that for each one, the “something happens.” When the list is
short, this may be a reasonable approach. However, as in this case, when the list is
infinite (or even just plain long), this approach is not practical.
We overcome this difficulty by using the choose-an-element method, where
we choose an arbitrary element with the given property. So in this case, we choose
an integer x that is a multiple of 6. We cannot use a specific multiple of 6 (such as
12 or 24), but rather the only thing we can assume is that the integer satisfies the
property that it is a multiple of 6. This is the key part of this method.
Whenever we choose an arbitrary element with a given property,
we are not selecting a specific element. Rather, the only thing we
can assume about the element is the given property.
It is important to realize that once we have chosen the arbitrary element, we have
added information to the forward process. So in the know-show table for this
proposition, we added the statement, “Let x 2 S” to the forward process. Follow-
ing is a completed proof of this propositionfollowing the outline of the know-show
table from Preview Activity 1.
Proposition 5.7. Let S be the set of all integers that are multiples of 6, and let T be
the set of all even integers. Then S is a subset of T.
Proof. Let S be the set of all integers that are multiples of 6, and let T be the set
of all even integers. We will show that S is a subset of T by showing that if an
integer x is an element of S, then it is also an element of T .
Let x 2 S. (Note: The use of the word “let” is often an indication that the we
are choosing an arbitrary element.) This means that x is a multiple of 6. Therefore,

there exists an integer m such that
x D 6m:
Since 6 D 2 3, this equation can be written in the form
x D 2.3m/:
By closure properties of the integers, 3m is an integer. Hence, this last equation
proves that x must be even. Therefore, we have shown that if x is an element of S,
then x is an element of T , and hence that S T .
Having proved that S is a subset of T , we can now ask if S is actually equal to
T . The work we did in Preview Activity 1 can help us answer this question. In that
preview activity, we should have found several elements that are in T but not in S.
For example, the integer 2 is in T since 2 is even but 2 … S since 2 is not a multiple
of 6. Therefore, S ¤ T and we can also conclude that S is a proper subset of T .
One reason we do this in a “two-step” process is that it is much easier to work
with the subset relation than the proper subset relation. The subset relation is de-
fined by a conditional statement and most of our work in mathematics deals with
proving conditional statements. In addition, the proper subset relation is a conjunc-
tion of two statements (S T and S ¤ T ) and so it is natural to deal with the two
parts of the conjunction separately.
Progress Check 5.8 (Subsets and Set Equality)
Let A D fx 2 Z j x is a multiple of 9g and let B D fx 2 Z j x is a multiple of 3g.
1. Is the set A a subset of B? Justify your conclusion.
2. Is the set A equal to the set B? Justify your conclusion.
Progress Check 5.9 (Using the Choose-an-Element Method)
The Venn diagram in Preview Activity 2 suggests that the following proposition is
true.
Proposition 5.10. Let A and B be subsets of the universal set U . If A B, then
Bc
Ac
.
1. The conclusion of the conditional statement is Bc
Ac
. Explain why we
should try the choose-an-element method to prove this proposition.

2. Complete the following know-show table for this proposition and explain
exactly where the choose-an-element method is used.
Step Know Reason
P A B Hypothesis
P1 Let x 2 Bc. Choose an arbitrary element
of Bc
.
P 2 If x 2 A, then x 2 B. Definition of “subset”
:
:
:
:
:
:
:
:
:
Q1 If x 2 Bc
, then x 2 Ac
.
Q Bc Ac Definition of “subset”
Step Show Reason
Proving Set Equality
One way to prove that two sets are equal is to use Theorem 5.2 and prove each of
the two sets is a subset of the other set. In particular, let A and B be subsets of some
universal set. Theorem 5.2 states that A D B if and only if A B and B A.
In Preview Activity 2, we created a Venn diagram that indicated that
A .A B/ D A B. Following is a proof of this result. Notice where the
choose-an-element method is used in each case.
Proposition 5.11. Let A and B be subsets of some universal set. Then
A .A B/ D A B.
Proof. Let A and B be subsets of some universal set. We will prove that
A .A B/ D A B by proving that A .A B/ A B and that
A B A .A B/.
First, let x 2 A .A B/. This means that
x 2 A and x … .A B/:
We know that an element is in .A B/ if and only if it is in A and not in B. Since
x … .A B/, we conclude that x … A or x 2 B. However, we also know that
x 2 A and so we conclude that x 2 B. This proves that
x 2 A and x 2 B:
This means that x 2 A B, and hence we have proved that A .A B/ A B.

Now choose y 2 A B. This means that
y 2 A and y 2 B:
We note that y 2 .A B/ if and only if y 2 A and y … B and hence, y … .A B/
if and only if y … A or y 2 B. Since we have proved that y 2 B, we conclude that
y … .A B/, and hence, we have established that y 2 A and y … .A B/. This
proves that if y 2 A B, then y 2 A .A B/ and hence, A B A .A B/.
Since we have proved that A .A B/ A B and A B A .A B/,
we conclude that A .A B/ D A B.
Progress Check 5.12 (Set Equality)
Prove the following proposition. To do so, prove each set is a subset of the other
set by using the choose-an-element method.
Proposition 5.13. Let A and B be subsets of some universal set. Then A B D
A Bc.
Disjoint Sets
Earlier in this section, we discussed the concept of set equality and the relation
of one set being a subset of another set. There are other possible relationships
between two sets; one is that the sets are disjoint. Basically, two sets are disjoint if
and only if they have nothing in common. We express this formally in the following
definition.
Definition. Let A and B be subsets of the universal set U . The sets A and B
are said to be disjoint provided that A B D ;.
For example, the Venn diagram in Figure 5.5 shows two sets A and B with
A B. The shaded region is the region that represents Bc. From the Venn
diagram, it appears that A Bc
D ;. This means that A and Bc
are disjoint. The
preceding example suggests that the following proposition is true:
If A B, then A Bc D ;.
If we would like to prove this proposition, a reasonable “backward question” is,
“How do we prove that a set .namely A Bc
/ is equal to the empty set?”

U
B
A
Figure 5.5: Venn Diagram with A B
This question seems difficult to answer since how do we prove that a set is
empty? This is an instance where proving the contrapositive or using a proof by
contradiction could be reasonable approaches. To illustrate these methods, let us
assume the proposition we are trying to prove is of the following form:
If P , then T D ;.
If we choose to prove the contrapositive or use a proof by contradiction, we will
assume that T ¤ ;. These methods can be outlined as follows:
The contrapositive of “If P , then T D ;” is, “If T ¤ ;, then :P .” So in
this case, we would assume T ¤ ; and try to prove :P .
Using a proof by contradiction, we would assume P and assume that T ¤ ;.
From these two assumptions, we would attempt to derive a contradiction.
One advantage of these methods is that when we assume that T ¤ ;, then we
know that there exists an element in the set T . We can then use that element
in the rest of the proof. We will prove one of th the conditional statements for
Proposition 5.14 by proving its contrapositive. The proof of the other conditional
statement associated with Proposition 5.14 is Exercise (10).
Proposition 5.14. Let A and B be subsets of some universal set. Then A B if
and only if A Bc D ;.
Proof. Let A and B be subsets of some universal set. We will first prove that if
A B, then A Bc
D ;, by proving its contrapositive. That is, we will prove

If A Bc ¤ ;, then A 6 B.
So assume that A Bc
¤ ;. We will prove that A 6 B by proving that there must
exist an element x such that x 2 A and x … B.
Since A Bc ¤ ;, there exists an element x that is in A Bc. This means
that
x 2 A and x 2 Bc
:
Now, the fact that x 2 Bc means that x … B. Hence, we can conclude that
x 2 A and x … B:
This means that A 6 B, and hence, we have proved that if A Bc ¤ ;, then
A 6 B, and therefore, we have proved that if A B, then A Bc
D ;.
The proof that if A Bc D ;, then A B is Exercise (10).
Progress Check 5.15 (Proving Two Sets Are Disjoint)
It has been noted that it is often possible to prove that two sets are disjoint by
using a proof by contradiction. In this case, we assume that the two sets are not
disjoint and hence, there intersection is not empty. Use this method to prove that
the following two sets are disjoint.
A D fx 2 Z j x 3 .mod 12/g and B D fy 2 Z j y 2 .mod 8/g:
A Final Comment
We have used the choose-an-element method to prove Propositions 5.7, 5.11,
and 5.14. Proofs involving sets that use this method are sometimes referred to as
element-chasing proofs. This name is used since the basic method is to choose an
arbitrary element from one set and “chase it” until you prove it must be in another
set.
? 1. Let A D
˚
x 2 Rj x2 4 and let B D fx 2 R j x 2g.
(a) Is A B? Justify your conclusion with a proof or a counterexample.

(b) Is B A? Justify your conclusion with a proof or a counterexample.
2. Let A, B, and C be subsets of a universal set U .
(a) Draw a Venn diagram with A B and B C . Does it appear that
A C ?
(b) Prove the following proposition:
If A B and B C , then A C .
Note: This may seem like an obvious result. However, one of the
reasons for this exercise is to provide practice at properly writing a
proof that one set is a subset of another set. So we should start the
proof by assuming that A B and B C . Then we should choose an
arbitrary element of A.
? 3. Let A D fx 2 Z j x 7 .mod 8/g and B D fx 2 Z j x 3 .mod 4/g.
(a) List at least five different elements of the set A and at least five elements
of the set B.
(b) Is A B? Justify your conclusion with a proof or a counterexample.
(c) Is B A? Justify your conclusion with a proof or a counterexample.
4. Let C D fx 2 Z j x 7 .mod 9/g and D D fx 2 Z j x 1 .mod 3/g.
(a) List at least five different elements of the set C and at least five ele-
ments of the set D.
(b) Is C D? Justify your conclusion with a proof or a counterexample.
(c) Is D C ? Justify your conclusion with a proof or a counterexample.
5. In each case, determine if A B, B A, A D B, or A B D ; or none
of these.
?
(a) A D fx 2 Z j x 2 .mod 3/g and B D fy 2 Z j 6 divides .2y 4/g.
(b) A D fx 2 Z j x 3 .mod 4/g and B D fy 2 Z j 3 divides .y 2/g.
? (c) A D fx 2 Z j x 1 .mod 5/g and B D fy 2 Z j y 7 .mod 10/g.
6. To prove the following set equalities, it may be necessary to use some of
the properties of positive and negative real numbers. For example, it may be
necessary to use the facts that:

The product of two real numbers is positive if and only if the two real
numbers are either both positive or are both negative.
The product of two real numbers is negative if and only if one of the
two numbers is positive and the other is negative.
For example, if x .x 2/ 0, then we can conclude that either (1) x 0
and x 2 0 or (2) x 0 and x 2 0. However, in the first case, we
must have x 0 and x 2, and this is impossible. Therefore, we conclude
that x 0 and x 2 0, which means that 0 x 2.
Use the choose-an-element method to prove each of the following:
(a)
˚
x 2 R j x2 3x 10 0 D fx 2 R j 2 x 5g
(b)
˚
x 2 R j x2
5x C 6 0 D fx 2 R j 2 x 3g
(c)
˚
x 2 R j x2
4 D fx 2 R j x 2g [ fx 2 R j x 2g
7. Let A and B be subsets of some universal set U . Prove each of the following:
?
(a) A B A
?
(b) A A [ B
(c) A A D A
(d) A [ A D A
?
(e) A ; D ;
(f) A [ ; D A
8. Let A and B be subsets of some universal set U . From Proposition 5.10, we
know that if A B, then Bc
Ac
. Now prove the following proposition:
For all sets A and B that are subsets of some universal set U , A B
if and only if Bc Ac.
For all sets A and B that are subsets of some universal set U , the sets
A B and A B are disjoint.
10. Complete the proof of Proposition 5.14 by proving the following conditional
statement:
Let A and B be subsets of some universal set. If A Bc
D ;, then
A B .
11. Let A, B, C , and D be subsets of some universal set U . Are the following
propositions true or false? Justify your conclusions.

(a) If A B and C D and A and C are disjoint, then B and D are
disjoint.
(b) If A B and C D and B and D are disjoint, then A and C are
disjoint.
12. Let A, B, and C be subsets of a universal set U . Prove:
?
(a) If A B, then A C B C .
(b) If A B, then A [ C B [ C .
13. Let A, B, and C be subsets of a universal set U . Are the following proposi-
tions true or false? Justify your conclusions.
(a) If A C B C , then A B.
(b) If A [ C B [ C , then A B.
(c) If A [ C D B [ C , then A D B.
(d) If A C D B [ C , then A D B.
(e) If A [ C D B [ C and A C D B C , then A D B.
For all sets A, B, and C that are subsets of some universal set, if
A B D A C and Ac B D Ac C , then B D C .
15. Are the following biconditional statements true or false? Justify your con-
clusion. If a biconditional statement is found to be false, you should clearly
determine if one of the conditional statements within it is true and provide a
proof of this conditional statement.
? (a) For all subsets A and B of some universal set U, A B if and only if
A Bc
D ;.
?
(b) For all subsets A and B of some universal set U, A B if and only if
A [ B D B.
(c) For all subsets A and B of some universal set U, A B if and only if
A B D A.
(d) For all subsets A, B, and C of some universal set U, A B [ C if and
only if A B or A C.
(e) For all subsets A, B, and C of some universal set U, A B C if and
only if A B and A C.
16. Let S, T , X, and Y be subsets of some universal set. Assume that

(i) S [ T X [ Y; (ii) S T D ;; and (iii) X S.
(a) Using assumption (i), what conclusion(s) can be made if it is known
that a 2 T ?
(b) Using assumption (ii), what conclusion(s) can be made if it is known
that a 2 T ?
(c) Using all three assumptions, either prove that T Y or explain why it
is not possible to do so.
17. Evaluation of Proofs
(a) Let A, B, and C be subsets of some universal set. If A 6 B and
B 6 C , then A 6 C .
Proof. We assume that A, B, and C be subsets of some universal set
and that A 6 B and B 6 C . This means that there exists an element x
in A that is not in B and there exists an element x that is in B and not in
C . Therefore, x 2 A and x … C , and we have proved that A 6 C .
(b) Let A, B, and C be subsets of some universal set. If A B D A C ,
then B D C .
Proof. We assume that A B D A C and will prove that B D C .
We will first prove that B C .
So let x 2 B. If x 2 A, then x 2 A B, and hence, x 2 A C .
From this we can conclude that x 2 C . If x … A, then x … A B,
and hence, x … A C . However, since x … A, we may conclude that
x 2 C . Therefore, B C .
The proof that C B may be done in a similar manner. Hence, B D
C .
(c) Let A, B, and C be subsets of some universal set. If A 6 B and
B C , then A 6 C .
Proof. Assume that A 6 B and B C . Since A 6 B, there exists
an element x such that x 2 A and x … B. Since B C , we may
conclude that x … C . Hence, x 2 A and x … C , and we have proved
that A 6 C .

18. Using the Choose-an-Element Method in a Different Setting. We have
used the choose-an-element method to prove results about sets. This method,
however, is a general proof technique and can be used in settings other than
set theory. It is often used whenever we encounter a universal quantifier in a
statement in the backward process. Consider the following proposition.
Proposition 5.16. Let a, b, and t be integers with t ¤ 0. If t divides a and t
divides b, then for all integers x and y, t divides (ax + by).
(a) Whenever we encounter a new proposition, it is a good idea to ex-
plore the proposition by looking at specific examples. For example, let
a D 20, b D 12, and t D 4. In this case, t j a and t j b. In each of the
following cases, determine the value of .ax C by/ and determine if t
divides .ax C by/.
i. x D 1; y D 1
ii. x D 1; y D 1
iii. x D 2; y D 2
iv. x D 2; y D 3
v. x D 2; y D 3
vi. x D 2; y D 5
(b) Repeat Part (18a) with a D 21, b D 6, and t D 3.
Notice that the conclusion of the conditional statement in this proposition
involves the universal quantifier. So in the backward process, we would
have
Q: For all integers x and y, t divides ax C by.
The “elements” in this sentence are the integers x and y. In this case, these
integers have no “given property” other than that they are integers. The
“something that happens” is that t divides .ax C by/. This means that in the
forward process, we can use the hypothesis of the proposition and choose
integers x and y. That is, in the forward process, we could have
P : a, b, and t are integers with t ¤ 0, t divides a and t divides b.
P1: Let x 2 Z and let y 2 Z.
(c) Complete the following proof of Proposition 5.16.

Proof. Let a, b, and t be integers with t ¤ 0, and assume that t divides
a and t divides b. We will prove that for all integers x and y, t divides
.ax C by/.
So let x 2 Z and let y 2 Z. Since t divides a, there exists an integer m
such that : : : :
5.3 Properties of Set Operations
Preview Activity 1 (Exploring a Relationship between Two Sets)
Let A and B be subsets of some universal set U .
1. Draw two general Venn diagrams for the sets A and B. On one, shade the
region that represents .A [ B/c
, and on the other, shade the region that rep-
resents Ac
Bc
. Explain carefully how you determined these regions.
2. Based on the Venn diagrams in Part (1), what appears to be the relationship
between the sets .A [ B/c
and Ac Bc?
Some of the properties of set operations are closely related to some of the logical
operators we studied in Section 2.1. This is due to the fact that set intersection is
defined using a conjunction (and), and set union is defined using a disjunction (or).
For example, if A and B are subsets of some universal set U , then an element x is
in A [ B if and only if x 2 A or x 2 B.
3. Use one of De Morgan’s Laws (Theorem 2.8 on page 48) to explain carefully
what it means to say that an element x is not in A [ B.
4. What does it mean to say that an element x is in Ac? What does it mean to
say that an element x is in Bc
?
5. Explain carefully what it means to say that an element x is in Ac Bc.
6. Compare your response in Part (3) to your response in Part (5). Are they
equivalent? Explain.
7. How do you think the sets .A [ B/c
and Ac
Bc
are related? Is this con-
sistent with the Venn diagrams from Part (1)?

5.3. Properties of Set Operations 245
Preview Activity 2 (Proving that Statements Are Equivalent)
1. Let X, Y , and Z be statements. Complete a truth table for
Œ.X ! Y / ^ .Y ! Z/ ! .X ! Z/.
2. Assume that P , Q, and R are statements and that we have proven that the
following conditional statements are true:
If P then Q .P ! Q/.
If Q then R .Q ! R/.
If R then P .R ! P /.
Explain why each of the following statements is true.
(a) P if and only if Q .P $ Q/.
(b) Q if and only if R .Q $ R/.
(c) R if and only if P .R $ P /.
Remember that X $ Y is logically equivalent to .X ! Y / ^ .Y ! X/.
Algebra of Sets – Part 1
This section contains many results concerning the properties of the set operations.
We have already proved some of the results. Others will be proved in this section or
in the exercises. The primary purpose of this section is to have in one place many
of the properties of set operations that we may use in later proofs. These results
are part of what is known as the algebra of sets or as set theory.
Theorem 5.17. Let A, B, and C be subsets of some universal set U . Then
A B A and A A [ B.
If A B, then A C B C and A [ C B [ C .
Proof. The first part of this theorem was included in Exercise (7) from Section 5.2.
The second part of the theorem was Exercise (12) from Section 5.2.
The next theorem provides many of the properties of set operations dealing
with intersection and union. Many of these results may be intuitively obvious, but
to be complete in the development of set theory, we should prove all of them. We
choose to prove only some of them and leave some as exercises.

Theorem 5.18 (Algebra of Set Operations). Let A, B, and C be subsets of some
universal set U . Then all of the following equalities hold.
Properties of the Empty Set
and the Universal Set
A ; D ;
A [ ; D A
A U D A
A [ U D U
Idempotent Laws A A D A A [ A D A
Commutative Laws A B D B A A [ B D B [ A
Associative Laws .A B/ C D A .B C /
.A [ B/ [ C D A [ .B [ C /
Distributive Laws A .B [ C / D .A B/ [ .A C /
A [ .B C / D .A [ B/ .A [ C /
Before proving some of these properties, we note that in Section 5.2, we learned
that we can prove that two sets are equal by proving that each one is a subset of the
other one. However, we also know that if S and T are both subsets of a universal
set U , then
S D T if and only if for each x 2 U , x 2 S if and only if x 2 T .
We can use this to prove that two sets are equal by choosing an element from
one set and chasing the element to the other set through a sequence of “if and only
if” statements. We now use this idea to prove one of the commutative laws.
Proof of One of the Commutative Laws in Theorem 5.18
Proof. We will prove that A B D B A. Let x 2 A B. Then
x 2 A B if and only if x 2 A and x 2 B: (1)
However, we know that if P and Q are statements, then P ^ Q is logically equiv-
alent to Q ^ P . Consequently, we can conclude that
x 2 A and x 2 B if and only if x 2 B and x 2 A: (2)
Now we know that
x 2 B and x 2 A if and only if x 2 B A: (3)
This means that we can use (1), (2), and (3) to conclude that
x 2 A B if and only if x 2 B A;
and, hence, we have proved that A B D B A.

Progress Check 5.19 (Exploring a Distributive Property)
We can use Venn diagrams to explore the more complicated properties in Theo-
rem 5.18, such as the associative and distributive laws. To that end, let A, B, and
C be subsets of some universal set U .
1. Draw two general Venn diagrams for the sets A, B, and C . On one, shade
the region that represents A [ .B C/, and on the other, shade the region
that represents .A [ B/ .A [ C /. Explain carefully how you determined
these regions.
2. Based on the Venn diagrams in Part (1), what appears to be the relationship
between the sets A [ .B C / and .A [ B/ .A [ C /?
Proof of One of the Distributive Laws in Theorem 5.18
We will now prove the distributive law explored in Progress Check 5.19. Notice
that we will prove two subset relations, and that for each subset relation, we will
begin by choosing an arbitrary element from a set. Also notice how nicely a proof
dealing with the union of two sets can be broken into cases.
Proof. Let A, B, and C be subsets of some universal set U . We will prove that
A [ .B C/ D .A [ B/ .A [ C/ by proving that each set is a subset of the
other set.
We will first prove that A [ .B C/ .A [ B/ .A [ C/. We let
x 2 A [ .B C /. Then x 2 A or x 2 B C .
So in one case, if x 2 A, then x 2 A [ B and x 2 A [ C . This means that
x 2 .A [ B/ .A [ C /.
On the other hand, if x 2 B C , then x 2 B and x 2 C . But x 2 B
implies that x 2 A [ B, and x 2 C implies that x 2 A [ C . Since x is in both
sets, we conclude that x 2 .A [ B/ .A [ C /. So in both cases, we see that
x 2 .A [ B/ .A [ C /, and this proves that A [ .B C/ .A [ B/ .A [ C/.
We next prove that .A [ B/ .A [ C / A [ .B C /. So let
y 2 .A [ B/ .A [ C /. Then, y 2 A [ B and y 2 A [ C . We must prove
that y 2 A [ .B C/. We will consider the two cases where y 2 A or y … A. In
the case where y 2 A, we see that y 2 A [ .B C /.
So we consider the case that y … A. It has been established that y 2 A[B and
y 2 A [ C . Since y … A and y 2 A [ B, y must be an element of B. Similarly,

since y … A and y 2 A [ C , y must be an element of C . Thus, y 2 B C and,
hence, y 2 A [ .B C/.
In both cases, we have proved that y 2 A [ .B C /. This proves that
.A [ B/.A [ C / A[.B C /. The two subset relations establish the equality
of the two sets. Thus, A [ .B C/ D .A [ B/ .A [ C/.
Important Properties of Set Complements
The three main set operations are union, intersection, and complementation. The-
orems 5.18 and 5.17 deal with properties of unions and intersections. The next
theorem states some basic properties of complements and the important relations
dealing with complements of unions and complements of intersections. Two rela-
tionships in the next theorem are known as De Morgan’s Laws for sets and are
closely related to De Morgan’s Laws for statements.
Theorem 5.20. Let A and B be subsets of some universal set U . Then the follow-
ing are true:
Basic Properties .Ac
/c
D A
A B D A Bc
Empty Set and Universal Set A ; D A and A U D ;
;c
D U and U c
D ;
De Morgan’s Laws .A B/c
D Ac
[ Bc
.A [ B/c
D Ac Bc
Subsets and Complements A B if and only if Bc Ac
Proof. We will only prove one of De Morgan’s Laws, namely, the one that was
explored in Preview Activity 1. The proofs of the other parts are left as exercises.
Let A and B be subsets of some universal set U . We will prove that .A [ B/c
D
Ac Bc by proving that an element is in .A [ B/c
if and only if it is in Ac Bc.
So let x be in the universal set U . Then
x 2 .A [ B/c
if and only if x … A [ B; (1)
and
x … A [ B if and only if x … A and x … B: (2)
Combining (1) and (2), we see that
x 2 .A [ B/c
if and only if x … A and x … B: (3)

In addition, we know that
x … A and x … B if and only if x 2 Ac
and x 2 Bc
; (4)
and this is true if and only if x 2 Ac
Bc
. So we can use (3) and (4) to conclude
that
x 2 .A [ B/c
if and only if x 2 Ac
Bc
;
and, hence, that .A [ B/c
D Ac
Bc
.
Progress Check 5.21 (Using the Algebra of Sets)
1. Draw two general Venn diagrams for the sets A, B, and C . On one, shade the
region that represents .A [ B/ C , and on the other, shade the region that
represents .A C/[.B C /. Explain carefully how you determined these
regions and why they indicate that .A [ B/ C D .A C/ [ .B C /.
It is possible to prove the relationship suggested in Part (1) by proving that each
set is a subset of the other set. However, the results in Theorems 5.18 and 5.20 can
be used to prove other results about set operations. When we do this, we say that
we are using the algebra of sets to prove the result. For example, we can start by
using one of the basic properties in Theorem 5.20 to write
.A [ B/ C D .A [ B/ C c
:
We can then use one of the commutative properties to write
.A [ B/ C D .A [ B/ C c
D C c
.A [ B/ :
2. Determine which properties from Theorems 5.18 and 5.20 justify each of
the last three steps in the following outline of the proof that .A [ B/ C D
.A C / [ .B C/.
.A [ B/ C D .A [ B/ C c
(Theorem 5.20)
D C c
.A [ B/ (Commutative Property)
D C c
A

[ C c
B

D A C c

[ B C c

D .A C / [ .B C /

Note: It is sometimes difficult to use the properties in the theorems when
the theorems use the same letters to represent the sets as those being used
in the current problem. For example, one of the distributive properties from
Theorems 5.18 can be written as follows: For all sets X, Y , and Z that are
subsets of a universal set U ,
X .Y [ Z/ D .X Y / [ .X Z/ :
Proving that Statements Are Equivalent
When we have a list of three statements P , Q, and R such that each statement in
the list is equivalent to the other two statements in the list, we say that the three
statements are equivalent. This means that each of the statements in the list implies
each of the other statements in the list.
The purpose of Preview Activity 2 was to provide one way to prove that three
(or more) statements are equivalent. The basic idea is to prove a sequence of con-
ditional statements so that there is an unbroken chain of conditional statements
from each statement to every other statement. This method of proof will be used in
Theorem 5.22.
Theorem 5.22. Let A and B be subsets of some universal set U . The following
are equivalent:
1. A B 2. A Bc D ; 3. Ac [ B D U
Proof. To prove that these are equivalent conditions, we will prove that (1) implies
(2), that (2) implies (3), and that (3) implies (1).
Let A and B be subsets of some universal set U . We have proved that (1)
implies (2) in Proposition 5.14.
To prove that (2) implies (3), we will assume that A Bc
D ; and use the fact
that ;c D U . We then see that
A Bc
c
D ;c
:
Then, using one of De Morgan’s Laws, we obtain
Ac
[ Bc
c
D U
Ac
[ B D U:

This completes the proof that (2) implies (3).
We now need to prove that (3) implies (1). We assume that Ac [ B D U and
will prove that A B by proving that every element of A must be in B.
So let x 2 A. Then we know that x … Ac. However, x 2 U and since
Ac
[ B D U , we can conclude that x 2 Ac
[ B. Since x … Ac
, we conclude that
x 2 B. This proves that A B and hence that (3) implies (1).
Since we have now proved that (1) implies (2), that (2) implies (3), and that (3)
implies (1), we have proved that the three conditions are equivalent.
1. Let A be a subset of some universal set U. Prove each of the following (from
Theorem 5.20):
? (a) .Ac/c
D A
(b) A ; D A
? (c) ;c D U
(d) U c D ;
? 2. Let A, B, and C be subsets of some universal set U. As part of Theorem 5.18,
we proved one of the distributive laws. Prove the other one. That is, prove
that
A .B [ C / D .A B/ [ .A C /:
3. Let A, B, and C be subsets of some universal set U . As part of Theo-
rem 5.20, we proved one of De Morgan’s Laws. Prove the other one. That
is, prove that
.A B/c
D Ac
[ Bc
:
4. Let A, B, and C be subsets of some universal set U.
? (a) Draw two general Venn diagrams for the sets A, B, and C . On one,
shade the region that represents A .B [ C/, and on the other, shade
the region that represents .A B/ .A C /. Based on the Venn
diagrams, make a conjecture about the relationship between the sets
A .B [ C / and .A B/ .A C /.
(b) Use the choose-an-element method to prove the conjecture from Exer-
cise (4a).

? (c) Use the algebra of sets to prove the conjecture from Exercise (4a).
(a) Draw two general Venn diagrams for the sets A, B, and C . On one,
shade the region that represents A .B C/, and on the other, shade
the region that represents .A B/ [ .A C /. Based on the Venn
A .B C/ and .A B/ [ .A C/.
(b) Use the choose-an-element method to prove the conjecture from Exer-
cise (5a).
(c) Use the algebra of sets to prove the conjecture from Exercise (5a).
6. Let A, B, and C be subsets of some universal set U. Prove or disprove each
of the following:
? (a) .A B/ C D .A C / .B C /
(b) .A [ B/ .A B/ D .A B/ [ .B A/
the region that represents .A B/ C . Based on the Venn diagrams,
make a conjecture about the relationship between the sets A .B C /
and .A B/ C . (Are the two sets equal? If not, is one of the sets a
subset of the other set?)
(b) Prove the conjecture from Exercise (7a).
the region that represents .A B/ [ .A C c
/. Based on the Venn
A .B C/ and .A B/ [ .A C c
/. (Are the two sets equal? If
not, is one of the sets a subset of the other set?)
(b) Prove the conjecture from Exercise (8a).
9. Let A and B be subsets of some universal set U.
?
(a) Prove that A and B A are disjoint sets.

(b) Prove that A [ B D A [ .B A/.
10. Let A and B be subsets of some universal set U .
(a) Prove that A B and A B are disjoint sets.
(b) Prove that A D .A B/ [ .A B/.
11. Let A and B be subsets of some universal set U. Prove or disprove each of
the following:
(a) A .A Bc/ D A B
(b) .Ac
[ B/c
A D A B
(c) .A [ B/ A D B A
(d) .A [ B/ B D A .A B/
(e) .A [ B/ .A B/ D .A B/ [ .B A/
(a) If A, B, and C are subsets of some universal set U , then A .B C / D
A .B [ C /.
Proof.
A .B C / D .A B/ .A C /
D A Bc

A C c

D A Bc
C c

D A .B [ C /c
D A .B [ C /
(b) If A, B, and C are subsets of some universal set U , then A .B [ C / D
.A B/ .A C /.
Proof. We first write A .B [ C / D A .B [ C /c
and then use one
of De Morgan’s Laws to obtain
A .B [ C / D A Bc
C c

:
We now use the fact that A D A A and obtain
A .B [ C / D A A Bc
C c
D .A Bc
/ .A C c
/ D
.A B/ .A C /.

13. (Comparison to Properties of the Real Numbers). The following are some
of the basic properties of addition and multiplication of real numbers.
Commutative Laws: a C b D b C a, for all a; b 2 R.
a b D b a, for all a; b 2 R.
Associative Laws: .a C b/Cc D aC.b C c/, for all a; b; c 2 R.
.a b/ c D a .b c/ , for all a; b; c 2 R.
Distributive Law: a.b C c/ D ab Cac, for all a; b; c 2 R.
Additive Identity: For all a 2 R, a C 0 D a D 0 C a.
Multiplicative Identity: For all a 2 R, a 1 D a D 1 a.
Additive Inverses: For all a 2 R, a C . a/ D 0 D . a/ C a.
Multiplicative Inverses: For all a 2 R with a ¤ 0, a a 1
D 1 D
a 1 a.
Discuss the similarities and differences among the properties of addition and
multiplication of real numbers and the properties of union and intersection
of sets.
5.4 Cartesian Products
Preview Activity 1 (An Equation with Two Variables)
In Section 2.3, we introduced the concept of the truth set of an open sentence with
one variable. This was defined to be the set of all elements in the universal set that
can be substituted for the variable to make the open sentence a true statement.
In previous mathematics courses, we have also had experience with open sen-
tences with two variables. For example, if we assume that x and y represent real
numbers, then the equation
2x C 3y D 12

5.4. Cartesian Products 255
is an open sentence with two variables. An element of the truth set of this open
sentence (also called a solution of the equation) is an ordered pair .a; b/ of real
numbers so that when a is substituted for x and b is substituted for y, the open
sentence becomes a true statement (a true equation in this case). For example, we
see that the ordered pair .6; 0/ is in the truth set for this open sentence since
2 6 C 3 D 12
is a true statement. On the other hand, the ordered pair .4; 1/ is not in the truth set
for this open sentence since
2 4 C 3 1 D 12
is a false statement.
Important Note: The order of the of the two numbers in the ordered pair is very
important. We are using the convention that the first number is to be substituted for
x and the second number is to be substituted for y. With this convention, .3; 2/ is a
solution of the equation 2x C 3y D 12, but .2; 3/ is not a solution of this equation.
1. List six different elements of the truth set (often called the solution set) of
the open sentence with two variables 2x C 3y D 12.
2. From previous mathematics courses, we know that the graph of the equation
2x C 3y D 12 is a straight line. Sketch the graph of the equation 2x C
3y D 12 in the xy-coordinate plane. What does the graph of the equation
2x C 3y D 12 show?
3. Write a description of the solution set S of the equation 2x C 3y D 12 using
set builder notation.
Preview Activity 2 (The Cartesian Product of Two Sets)
In Preview Activity 1, we worked with ordered pairs without providing a formal
definition of an ordered pair. We instead relied on your previous work with ordered
pairs, primarily from graphing equations with two variables. Following is a formal
definition of an ordered pair.

Definition. Let A and B be sets. An ordered pair (with first element from
A and second element from B) is a single pair of objects, denoted by .a; b/,
with a 2 A and b 2 B and an implied order. This means that for two ordered
pairs to be equal, they must contain exactly the same objects in the same order.
That is, if a; c 2 A and b; d 2 B, then
.a; b/ D .c; d/ if and only if a D c and b D d:
The objects in the ordered pair are called the coordinates of the ordered pair.
In the ordered pair .a; b/, a is the first coordinate and b is the second coor-
dinate.
We will now introduce a new set operation that gives a way of combining el-
ements from two given sets to form ordered pairs. The basic idea is that we will
create a set of ordered pairs.
Definition. If A and B are sets, then the Cartesian product, A B, of A and
B is the set of all ordered pairs .x; y/ where x 2 A and y 2 B. We use the
notation A B for the Cartesian product of A and B, and using set builder
notation, we can write
A B D f.x; y/ j x 2 A and y 2 Bg :
We frequently read A B as “A cross B.” In the case where the two sets are
the same, we will write A2
for A A. That is,
A2
D A A D f.a; b/ j a 2 A and b 2 Ag :
Let A D f1; 2; 3g and B D fa; bg.
1. Is the ordered pair .3; a/ in the Cartesian product A B? Explain.
2. Is the ordered pair .3; a/ in the Cartesian product A A? Explain.
3. Is the ordered pair .3; 1/ in the Cartesian product A A? Explain.
4. Use the roster method to specify all the elements of A B. (Remember that
the elements of A B will be ordered pairs.
5. Use the roster method to specify all of the elements of the set A A D A2
.

6. For any sets C and D, explain carefully what it means to say that the ordered
pair .x; y/ is not in the Cartesian product C D.
Cartesian Products
When working with Cartesian products, it is important to remember that the Carte-
sian product of two sets is itself a set. As a set, it consists of a collection of el-
ements. In this case, the elements of a Cartesian product are ordered pairs. We
should think of an ordered pair as a single object that consists of two other objects
in a specified order. For example,
If a ¤ 1, then the ordered pair .1; a/ is not equal to the ordered pair .a; 1/.
That is, .1; a/ ¤ .a; 1/.
If A D f1; 2; 3g and B D fa; bg, then the ordered pair .3; a/ is an element
of the set A B. That is, .3; a/ 2 A B.
If A D f1; 2; 3g and B D fa; bg, then the ordered pair .5; a/ is not an
element of the set A B since 5 … A. That is, .5; a/ … A B.
In Section 5.3, we studied certain properties of set union, set intersection, and
set complements, which we called the algebra of sets. We will now begin some-
thing similar for Cartesian products. We begin by examining some specific exam-
ples in Progress Check 5.23 and a little later in Progress Check 5.24.
Progress Check 5.23 (Relationships between Cartesian Products)
Let A D f1; 2; 3g, T D f1; 2g, B D fa; bg , and C D fa; cg. We can then form
new sets from all of the set operations we have studied. For example, BC D fag,
and so
A .B C/ D f.1; a/ ; .2; a/ ; .3; a/g :
1. Use the roster method to list all of the elements (ordered pairs) in each of the
following sets:
(a) A B
(b) T B
(c) A C
(d) A .B C/
(e) .A B/ .A C /
(f) A .B [ C/
(g) .A B/ [ .A C /
(h) A .B C/
(i) .A B/ .A C /
(j) B A

2. List all the relationships between the sets in Part (1) that you observe.
The Cartesian Plane
In Preview Activity 1, we sketched the graph of the equation 2x C 3y D 12 in the
xy-plane. This xy-plane, with which you are familiar, is a representation of the set
R R or R2
. This plane is called the Cartesian plane.
The basic idea is that each ordered pair of real numbers corresponds to a point
in the plane, and each point in the plane corresponds to an ordered pair of real
numbers. This geometric representation of R2
is an extension of the geometric
representation of R as a straight line whose points correspond to real numbers.
Since the Cartesian product R2 corresponds to the Cartesian plane, the Carte-
sian product of two subsets of R corresponds to a subset of the Cartesian plane.
For example, if A is the interval Œ1; 3, and B is the interval Œ2; 5, then
A B D
˚
.x; y/ 2 R2
j 1 x 3 and 2 y 5 :
A graph of the set A B can then be drawn in the Cartesian plane as shown in
Figure 5.6.
1 2 3 4
1
2
3
4
5
6
x
y
Figure 5.6: Cartesian Product A B
This illustrates that the graph of a Cartesian product of two intervals of finite
length in R corresponds to the interior of a rectangle and possibly some or all of its

boundary. The solid line for the boundary in Figure 5.6 indicates that the boundary
is included. In this case, the Cartesian product contained all of the boundary of the
rectangle. When the graph does not contain a portion of the boundary, we usually
draw that portion of the boundary with a dotted line.
Note: A Caution about Notation. The standard notation for an open interval in R
is the same as the notation for an ordered pair, which is an element of R R. We
need to use the context in which the notation is used to determine which interpre-
tation is intended. For example,
If we write
p
2; 7

2 R R, then we are using
p
2; 7

to represent an
ordered pair of real numbers.
If we write .1; 2/ f4g, then we are interpreting .1; 2/ as an open interval.
We could write
.1; 2/ f4g D f.x; 4/ j 1 x 2g:
The following progress check explores some of the same ideas explored in Progress
Check 5.23 except that intervals of real numbers are used for the sets.
Progress Check 5.24 (Cartesian Products of Intervals)
We will use the following intervals that are subsets of R.
A D Œ0; 2 T D .1; 2/ B D Œ2; 4/ C D .3; 5
1. Draw a graph of each of the following subsets of the Cartesian plane and
write each subset using set builder notation.
(a) A B
(b) T B
(c) A C
(d) A .B C/
(e) .A B/ .A C /
(f) A .B [ C/
(g) .A B/ [ .A C /
(h) A .B C/
(i) .A B/ .A C /
(j) B A
2. List all the relationships between the sets in Part (1) that you observe.
One purpose of the work in Progress Checks 5.23 and 5.24 was to indicate the
plausibility of many of the results contained in the next theorem.

Theorem 5.25. Let A, B, and C be sets. Then
1. A .B C / D .A B/ .A C /
2. A .B [ C / D .A B/ [ .A C /
3. .A B/ C D .A C / .B C/
4. .A [ B/ C D .A C / [ .B C/
5. A .B C / D .A B/ .A C /
6. .A B/ C D .A C / .B C/
7. If T A, then T B A B.
8. If Y B, then A Y A B.
We will not prove all these results; rather, we will prove Part (2) of Theo-
rem 5.25 and leave some of the rest to the exercises. In constructing these proofs,
we need to keep in mind that Cartesian products are sets, and so we follow many of
the same principles to prove set relationships that were introduced in Sections 5.2
and 5.3.
The other thing to remember is that the elements of a Cartesian product are
ordered pairs. So when we start a proof of a result such as Part (2) of Theorem 5.25,
the primary goal is to prove that the two sets are equal. We will do this by proving
that each one is a subset of the other one. So if we want to prove that A.B [ C/
.A B/[.A C /, we can start by choosing an arbitrary element of A.B [ C/.
The goal is then to show that this element must be in .A B/[.A C/. When we
start by choosing an arbitrary element of A .B [ C/, we could give that element
a name. For example, we could start by letting
u be an element of A .B [ C/: (1)
We can then use the definition of “ordered pair” to conclude that
there exists x 2 A and there exists y 2 B [ C such that u D .x; y/ : (2)
In order to prove that A .B [ C / .A B/ [ .A C /, we must now show that
the ordered pair u from (1) is in .A B/ [ .A C/. In order to do this, we can
use the definition of set union and prove that
u 2 .A B/ or u 2 .A C /: (3)

Since u D .x; y/, we can prove (3) by proving that
.x 2 A and y 2 B/ or .x 2 A and y 2 C / : (4)
If we look at the sentences in (2) and (4), it would seem that we are very close to
proving that A .B [ C / .A B/ [ .A C /. Following is a proof of Part (2)
of Theorem 5.25.
Theorem 5.25 (Part (2)). Let A, B, and C be sets. Then
A .B [ C / D .A B/ [ .A C / :
Proof. Let A, B, and C be sets. We will prove that A .B [ C/ is equal to
.A B/ [ .A C / by proving that each set is a subset of the other set.
To prove that A .B [ C / .A B/ [ .A C /, we let u 2 A .B [ C/.
Then there exists x 2 A and there exists y 2 B [ C such that u D .x; y/. Since
y 2 B [ C , we know that y 2 B or y 2 C .
In the case where y 2 B, we have u D .x; y/, where x 2 A and y 2 B. So
in this case, u 2 A B, and hence u 2 .A B/ [ .A C/. Similarly, in the case
where y 2 C , we have u D .x; y/, where x 2 A and y 2 C . So in this case,
u 2 A C and, hence, u 2 .A B/ [ .A C/.
In both cases, u 2 .A B/ [ .A C /. Hence, we may conclude that if u is an
element of A .B [ C /, then u 2 .A B/ [ .A C/, and this proves that
A .B [ C / .A B/ [ .A C/: (1)
We must now prove that .A B/ [ .A C / A .B [ C /. So we let
v 2 .A B/ [ .A C /. Then v 2 .A B/ or v 2 .A C/.
In the case where v 2 .A B/, we know that there exists s 2 A and there
exists t 2 B such that v D .s; t/. But since t 2 B, we know that t 2 B [ C , and
hence v 2 A .B [ C/. Similarly, in the case where v 2 .A C /, we know that
there exists s 2 A and there exists t 2 C such that v D .s; t/. But because t 2 C ,
we can conclude that t 2 B [ C and, hence, v 2 A .B [ C /.
In both cases, v 2 A .B [ C /. Hence, we may conclude that if
v 2 .A B/ [ .A C /, then v 2 A .B [ C /, and this proves that
.A B/ [ .A C / A .B [ C /: (2)
The relationshipsin (1) and (2) prove that A.B [ C/ D .A B/[.A C /.

Final Note. The definition of an ordered pair in Preview Activity 2 may have
seemed like a lengthy definition, but in some areas of mathematics, an even more
formal and precise definition of “ordered pair” is needed. This definition is ex-
plored in Exercise (10).
? 1. Let A D f1; 2g, B D fa; b; c; dg, and C D f1; a; bg. Use the roster method
to list all of the elements of each of the following sets:
(a) A B
(b) B A
(c) A C
(d) A2
(e) A .B C /
(f) .A B/ .A C/
(g) A ;
(h) B f2g
2. Sketch a graph of each of the following Cartesian products in the Cartesian
plane.
(a) Œ0; 2 Œ1; 3
(b) .0; 2/ .1; 3
(c) Œ2; 3 f1g
(d) f1g Œ2; 3
(e) R .2; 4/
(f) .2; 4/ R
(g) R f 1g
(h) f 1g Œ1; C1/
? 3. Prove Theorem 5.25, Part (1): A .B C/ D .A B/ .A C /.
?
4. Prove Theorem 5.25, Part (4): .A [ B/ C D .A C/ [ .B C /.
5. Prove Theorem 5.25, Part (5): A .B C/ D .A B/ .A C /.
6. Prove Theorem 5.25, Part (7): If T A, then T B A B.
7. Let A D f1g; B D f2g; and C D f3g.
(a) Explain why A B ¤ B A.
(b) Explain why .A B/ C ¤ A .B C/.
8. Let A and B be nonempty sets. Prove that A B D B A if and only if
A D B.

Let A, B, and C be sets with A ¤ ;. If A B D A C, then B D C.
Explain where the assumption that A ¤ ; is needed.
10. (A Set Theoretic Definition of an Ordered Pair) In elementary mathemat-
ics, the notion of an ordered pair introduced at the beginning of this section
will suffice. However, if we are interested in a formal development of the
Cartesian product of two sets, we need a more precise definition of ordered
pair. Following is one way to do this in terms of sets. This definition is
credited to Kazimierz Kuratowski (1896 – 1980). Kuratowski was a famous
Polish mathematician whose main work was in the areas of topology and set
theory. He was appointed the Director of the Polish Academy of Sciences
and served in that position for 19 years.
Let x be an element of the set A, and let y be an element of the set B. The
ordered pair .x; y/ is defined to be the set ffxg ; fx; ygg. That is,
.x; y/ D ffxg ; fx; ygg:
(a) Explain how this definition allows us to distinguish between the or-
dered pairs .3; 5/ and .5; 3/.
(b) Let A and B be sets and let a; c 2 A and b; d 2 B. Use this definition
of an ordered pair and the concept of set equality to prove that .a; b/ D
.c; d/ if and only if a D c and b D d.
An ordered triple can be thought of as a single triple of objects, denoted
by .a; b; c/, with an implied order. This means that in order for two ordered
triples to be equal, they must contain exactly the same objects in the same
order. That is, .a; b; c/ D .p; q; r/ if and only if a D p, b D q and c D r.
(c) Let A, B, and C be sets, and let x 2 A; y 2 B; and z 2 C . Write a
set theoretic definition of the ordered triple .x; y; z/ similar to the set
theoretic definition of “ordered pair.”

5.5 Indexed Families of Sets
Preview Activity 1 (The Union and Intersection of a Family of Sets)
In Section 5.3, we discussed various properties of set operations. We will now
focus on the associative properties for set union and set intersection. Notice that
the definition of “set union” tells us how to form the union of two sets. It is the
associative law that allows us to discuss the union of three sets. Using the associate
law, if A, B, and C are subsets of some universal set, then we can define A[B [C
to be .A [ B/ [ C or A [ .B [ C /. That is,
A [ B [ C D .A [ B/ [ C D A [ .B [ C /:
For this activity, the universal set is N and we will use the following four sets:
A D f1; 2; 3; 4; 5g
B D f2; 3; 4; 5; 6g
C D f3; 4; 5; 6; 7g
D D f4; 5; 6; 7; 8g.
1. Use the roster method to specify the sets A[B [C , B [C [D, AB C ,
and B C D.
2. Use the roster method to specify each of the following sets. In each case, be
sure to follow the order specified by the parentheses.
(a) .A [ B [ C / [ D
(b) A [ .B [ C [ D/
(c) A [ .B [ C / [ D
(d) .A [ B/ [ .C [ D/
(e) .A B C / D
(f) A .B C D/
(g) A .B C / D
(h) .A B/ .C D/
3. Based on the work in Part (2), does the placement of the parentheses matter
when determining the union (or intersection) of these four sets? Does this
make it possible to define A [ B [ C [ D and A B C D?
We have already seen that the elements of a set may themselves be sets. For
example, the power set of a set T , P.T /, is the set of all subsets of T . The phrase,
“a set of sets” sounds confusing, and so we often use the terms collection and
family when we wish to emphasize that the elements of a given set are themselves
sets. We would then say that the power set of T is the family (or collection) of sets
that are subsets of T .

5.5. Indexed Families of Sets 265
One of the purposes of the work we have done so far in this preview activity
was to show that it is possible to define the union and intersection of a family of
sets.
Definition. Let C be a family of sets. The union over C is defined as the set
of all elements that are in at least one of the sets in C. We write
[
X2C
X D fx 2 U j x 2 X for some X 2 Cg
The intersection over C is defined as the set of all elements that are in all of
the sets in C. That is,

X2C
X D fx 2 U j x 2 X for all X 2 Cg
For example, consider the four sets A, B, C , and D used earlier in this preview
activity and the sets
S D f5; 6; 7; 8; 9g and T D f6; 7; 8; 9; 10g:
We can then consider the following families of sets: A D fA; B; C; Dg and
B D fA; B; C; D; S; Tg.
4. Explain why
[
X2A
X D A [ B [ C [ D and

X2A
X D A B C D;
and use your work in (1), (2), and (3) to determine
S
X2A
X and
T
X2A
X.
5. Use the roster method to specify
S
X2B
X and
T
X2B
X.
6. Use the roster method to specify the sets

S
X2A
X
c
and
T
X2A
Xc
. Remember
that the universal set is N.

Preview Activity 2 (An Indexed Family of Sets)
We often use subscripts to identify sets. For example, in Preview Activity 1, instead
of using A, B, C , and D as the names of the sets, we could have used A1, A2, A3,
and A4. When we do this, we are using the subscript as an identifying tag, or
index, for each set. We can also use this idea to specify an infinite family of sets.
For example, for each natural number n, we define
Cn D fn; n C 1; n C 2; n C 3; n C 4g:
So if we have a family of sets C D fC1; C2; C3; C4g, we use the notation
4
S
j D1
Cj to
mean the same thing as
S
X2C
X.
1. Determine
4
S
j D1
Cj and
4
T
j D1
Cj.
We can see that with the use of subscripts, we do not even have to define the family
of sets A. We can work with the infinite family of sets
C
D fAn j n 2 Ng
and use the subscripts to indicate which sets to use in a union or an intersection.
2. Use the roster method to specify each of the following pairs of sets. The
universal set is N.
(a)
6
S
j D1
Cj and
6
T
j D1
Cj
(b)
8
S
j D1
Cj and
8
T
j D1
Cj
(c)
8
S
j D4
Cj and
8
T
j D4
Cj
(d)
4
T
j D1
Cj
!c
and
4
S
j D1
C c
j
The Union and Intersection over an Indexed Family of Sets
One of the purposes of the preview activities was to show that we often encounter
situations in which more than two sets are involved, and it is possible to define the
union and intersection of more than two sets. In Preview Activity 2, we also saw
that it is often convenient to “index” the sets in a family of sets. In particular, if n

is a natural number and A D fA1; A2; : : :; Ang is a family of n sets, then the union
of these n sets, denoted by A1 [ A2 [ [ An or
n
S
j D1
Aj , is defined as
n
[
j D1
Aj D
˚
x 2 U j x 2 Aj ; for some j with 1 j n :
We can also define the intersection of these n sets, denoted by A1 A2 An
or
n
T
j D1
Aj , as
n

j D1
Aj D
˚
x 2 U j x 2 Aj ; for all j with 1 j n :
We can also extend this idea to define the union and intersection of a family that
consists of infinitely many sets. So if B D fB1; B2; : : :; Bn; : : : g, then
1
[
j D1
Bj D
˚
x 2 U j x 2 Bj ; for some j with j 1 ; and
1

j D1
Bj D
˚
x 2 U j x 2 Bj ; for all j with j 1 :
Progress Check 5.26 (An Infinite Family of Sets)
For each natural number n, let An D
˚
1; n; n2 . For example,
A1 D f1g A2 D f1; 2; 4g, A3 D f1; 3; 9g,
and
3
S
j D1
Aj D f1; 2; 3; 4; 9g,
3
T
j D1
Aj D f1g.
Determine each of the following sets:
1.
6
S
j D1
Aj 2.
6
T
j D1
Aj 3.
6
S
j D3
Aj

4.
6
T
j D3
Aj 5.
1
S
j D1
Aj 6.
1
T
j D1
Aj
In all of the examples we have studied so far, we have used N or a subset of N
to index or label the sets in a family of sets. We can use other sets to index or label
sets in a family of sets. For example, for each real number x, we can define Bx to
be the closed interval Œx; x C 2. That is,
Bx D fy 2 R j x y x C 2g :
So we make the following definition. In this definition, ƒ is the uppercase Greek
letter lambda and ˛ is the lowercase Greek letter alpha.
Definition. Let ƒ be a nonempty set and suppose that for each ˛ 2 ƒ, there is
a corresponding set A˛. The family of sets fA˛ j ˛ 2 ƒg is called an indexed
family of sets indexed by ƒ. Each ˛ 2 ƒ is called an index and ƒ is called
an indexing set.
Progress Check 5.27 (Indexed Families of Sets)
In each of the indexed families of sets that we seen so far, if the indices were
different, then the sets were different. That is, if ƒ is an indexing set for the family
of sets A D fA˛ j ˛ 2 ƒg, then if ˛; ˇ 2 ƒ and ˛ ¤ ˇ, then A˛ ¤ Aˇ . (Note:
The letter ˇ is the Greek lowercase beta.)
1. Let ƒ D f1; 2; 3; 4g, and for each n 2 ƒ, let An D f2n C 6; 16 2ng, and
let A D fA1; A2; A3; A4g. Determine A1, A2, A3, and A4.
2. Is the following statement true or false for the indexed family A in (1)?
For all m; n 2 ƒ, if m ¤ n, then Am ¤ An.
3. Now let ƒ D R. For each x 2 R, define Bx D
˚
0; x2; x4 . Is the following
statement true for the indexed family of sets B D fBx j x 2 Rg?
For all x; y 2 R, if x ¤ y, then Bx ¤ By .
We now restate the definitions of the union and intersection of a family of sets
for an indexed family of sets.

Definition. Let ƒ be a nonempty indexing set and let A D fA˛ j ˛ 2 ƒg
be an indexed family of sets. The union over A is defined as the set of all
elements that are in at least one of sets A˛, where ˛ 2 ƒ. We write
[
˛2ƒ
A˛ D fx 2 U j there exists an ˛ 2 ƒ with x 2 A˛g:
The intersection over A is the set of all elements that are in all of the sets A˛
for each ˛ 2 ƒ. That is,

˛2ƒ
A˛ D fx 2 U j for all ˛ 2 ƒ; x 2 A˛g:
Example 5.28 (A Family of Sets Indexed by the Positive Real Numbers)
For each positive real number ˛, let A˛ be the interval . 1; ˛. That is,
A˛ D fx 2 R j 1 x ˛g:
If we let RC be the set of positive real numbers, then we have a family of sets
indexed by RC
. We will first determine the union of this family of sets. Notice
that for each ˛ 2 RC, ˛ 2 A˛, and if y is a real number with 1 y 0, then
y 2 A˛. Also notice that if y 2 R and y 1, then for each ˛ 2 RC
, y … A˛.
With these observations, we conclude that
[
˛2RC
A˛ D . 1; 1/ D fx 2 R j 1 xg:
To determine the intersection of this family, notice that
if y 2 R and y 1, then for each ˛ 2 RC
, y … A˛;
if y 2 R and 1 y 0, then y 2 A˛ for each ˛ 2 RC; and
if y 2 R and y 0, then if we let ˇ D
y
2
, y ˇ and y … Aˇ .
From these observations, we conclude that

˛2RC
A˛ D . 1; 0 D fx 2 R j 1 x 0g:

Progress Check 5.29 (A Continuation of Example 5.28)
Using the family of sets from Example 5.28, for each ˛ 2 RC
, we see that
Ac
˛ D . 1; 1 [ .˛; 1/:
Use the results from Example 5.28 to help determine each of the following sets.
For each set, use either interval notation or set builder notation.
1.
S
˛2RC
A˛
!c
2.
T
˛2RC
Ac
˛
3.
T
˛2RC
A˛
!c
4.
S
˛2RC
Ac
˛
Properties of Union and Intersection
In Theorem 5.30, we will prove some properties of set operations for indexed fami-
lies of sets. Some of these properties are direct extensions of corresponding proper-
ties for two sets. For example, we have already proved De Morgan’s Laws for two
sets in Theorem 5.20 on page 248. The work in the preview activities and Progress
Check 5.29 suggests that we should get similar results using set operations with an
indexed family of sets. For example, in Preview Activity 2, we saw that
0
@
4

j D1
Aj
1
A
c
D
4
[
j D1
Ac
j :
Theorem 5.30. Let ƒ be a nonempty indexing set and let A D fA˛ j ˛ 2 ƒg be
an indexed family of sets. Then
1. For each ˇ 2 ƒ,
T
˛2ƒ
A˛ Aˇ .
2. For each ˇ 2 ƒ, Aˇ
S
˛2ƒ
A˛.
3.

T
˛2ƒ
A˛
c
D
S
˛2ƒ
Ac
˛
4.

S
˛2ƒ
A˛
c
D
T
˛2ƒ
Ac
˛

Parts (3) and (4) are known as De Morgan’s Laws.
Proof. We will prove Parts (1) and (3). The proofs of Parts (2) and (4) are included
in Exercise (4). So we let ƒ be a nonempty indexing set and let A D fA˛ j ˛ 2 ƒg
be an indexed family of sets. To prove Part (1), we let ˇ 2 ƒ and note that if
x 2
T
˛2ƒ
A˛, then x 2 A˛, for all ˛ 2 ƒ. Since ˇ is one element in ƒ, we may
conclude that x 2 Aˇ . This proves that
T
˛2ƒ
A˛ Aˇ .
To prove Part (3), we will prove that each set is a subset of the other set. We
first let x 2

T
˛2ƒ
A˛
c
. This means that x …

T
˛2ƒ
A˛

, and this means that
there exists a ˇ 2 ƒ such that x … Aˇ .
Hence, x 2 Ac
ˇ
, which implies that x 2
S
˛2ƒ
Ac
˛. Therefore, we have proved that

˛2ƒ
A˛
!c

[
˛2ƒ
Ac
˛: (1)
We now let y 2
S
˛2ƒ
Ac
˛. This means that there exists a ˇ 2 ƒ such that
y 2 Ac
ˇ
or y … Aˇ . However, since y … Aˇ , we may conclude that y …
T
˛2ƒ
A˛
and, hence, y 2

T
˛2ƒ
A˛
c
. This proves that
[
˛2ƒ
Ac
˛

˛2ƒ
A˛
!c
: (2)
Using the results in (1) and (2), we have proved that

T
˛2ƒ
A˛
c
D
S
˛2ƒ
Ac
˛.
Many of the other properties of set operations are also true for indexed families
of sets. Theorem 5.31 states the distributive laws for set operations.
Theorem 5.31. Let ƒ be a nonempty indexing set, let A D fA˛ j ˛ 2 ƒg be an
indexed family of sets, and let B be a set. Then

1. B

S
˛2ƒ
A˛

D
S
˛2ƒ
.B A˛/, and
2. B [

T
˛2ƒ
A˛

D
T
˛2ƒ
.B [ A˛/.
The proof of Theorem 5.31 is Exercise (5).
Pairwise Disjoint Families of Sets
In Section 5.2, we defined two sets A and B to be disjoint provided that AB D ;.
In a similar manner, if ƒ is a nonempty indexing set and A D fA˛ j ˛ 2 ƒg is
an indexed family of sets, we can say that this indexed family of sets is disjoint
provided that
T
˛2ƒ
A˛ D ;. However, we can use the concept of two disjoint sets
to define a somewhat more interesting type of “disjointness” for an indexed family
of sets.
Definition. Let ƒ be a nonempty indexing set, and let A D fA˛ j ˛ 2 ƒg be
an indexed family of sets. We say that A is pairwise disjoint provided that
for all ˛ and ˇ in ƒ, if A˛ ¤ Aˇ , then A˛ Aˇ D ;.
Progress Check 5.32 (Disjoint Families of Sets)
Figure 5.7 shows two families of sets,
A D fA1; A2; A3; A4g and B D fB1; B2; B3; B4g.
U
A
1
A2
A3
4
A
U
2
B
3
B
4
B
B1
Figure 5.7: Two Families of Indexed Sets

1. Is the family of sets A a disjoint family of sets? A pairwise disjoint family
of sets?
2. Is the family of sets B a disjoint family of sets? A pairwise disjoint family
of sets?
Now let the universal be R. For each n 2 N, let Cn D .n; 1/, and let
C D fCn j n 2 Ng.
3. Is the family of sets C a disjoint family of sets? A pairwise disjoint family of
sets?
1. For each natural number n, let An D fn; n C 1; n C 2; n C 3g. Use the
roster method to specify each of the following sets:
?
(a)
3
T
j D1
Aj
(b)
3
S
j D1
Aj
(c)
7
T
j D3
Aj
?
(d)
7
S
j D3
Aj
(e) A9
7
S
j D3
Aj
!
(f)
7
S
j D3
A9 Aj

2. For each natural number n, let An D fk 2 N j k ng. Use the roster method
or set builder notation to specify each of the following sets:
?
(a)
5
T
j D1
Aj
(b)
5
T
j D1
Aj
!c
? (c)
5
T
j D1
Ac
j
?
(d)
5
S
j D1
Ac
j
(e)
5
S
j D1
Aj
? (f)
5
S
j D1
Aj
!c
(g)
T
j 2N
Aj
(h)
S
j 2N
Aj

3. For each positive real number r, define Tr to be the closed interval

r2; r2

.
That is,
Tr D
˚
x 2 R
ˇ
ˇ r2
x r2
:
Let ƒ D fm 2 N j1 m 10g. Use either interval notation or set builder
notation to specify each of the following sets:
? (a)
S
k2ƒ
Tk
?
(b)
T
k2ƒ
Tk
(c)
S
r2RC
Tr
(d)
T
r2RC
Tr
(e)
S
k2N
Tk
(f)
T
k2N
Tk
4. Prove Parts (2) and (4) of Theorem 5.30. Let ƒ be a nonempty indexing set
and let A D fA˛ j ˛ 2 ƒg be an indexed family of sets.
?
(a) For each ˇ 2 ƒ, Aˇ
S
˛2ƒ
A˛.
(b)

S
˛2ƒ
A˛
c
D
T
˛2ƒ
Ac
˛
5. Prove Theorem 5.31. Let ƒ be a nonempty indexing set, let A D fA˛ j ˛ 2 ƒg
be an indexed family of sets, and let B be a set. Then
?
(a) B

S
˛2ƒ
A˛

D
S
˛2ƒ
.B A˛/, and
(b) B [

T
˛2ƒ
A˛

D
T
˛2ƒ
.B [ A˛/.
6. Let ƒ and € be nonempty indexing sets and let A D fA˛ j ˛ 2 ƒg and
B D
˚
Bˇ j ˇ 2 € be indexed families of sets. Use the distributive laws in
Exercise (5) to:
(a) Write

S
˛2ƒ
A˛

S
ˇ2€
Bˇ
!
as a union of intersections of two sets.
(b) Write

T
˛2ƒ
A˛

[
T
ˇ2€
Bˇ
!
as an intersection of unions of two
sets.
7. Let ƒ be a nonempty indexing set and let A D fA˛ j ˛ 2 ƒg be an indexed
family of sets. Also, assume that € ƒ and € ¤ ;. (Note: The letter € is
the uppercase Greek letter gamma.) Prove that

(a)
S
˛2€
A˛
S
˛2ƒ
A˛ (b)
T
˛2ƒ
A˛
T
˛2€
A˛
8. Let ƒ be a nonempty indexing set and let A D fA˛ j ˛ 2 ƒg be an indexed
family of sets.
? (a) Prove that if B is a set such that B A˛ for every ˛ 2 ƒ, then
B
T
˛2ƒ
A˛.
(b) Prove that if C is a set such that A˛ C for every ˛ 2 ƒ, then
S
˛2ƒ
A˛ C .
9. For each natural number n, let An D fx 2 R j n 1 x ng. Prove that
fAn jn 2 Ng is a pairwise disjoint family of sets and that
S
n2N
An D RC
N

.
10. For each natural number n, let An D fk 2 N j k ng. Determine if the
following statements are true or false. Justify each conclusion.
(a) For all j; k 2 N, if j ¤ k, then Aj Ak ¤ ;.
(b)
T
k2N
Ak D ;
11. Give an example of an indexed family of sets fAn jn 2 Ng such all three of
the following conditions are true:
(i) For each m 2 N, Am .0; 1/;
(ii) For each j; k 2 N, if j ¤ k, then Aj Ak ¤ ;; and
(iii)
T
k2N
Ak D ;.
12. Let ƒ be a nonempty indexing set, let A D fA˛ j ˛ 2 ƒg be an indexed
family of sets, and let B be a set. Use the results of Theorem 5.30 and
Theorem 5.31 to prove each of the following:
? (a)

S
˛2ƒ
A˛

B D
S
˛2ƒ
.A˛ B/
(b)

T
˛2ƒ
A˛

B D
T
˛2ƒ
.A˛ B/
(c) B

S
˛2ƒ
A˛

D
T
˛2ƒ
.B A˛/

(d) B

T
˛2ƒ
A˛

D
S
˛2ƒ
.B A˛/
13. An Indexed Family of Subsets of the Cartesian Plane. Let R be the set
of nonnegative real numbers, and for each r 2 R
, let
Cr D
˚
.x; y/ 2 R R j x2
C y2
D r2
Dr D
˚
.x; y/ 2 R R j x2
C y2
r2
Tr D
˚
.x; y/ 2 R R j x2
C y2
r2
D Dc
r :
If r 0, then the set Cr is the circle of radius r with center at the origin
as shown in Figure 5.8, and the set Dr is the shaded disk (including the
boundary) shown in Figure 5.8.
x
y
r
r
x
y
r
r
r r
D
C
Figure 5.8: Two Sets for Activity 13
(a) Determine
S
r2R
Cr and
T
r2R
Cr .
(b) Determine
S
r2R
Dr and
T
r2R
Dr .
(c) Determine
S
r2R
Tr and
T
r2R
Tr.
(d) Let C D fCr j r 2 Rg, D D fDr j r 2 Rg, and T D fTr j r 2 Rg.
Are any of these indexed families of sets pairwise disjoint? Explain.
Now let I be the closed interval Œ0; 2 and let J be the closed interval Œ1; 2.

(e) Determine
S
r2I
Cr ,
T
r2I
Cr ,
S
r2J
Cr , and
T
r2J
Cr.
(f) Determine
S
r2I
Dr ,
T
r2I
Dr ,
S
r2J
Dr , and
T
r2J
Dr .
(g) Determine

S
r2I
Dr
c
,

T
r2I
Dr
c
,

S
r2J
Dr
c
, and

T
r2J
Dr
c
.
(h) Determine
S
r2I
Tr ,
T
r2I
Tr ,
S
r2J
Tr, and
T
r2J
Tr .
(i) Use De Morgan’s Laws to explain the relationship between your an-
swers in Parts (13g) and (13h).
Equal sets, page 55
Subset, page 55
Proper subset, page 218
Power set, page 222
Cardinality of a finite set,
page 223
Intersection of two sets, page 216
Union of two sets, page 216
Set difference, page 216
Complement of a set, page 216
Disjoint sets, page 236
Cartesian product of two sets,
pages 256
Ordered pair, page 256
Union over a family of sets,
page 265
Intersection over a family of sets,
page 265
Indexing set, page 268
Indexed family of sets, page 268
Union over an indexed family of
sets, page 269
Intersection over an indexed fam-
ily of sets, page 269
Pairwise disjoint family of sets,
page 272

Important Theorems and Results about Sets
Theorem 5.5. Let n be a nonnegative integer and let A be a subset of some
universal set. If A is a finite set with n elements, then A has 2n
subsets. That
is, if jAj D n, then jP .A/j D 2n.
Theorem 5.18. Let A, B, and C be subsets of some universal set U. Then
all of the following equalities hold.
Properties of the Empty Set
and the Universal Set
A ; D ;
A [ ; D A
A U D A
A [ U D U
Idempotent Laws A A D A A [ A D A
Commutative Laws A B D B A A [ B D B [ A
Associative Laws .A B/ C D A .B C /
.A [ B/ [ C D A [ .B [ C /
Distributive Laws A .B [ C / D .A B/ [ .A C /
A [ .B C / D .A [ B/ .A [ C /
Theorem 5.20. Let A and B be subsets of some universal set U . Then the
following are true:
Basic Properties .Ac/c
D A
A B D A Bc
Empty Set, Universal Set A ; D A and A U D ;
;c
D U and U c
D ;
De Morgan’s Laws .A B/c
D Ac [ Bc
.A [ B/c
D Ac
Bc
Subsets and Complements A B if and only if Bc Ac.

Theorem 5.25. Let A, B, and C be sets. Then
1. A .B C/ D .A B/ .A C /
2. A .B [ C/ D .A B/ [ .A C /
3. .A B/ C D .A C/ .B C /
4. .A [ B/ C D .A C/ [ .B C /
5. A .B C/ D .A B/ .A C /
6. .A B/ C D .A C/ .B C /
7. If T A, then T B A B.
8. If Y B, then A Y A B.
Theorem 5.30. Let ƒ be a nonempty indexing set and let A D fA˛ j ˛ 2 ƒg
be an indexed family of sets. Then
1. For each ˇ 2 ƒ,
T
˛2ƒ
A˛ Aˇ .
2. For each ˇ 2 ƒ, Aˇ
T
˛2ƒ
A˛.
3.

T
˛2ƒ
A˛
c
D
S
˛2ƒ
Ac
˛
4.

S
˛2ƒ
A˛
c
D
T
˛2ƒ
Ac
˛
Parts (3) and (4) are known as De Morgan’s Laws.
Theorem 5.31. Let ƒ be a nonempty indexing set, let A D fA˛ j ˛ 2 ƒg be
an indexed family of sets, and let B be a set. Then
1. B

S
˛2ƒ
A˛

D
S
˛2ƒ
.B A˛/, and
2. B [

T
˛2ƒ
A˛

D
T
˛2ƒ
.B [ A˛/.
Important Proof Method
The Choose-an-Element Method
The choose-an-element method is frequently used when we encounter a universal
quantifier in a statement in the backward process of a proof. This statement often
has the form

For each element with a given property, something happens.
In the forward process of the proof, we then we choose an arbitrary element with
the given property.
Whenever we choose an arbitrary element with a given property,
we are not selecting a specific element. Rather, the only thing we
can assume about the element is the given property.
For more information, see page 232.

Chapter 6
Functions
6.1 Introduction to Functions
Preview Activity 1 (Functions from Previous Courses)
One of the most important concepts in modern mathematics is that of a function.
In previous mathematics courses, we have often thought of a function as some sort
of input-output rule that assigns exactly one output to each input. So in this con-
text, a function can be thought of as a procedure for associating with each element
of some set, called the domain of the function, exactly one element of another
set, called the codomain of the function. This procedure can be considered an
input-output-rule. The function takes the input, which is an element of the domain,
and produces an output, which is an element of the codomain. In calculus and pre-
calculus, the inputs and outputs were almost always real numbers. So the notation
f.x/ D x2
sin x means the following:
f is the name of the function.
f.x/ is a real number. It is the output of the function when the input is the
real number x. For example,
f

2

D

2
2
sin

2

D
2
4
1
D
2
4
:
281

282 Chapter 6. Functions
For this function, it is understood that the domain of the function is the set R of
all real numbers. In this situation, we think of the domain as the set of all possible
inputs. That is, the domain is the set of all possible real numbers x for which a real
number output can be determined.
This is closely related to the equation y D x2
sin x. With this equation, we
frequently think of x as the input and y as the output. In fact, we sometimes write
y D f.x/. The key to remember is that a function must have exactly one output
for each input. When we write an equation such as
y D
1
2
x3
1;
we can use this equation to define y as a function of x. This is because when we
substitute a real number for x (the input), the equation produces exactly one real
number for y (the output). We can give this function a name, such as g, and write
y D g.x/ D
1
2
x3
1:
However, as written, an equation such as
y2
D x C 3
cannot be used to define y as a function of x since there are real numbers that
can be substituted for x that will produce more than one possible value of y. For
example, if x D 1, then y2
D 4, and y could be 2 or 2.
Which of the following equations can be used to define a function with x 2 R
as the input and y 2 R as the output?
1. y D x2 2
2. y2
D x C 3
3. y D
1
2
x3
1
4. y D
1
2
x sin x
5. x2 C y2 D 4
6. y D 2x 1
7. y D
x
x 1
Preview Activity 2 (Some Other Types of Functions)
The domain and codomain of the functions in Preview Activity 1 is the set R of all
real numbers, or some subset of R. In most of these cases, the way in which the
function associates elements of the domain with elements of the codomain is by a

6.1. Introduction to Functions 283
rule determined by some mathematical expression. For example, when we say that
f is the function such that
f.x/ D
x
x 1
;
then the algebraic rule that determines the output of the function f when the input
is x is
x
x 1
. In this case, we would say that the domain of f is the set of all real
numbers not equal to 1 since division by zero is not defined.
However, the concept of a function is much more general than this. The do-
main and codomain of a function can be any set, and the way in which a function
associates elements of the domain with elements of the codomain can have many
different forms. The input-output rule for a function can be a formula, a graph,
a table, a random process, or a verbal description. We will explore two different
examples in this preview activity.
1. Let b be the function that assigns to each person his or her birthday (month
and day). The domain of the function b is the set of all people and the
codomain of b is the set of all days in a leap year (i.e., January 1 through
December 31, including February 29).
(a) Explain why b really is a function. We will call this the birthday
function.
(b) In 1995, Andrew Wiles became famous for publishing a proof of Fer-
mat’s Last Theorem. (See A. D. Aczel, Fermat’s Last Theorem: Un-
locking the Secret of an Ancient Mathematical Problem, Dell Publish-
ing, New York, 1996.) Andrew Wiles’s birthday is April 11, 1953.
Translate this fact into functional notation using the “birthday func-
tion” b. That is, fill in the spaces for the following question marks:
b. ‹ / D ‹:
(c) Is the following statement true or false? Explain.
For each day D of the year, there exists a person x such that
b.x/ D D.
(d) Is the following statement true or false? Explain.
For any people x and y, if x and y are different people, then
b.x/ ¤ b.y/.
2. Let s be the function that associates with each natural number the sum of
its distinct natural number divisors. This is called the sum of the divisors

function. For example, the natural number divisors of 6 are 1, 2, 3, and 6,
and so
s.6/ D 1 C 2 C 3 C 6
D 12:
(a) Calculate s.k/ for each natural number k from 1 through 15.
(b) Does there exist a natural number n such that s.n/ D 5? Justify your
conclusion.
(c) Is it possible to find two different natural numbers m and n such that
s.m/ D s.n/? Explain.
(d) Use your responses in (b) and (c) to determine whether the following
statements true or false.
i. For each m 2 N, there exists a natural number n such that s.n/ D
m.
ii. For all m; n 2 N, if m ¤ n, then s.m/ ¤ s.n/.
The Definition of a Function
The concept of a function is much more general than the idea of a function used
in calculus or precalculus. In particular, the domain and codomain do not have to
be subsets of R. In addition, the way in which a function associates elements of
the domain with elements of the codomain can have many different forms. This
input-output rule can be a formula, a graph, a table, a random process, a computer
algorithm, or a verbal description. Two such examples were introduced in Preview
Activity 2.
For the birthday function, the domain would be the set of all people and the
codomain would be the set of all days in a leap year. For the sum of the divisors
function, the domain is the set N of natural numbers, and the codomain could also
be N. In both of these cases, the input-output rule was a verbal description of how
to assign an element of the codomain to an element of the domain.
We formally define the concept of a function as follows:
Definition. A function from a set A to a set B is a rule that associates with
each element x of the set A exactly one element of the set B. A function from
A to B is also called a mapping from A to B.

Function Notation. When we work with a function, we usually give it a name.
The name is often a single letter, such as f or g. If f is a function from the set A
to be the set B, we will write f WA ! B. This is simply shorthand notation for the
fact that f is a function from the set A to the set B. In this case, we also say that
f maps A to B.
Definition. Let f W A ! B. (This is read, “Let f be a function from A
to B.”) The set A is called the domain of the function f , and we write
A D dom.f /. The set B is called the codomain of the function f , and we
write B D codom.f /.
If a 2 A, then the element of B that is associated with a is denoted by f.a/
and is called the image of a under f. If f.a/ D b, with b 2 B, then a is
called a preimage of b under f.
Some Function Terminology with an Example. When we have a function
f W A ! B, we often write y D f.x/. In this case, we consider x to be an
unspecified object that can be chosen from the set A, and we would say that x is
the independent variable of the function f and y is the dependent variable of
the function f .
For a specific example, consider the function gWR ! R, where g.x/ is defined
by the formula
g.x/ D x2
2:
Note that this is indeed a function since given any input x in the domain, R, there
is exactly one output g.x/ in the codomain, R. For example,
g. 2/ D . 2/2
2 D 2;
g.5/ D 52
2 D 23;
g.
p
2/ D
p
2
2
2 D 0;
g.
p
2/ D
p
2
2
2 D 0:
So we say that the image of 2 under g is 2, the image of 5 under g is 23, and so
on.
Notice in this case that the number 0 in the codomain has two preimages,
p
2 and
p
2. This does not violate the mathematical definition of a function
since the definition only states that each input must produce one and only one out-
put. That is, each element of the domain has exactly one image in the codomain.

Nowhere does the definition stipulate that two different inputs must produce dif-
ferent outputs.
Finding the preimages of an element in the codomain can sometimes be diffi-
cult. In general, if y is in the codomain, to find its preimages, we need to ask, “For
which values of x in the domain will we have y D g.x/?” For example, for the
function g, to find the preimages of 5, we need to find all x for which g.x/ D 5.
In this case, since g.x/ D x2
2, we can do this by solving the equation
x2
2 D 5:
The solutions of this equation are
p
7 and
p
7. So for the function g, the preim-
ages of 5 are
p
7 and
p
7. We often use set notation for this and say that the set
of preimages of 5 for the function g is
n p
7;
p
7
o
.
Also notice that for this function, not every element in the codomain has a
preimage. For example, there is no input x such that g .x/ D 3. This is true
since for all real numbers x, x2
0 and hence x2
2 2. This means that for
all x in R, g .x/ 2.
Finally, note that we introduced the function g with the sentence, “Consider the
function gWR ! R, where g.x/ is defined by the formula g.x/ D x2
2.” This is
one correct way to do this, but we will frequently shorten this to, “Let g WR ! R
be defined by g.x/ D x2
2”, or “Let gWR ! R, where g.x/ D x2
2.”
Progress Check 6.1 (Images and Preimages)
Let f WR ! R be defined by f.x/ D x2
5x for all x 2 R, and let gWZ ! Z be
defined by g.m/ D m2 5m for all m 2 Z.
1. Determine f. 3/ and f
p
8

.
2. Determine g.2/ and g. 2/.
3. Determine the set of all preimages of 6 for the function f.
4. Determine the set of all preimages of 6 for the function g.
5. Determine the set of all preimages of 2 for the function f.
6. Determine the set of all preimages of 2 for the function g.

The Codomain and Range of a Function
Besides the domain and codomain, there is another important set associated with
a function. The need for this was illustrated in the example of the function g on
page 285. For this function, it was noticed that there are elements in the codomain
that have no preimage or, equivalently, there are elements in the codomain that are
not the image of any element in the domain. The set we are talking about is the
subset of the codomain consisting of all images of the elements of the domain of
the function, and it is called the range of the function.
Definition. Let f WA ! B. The set ff.x/ j x 2 Ag is called the range of the
function f and is denoted by range .f /. The range of f is sometimes called
the image of the function f (or the image of A under f ).
The range of f WA ! B could equivalently be defined as follows:
range.f / D fy 2 B j y D f .x/ for some x 2 Ag:
Notice that this means that range.f / codom.f / but does not necessarily mean
that range.f / D codom.f /. Whether we have this set equality or not depends on
the function f . More about this will be explored in Section 6.3.
Progress Check 6.2 (Codomain and Range)
1. Let b be the function that assigns to each person his or her birthday (month
and day).
(a) What is the domain of this function?
(b) What is a codomain for this function?
(c) In Preview Activity 2, we determined that the following statement is
true: For each day D of the year, there exists a person x such that
b.x/ D D. What does this tell us about the range of the function b?
Explain.
2. Let s be the function that associates with each natural number the sum of its
distinct natural number factors.
(a) What is the domain of this function?
(b) What is a codomain for this function?
(c) In Preview Activity 2, we determined that the following statement is
false:

For each m 2 N, there exists a natural number n such that
s.n/ D m.
Give an example of a natural number m that shows this statement is
false, and explain what this tells us about the range of the function s.
The Graph of a Real Function
We will finish this section with methods to visually communicate information
about two specific types of functions. The first is the familiar method of graph-
ing functions that was a major part of some previous mathematics courses. For
example, consider the function gWR ! R defined by g.x/ D x2
2x 1.
Figure 6.1: Graph of y D g.x/, where g.x/ D x2 2x 1
Every point on this graph corresponds to an ordered pair .x; y/ of real num-
bers, where y D g.x/ D x2
2x 1. Because we use the Cartesian plane when
drawing this type of graph, we can only use this type of graph when both the do-
main and the codomain of the function are subsets of the real numbers R. Such
a function is sometimes called a real function. The graph of a real function is a
visual way to communicate information about the function. For example, the range
of g is the set of all y-values that correspond to points on the graph. In this case,
the graph of g is a parabola and has a vertex at the point .1; 2/. (Note: The x-
coordinate of the vertex can be found by using calculus and solving the equation
f 0
.x/ D 0.) Since the graph of the function g is a parabola, we know that pattern
shown on the left end and the right end of the graph continues and we can conclude

that the range of g is the set of all y 2 R such that y 2. That is,
range.g/ D fy 2 R j y 2g:
Progress Check 6.3 (Using the Graph of a Real Function)
The graph in Figure 6.2 shows the graph of (slightly more than) two complete
periods for a function f WR ! R, where f.x/ D A sin.Bx/ for some positive real
number constants A and B.
Figure 6.2: Graph of y D f.x/
1. We can use the graph to estimate the output for various inputs. This is done
by estimating the y-coordinate for the point on the graph with a specified
x-coordinate. On the graph, draw vertical lines at x D 1 and x D 2 and
estimate the values of f. 1/ and f.2/.
2. Similarly, we can estimate inputs of the function that produce a specified
output. This is done by estimating the x-coordinates of the points on the
graph that have a specified y-coordinate. Draw a horizontal line at y D 2
and estimate at least two values of x such that f.x/ D 2.
3. Use the graph Figure 6.2 to estimate the range of the function f .

Arrow Diagrams
Sometimes the domain and codomain of a function are small, finite sets. When
this is the case, we can define a function simply by specifying the outputs for each
input in the domain. For example, if we let A D f1; 2; 3g and let B D fa; bg, we
can define a function F WA ! B by specifying that
F.1/ D a; F.2/ D a; and F.3/ D b:
This is a function since each element of the domain is mapped to exactly one el-
ement in B. A convenient way to illustrate or visualize this type of function is
with a so-called arrow diagram as shown in Figure 6.3. An arrow diagram can
3
2
1
b
a
A B
F
Figure 6.3: Arrow Diagram for a Function
be used when the domain and codomain of the function are finite (and small). We
represent the elements of each set with points and then use arrows to show how
the elements of the domain are associated with elements of the codomain. For ex-
ample, the arrow from the point 2 in A to the point a in B represents the fact that
F.2/ D a. In this case, we can use the arrow diagram in Figure 6.3 to conclude
that range.F / D fa; bg.
Progress Check 6.4 (Working with Arrow Diagrams)
Let A D f1; 2; 3; 4g and let B D fa; b; cg.
1. Which of the arrow diagrams in Figure 6.4 can be used to represent a func-
tion from A to B? Explain.
2. For those arrow diagrams that can be used to represent a function from A to
B, determine the range of the function.

1
2
3
4
a
b
c
A B
1
2
3
4
a
b
c
A B
(a) (b)
1
2
3
4
a
b
c
A B
(c)
Figure 6.4: Arrow Diagram for a Function
Exercises 6.1
?
1. Let f WR ! R be defined by f .x/ D x2
2x.
(a) Evaluate f. 3/; f. 1/; f.1/; and f.3/.
(b) Determine the set of all of the preimages of 0 and the set of all of the
preimages of 4.
(c) Sketch a graph of the function f.
(d) Determine the range of the function f.
2. Let R D fx 2 R j x 0g, and let sWR ! R be defined by s.x/ D x2.
(a) Evaluate s. 3/; s. 1/; s.1/; and s.3/.
(b) Determine the set of all of the preimages of 0 and the set of all preim-
ages of 2.
(c) Sketch a graph of the function s.
(d) Determine the range of the function s.
?
3. Let f WZ ! Z be defined by f.m/ D 3 m.
(b) Determine the set of all of the preimages of 5 and the set of all of the
preimages of 4.
(c) Determine the range of the function f.
(d) This function can be considered a real function since Z R. Sketch a
graph of this function. Note: The graph will be an infinite set of points
that lie on a line. However, it will not be a line since its domain is not
R but is Z.

4. Let f WZ ! Z be defined by f.m/ D 2m C 1.
? (b) Determine the set of all of the preimages of 5 and the set of all of the
preimages 4.
?
(c) Determine the range of the function f.
? (d) Sketch a graph of the function f . See the comments in Exercise (3d).
5. Recall that a real function is a function whose domain and codomain are
subsets of the real numbers R. (See page 288.) Most of the functions used in
calculus are real functions. Quite often, a real function is given by a formula
or a graph with no specific reference to the domain or the codomain. In these
cases, the usual convention is to assume that the domain of the real function
f is the set of all real numbers x for which f.x/ is a real number, and that
the codomain is R. For example, if we define the (real) function f by
f.x/ D
x
x 2
;
we would be assuming that the domain is the set of all real numbers that are
not equal to 2 and that the codomain in R.
Determine the domain and range of each of the following real functions. It
might help to use a graphing calculator to plot a graph of the function.
(a) The function k defined by k.x/ D
p
x 3
?
(b) The function F defined by F.x/ D ln .2x 1/
(c) The function f defined by f.x/ D 3 sin.2x/
?
(d) The function g defined by g.x/ D
4
x2 4
(e) The function G defined by G.x/ D 4 cos .x/ C 8
? 6. The number of divisors function. Let d be the function that associates
with each natural number the number of its natural number divisors. That is,
d W N ! N where d.n/ is the number of natural number divisors of n. For
example, d.6/ D 4 since 1, 2, 3, and 6 are the natural number divisors of 6.
(a) Calculate d.k/ for each natural number k from 1 through 12.
(b) Does there exist a natural number n such that d.n/ D 1? What is the
set of preimages of the natural number 1.

(c) Does there exist a natural number n such that d.n/ D 2? If so, deter-
mine the set of all preimages of the natural number 2.
(d) Is the following statement true or false? Justify your conclusion.
For all m; n 2 N, if m ¤ n, then d.m/ ¤ d.n/.
(e) Calculate d

2k

for k D 0 and for each natural number k from 1
through 6.
(f) Based on your work in Exercise (6e), make a conjecture for a formula
for d.2n
/ where n is a nonnegative integer. Then explain why your
conjecture is correct.
(g) Is the following statement is true or false?
For each n 2 N, there exists a natural number m such that
d.m/ D n.
7. In Exercise (6), we introduced the number of divisors function d. For this
function, d W N ! N, where d.n/ is the number of natural number divisors
of n.
A function that is related to this function is the so-called set of divisors
function. This can be defined as a function S that associates with each
natural number the set of its distinct natural number factors. For example,
S.6/ D f1; 2; 3; 6g and S.10/ D f1; 2; 5; 10g.
?
(a) Discuss the function S by carefully stating its domain, codomain, and
its rule for determining outputs.
?
(b) Determine S.n/ for at least five different values of n.
?
(c) Determine S.n/ for at least three different prime number values of n.
(d) Does there exist a natural number n such that card .S.n// D 1? Ex-
plain. [Recall that card .S.n// is the number of elements in the set
S.n/.]
(e) Does there exist a natural number n such that card .S.n// D 2? Ex-
plain.
(f) Write the output for the function d in terms of the output for the func-
tion S. That is, write d.n/ in terms of S.n/.
(g) Is the following statement true or false? Justify your conclusion.
For all natural numbers m and n, if m ¤ n, then S.m/ ¤ S.n/.
(h) Is the following statement true or false? Justify your conclusion.
For all sets T that are subsets of N, there exists a natural number
n such that S.n/ D T .

8. Creating Functions with Finite Domains.Let A D fa; b; c; dg, B D fa; b; cg,
and C D fs; t; u; vg. In each of the following exercises, draw an arrow dia-
gram to represent your function when it is appropriate.
(a) Create a function f WA ! C whose range is the set C or explain why
it is not possible to construct such a function.
(b) Create a function f W A ! C whose range is the set fu; vg or explain
why it is not possible to construct such a function.
(c) Create a function f WB ! C whose range is the set C or explain why
(d) Create a function f WA ! C whose range is the set fug or explain why
(e) If possible, create a function f W A ! C that satisfies the following
condition:
For all x; y 2 A, if x ¤ y, then f.x/ ¤ f.y/.
If it is not possible to create such a function, explain why.
(f) If possible, create a function f WA ! fs; t; ug that satisfies the follow-
ing condition:
If it is not possible to create such a function, explain why.
6.2 More about Functions
In Section 6.1, we have seen many examples of functions. We have also seen
various ways to represent functions and to convey information about them. For
example, we have seen that the rule for determining outputs of a function can be
given by a formula, a graph, or a table of values. We have also seen that sometimes
it is more convenient to give a verbal description of the rule for a function. In cases
where the domain and codomain are small, finite sets, we used an arrow diagram to
convey information about how inputs and outputs are associated without explicitly
stating a rule. In this section, we will study some types of functions, some of which
we may not have encountered in previous mathematics courses.
Preview Activity 1 (The Number of Diagonals of a Polygon)
A polygon is a closed plane figure formed by the joining of three or more straight

6.2. More about Functions 295
lines. For example, a triangle is a polygon that has three sides; a quadrilateral is
a polygon that has four sides and includes squares, rectangles, and parallelograms;
a pentagon is a polygon that has five sides; and an octagon is a polygon that has
eight sides. A regular polygon is one that has equal-length sides and congruent
interior angles.
A diagonal of a polygon is a line segment that connects two nonadjacent ver-
tices of the polygon. In this activity, we will assume that all polygons are convex
polygons so that, except for the vertices, each diagonal lies inside the polygon.
For example, a triangle (3-sided polygon) has no diagonals and a rectangle has two
diagonals.
1. How many diagonals does any quadrilateral (4-sided polygon) have?
2. Let D D N f1; 2g. Define d WD ! N [ f0g so that d.n/ is the number of
diagonals of a convex polygon with n sides. Determine the values of d.3/,
d.4/, d.5/, d.6/, d.7/, and d.8/. Arrange the results in the form of a table
of values for the function d.
3. Let f WR ! R be defined by
f.x/ D
x .x 3/
2
:
Determine the values of f.0/, f.1/, f.2/, f.3/, f.4/, f.5/, f.6/, f.7/,
f.8/, and f.9/. Arrange the results in the form of a table of values for the
function f.
4. Compare the functions in Parts (2) and (3). What are the similarities between
the two functions and what are the differences? Should these two functions
be considered equal functions? Explain.
Preview Activity 2 (Derivatives)
In calculus, we learned how to find the derivatives of certain functions. For exam-
ple, if f.x/ D x2.sin x/, then we can use the product rule to obtain
f 0
.x/ D 2x.sin x/ C x2
.cos x/:
1. If possible, find the derivative of each of the following functions:

(a) f.x/ D x4 5x3 C 3x 7
(b) g.x/ D cos.5x/
(c) h.x/ D
sin x
x
(d) k.x/ D e x2
(e) r.x/ D jxj
2. Is it possible to think of differentiation as a function? Explain. If so, what
would be the domain of the function, what could be the codomain of the
function, and what is the rule for computing the element of the codomain
(output) that is associated with a given element of the domain (input)?
Functions Involving Congruences
Theorem 3.31 and Corollary 3.32 (see page 150) state that an integer is congruent
(mod n) to its remainder when it is divided by n. (Recall that we always mean the
remainder guaranteed by the Division Algorithm, which is the least nonnegative
remainder.) Since this remainder is unique and since the only possible remainders
for division by n are 0; 1; 2; : : :; n 1, we then know that each integer is congruent,
modulo n, to precisely one of the integers 0; 1; 2; : : :; n 1. So for each natural
number n, we will define a new set Zn as follows:
Zn D f0; 1; 2; : : :; n 1g:
For example, Z4 D f0; 1; 2; 3g and Z6 D f0; 1; 2; 3; 4; 5g. We will now explore a
method to define a function from Z6 to Z6.
For each x 2 Z6, we can compute x2 C 3 and then determine the value of r in
Z6 so that
x2
C 3

r .mod 6/ :
Since r must be in Z6, we must have 0 r 6. The results are shown in the
following table.
r where r where
x .x2 C 3/ r.mod6/ x .x2 C 3/ r.mod6/
0 3 3 0
1 4 4 1
2 1 5 4
Table 6.1: Table of Values Defined by a Congruence

The value of x in the first column can be thought of as the input for a function
with the value of r in the second column as the corresponding output. Each input
produces exactly one output. So we could write
f W Z6 ! Z6 by f.x/ D r where .x2
C 3/ r.mod6/.
This description and the notation for the outputs of this function are quite cumber-
some. So we will use a more concise notation. We will, instead, write
Let f WZ6 ! Z6 by f.x/ D x2 C 3

.mod 6/.
Progress Check 6.5 (Functions Defined by Congruences)
Let Z5 D f0; 1; 2; 3; 4g. Define
f WZ5 ! Z5 by f.x/ D x4
.mod 5/, for each x 2 Z5;
gWZ5 ! Z5 by g.x/ D x5
.mod 5/, for each x 2 Z5.
1. Determine f.0/, f.1/, f.2/, f.3/, and f.4/ and represent the function f
with an arrow diagram.
2. Determine g.0/, g.1/, g.2/, g.3/, and g.4/ and represent the function g with
an arrow diagram.
Equality of Functions
The idea of equality of functions has been in the background of our discussion of
functions, and it is now time to discuss it explicitly. The preliminary work for this
discussion was Preview Activity 1, in which D D N f1; 2g, and there were two
functions:
d WD ! N[f0g, where d.n/ is the number of diagonals of a convex polygon
with n sides
f WR ! R, where f.x/ D
x .x 3/
2
, for each real number x.
In Preview Activity 1, we saw that these two functions produced the same
outputs for certain values of the input (independent variable). For example, we can

verify that
d.3/ D f.3/ D 0; d.4/ D f.4/ D 2;
d.5/ D f.5/ D 5; and d.6/ D f.6/ D 9:
Although the functions produce the same outputs for some inputs, these are two
different functions. For example, the outputs of the function f are determined by a
formula, and the outputs of the function d are determined by a verbal description.
This is not enough, however, to say that these are two different functions. Based
on the evidence from Preview Activity 1, we might make the following conjecture:
For n 3, d.n/ D
n .n 3/
2
.
Although we have not proved this statement, it is a true statement. (See Exercise 6.)
However, we know the function d and the function f are not the same function.
For example,
f.0/ D 0, but 0 is not in the domain of d;
f./ D
. 3/
2
, but is not in the domain of d.
We thus see the importance of considering the domain and codomain of each of
the two functions in determining whether the two functions are equal or not. This
motivates the following definition.
Definition. Two functions f and g are equal provided that
The domain of f equals the domain of g. That is,
dom.f / D dom.g/.
The codomain of f equals the codomain of g. That is,
codom.f / D codom.g/.
For each x in the domain of f (which equals the domain of g),
f.x/ D g.x/.
Progress Check 6.6 (Equality of Functions)
Let A be a nonempty set. The identity function on the set A, denoted by IA , is

the function IA W A ! A defined by IA.x/ D x for every x in A. That is, for the
identity map, the output is always equal to the input.
For this progress check, we will use the functions f and g from Progress
Check 6.5. The identity function on the set Z5 is
IZ5 W Z5 ! Z5 by IZ5 .x/ D x .mod 5/, for each x 2 Z5.
Is the identity function on Z5 equal to either of the functions f or g from Progress
Check 6.5? Explain.
Mathematical Processes as Functions
Certain mathematical processes can be thought of as functions. In Preview Activ-
ity 2, we reviewed how to find the derivatives of certain functions, and we con-
sidered whether or not we could think of this differentiation process as a function.
If we use a differentiable function as the input and consider the derivative of that
function to be the output, then we have the makings of a function. Computer alge-
bra systems such as Maple and Mathematica have this derivative function as one
of their predefined operators.
Following is some Maple code (using the Classic Worksheet version of Maple)
that can be used to find the derivative function of the function given by f.x/ D
x2
.sin x/. The lines that start with the Maple prompt, Œ, are the lines typed by
the user. The centered lines following these show the resulting Maple output. The
first line defines the function f, and the second line uses the derivative function D
to produce the derivative of the function f.
Œ f := x ! xb2 sin(x);
f WD x ! x2
sin.x/
Œ f1 := D(f);
f 1 WD x ! 2x sin .x/ C x2
cos .x/
We must be careful when determining the domain for the derivative function since
there are functions that are not differentiable. To make things reasonably easy, we
will let F be the set of all real functions that are differentiable and call this the
domain of the derivative function D. We will use the set T of all real functions as
the codomain. So our function D is
DWF ! T by D.f / D f 0
:

Progress Check 6.7 (Average of a Finite Set of Numbers)
Let A D fa1; a2; : : :ang be a finite set whose elements are the real numbers
a1; a2; : : :an. We define the average of the set A to be the real number N
A, where
N
A D
a1 C a2 C C an
n
:
1. Find the average of A D f3; 7; 1; 5g.
2. Find the average of B D f7; 2; 3:8; 4:2; 7:1g.
3. Find the average of C D
np
2;
p
3;
p
3
o
.
4. Now let F.R/ be the set of all finite subsets of R. That is, a subset A of R is
in F.R/ if and only if A contains only a finite number of elements. Carefully
explain how the process of finding the average of a finite subset of R can be
thought of as a function. In doing this, be sure to specify the domain of the
function and the codomain of the function.
Sequences as Functions
A sequence can be considered to be an infinite list of objects that are indexed
(subscripted) by the natural numbers (or some infinite subset of N [ f0g). Using
this idea, we often write a sequence in the following form:
a1; a2; : : :; an; : : ::
In order to shorten our notation, we will often use the notation hani to represent this
sequence. Sometimes a formula can be used to represent the terms of a sequence,
and we might include this formula as the nth term in the list for a sequence such as
in the following example:
1;
1
2
;
1
3
; : : :;
1
n
; : : ::
In this case, the nth term of the sequence is
1
n
. If we know a formula for the nth
term, we often use this formula to represent the sequence. For example, we might
say
Define the sequence hani by an D
1
n
for each n 2 N.

This shows that this sequence is a function with domain N. If it is understood that
the domain is N, we could refer to this as the sequence

1
n

. Given an element
of the domain, we can consider an to be the output. In this case, we have used
subscript notation to indicate the output rather than the usual function notation.
We could just as easily write
a.n/ D
1
n
instead of an D
1
n
:
We make the following formal definition.
Definition. An (infinite) sequence is a function whose domain is N or some
infinite subset of N [ f0g.
Progress Check 6.8 (Sequences)
Find the sixth and tenth terms of each of the following sequences:
1.
1
3
;
1
6
;
1
9
;
1
12
; : : :
2. hani, where an D
1
n2
for each n 2 N
3. h. 1/ni
Functions of Two Variables
In Section 5.4, we learned how to form the Cartesian product of two sets. Recall
that a Cartesian product of two sets is a set of ordered pairs. For example, the set
Z Z is the set of all ordered pairs, where each coordinate of an ordered pair is
an integer. Since a Cartesian product is a set, it could be used as the domain or
codomain of a function. For example, we could use Z Z as the domain of a
function as follows:
Let f WZ Z ! Z be defined by f.m; n/ D 2m C n.
Technically, an element of Z Z is an ordered pair, and so we should write
f..m; n// for the output of the function f when the input is the ordered pair
.m; n/. However, the double parentheses seem unnecessary in this context

and there should be no confusion if we write f.m; n/ for the output of the
function f when the input is .m; n/. So, for example, we simply write
f.3; 2/ D 2 3 C 2 D 8; and
f. 4; 5/ D 2 . 4/ C 5 D 3:
Since the domain of this function is Z Z and each element of Z Z is an
ordered pair of integers, we frequently call this type of function a function
of two variables.
Finding the preimages of an element of the codomain for the function f , Z,
usually involves solving an equation with two variables. For example, to find the
preimages of 0 2 Z, we need to find all ordered pairs .m; n/ 2 Z Z such that
f.m; n/ D 0. This means that we must find all ordered pairs .m; n/ 2 Z Z such
that
2m C n D 0:
Three such ordered pairs are .0; 0/, .1; 2/, and . 1; 2/. In fact, whenever we
choose an integer value for m , we can find a corresponding integer n such that
2m C n D 0. This means that 0 has infinitely many preimages, and it may be
difficult to specify the set of all of the preimages of 0 using the roster method. One
way that can be used to specify this set is to use set builder notation and say that
the following set consists of all of the preimages of 0:
f.m; n/ 2 Z Z j 2m C n D 0g D f.m; n/ 2 Z Z j n D 2mg:
The second formulation for this set was obtained by solving the equation
2m C n D 0 for n.
Progress Check 6.9 (Working with a Function of Two Variables)
Let gWZ Z ! Z be defined by g.m; n/ D m2
n for all .m; n/ 2 Z Z.
1. Determine g.0; 3/, g.3; 2/, g. 3; 2/, and g.7; 1/.
2. Determine the set of all preimages of the integer 0 for the function g. Write
your answer using set builder notation.
3. Determine the set of all preimages of the integer 5 for the function g. Write
your answer using set builder notation.

Exercises 6.2
? 1. Let Z5 D f0; 1; 2; 3; 4g. Define f WZ5 ! Z5 by f.x/ D x2 C 4 .mod 5/,
and define gWZ5 ! Z5 by g.x/ D .x C 1/.x C 4/ .mod 5/.
(a) Calculate f.0/, f.1/; f.2/, f.3/, and f.4/.
(b) Calculate g.0/, g.1/; g.2/, g.3/, and g.4/.
(c) Is the function f equal to the function g? Explain.
2. Let Z6 D f0; 1; 2; 3; 4; 5g. Define f WZ6 ! Z6 by f.x/ D x2 C4 .mod 6/,
and define gWZ6 ! Z6 by g.x/ D .x C 1/.x C 4/ .mod 6/
(a) Calculate f.0/, f.1/; f.2/, f.3/, f.4/, and f.5/.
(b) Calculate g.0/, g.1/; g.2/, g.3/, g.4/, and g.5/.
? 3. Let f W .R f0g/ ! R by f.x/ D
x3
C 5x
x
and let g W R ! R by
g.x/ D x2
C 5.
(a) Calculate f.2/; f. 2/, f.3/, and f.
p
2/.
(b) Calculate g.0/, g.2/; g. 2/, g.3/, and g.
p
2/.
(d) Now let hW.R f0g/ ! R by h.x/ D x2
C 5. Is the function f equal
to the function h? Explain.
4. Represent each of the following sequences as functions. In each case, state
the domain, codomain, and rule for determining the outputs of the function.
Also, determine if any of the sequences are equal.
? (a) 1;
1
4
;
1
9
;
1
16
; : : :
(b)
1
3
;
1
9
;
1
27
;
1
81
; : : :
(c) 1; 1; 1; 1; 1; 1; : : :
? (d) cos.0/; cos./; cos.2/; cos.3/; cos.4/; : : :

5. Let A and B be two nonempty sets. There are two projection functions
with domain A B, the Cartesian product of A and B. One projection func-
tion will map an ordered pair to its first coordinate, and the other projection
function will map the ordered pair to its second coordinate. So we define
p1 WA B ! A by p1.a; b/ D a for every .a; b/ 2 A B; and
p2 WA B ! B by p2.a; b/ D b for every .a; b/ 2 A B.
Let A D f1; 2g and let B D fx; y; zg.
? (a) Determine the outputs for all possibleinputs for the projection function
p1 WA B ! A.
(b) Determine the outputs for all possibleinputs for the projection function
p2 WA B ! A.
? (c) What is the range of these projection functions?
(d) Is the following statement true or false? Explain.
For all .m; n/; .u; v/ 2 A B, if .m; n/ ¤ .u; v/, then
p1.m; n/ ¤ p1.u; v/.
?
6. Let D D N f1; 2g and define d W D ! N [ f0g by d.n/ D the number
of diagonals of a convex polygon with n sides. In Preview Activity 1, we
showed that for values of n from 3 through 8,
d.n/ D
n .n 3/
2
:
Use mathematical induction to prove that for all n 2 D,
d.n/ D
n .n 3/
2
.
Hint: To get an idea of how to handle the inductive step, use a pentagon.
First, form all the diagonals that can be made from four of the vertices. Then
consider how to make new diagonals when the fifth vertex is used. This may
generate an idea of how to proceed from a polygon with k sides to a polygon
with k C 1 sides.
? 7. Let f WZ Z ! Z be defined by f.m; n/ D m C 3n.
(a) Calculate f. 3; 4/ and f. 2; 7/.
(b) Determine the set of all the preimages of 4 by using set builder notation
to describe the set of all .m; n/ 2 Z Z such that f.m; n/ D 4.

8. Let gWZ Z ! Z Z be defined by g.m; n/ D .2m; m n/.
? (a) Calculate g.3; 5/ and g. 1; 4/.
(b) Determine all the preimages of .0; 0/. That is, find all .m; n/ 2 Z Z
such that g.m; n/ D .0; 0/.
?
(c) Determine the set of all the preimages of .8; 3/.
(d) Determine the set of all the preimages of .1; 1/.
(e) Is the following proposition true or false? Justify your conclusion.
For each .s; t/ 2 Z Z, there exists an .m; n/ 2 Z Z such that
g.m; n/ D .s; t/.
9. A 2 by 2 matrix over R is a rectangular array of four real numbers arranged
in two rows and two columns. We usually write this array inside brackets (or
parentheses) as follows:
A D

a b
c d

;
where a, b, c, and d are real numbers. The determinant of the 2 by 2 matrix
A, denoted by det.A/ , is defined as
det.A/ D ad bc:
? (a) Calculate the determinant of each of the following matrices:

3 5
4 1

;

1 0
0 7

; and

3 2
5 0

:
(b) Let M2.R/ be the set of all 2 by 2 matrices over R. The mathematical
process of finding the determinant of a 2 by 2 matrix over R can be
thought of as a function. Explain carefully how to do so, including a
clear statement of the domain and codomain of this function.
10. Using the notation from Exercise (9), let
A D

a b
c d

be a 2 by 2 matrix over R. The transpose of the matrix A, denoted by AT ,
is the 2 by 2 matrix over R defined by
AT
D

a c
b d

:

(a) Calculate the transpose of each of the following matrices:

3 5
4 1

;

1 0
0 7

; and

3 2
5 0

:
(b) Let M2.R/ be the set of all 2 by 2 matrices over R. The mathemati-
cal process of finding the transpose of a 2 by 2 matrix over R can be
thought of as a function. Carefully explain how to do so, including a
clear statement of the domain and codomain of this function.
11. Integration as a Function. In calculus, we learned that if f is real func-
tion that is continuous on the closed interval Œa; b, then the definite integral
R b
a f.x/ dx is a real number. In fact, one form of the Fundamental Theo-
rem of Calculus states that
Z b
a
f.x/ dx D F.b/ F.a/;
where F is any antiderivative of f , that is, where F 0 D f.
(a) Let Œa; b be a closed interval of real numbers and let C Œa; b be the set
of all real functions that are continuous on Œa; b. That is,
C Œa; b D ff WŒa; b ! R j f is continuous on Œa; bg :
i. Explain how the definite integral
R b
a f.x/ dx can be used to define
a function I from C Œa; b to R.
ii. Let Œa; b D Œ0; 2. Calculate I.f /, where f.x/ D x2 C 1.
iii. Let Œa; b D Œ0; 2. Calculate I.g/, where g.x/ D sin.x/.
In calculus, we also learned how to determine the indefiniteintegral
R
f.x/ dx
of a continuous function f .
(b) Let f.x/ D x2 C 1 and g.x/ D cos.2x/. Determine
R
f.x/ dx and
R
g.x/ dx.
(c) Let f be a continuous function on the closed interval Œ0; 1 and let T
be the set of all real functions. Can the process of determining the
indefinite integral of a continuous function be used to define a function
from C Œ0; 1 to T ? Explain.

6.3. Injections, Surjections, and Bijections 307
(d) Another form of the Fundamental Theorem of Calculus states that if f
is continuous on the interval Œa; b and if
g.x/ D
Z x
a
f.t/ dt
for each x in Œa; b, then g0 .x/ D f .x/. That is, g is an antiderivative
of f . Explain how this theorem can be used to define a function from
C Œa; b to T, where the output of the function is an antiderivative of
the input. (Recall that T is the set of all real functions.)
6.3 Injections, Surjections, and Bijections
Functions are frequently used in mathematics to define and describe certain rela-
tionships between sets and other mathematical objects. In addition, functions can
be used to impose certain mathematical structures on sets. In this section, we will
study special types of functions that are used to describe these relationships that
are called injections and surjections. Before defining these types of functions, we
will revisit what the definition of a function tells us and explore certain functions
with finite domains.
Preview Activity 1 (Functions with Finite Domains)
Let A and B be sets. Given a function f WA ! B, we know the following:
For every x 2 A, f.x/ 2 B. That is, every element of A is an input for
the function f . This could also be stated as follows: For each x 2 A, there
exists a y 2 B such that y D f.x/.
For a given x 2 A, there is exactly one y 2 B such that y D f.x/.
The definition of a function does not require that different inputs produce dif-
ferent outputs. That is, it is possible to have x1; x2 2 A with x1 ¤ x2 and
f.x1/ D f.x2/. The arrow diagram for the function f in Figure 6.5 illustrates
such a function.
Also, the definition of a function does not require that the range of the function
must equal the codomain. The range is always a subset of the codomain, but these
two sets are not required to be equal. That is, if g W A ! B, then it is possible
to have a y 2 B such that g.x/ ¤ y for all x 2 A. The arrow diagram for the
function g in Figure 6.5 illustrates such a function.

3
2
1
b
a
B
3
2
1
c
b
a
A B
A
f g
Figure 6.5: Arrow Diagram for Two Functions
Now let A D f1; 2; 3g, B D fa; b; c; dg, and C D fs; tg. Define
f WA ! B by gWA ! B by hWA ! C by
f.1/ D a g.1/ D a h.1/ D s
f.2/ D b g.2/ D b h.2/ D t
f.3/ D c g.3/ D a h.3/ D s
1. Which of these functions satisfy the following property for a function F ?
For all x; y 2 dom.F /, if x ¤ y, then F.x/ ¤ F.y/.
2. Which of these functions satisfy the following property for a function F ?
For all x; y 2 dom.F /, if F.x/ D F.y/, then x D y.
3. Determine the range of each of these functions.
4. Which of these functions have their range equal to their codomain?
5. Which of the these functions satisfy the following property for a function F ?
For all y in the codomain of F , there exists an x 2 dom.F / such that
F.x/ D y.
Preview Activity 2 (Statements Involving Functions)
Let A and B be nonempty sets and let f W A ! B. In Preview Activity 1, we
determined whether or not certain functions satisfied some specified properties.
These properties were written in the form of statements, and we will now examine
these statements in more detail.

1. Consider the following statement:
(a) Write the contrapositive of this conditional statement.
(b) Write the negation of this conditional statement.
2. Now consider the statement:
For all y 2 B, there exists an x 2 A such that f.x/ D y.
Write the negation of this statement.
3. Let g W R ! R be defined by g.x/ D 5x C 3, for all x 2 R. Complete the
following proofs of the following propositions about the function g.
Proposition 1. For all a; b 2 R, if g.a/ D g.b/, then a D b.
Proof. We let a; b 2 R, and we assume that g.a/ D g.b/ and will prove
that a D b. Since g.a/ D g.b/, we know that
5a C 3 D 5b C 3:
(Now prove that in this situation, a D b.)
Proposition 2. For all b 2 R, there exists an a 2 R such that g.a/ D b.
Proof. We let b 2 R. We will prove that there exists an a 2 R such that
g.a/ D b by constructing such an a in R. In order for this to happen, we
need g.a/ D 5a C 3 D b.
(Now solve the equation for a and then show that for this real number a,
g.a/ D b.)
Injections
In previous sections and in Preview Activity 1, we have seen examples of functions
for which there exist different inputs that produce the same output. Using more
formal notation, this means that there are functions f W A ! B for which there
exist x1; x2 2 A with x1 ¤ x2 and f.x1/ D f.x2/. The work in the preview
activities was intended to motivate the following definition.

Definition. Let f W A ! B be a function from the set A to the set B. The
function f is called an injection provided that
for all x1; x2 2 A, if x1 ¤ x2, then f.x1/ ¤ f.x2/.
When f is an injection, we also say that f is a one-to-one function, or that
f is an injective function.
Notice that the condition that specifies that a function f is an injection is given
in the form of a conditional statement. As we shall see, in proofs, it is usually
easier to use the contrapositive of this conditional statement. Although we did not
define the term then, we have already written the contrapositive for the conditional
statement in the definition of an injection in Part (1) of Preview Activity 2. In
that preview activity, we also wrote the negation of the definition of an injection.
Following is a summary of this work giving the conditions for f being an injection
or not being an injection.
Let f WA ! B.
“The function f is an injection” means that
For all x1; x2 2 A, if x1 ¤ x2, then f.x1/ ¤ f.x2/; or
For all x1; x2 2 A, if f.x1/ D f.x2/, then x1 D x2.
“The function f is not an injection” means that
There exist x1; x2 2 A such that x1 ¤ x2 and f.x1/ D f.x2/.
Progress Check 6.10 (Working with the Definition of an Injection)
Now that we have defined what it means for a function to be an injection, we can
see that in Part (3) of Preview Activity 2, we proved that the function g W R ! R
is an injection, where g.x/ D 5x C 3 for all x 2 R. Use the definition (or its
negation) to determine whether or not the following functions are injections.
1. k W A ! B, where A D fa; b; cg, B D f1; 2; 3; 4g, and k.a/ D 4, k.b/ D 1,
and k.c/ D 3.
2. f W A ! C , where A D fa; b; cg, C D f1; 2; 3g, and f.a/ D 2,f.b/ D 3,
and f.c/ D 2.

3. F W Z ! Z defined by F.m/ D 3m C 2 for all m 2 Z
4. h W R ! R defined by h.x/ D x2
3x for all x 2 R
5. s W Z5 ! Z5 defined by s.x/ D x3 for all x 2 Z5.
Surjections
In previous sections and in Preview Activity 1, we have seen that there exist func-
tions f WA ! B for which range.f / D B. This means that every element of B is
an output of the function f for some input from the set A. Using quantifiers, this
means that for every y 2 B, there exists an x 2 A such that f.x/ D y . One of the
objectives of the preview activities was to motivate the following definition.
Definition. Let f W A ! B be a function from the set A to the set B. The
function f is called a surjection provided that the range of f equals the
codomain of f . This means that
for every y 2 B, there exists an x 2 A such that f.x/ D y.
When f is a surjection, we also say that f is an onto function or that f maps
A onto B. We also say that f is a surjective function.
One of the conditions that specifies that a function f is a surjection is given in
the form of a universally quantified statement, which is the primary statement used
in proving a function is (or is not) a surjection. Although we did not define the term
then, we have already written the negation for the statement defining a surjection
in Part (2) of Preview Activity 2. We now summarize the conditions for f being a
surjection or not being a surjection.

Let f WA ! B.
“The function f is a surjection” means that
range.f / D codom.f / D B; or
For every y 2 B, there exists an x 2 A such that f.x/ D y.
“The function f is not a surjection” means that
range.f / ¤ codom.f /; or
There exists a y 2 B such that for all x 2 A, f.x/ ¤ y.
One other important type of function is when a function is both an injection and
surjection. This type of function is called a bijection.
Definition. A bijection is a function that is both an injection and a surjection.
If the function f is a bijection, we also say that f is one-to-one and onto
and that f is a bijective function.
Progress Check 6.11 (Working with the Definition of a Surjection)
Now that we have defined what it means for a function to be a surjection, we can
see that in Part (3) of Preview Activity 2, we proved that the function g W R ! R
is a surjection, where g.x/ D 5x C 3 for all x 2 R. Determine whether or not the
following functions are surjections.
1. k WA ! B, where A D fa; b; cg, B D f1; 2; 3; 4g, and k.a/ D 4, k.b/ D 1,
and k.c/ D 3.
2. f WR ! R defined by f.x/ D 3x C 2 for all x 2 R.
3. F WZ ! Z defined by F.m/ D 3m C 2 for all m 2 Z.
4. s W Z5 ! Z5 defined by s.x/ D x3
for all x 2 Z5.
The Importance of the Domain and Codomain
The functions in the next two examples will illustrate why the domain and the
codomain of a function are just as important as the rule defining the outputs of a
function when we need to determine if the function is a surjection.

Example 6.12 (A Function that Is Neither an Injection nor a Surjection)
C 1. Notice that
f.2/ D 5 and f. 2/ D 5:
This is enough to prove that the function f is not an injection since this shows that
there exist two different inputs that produce the same output.
Since f.x/ D x2 C 1, we know that f.x/ 1 for all x 2 R. This implies that
the function f is not a surjection. For example, 2 is in the codomain of f and
f.x/ ¤ 2 for all x in the domain of f.
Example 6.13 (A Function that Is Not an Injection but Is a Surjection)
Let T D fy 2 R j y 1g, and define F W R ! T by F.x/ D x2 C 1. As in
Example 6.12, the function F is not an injection since F.2/ D F. 2/ D 5.
Is the function F a surjection? That is, does F map R onto T ? As in Exam-
ple 6.12, we do know that F.x/ 1 for all x 2 R.
To see if it is a surjection, we must determine if it is true that for every y 2 T ,
there exists an x 2 R such that F.x/ D y. So we choose y 2 T. The goal is to
determine if there exists an x 2 R such that
F.x/ D y, or
x2
C 1 D y:
One way to proceed is to work backward and solve the last equation (if possible)
for x. Doing so, we get
x2
D y 1
x D
p
y 1 or x D
p
y 1:
Now, since y 2 T , we know that y 1 and hence that y 1 0. This means that
p
y 1 2 R. Hence, if we use x D
p
y 1, then x 2 R, and
F.x/ D F
p
y 1

D
p
y 1
2
C 1
D .y 1/ C 1
D y:
This proves that F is a surjection since we have shown that for all y 2 T, there
exists an x 2 R such that F.x/ D y. Notice that for each y 2 T, this was a
constructive proof of the existence of an x 2 R such that F.x/ D y.

An Important Lesson. In Examples 6.12 and 6.13, the same mathematical
formula was used to determine the outputs for the functions. However, one
function was not a surjection and the other one was a surjection. This illus-
trates the important fact that whether a function is surjective not only depends
on the formula that defines the output of the function but also on the domain
and codomain of the function.
The next example will show that whether or not a function is an injection also
depends on the domain of the function.
Example 6.14 (A Function that Is a Injection but Is Not a Surjection)
Let Z
D fx 2 Z j x 0g D N [ f0g. Define g W Z
! N by g.x/ D x2
C 1.
(Notice that this is the same formula used in Examples 6.12 and 6.13.) Following
is a table of values for some inputs for the function g.
x g.x/ x g.x/
0 1 3 10
1 2 4 17
2 5 5 26
Notice that the codomain is N, and the table of values suggests that some natural
numbers are not outputs of this function. So it appears that the function g is not a
surjection.
To prove that g is not a surjection, pick an element of N that does not appear
to be in the range. We will use 3, and we will use a proof by contradiction to prove
that there is no x in the domain .Z
/ such that g.x/ D 3. So we assume that there
exists an x 2 Z with g.x/ D 3. Then
x2
C 1 D 3
x2
D 2
x D ˙
p
2:
But this is not possible since
p
2 … Z. Therefore, there is no x 2 Z with
g.x/ D 3. This means that for every x 2 Z
, g.x/ ¤ 3. Therefore, 3 is not in the
range of g, and hence g is not a surjection.
The table of values suggests that different inputs produce different outputs, and
hence that g is an injection. To prove that g is an injection, assume that s; t 2 Z

(the domain) with g.s/ D g.t/. Then
s2
C 1 D t2
C 1
s2
D t2
:
Since s; t 2 Z, we know that s 0 and t 0. So the preceding equation implies
that s D t. Hence, g is an injection.
An Important Lesson. The functions in the three preceding examples all
used the same formula to determine the outputs. The functions in Exam-
ples 6.12 and 6.13 are not injections but the function in Example 6.14 is an
injection. This illustrates the important fact that whether a function is injec-
tive not only depends on the formula that defines the output of the function
but also on the domain of the function.
Progress Check 6.15 (The Importance of the Domain and Codomain)
Let RC
D fy 2 R j y 0g. Define
f WR ! R by f.x/ D e x
, for each x 2 R, and
gWR ! RC
by g.x/ D e x
, for each x 2 R.
Determine if each of these functions is an injection or a surjection. Justify
your conclusions. Note: Before writing proofs, it might be helpful to draw the
graph of y D e x
. A reasonable graph can be obtained using 3 x 3
and 2 y 10. Please keep in mind that the graph is does not prove your
conclusions, but may help you arrive at the correct conclusions, which will still
need proof.
Working with a Function of Two Variables
It takes time and practice to become efficient at working with the formal definitions
of injection and surjection. As we have seen, all parts of a function are important
(the domain, the codomain, and the rule for determining outputs). This is especially
true for functions of two variables.
For example, we define f WR R ! R R by
f.a; b/ D .2a C b; a b/ for all .a; b/ 2 R R:

Notice that both the domain and the codomain of this function is the set R R.
Thus, the inputs and the outputs of this function are ordered pairs of real numbers.
For example,
f.1; 1/ D .3; 0/ and f. 1; 2/ D .0; 3/:
To explore whether or not f is an injection, we assume that .a; b/ 2 R R,
.c; d/ 2 R R, and f.a; b/ D f.c; d/. This means that
.2a C b; a b/ D .2c C d; c d/:
Since this equation is an equality of ordered pairs, we see that
2a C b D 2c C d, and
a b D c d:
By adding the corresponding sides of the two equations in this system, we obtain
3a D 3c and hence, a D c. Substituting a D c into either equation in the system
give us b D d. Since a D c and b D d, we conclude that
.a; b/ D .c; d/:
Hence, we have shown that if f.a; b/ D f.c; d/, then .a; b/ D .c; d/. Therefore,
f is an injection.
Now, to determine if f is a surjection, we let .r; s/ 2 R R, where .r; s/ is
considered to be an arbitrary element of the codomain of the function f . Can we
find an ordered pair .a; b/ 2 RR such that f.a; b/ D .r; s/? Working backward,
we see that in order to do this, we need
.2a C b; a b/ D .r; s/:
That is, we need
2a C b D r and a b D s:
Solving this system for a and b yields
a D
r C s
3
and b D
r 2s
3
:
Since r; s 2 R, we can conclude that a 2 R and b 2 R and hence that .a; b/ 2 R R.

We now need to verify that for these values of a and b, we get f.a; b/ D .r; s/. So
f.a; b/ D f

r C s
3
;
r 2s
3

D

2

r C s
3

C
r 2s
3
;
r C s
3
r 2s
3

D

2r C 2s C r 2s
3
;
r C s r C 2s
3

D .r; s/:
This proves that for all .r; s/ 2 R R, there exists .a; b/ 2 R R such that
f.a; b/ D .r; s/. Hence, the function f is a surjection. Since f is both an injection
and a surjection, it is a bijection.
Progress Check 6.16 (A Function of Two Variables)
Let gWR R ! R be defined by g.x; y/ D 2x C y, for all .x; y/ 2 R R.
Note: Be careful! One major difference between this function and the previous
example is that for the function g, the codomain is R, not R R. It is a good idea
to begin by computing several outputs for several inputs (and remember that the
inputs are ordered pairs).
1. Notice that the ordered pair .1; 0/ 2 R R. That is .1; 0/ is in the domain
of g. Also notice that g.1; 0/ D 2. Is it possible to find another ordered pair
.a; b/ 2 R R such that g.a; b/ D 2?
2. Let z 2 R. Then .0; z/ 2 R R and so .0; z/ 2 dom .g/. Now determine
g.0; z/.
3. Is the function g an injection? Is the function g a surjection? Justify your
conclusions.
Exercises 6.3
1. (a) Draw an arrow diagram that represents a function that is an injection
but is not a surjection.
(b) Draw an arrow diagram that represents a function that is an injection
and is a surjection.

(c) Draw an arrow diagram that represents a function that is not an injec-
tion and is not a surjection.
(d) Draw an arrow diagram that represents a function that is not an injec-
tion but is a surjection.
(e) Draw an arrow diagram that represents a function that is not a bijection.
2. Let Z5 D f0; 1; 2; 3; 4g and let Z6 D f0; 1; 2; 3; 4; 5g. For each of the fol-
lowing functions, determine if the function is an injection and determine if
the function is a surjection. Justify all conclusions.
?
(a) f WZ5 ! Z5 by f.x/ D x2
C 4 .mod 5/, for all x 2 Z5
(b) gWZ6 ! Z6 by g.x/ D x2
C 4 .mod 6/, for all x 2 Z6
? (c) F WZ5 ! Z5 by F.x/ D x3 C 4 .mod 5/, for all x 2 Z5
3. For each of the following functions, determine if the function is an injection
and determine if the function is a surjection. Justify all conclusions.
? (a) f WZ ! Z defined by f.x/ D 3x C 1, for all x 2 Z.
? (b) F WQ ! Q defined by F.x/ D 3x C 1, for all x 2 Q.
(c) gWR ! R defined by g.x/ D x3, for all x 2 R.
(d) G WQ ! Q defined by G.x/ D x3, for all x 2 Q.
(e) kWR ! R defined by k.x/ D e x2
, for all x 2 R.
(f) K WR ! R defined by K.x/ D e x2
, for all x 2 R.
Note: R
D fx 2 R j x 0g.
(g) K1 W R ! T defined by K1.x/ D e x2
, for all x 2 R, where
T D fy 2 R j 0 y 1g.
? (h) hWR ! R defined by h.x/ D
2x
x2 C 4
, for all x 2 R.
(i) H W fx 2 R j x 0g !

y 2 R
ˇ
ˇ
ˇ
ˇ0 y
1
2

defined by
H.x/ D
2x
x2 C 4
, for all x 2 fx 2 R j x 0g.
4. For each of the following functions, determine if the function is a bijection.
Justify all conclusions.
?
(a) F WR ! R defined by F.x/ D 5x C 3, for all x 2 R.
?
(b) G WZ ! Z defined by G.x/ D 5x C 3, for all x 2 Z.

(c) f W.R f4g/ ! R defined by f.x/ D
3x
x 4
, for all x 2 .R f4g/.
(d) g W .R f4g/ ! .R f3g/ defined by g.x/ D
3x
x 4
, for all
x 2 .R f4g/.
5. Let s WN ! N, where for each n 2 N, s.n/ is the sum of the distinct natural
number divisors of n. This is the sum of the divisors function that was
introduced in Preview Activity 2 from Section 6.1. Is s an injection? Is s a
surjection? Justify your conclusions.
6. Let d W N ! N, where d.n/ is the number of natural number divisors of
n. This is the number of divisors function introduced in Exercise (6) from
Section 6.1. Is the function d an injection? Is the function d a surjection?
Justify your conclusions.
? 7. In Preview Activity 2 from Section 6.1 , we introduced the birthday func-
tion. Is the birthday function an injection? Is it a surjection? Justify your
conclusions.
8. (a) Let f WZZ ! Z be defined by f.m; n/ D 2mCn. Is the function f
an injection? Is the function f a surjection? Justify your conclusions.
(b) Let gWZZ ! Z be defined by g.m; n/ D 6mC3n. Is the function g
an injection? Is the function g a surjection? Justify your conclusions.
? 9. (a) Let f WR R ! R R be defined by f.x; y/ D .2x; x C y/. Is the
function f an injection? Is the function f a surjection? Justify your
conclusions.
(b) Let g W Z Z ! Z Z be defined by g.x; y/ D .2x; x C y/. Is the
function g an injection? Is the function g a surjection? Justify your
conclusions.
10. Let f W R R ! R be the function defined by f.x; y/ D x2y C 3y, for
all .x; y/ 2 R R. Is the function f and injection? Is the function f a
11. Let g W R R ! R be the function defined by g.x; y/ D .x3
C 2/ siny,
for all .x; y/ 2 R R. Is the function g and injection? Is the function g a
12. Let A be a nonempty set. The identity function on the set A, denoted by
IA, is the function IA WA ! A defined by IA.x/ D x for every x in A. Is IA
an injection? Is IA a surjection? Justify your conclusions.

13. Let A and B be two nonempty sets. Define
p1 WA B ! A by p1.a; b/ D a
for every .a; b/ 2 A B. This is the first projection function introduced in
Exercise (5) in Section 6.2.
(a) Is the function p1 a surjection? Justify your conclusion.
(b) If B D fbg, is the function p1 an injection? Justify your conclusion.
(c) Under what condition(s) is the function p1 not an injection? Make a
conjecture and prove it.
14. Define f W N ! Z be defined as follows: For each n 2 N,
f.n/ D
1 C . 1/n.2n 1/
4
:
Is the function f an injection? Is the function f a surjection? Justify your
conclusions.
Suggestions. Start by calculating several outputs for the function before
you attempt to write a proof. In exploring whether or not the function is an
injection, it might be a good idea to uses cases based on whether the inputs
are even or odd. In exploring whether f is a surjection, consider using cases
based on whether the output is positive or less than or equal to zero.
15. Let C be the set of all real functions that are continuous on the closed interval
Œ0; 1. Define the function AWC ! R as follows: For each f 2 C ,
A.f / D
Z 1
0
f.x/ dx:
Is the function A an injection? Is it a surjection? Justify your conclusions.
16. Let A D f.m; n/ j m 2 Z; n 2 Z; and n ¤ 0g. Define f W A ! Q as fol-
lows:
For each .m; n/ 2 A, f.m; n/ D
m C n
n
.
(a) Is the function f an injection? Justify your conclusion.
(b) Is the function f a surjection? Justify your conclusion.

(a) Proposition. The function f WR R ! R R defined by f.x; y/ D
.2x C y; x y/ is an injection.
Proof. For each .a; b/ and .c; d/ in R R, if f.a; b/ D f.c; d/, then
.2a C b; a b/ D .2c C d; c d/:
We will use systems of equations to prove that a D c and b D d.
2a C b D 2c C d
a b D c d
3a D 3c
a D c
Since a D c, we see that
.2c C b; c b/ D .2c C d; c d/:
So b D d. Therefore, we have proved that the function f is an injec-
tion.
(b) Proposition. The function f WR R ! R R defined by f.x; y/ D
.2x C y; x y/ is a surjection.
Proof. We need to find an ordered pair such that f.x; y/ D .a; b/ for
each .a; b/ in R R. That is, we need .2x C y; x y/ D .a; b/, or
2x C y D a and x y D b:
Treating these two equations as a system of equations and solving for
x and y, we find that
x D
a C b
3
and y D
a 2b
3
:

Hence, x and y are real numbers, .x; y/ 2 R R, and
f.x; y/ D f

a C b
3
;
a 2b
3

D

2

a C b
3

C
a 2b
3
;
a C b
3
a 2b
3

D

2a C 2b C a 2b
3
;
a C b a C 2b
3

D

3a
3
;
3b
3

D .a; b/:
Therefore, we have proved that for every .a; b/ 2 R R, there exists
an .x; y/ 2 R R such that f.x; y/ D .a; b/. This proves that the
function f is a surjection.
18. Piecewise Defined Functions. We often say that a function is a piecewise
defined function if it has different rules for determining the output for dif-
ferent parts of its domain. For example, we can define a function f WR ! R
by giving a rule for calculating f.x/ when x 0 and giving a rule for
calculating f.x/ when x 0 as follows:
f.x/ D
(
x2
C 1; if x 0;
x 1 if x 0.
(a) Sketch a graph of the function f. Is the function f and injection? Is
the function f a surjection? Justify your conclusions.
For each of the following functions, determine if the function is an injection
and determine if the function is a surjection. Justify all conclusions.
(b) gWŒ0; 1 ! .0; 1/ by
g.x/ D
8̂

:̂
0:8; if x D 0;
0:5x; if 0 x 1;
0:6 if x D 1.
(c) hWZ ! f0; 1g by
h.x/ D
(
0; if x is even;
1; if x is odd.

6.4. Composition of Functions 323
19. Functions Whose Domain is M2.R/. Let M2.R/ represent the set of all
2 by 2 matrices over R.
(a) Define detWM2.R/ ! R by
det

a b
c d

D ad bc:
This is the determinant function introduced in Exercise (9) from Sec-
tion 6.2. Is the determinant function an injection? Is the determinant
function a surjection? Justify your conclusions.
(b) Define tranWM2.R/ ! M2.R/ by
tran

a b
c d

D AT
D

a c
b d

:
This is the transpose function introduced in Exercise (10) from Sec-
tion 6.2. Is the transpose function an injection? Is the transpose func-
tion a surjection? Justify your conclusions.
(c) Define F WM2.R/ ! R by
F

a b
c d

D a2
C d2
b2
c2
:
Is the function F an injection? Is the function F a surjection? Justify
your conclusions.
6.4 Composition of Functions
Preview Activity 1 (Constructing a New Function)
Let A D fa; b; c; dg, B D fp; q; rg, and C D fs; t; u; vg. The arrow diagram in
Figure 6.6 shows two functions: f WA ! B and gWB ! C . Notice that if x 2 A,
then f.x/ 2 B. Since f.x/ 2 B, we can apply the function g to f.x/, and we
obtain g.f.x//, which is an element of C .
Using this process, determine g.f.a//, g.f.b//, g.f.c//, and g.f.d//. Then
explain how we can use this information to define a function from A to C .

a
b
c
d
p
q
r
s
t
u
v
A B C
f g
Figure 6.6: Arrow Diagram Showing Two Functions
Preview Activity 2 (Verbal Descriptions of Functions)
The outputs of most real functions we have studiedin previous mathematics courses
have been determined by mathematical expressions. In many cases, it is possible
to use these expressions to give step-by-step verbal descriptions of how to compute
the outputs. For example, if
f WR ! R is defined by f.x/ D .3x C 2/3
;
we could describe how to compute the outputs as follows:
Step Verbal Description Symbolic Result
1 Choose an input. x
2 Multiply by 3. 3x
3 Add 2. 3x C 2
4 Cube the result. .3x C 2/3
Complete step-by-step verbal descriptions for each of the following functions.
1. f WR ! R by f.x/ D
p
3x2 C 2, for each x 2 R.
2. gWR ! R by g.x/ D sin 3x2
C 2

, for each x 2 R.
3. hWR ! R by h.x/ D e3x2C2
, for each x 2 R.
4. G WR ! R by G.x/ D ln.x4
C 3/, for each x 2 R.

5. kWR ! R by k.x/ D 3
r
sin.4x C 3/
x2 C 1
, for each x 2 R.
Composition of Functions
There are several ways to combine two existing functions to create a new function.
For example, in calculus, we learned how to form the product and quotient of two
functions and then how to use the product rule to determine the derivative of a
product of two functions and the quotient rule to determine the derivative of the
quotient of two functions.
The chain rule in calculus was used to determine the derivative of the compo-
sition of two functions, and in this section, we will focus only on the composition
of two functions. We will then consider some results about the compositions of
injections and surjections.
The basic idea of function composition is that when possible, the output of a
function f is used as the input of a function g. This can be referred to as “f
followed by g” and is called the composition of f and g. In previous mathematics
courses, we used this idea to determine a formula for the composition of two real
functions.
For example, if
f.x/ D 3x2
C 2 and g.x/ D sinx;
then we can compute g.f.x// as follows:
g.f.x// D g 3x2
C 2

D sin 3x2
C 2

:
In this case, f.x/, the output of the function f , was used as the input for the
function g. We now give the formal definition of the composition of two functions.
Definition. Let A, B, and C be nonempty sets, and let f W A ! B and
g W B ! C be functions. The composition of f and g is the function
g ı f WA ! C defined by
.g ı f /.x/ D g .f.x//
for all x 2 A. We often refer to the function g ı f as a composite function.
It is helpful to think of the composite function g ı f as “f followed by g.”
We then refer to f as the inner function and g as the outer function.

Composition and Arrow Diagrams
The concept of the composition of two functions can be illustrated with arrow
diagrams when the domain and codomain of the functions are small, finite sets.
Although the term “composition” was not used then, this was done in Preview
Activity 1, and another example is given here.
Let A D fa; b; c; dg, B D fp; q; rg, and C D fs; t; u; vg. The arrow diagram
in Figure 6.7 shows two functions: f WA ! B and gWB ! C .
a
b
c
d
p
q
r
s
t
u
v
A B C
f g
Figure 6.7: Arrow Diagram for Two Functions
If we follow the arrows from the set A to the set C , we will use the outputs of
f as inputs of g, and get the arrow diagram from A to C shown in Figure 6.8. This
diagram represents the composition of f followed by g.
Figure 6.8: Arrow Diagram for g ı f WA ! C

Progress Check 6.17 (The Composition of Two Functions)
Let A D fa; b; c; dg and B D f1; 2; 3g. Define the functions f and g as follows:
f WA ! B defined by f.a/ D 2, f.b/ D 3, f.c/ D 1, and f.d/ D 2.
gWB ! B defined by g.1/ D 3, g.2/ D 1, and g.3/ D 2.
Create arrow diagrams for the functions f , g, g ı f , and g ı g.
Decomposing Functions
We use the chain rule in calculus to find the derivative of a composite function.
The first step in the process is to recognize a given function as a composite function.
This can be done in many ways, but the work in Preview Activity 2 can be used
to decompose a function in a way that works well with the chain rule. The use of
the terms “inner function” and “outer function” can also be helpful. The idea is
that we use the last step in the process to represent the outer function, and the steps
prior to that to represent the inner function. So for the function,
f WR ! R by f.x/ D .3x C 2/3
;
the last step in the verbal description table was to cube the result. This means that
we will use the function g (the cubing function) as the outer function and will use
the prior steps as the inner function. We will denote the inner function by h. So we
let hWR ! R by h.x/ D 3x C 2 and gWR ! R by g.x/ D x3. Then
.g ı h/.x/ D g .h.x//
D g.3x C 2/
D .3x C 2/3
D f.x/:
We see that gıh D f and, hence, we have “decomposed” the function f . It should
be noted that there are other ways to write the function f as a composition of two
functions, but the way just described is the one that works well with the chain rule.
In this case, the chain rule gives
f 0
.x/ D .g ı h/0
.x/
D g0
.h.x// h0
.x/
D 3.h.x//2
3
D 9.3x C 2/2

Progress Check 6.18 (Decomposing Functions)
Write each of the following functions as the composition of two functions.
1. F WR ! R by F.x/ D .x2 C 3/3
2. G WR ! R by G.x/ D ln.x2
C 3/
3. f WZ ! Z by f.x/ D jx2
3j
4. gWR ! R by g.x/ D cos

2x 3
x2 C 1

Theorems about Composite Functions
If f W A ! B and g W B ! C , then we can form the composite function g ı f W
A ! C . In Section 6.3, we learned about injections and surjections. We now
explore what type of function g ı f will be if the functions f and g are injections
(or surjections).
Progress Check 6.19 (Compositions of Injections and Surjections)
Although other representations of functions can be used, it will be helpful to use
arrow diagrams to represent the functions in this progress check. We will use the
following sets:
A D fa; b; cg; B D fp; q; rg; C D fu; v; w; xg; and D D fu; vg:
1. Draw an arrow diagram for a function f WA ! B that is an injection and an
arrow diagram for a function g WB ! C that is an injection. In this case, is
the composite function g ı f WA ! C an injection? Explain.
2. Draw an arrow diagram for a function f WA ! B that is a surjection and an
arrow diagram for a function g WB ! D that is a surjection. In this case, is
the composite function g ı f WA ! D a surjection? Explain.
3. Draw an arrow diagram for a function f WA ! B that is a bijection and an
arrow diagram for a function g W B ! A that is a bijection. In this case, is
the composite function g ı f WA ! A bijection? Explain.
In Progress Check 6.19, we explored some properties of composite functions
related to injections, surjections, and bijections. The following theorem contains

results that these explorations were intended to illustrate. Some of the proofs will
be included in the exercises.
Theorem 6.20. Let A, B, and C be nonempty sets and assume that f W A ! B
and gWB ! C .
1. If f and g are both injections, then .g ı f /WA ! C is an injection.
2. If f and g are both surjections, then .g ı f /WA ! C is a surjection.
3. If f and g are both bijections, then .g ı f /WA ! C is a bijection.
The proof of Part (1) is Exercise (6). Part (3) is a direct consequence of the first
two parts. We will discuss a process for constructing a proof of Part (2). Using
the forward-backward process, we first look at the conclusion of the conditional
statement in Part (2). The goal is to prove that g ı f is a surjection. Since g ı f W
A ! C , this is equivalent to proving that
For all c 2 C , there exists an a 2 A such that .g ı f /.a/ D c.
Since this statement in the backward process uses a universal quantifier, we
will use the choose-an-element method and choose an arbitrary element c in the
set C . The goal now is to find an a 2 A such that .g ı f /.a/ D c.
Now we can look at the hypotheses. In particular, we are assuming that both
f W A ! B and g W B ! C are surjections. Since we have chosen c 2 C , and
gWB ! C is a surjection, we know that
there exists a b 2 B such that g.b/ D c.
Now, b 2 B and f WA ! B is a surjection. Hence
there exists an a 2 A such that f.a/ D b.
If we now compute .g ı f /.a/, we will see that
.g ı f /.a/ D g .f.a// D g.b/ D c:
We can now write the proof as follows:

Proof of Theorem 6.20, Part (2). Let A, B, and C be nonempty sets and assume
that f W A ! B and g W B ! C are both surjections. We will prove that g ı f W
A ! C is a surjection.
Let c be an arbitrary element of C . We will prove there exists an a 2 A such
that .g ı f /.a/ D c. Since gWB ! C is a surjection, we conclude that
there exists a b 2 B such that g.b/ D c.
Now, b 2 B and f WA ! B is a surjection. Hence
there exists an a 2 A such that f.a/ D b.
We now see that
.g ı f /.a/ D g .f.a//
D g.b/
D c:
We have now shown that for every c 2 C , there exists an a 2 A such that
.g ı f /.a/ D c, and this proves that g ı f is a surjection.
Theorem 6.20 shows us that if f and g are both special types of functions, then
the composition of f followed by g is also that type of function. The next question
is, “If the composition of f followed by g is an injection (or surjection), can we
make any conclusions about f or g?” A partial answer to this question is provided
in Theorem 6.21. This theorem will be investigated and proved in the Explorations
and Activities for this section. See Exercise (10).
Theorem 6.21. Let A, B, and C be nonempty sets and assume that f W A ! B
and gWB ! C .
1. If g ı f WA ! C is an injection, then f WA ! B is an injection.
2. If g ı f WA ! C is a surjection, then gWB ! C is a surjection.
Exercises 6.4
1. In our definition of the composition of two functions, f and g, we re-
quired that the domain of g be equal to the codomain of f . However, it
is sometimes possible to form the composite function g ı f even though
dom.g/ ¤ codom.f /. For example, let

f WR ! R be defined by f.x/ D x2 C 1, and let
gWR f0g ! R be defined by g.x/ D
1
x
.
(a) Is it possible to determine .g ı f /.x/ for all x 2 R? Explain.
(b) In general, let f W A ! T and g W B ! C . Find a condition on the
domain of g (other than B D T ) that results in a meaningful definition
of the composite function g ı f WA ! C .
? 2. Let h W R ! R be defined by h.x/ D 3x C 2 and g W R ! R be defined by
g.x/ D x3
. Determine formulas for the composite functions g ı h and hı g.
Is the function g ı h equal to the function h ı g? Explain . What does this
tell you about the operation of composition of functions?
? 3. Following are formulas for certain real functions. Write each of these real
functions as the composition of two functions. That is, decompose each of
the functions.
(a) F.x/ D cos.ex
/
(b) G.x/ D ecos.x/
(c) H.x/ D
1
sin x
(d) K.x/ D cos

e x2

4. The identity function on a set S, denoted by IS , is defined as follows:
IS WS ! S by IS .x/ D x for each x 2 S. Let f WA ! B.
?
(a) For each x 2 A, determine .f ı IA/.x/ and use this to prove that f ı
IA D f .
(b) Prove that IB ı f D f.
5. ? (a) Let f WR ! R be defined by f.x/ D x2, let g WR ! R be defined by
g.x/ D sin x, and let hWR ! R be defined by h.x/ D 3
p
x.
Determine formulas for Œ.h ı g/ ı f .x/ and Œh ı .g ı f / .x/.
Does this prove that .h ı g/ ı f D h ı .g ı f / for these particular
functions? Explain.
(b) Now let A, B, and C be sets and let f W A ! B, g W B ! C , and
hWC ! D. Prove that .h ı g/ ı f D h ı .g ı f /. That is, prove that
function composition is an associative operation.

? 6. Prove Part (1) of Theorem 6.20.
Let A, B, and C be nonempty sets and let f WA ! B and g WB ! C . If f
and g are both injections, then g ı f is an injection.
7. For each of the following, give an example of functions f W A ! B and
gWB ! C that satisfy the stated conditions, or explain why no such example
exists.
? (a) The function f is a surjection, but the function gıf is not a surjection.
?
(b) The function f is an injection, but the function gıf is not an injection.
(c) The function g is a surjection, but the function gıf is not a surjection.
(d) The function g is an injection, but the function gıf is not an injection.
(e) The function f is not a surjection, but the function gıf is a surjection.
?
(f) The function f is not an injection, but the function gıf is an injection.
(g) The function f is not an injection, but the function gıf is an injection.
(h) The function g is not an injection, but the function gıf is an injection.
8. Let A be a nonempty set and let f W A ! A. For each n 2 N, define
a function f n
W A ! A recursively as follows: f 1
D f and for each
n 2 N, f nC1 D f ı f n. For example, f 2 D f ı f 1 D f ı f and
f 3
D f ı f 2
D f ı .f ı f /.
(a) Let f WR ! R by f.x/ D x C 1 for each x 2 R. For each n 2 N and
for each x 2 R, determine a formula for f n.x/ and use induction to
prove that your formula is correct.
(b) Let a; b 2 R and let f W R ! R by f.x/ D ax C b for each x 2 R.
For each n 2 N and for each x 2 R, determine a formula for f n
.x/
and use induction to prove that your formula is correct.
(c) Now let A be a nonempty set and let f W A ! A. Use induction to
prove that for each n 2 N, f nC1
D f n
ı f . (Note: You will need to
use the result in Exercise (5).)
9. Exploring Composite Functions. Let A, B, and C be nonempty sets and
let f W A ! B and g W B ! C . For this activity, it may be useful to draw
your arrow diagrams in a triangular arrangement as follows:

A B
C
f
g
g f
It might be helpful to consider examples where the sets are small. Try con-
structing examples where the set A has 2 elements, the set B has 3 elements,
and the set C has 2 elements.
(a) Is it possible to construct an example where g ı f is an injection, f is
an injection, but g is not an injection? Either construct such an example
or explain why it is not possible.
(b) Is it possibleto construct an example where gıf is an injection, g is an
injection, but f is not an injection? Either construct such an example
(c) Is it possible to construct an example where gıf is a surjection, f is a
surjection, but g is not a surjection? Either construct such an example
(d) Is it possible to construct an example where g ı f is surjection, g is a
surjection, but f is not a surjection? Either construct such an example
10. The Proof of Theorem 6.21. Use the ideas from Exercise (9) to prove The-
orem 6.21. Let A, B, and C be nonempty sets and let f W A ! B and
gWB ! C .
(a) If g ı f WA ! C is an injection, then f WA ! B is an injection.
(b) If g ı f WA ! C is a surjection, then gWB ! C is a surjection.
Hint: For part (a), start by asking, “What do we have to do to prove that f
is an injection? ” Start with a similar question for part (b).

6.5 Inverse Functions
For this section, we will use the concept of Cartesian product of two sets A and B,
denoted by A B, which is the set of all ordered pairs .x; y/ where x 2 A and
y 2 B. That is,
A B D f.x; y/ j x 2 A and y 2 Bg :
See Preview Activity 2 in Section 5.4 for a more thorough discussion of this con-
cept.
Preview Activity 1 (Functions and Sets of Ordered Pairs)
When we graph a real function, we plot ordered pairs in the Cartesian plane where
the first coordinate is the input of the function and the second coordinate is the
output of the function. For example, if gWR ! R, then every point on the graph of
g is an ordered pair .x; y/ of real numbers where y D g.x/. This shows how we
can generate ordered pairs from a function. It happens that we can do this with any
function. For example, let
A D f1; 2; 3g and B D fa; bg :
Define the function F WA ! B by
F.1/ D a; F.2/ D b; and F.3/ D b:
We can convert each of these to an ordered pair in A B by using the input as
the first coordinate and the output as the second coordinate. For example, F.1/ D
a is converted to .1; a/, F.2/ D b is converted to .2; b/, and F.3/ D b is converted
to .3; b/. So we can think of this function as a set of ordered pairs, which is a subset
of A B, and write
F D f.1; a/; .2; b/; .3; b/g:
Note: Since F is the name of the function, it is customary to use F as the name
for the set of ordered pairs.
1. Let A D f1; 2; 3g and let C D fa; b; c; dg. Define the function g W A ! C
by g.1/ D a, g.2/ D b, and g.3/ D d. Write the function g as a set of
ordered pairs in A C .
For another example, if we have a real function, such as g W R ! R by g.x/ D
x2 2, then we can think of g as the following infinite subset of R R:
g D
˚
.x; y/ 2 R R j y D x2
2 :

6.5. Inverse Functions 335
We can also write this sometimes write this as g D
˚
.x; x2 2/ j x 2 R .
2. Let f WZ ! Z be defined by f.m/ D 3mC5, for all m 2 Z. Use set builder
notation to write the function f as a set of ordered pairs, and then use the
roster method to write the function f as a set of ordered pairs.
So any function f W A ! B can be thought of as a set of ordered pairs that is a
subset of A B. This subset is
f D f.a; f.a// j a 2 Ag or f D f.a; b/ 2 A B j b D f.a/g:
On the other hand, if we started with A D f1; 2; 3g, B D fa; bg, and
G D f.1; a/; .2; a/; .3; b/g A B;
then we could think of G as a function from A to B with G.1/ D a, G.2/ D a,
and G.3/ D b: The idea is to use the first coordinate of each ordered pair as the
input, and the second coordinate as the output. However, not every subset of AB
can be used to define a function from A to B. This is explored in the following
questions.
3. Let f D f.1; a/; .2; a/; .3; a/; .1; b/g. Could this set of ordered pairs be
used to define a function from A to B? Explain.
4. Let g D f.1; a/; .2; b/; .3; a/g. Could this set of ordered pairs be used to
define a function from A to B? Explain.
5. Let h D f.1; a/; .2; b/g. Could this set of ordered pairs be used to define a
function from A to B? Explain.
Preview Activity 2 (A Composition of Two Specific Functions)
Let A D fa; b; c; dg and let B D fp; q; r; sg.
1. Construct an example of a function f WA ! B that is a bijection. Draw an
arrow diagram for this function.
2. On your arrow diagram, draw an arrow from each element of B back to its
corresponding element in A. Explain why this defines a function from B to
A.

3. If the name of the function in Part (2) is g, so that gWB ! A, what are g.p/,
g.q/, g.r/, and g.s/?
4. Construct a table of values for each of the functions g ı f W A ! A and
f ı gWB ! B. What do you observe about these tables of values?
The Ordered Pair Representation of a Function
In Preview Activity 1, we observed that if we have a function f WA ! B, we can
generate a set of ordered pairs f that is a subset of A B as follows:
f D f.a; f.a// j a 2 Ag or f D f.a; b/ 2 A B j b D f.a/g:
However, we also learned that some sets of ordered pairs cannot be used to define
a function. We now wish to explore under what conditions a set of ordered pairs
can be used to define a function. Starting with a function f W A ! B, since
dom.f / D A, we know that
For every a 2 A; there exists a b 2 B such that .a; b/ 2 f: (1)
Specifically, we use b D f.a/. This says that every element of A can be used as
an input. In addition, to be a function, each input can produce only one output. In
terms of ordered pairs, this means that there will never be two ordered pairs .a; b/
and .a; c/ in the function f where a 2 A, b; c 2 B, and b ¤ c. We can formulate
this as a conditional statement as follows:
For every a 2 A and every b; c 2 B;
if .a; b/ 2 f and .a; c/ 2 f; then b D c: (2)
This also means that if we start with a subset f of A B that satisfies condi-
tions (1) and (2), then we can consider f to be a function from A to B by using
b D f.a/ whenever .a; b/ is in f . This proves the following theorem.
Theorem 6.22. Let A and B be nonempty sets and let f be a subset of A B that
satisfies the following two properties:
For every a 2 A, there exists b 2 B such that .a; b/ 2 f; and
For every a 2 A and every b; c 2 B, if .a; b/ 2 f and .a; c/ 2 f , then
b D c.
If we use f.a/ D b whenever .a; b/ 2 f , then f is a function from A to B.

A Note about Theorem 6.22. The first condition in Theorem 6.22 means that
every element of A is an input, and the second condition ensures that every input
has exactly one output. Many texts will use Theorem 6.22 as the definition of a
function. Many mathematicians believe that this ordered pair representation of a
function is the most rigorous definition of a function. It allows us to use set theory
to work with and compare functions. For example, equality of functions becomes
a question of equality of sets. Therefore, many textbooks will use the ordered pair
representation of a function as the definition of a function.
Progress Check 6.23 (Sets of Ordered Pairs that Are Not Functions)
Let A D f1; 2; 3g and let B D fa; bg. Explain why each of the following subsets
of A B cannot be used to define a function from A to B.
1. F D f.1; a/; .2; a/g. 2. G D f.1; a/; .2; b/; .3; c/; .2; c/g.
The Inverse of a Function
In previous mathematics courses, we learned that the exponential function (with
base e) and the natural logarithm functions are inverses of each other. This was
often expressed as follows:
For each x 2 R with x 0 and for each y 2 R;
y D ln x if and only if x D ey
:
Notice that this means that x is the input and y is the output for the natural loga-
rithm function if and only if y is the input and x is the output for the exponential
function. In essence, the inverse function (in this case, the exponential function)
reverses the action of the original function (in this case, the natural logarithm func-
tion). In terms of ordered pairs (input-output pairs), this means that if .x; y/ is
an ordered pair for a function, then .y; x/ is an ordered pair for its inverse. This
idea of reversing the roles of the first and second coordinates is the basis for our
definition of the inverse of a function.

Definition. Let f WA ! B be a function. The inverse of f, denoted by f 1
,
is the set of ordered pairs f.b; a/ 2 B A j f.a/ D bg. That is,
f 1
D f.b; a/ 2 B A j f.a/ D bg:
If we use the ordered pair representation for f, we could also write
f 1
D f.b; a/ 2 B A j .a; b/ 2 f g:
Notice that this definition does not state that f 1
is a function. It is simply a
subset of B A. After we study the material in Chapter 7, we will say that this
means that f 1
is a relation from B to A. This fact, however, is not important to
us now. We are mainly interested in the following question:
Under what conditions will the inverse of the function f WA ! B be a
function from B to A?
Progress Check 6.24 (Exploring the Inverse of a Function)
Let A D fa; b; cg, B D fa; b; c; dg, and C D fp; q; rg. Define
f WA ! C by gWA ! C by hWB ! C by
f.a/ D r g.a/ D p h.a/ D p
f.b/ D p g.b/ D q h.b/ D q
f.c/ D q g.c/ D p h.c/ D r
h.d/ D q
1. Draw an arrow diagram for each function.
2. Determine the inverse of each function as a set of ordered pairs.
3. (a) Is f 1 a function from C to A? Explain.
(b) Is g 1
a function from C to A? Explain.
(c) Is h 1
a function from C to B? Explain.
4. Draw an arrow diagram for each inverse from Part (3) that is a function. Use
your existing arrow diagram from Part (1) to draw this arrow diagram.
5. Make a conjecture about what conditions on a function F W S ! T will
ensure that its inverse is a function from T to S.

We will now consider a general argument suggested by the explorations in
Progress Check 6.24. By definition, if f W A ! B is a function, then f 1
is a
subset of B A. However, f 1 may or may not be a function from B to A. For
example, suppose that s; t 2 A with s ¤ t and f.s/ D f.t/. This is represented in
Figure 6.9.
t
s
y
z
A B
f
Figure 6.9: The Inverse Is Not a Function
In this case, if we try to reverse the arrows, we will not get a function from B
to A. This is because .y; s/ 2 f 1
and .y; t/ 2 f 1
with s ¤ t. Consequently,
f 1 is not a function. This suggests that when f is not an injection, then f 1 is
not a function.
Also, if f is not a surjection, then there exists a z 2 B such that f.a/ ¤ z for
all a 2 A, as in the diagram in Figure 6.9. In other words, there is no ordered pair
in f with z as the second coordinate. This means that there would be no ordered
pair in f 1 with z as a first coordinate. Consequently, f 1 cannot be a function
from B to A.
This motivates the statement in Theorem 6.25. In the proof of this theorem,
we will frequently change back and forth from the input-output representation of
a function and the ordered pair representation of a function. The idea is that if
G WS ! T is a function, then for s 2 S and t 2 T,
G.s/ D t if and only if .s; t/ 2 G:
When we use the ordered pair representation of a function, we will also use the
ordered pair representation of its inverse. In this case, we know that
.s; t/ 2 G if and only if .t; s/ 2 G 1
:
Theorem 6.25. Let A and B be nonempty sets and let f WA ! B. The inverse of
f is a function from B to A if and only if f is a bijection.
Proof. Let A and B be nonempty sets and let f W A ! B. We will first assume

that f is a bijection and prove that f 1 is a function from B to A. To do this, we
will show that f 1
satisfies the two conditions of Theorem 6.22.
We first choose b 2 B. Since the function f is a surjection, there exists an
a 2 A such that f.a/ D b. This implies that .a; b/ 2 f and hence that .b; a/ 2
f 1
. Thus, each element of B is the first coordinate of an ordered pair in f 1
,
and hence f 1 satisfies the first condition of Theorem 6.22.
To prove that f 1 satisfies the second condition of Theorem 6.22, we must
show that each element of B is the first coordinate of exactly one ordered pair in
f 1. So let b 2 B, a1; a2 2 A and assume that
.b; a1/ 2 f 1
and .b; a2/ 2 f 1
:
This means that .a1; b/ 2 f and .a2; b/ 2 f . We can then conclude that
f.a1/ D b and f.a2/ D b:
But this means that f.a1/ D f.a2/. Since f is a bijection, it is an injection, and
we can conclude that a1 D a2. This proves that b is the first element of only
one ordered pair in f 1
. Consequently, we have proved that f 1
satisfies both
conditions of Theorem 6.22 and hence that f 1 is a function from B to A.
We now assume that f 1 is a function from B to A and prove that f is a
bijection. First, to prove that f is an injection, we assume that a1; a2 2 A and that
f.a1/ D f.a2/. We wish to show that a1 D a2. If we let b D f.a1/ D f.a2/,
we can conclude that
.a1; b/ 2 f and .a2; b/ 2 f:
But this means that
.b; a1/ 2 f 1
and .b; a2/ 2 f 1
:
Since we have assumed that f 1 is a function, we can conclude that a1 D a2.
Hence, f is an injection.
Now to prove that f is a surjection, we choose b 2 B and will show that there
exists an a 2 A such that f.a/ D b. Since f 1 is a function, b must be the first
coordinate of some ordered pair in f 1
. Consequently, there exists an a 2 A such
that
.b; a/ 2 f 1
:
Now this implies that .a; b/ 2 f and hence that f.a/ D b. This proves that f is a
surjection. Since we have also proved that f is an injection, we conclude that f is
a bijection.

Inverse Function Notation
In the situation where f WA ! B is a bijection and f 1
is a function from B to A,
we can write f 1 WB ! A. In this case, we frequently say that f is an invertible
function, and we usually do not use the ordered pair representation for either f or
f 1. Instead of writing .a; b/ 2 f , we write f.a/ D b, and instead of writing
.b; a/ 2 f 1
, we write f 1
.b/ D a. Using the fact that .a; b/ 2 f if and only
if .b; a/ 2 f 1, we can now write f.a/ D b if and only if f 1.b/ D a. We
summarize this in Theorem 6.26.
Theorem 6.26. Let A and B be nonempty sets and let f WA ! B be a bijection.
Then f 1
WB ! A is a function, and for every a 2 A and b 2 B,
f.a/ D b if and only if f 1
.b/ D a:
Example 6.27 (Inverse Function Notation)
For an example of the use of the notationin Theorem 6.26, let RC
D fx 2 R j x 0g.
Define
f WR ! R by f.x/ D x3; and gWR ! RC by g.x/ D ex.
Notice that RC
is the codomain of g. We can then say that both f and g are
bijections. Consequently, the inverses of these functions are also functions. In fact,
f 1
WR ! R by f 1
.y/ D 3
p
y; and g 1
WRC
! R by g 1
.y/ D ln y.
For each function (and its inverse), we can write the result of Theorem 6.26 as
follows:
Theorem 6.26 Translates to:
For x; y 2 R, f.x/ D y For x; y 2 R, x3 D y
if and only if f 1
.y/ D x. if and only if 3
p
y D x.
For x 2 R; y 2 RC
, g.x/ D y For x 2 R; y 2 RC
, ex
D y
if and only if g 1
.y/ D x. if and only if ln y D x.

Theorems about Inverse Functions
The next two results in this section are two important theorems about inverse func-
tions. The first is actually a corollary of Theorem 6.26.
Corollary 6.28. Let A and B be nonempty sets and let f WA ! B be a bijection.
Then
1. For every x in A, f 1 ı f

.x/ D x.
2. For every y in B, f ı f 1

.y/ D y.
Proof. Let A and B be nonempty sets and assume that f WA ! B is a bijection. So
let x 2 A and let f.x/ D y. By Theorem 6.26, we can conclude that f 1.y/ D x.
Therefore,
f 1
ı f

.x/ D f 1
.f.x//
D f 1
.y/
D x:
Hence, for each x 2 A, f 1
ı f

.x/ D x.
The proof that for each y in B, f ı f 1

.y/ D y is Exercise (4).
Example 6.27 (continued)
For the cubing function and the cube root function, we have seen that
For x; y 2 R, x3 D y if and only if 3
p
y D x.
Notice that
If we substitute x3
D y into the equation 3
p
y D x, we obtain
3
p
x3 D x.
If we substitute 3
p
y D x into the equation x3
D y, we obtain 3
p
y
3
D y.
This is an illustration of Corollary 6.28. We can see this by using f W R ! R
defined by f.x/ D x3 and f 1 W R ! R defined by f 1.y/ D 3
p
y. Then
f 1
ı f WR ! R and f 1
ı f D IR. So for each x 2 R,
f 1
ı f

.x/ D x
f 1
.f.x// D x
f 1
x3

D x
3
p
x3 D x:

Similarly, the equation 3
p
y
3
D y for each y 2 R can be obtained from the fact
that for each y 2 R, .f ı f 1/.y/ D y.
We will now consider the case where f W A ! B and g W B ! C are both
bijections. In this case, f 1 WB ! A and g 1 WC ! B. Figure 6.10 can be used
to illustrate this situation.
A B C
f g
f 1 g 1
g f
(g f ) 1
Figure 6.10: Composition of Two Bijections
By Theorem 6.20, g ı f WA ! C is also a bijection. Hence, by Theorem 6.25,
.g ı f / 1 is a function and, in fact, .g ı f / 1 W C ! A. Notice that we can
also form the composition of g 1
followed by f 1
to get f 1
ı g 1
W C ! A.
Figure 6.10 helps illustrate the result of the next theorem.
Theorem 6.29. Let f W A ! B and g W B ! C be bijections. Then g ı f is a
bijection and .g ı f / 1 D f 1 ı g 1.
Proof. Let f W A ! B and g W B ! C be bijections. Then f 1
W B ! A and
g 1 WC ! B. Hence, f 1 ıg 1 WC ! A. Also, by Theorem 6.20, g ıf WA ! C
is a bijection, and hence .g ı f / 1
W C ! A. We will now prove that for each
z 2 C , .g ı f / 1.z/ D .f 1 ı g 1/.z/.
Let z 2 C . Since the function g is a surjection, there exists a y 2 B such that
g.y/ D z: (1)
Also, since f is a surjection, there exists an x 2 A such that
f.x/ D y: (2)

Now these two equations can be written in terms of the respective inverse functions
as
g 1
.z/ D yI and (3)
f 1
.y/ D x: (4)
Using equations (3) and (4), we see that
f 1
ı g 1

.z/ D f 1
g 1
.z/

D f 1
.y/
D x: (5)
Using equations (1) and (2) again, we see that .g ı f /.x/ D z. However, in terms
of the inverse function, this means that
.g ı f / 1
.z/ D x: (6)
Comparing equations (5) and (6), we have shown that for all z 2 C ,
.g ı f / 1
.z/ D .f 1
ı g 1
/.z/. This proves that .g ı f / 1
D f 1
ı g 1
.
Exercises 6.5
1. Let A D f1; 2; 3g and B D fa; b; cg.
(a) Construct an example of a function f W A ! B that is not a bijection.
Write the inverse of this function as a set of ordered pairs. Is the inverse
of f a function? Explain. If so, draw an arrow diagram for f and f 1
.
(b) Construct an example of a function gWA ! B that is a bijection. Write
the inverse of this function as a set of ordered pairs. Is the inverse of g
a function? Explain. If so, draw an arrow diagram for g and g 1.
? 2. Let S D fa; b; c; dg. Define f W S ! S by defining f to be the following
set of ordered pairs.
f D f.a; c/; .b; b/; .c; d/; .d; a/g
(a) Draw an arrow diagram to represent the function f . Is the function f
a bijection?

(b) Write the inverse of f as a set of ordered pairs. Is f 1 a function?
Explain.
(c) Draw an arrow diagram for f 1
using the arrow diagram from Exer-
cise (2a).
(d) Compute f 1 ı f

.x/ and f ı f 1

.x/ for each x in S. What
theorem does this illustrate?
? 3. Inverse functions can be used to help solve certain equations. The idea is to
use an inverse function to undo the function.
(a) Since the cube root function and the cubing function are inverses of
each other, we can often use the cube root function to help solve an
equation involving a cube. For example, the main step in solving the
equation
.2t 1/3
D 20
is to take the cube root of each side of the equation. This gives
3
q
.2t 1/3 D
3
p
20
2t 1 D
3
p
20:
Explain how this step in solving the equation is a use of Corollary 6.28.
(b) A main step in solving the equation e2t 1 D 20 is to take the natural
logarithm of both sides of this equation. Explain how this step is a
use of Corollary 6.28, and then solve the resulting equation to obtain a
solution for t in terms of the natural logarithm function.
(c) How are the methods of solving the equations in Exercise (3a) and
Exercise (3b) similar?
?
4. Prove Part (2) of Corollary 6.28. Let A and B be nonempty sets and let
f WA ! B be a bijection. Then for every y in B, f ı f 1

.y/ D y.
5. In Progress Check 6.6 on page 298, we defined the identity function on a
set. The identity function on the set T , denoted by IT , is the function
IT W T ! T defined by IT .t/ D t for every t in T . Explain how Corol-
lary 6.28 can be stated using the concept of equality of functions and the
identity functions on the sets A and B.
6. Let f WA ! B and g WB ! A. Let IA and IB be the identity functions on
the sets A and B, respectively. Prove each of the following:

? (a) If g ı f D IA, then f is an injection.
? (b) If f ı g D IB , then f is a surjection.
(c) If g ı f D IA and f ı g D IB , then f and g are bijections and
g D f 1
.
?
7. (a) Define f W R ! R by f.x/ D e x2
. Is the inverse of f a function?
(b) Let R D fx 2 R j x 0g. Define g W R ! .0; 1 by g.x/ D e x2
.
Is the inverse of g a function? Justify your conclusion.
8. (a) Let f WR ! R be defined by f.x/ D x2
. Explain why the inverse of
f is not a function.
(b) Let R D ft 2 R j t 0g. Define g W R ! R by g.x/ D x2. Ex-
plain why this squaring function (with a restricted domain and codomain)
is a bijection.
(c) Explain how to define the square root function as the inverse of the
function in Exercise (8b).
(d) True or false:
p
x
2
D x for all x 2 R such that x 0.
(e) True or false:
p
x2 D x for all x 2 R.
9. Prove the following:
If f WA ! B is a bijection, then f 1
WB ! A is also a bijection.
10. For each natural number k, let Ak be a set, and for each natural number n,
let fn WAn ! AnC1.
For example, f1 WA1 ! A2, f2 WA2 ! A3, f3 WA3 ! A4, and so on.
Use mathematical induction to prove that for each natural number n with
n 2, if f1; f2; : : :; fn are all bijections, then fn ı fn 1 ı ı f2 ı f1 is a
bijection and
.fn ı fn 1 ı ı f2 ı f1/ 1
D f 1
1 ı f 1
2 ı ı f 1
n 1 ı f 1
n :
Note: This is an extension of Theorem 6.29. In fact, Theorem 6.29 is the
basis step of this proof for n D 2.
11. (a) Define f WR ! R by f.x/ D x2
4 for all x 2 R. Explain why the
inverse of the function f is not a function.

(b) Let R D fx 2 R j x 0g and let T D fy 2 R j y 4g. Define
F W R
! T by F.x/ D x2
4 for all x 2 R
. Explain why the
inverse of the function F is a function and find a formula for F 1.y/,
where y 2 T .
12. Let Z5 D f0; 1; 2; 3; 4g.
(a) Define f WZ5 ! Z5 by f.x/ D x2
C 4 .mod 5/ for all x 2 Z5. Write
the inverse of f as a set of ordered pairs and explain why f 1 is not a
function.
(b) Define g W Z5 ! Z5 by g.x/ D x3 C 4 .mod 5/ for all x 2 Z5.
Write the inverse of g as a set of ordered pairs and explain why g 1
is
a function.
(c) Is it possibleto write a formula for g 1.y/, where y 2 Z5? The answer
to this question depends on whether or not is possible to define a cube
root of elements of Z5. Recall that for a real number x, we define the
cube root of x to the real number y such that y3
D x. That is,
y D 3
p
x if and only if y3 D x.
Using this idea, is it possible to define the cube root of each number in
Z5? If so, what is 3
p
0, 3
p
1, 3
p
2, 3
p
3, and 3
p
4.
(d) Now answer the question posed at the beginning of Part (c). If possible,
determine a formula for g 1.y/ where g 1 WZ5 ! Z5.
13. Constructing an Inverse Function. If f W A ! B is a bijection, then
we know that its inverse is a function. If we are given a formula for the
function f , it may be desirable to determine a formula for the function f 1
.
This can sometimes be done, while at other times it is very difficult or even
impossible.
Let f W R ! R be defined by f.x/ D 2x3
7. A graph of this function
would suggest that this function is a bijection.
(a) Prove that the function f is an injection and a surjection.
Let y 2 R. One way to prove that f is a surjection is to set y D f.x/
and solve for x. If this can be done, then we would know that there exists
an x 2 R such that f.x/ D y. For the function f , we are using x for the

input and y for the output. By solving for x in terms of y, we are attempting
to write a formula where y is the input and x is the output. This formula
represents the inverse function.
(b) Solve the equation y D 2x3
7 for x. Use this to write a formula for
f 1.y/, where f 1 WR ! R.
(c) Use the result of Part (13b) to verify that for each x 2 R, f 1.f.x// D
x and for each y 2 R, f f 1.y/

D y.
Now let RC D fy 2 R j y 0g. Define gWR ! RC by g.x/ D e2x 1.
(d) Set y D e2x 1 and solve for x in terms of y.
(e) Use your work in Exercise (13d) to define a function hWRC
! R.
(f) For each x 2 R, determine .h ı g/.x/ and for each y 2 RC
, determine
.g ı h/.y/.
(g) Use Exercise (6) to explain why h D g 1.
14. The Inverse Sine Function. We have seen that in order to obtain an inverse
function, it is sometimes necessary to restrict the domain (or the codomain)
of a function.
(a) Let f W R ! R be defined by f.x/ D sinx. Explain why the inverse
of the function f is not a function. (A graph may be helpful.)
Notice that if we use the ordered pair representation, then the sine function
can be represented as
f D f.x; y/ 2 R R j y D sinxg:
If we denote the inverse of the sine function by sin 1
, then
f 1
D f.y; x/ 2 R R j y D sin xg:
Part (14a) proves that f 1 is not a function. However, in previous mathe-
matics courses, we frequently used the “inverse sine function.” This is not
really the inverse of the sine function as defined in Part (14a) but, rather, it is
the inverse of the sine function restricted to the domain
h
2
;

2
i
.
(b) Explain why the function F W
h
2
;

2
i
! Œ 1; 1 defined by F.x/ D
sinx is a bijection.

6.6. Functions Acting on Sets 349
The inverse of the function in Part (14b) is itself a function and is called the
inverse sine function (or sometimes the arcsine function).
(c) What is the domain of the inverse sine function? What are the range
and codomain of the inverse sine function?
Let us now use F.x/ D Sin.x/ to represent the restricted sine function in
Part (14b). Therefore, F 1.x/ D Sin 1
.x/ can be used to represent the
inverse sine function. Observe that
F W
h
2
;

2
i
! Œ 1; 1 and F 1
WŒ 1; 1 !
h
2
;

2
i
:
(d) Using this notation, explain why
Sin 1
y D x if and only if
h
y D sin x and

2
x

2
i
;
Sin Sin 1
.y/

D y for all y 2 Œ 1; 1; and
Sin 1
.Sin.x// D x for all x 2
h
2
;

2
i
.
6.6 Functions Acting on Sets
Preview Activity 1 (Functions and Sets)
Let S D fa; b; c; dg and T D fs; t; ug. Define f WS ! T by
f.a/ D s f.b/ D t f.c/ D t f.d/ D s:
1. Let A D fa; cg and B D fa; dg. Notice that A and B are subsets of S. Use
the roster method to specify the elements of the following two subsets of T :
(a) ff.x/ j x 2 Ag (b) ff.x/ j x 2 Bg
2. Let C D fs; tg and D D fs; ug. Notice that C and D are subsets of T . Use
the roster method to specify the elements of the following two subsets of S:
(a) fx 2 S j f.x/ 2 C g (b) fx 2 S j f.x/ 2 Dg
Now let gWR ! R be defined by g.x/ D x2, for each x 2 R.
3. Let A D f1; 2; 3; 1g. Use the roster method to specify the elements of the
set fg.x/ j x 2 Ag.
4. Use the roster method to specify the elements of each of the following sets:

(a) fx 2 R j g.x/ D 1g
(b) fx 2 R j g.x/ D 9g
(c) fx 2 R j g.x/ D 15g
(d) fx 2 R j g.x/ D 1g
5. Let B D f1; 9; 15; 1g. Use the roster method to specify the elements of the
set
fx 2 R j g.x/ 2 Bg.
Preview Activity 2 (Functions and Intervals)
Let gWR ! R be defined by g.x/ D x2, for each x 2 R.
1. We will first determine where g maps the closed interval Œ1; 2. (Recall that
Œ1; 2 D fx 2 R j 1 x 2g.) That is, we will describe, in simpler terms,
the set fg.x/ j x 2 Œ1; 2g. This is the set of all images of the real numbers
in the closed interval Œ1; 2.
(a) Draw a graph of the function g using 3 x 3.
(b) On the graph, draw the vertical lines x D 1 and x D 2 from the x-axis
to the graph. Label the points P .1; f.1// and Q.2; f.2// on the graph.
(c) Now draw horizontal lines from the points P and Q to the y-axis. Use
this information from the graph to describe the set fg.x/ j x 2 Œ1; 2g
in simpler terms. Use interval notation or set builder notation.
2. We will now determine all real numbers that g maps into the closed interval
Œ1; 4. That is, we will describe the set fx 2 R j g.x/ 2 Œ1; 4g in simpler
terms. This is the set of all preimages of the real numbers in the closed
interval Œ1; 4.
(a) Draw a graph of the function g using 3 x 3.
(b) On the graph, draw the horizontal lines y D 1 and y D 4 from the
y-axis to the graph. Label all points where these two lines intersect the
graph.
(c) Now draw vertical lines from the points in Part (2) to the x-axis, and
then use the resulting information to describe the set
fx 2 R j g.x/ 2 Œ1; 4g in simpler terms. (You will need to describe
this set as a union of two intervals. Use interval notation or set builder
notation.)

Functions Acting on Sets
In our study of functions, we have focused on how a function “maps” individ-
ual elements of its domain to the codomain. We also studied the preimage of an
individual element in its codomain. For example, if f W R ! R is defined by
f.x/ D x2
, for each x 2 R, then
f.2/ D 4. We say that f maps 2 to 4 or that 4 is the image of 2 under the
function f .
Since f.x/ D 4 implies that x D 2 or x D 2, we say that the preimages of
4 are 2 and 2 or that the set of preimages of 4 is f 2; 2g.
For a function f W S ! T , the next step is to consider subsets of S or T and
what corresponds to them in the other set. We did this in the Preview Activities. We
will give some definitions and then revisit the examples in the Preview Activities
in light of these definitions. We will first consider the situation where A is a subset
of S and consider the set of outputs whose inputs are from A. This will be a subset
of T .
Definition. Let f W S ! T. If A S, then the image of A under f is the
set f.A/, where
f.A/ D ff.x/ j x 2 Ag:
If there is no confusion as to which function is being used, we call f.A/ the
image of A.
We now consider the situation in which C is a subset of T and consider the
subset of A consisting of all elements of T whose outputs are in C .
Definition. Let f W S ! T . If C T , then the preimage of C under f is
the set f 1.C /, where
f 1
.C / D fx 2 S j f.x/ 2 C g :
If there is no confusion as to which function is being used, we call f 1.C /
the preimage of C. The preimage of the set C under f is also called the
inverse image of C under f.
Notice that the set f 1
.C / is defined whether or not f 1
is a function.

Progress Check 6.30 (Preview Activity 1 Revisited)
Let S D fa; b; c; dg and T D fs; t; ug. Define f WS ! T by
f.a/ D s f.b/ D t f.c/ D t f.d/ D s:
Let A D fa; cg ; B D fa; dg ; C D fs; tg; and D D fs; ug.
Use your work in Preview Activity 1 to determine each of the following sets:
1. f.A/ 2. f.B/ 3. f 1.C / 4. f 1.D/
Example 6.31 (Images and Preimages of Sets)
Let gWR ! R be defined by g.x/ D x2
, for each x 2 R. The following results are
based on the examples in Preview Activity 1 and Preview Activity 2.
Let A D f1; 2; 3; 1g. Then f.A/ D f1; 4; 9g.
Let B D f1; 9; 15; 1g. Then f 1
.B/ D
n p
15; 3; 1; 1; 3;
p
15
o
.
The graphs from Preview Activity 2 illustrate the following results:
If T is the closed interval Œ1; 2, then the image of the set T is
f.T / D ff.x/ j x 2 Œ1; 2g
D Œ1; 4 :
If C is the closed interval Œ1; 4, then the preimage of the set C is
f 1
.C / D fx 2 R j f.x/ 2 Œ1; 4g D Œ 2; 1 [ Œ1; 2:
Set Operations and Functions Acting on Sets
We will now consider the following situation: Let S and T be sets and let f be a
function from S to T . Also, let A and B be subsets of S and let C and D be subsets
of T . In the remainder of this section, we will consider the following situations and
answer the questions posed in each case.

The set AB is a subset of S and so f.AB/ is a subset of T . In addition,
f.A/ and f.B/ are subsets of T . Hence, f.A/ f.B/ is a subset of T .
Is there any relationship between f.A B/ and f.A/ f.B/?
The set A[B is a subset of S and so f.A[B/ is a subset of T . In addition,
f.A/ and f.B/ are subsets of T . Hence, f.A/ [ f.B/ is a subset of T .
Is there any relationship between f.A [ B/ and f.A/ [ f.B/?
The set C D is a subset of T and so f 1
.C D/ is a subset of S. In
addition, f 1.C / and f 1.D/ are subsets of S. Hence, f 1.C /f 1.D/
is a subset of S.
Is there any relationship between the sets f 1
.C D/ and
f 1.C / f 1.D/?
The set C [ D is a subset of T and so f 1
.C [ D/ is a subset of S. In
addition, f 1.C / and f 1.D/ are subsets of S. Hence, f 1.C /[f 1.D/
is a subset of S.
Is there any relationship between the sets f 1
.C [ D/ and
f 1.C / [ f 1.D/?
These and other questions will be explored in the next progress check.
Progress Check 6.32 (Set Operations and Functions Acting on Sets)
In Section 6.2, we introduced functions involving congruences. For example, if we
let
Z8 D f0; 1; 2; 3; 4; 5; 6; 7g;
then we can define f WZ8 ! Z8 by f.x/ D r, where x2 C 2

r .mod 8/ and
r 2 Z8. Moreover, we shortened this notation to
f.x/ D x2
C 2

.mod 8/ :
We will use the following subsets of Z8:
A D f1; 2; 4g B D f3; 4; 6g C D f1; 2; 3g D D f3; 4; 5g.
1. Verify that f.0/ D 2, f.1/ D 3, f.2/ D 6, and f.3/ D 3. Then determine
f.4/, f.5/, f.6/, and f.7/.
2. Determine f.A/, f.B/, f 1
.C /, and f 1
.D/.

3. For each of the following, determine the two subsets of Z8 and then deter-
mine if there is a relationship between the two sets. For example, A B D
f4g and since f.4/ D 2, we see that f.A B/ D f2g.
(a) f.A B/ and f.A/ f.B/
(b) f.A [ B/ and f.A/ [ f.B/
(c) f 1.C D/ and f 1.C / f 1.D/
(d) f 1.C [ D/ and f 1.C / [ f 1.D/
4. Notice that f.A/ is a subset of the codomain, Z8. Consequently,
f 1.f.A// is a subset of the domain, Z8. Is there any relation between
A and f 1
.f.A// in this case?
5. Notice that f 1
.C / is a subset of the domain, Z8. Consequently,
f f 1.C /

is a subset of the codomain, Z8. Is there any relation between
C and f f 1
.C /

in this case?
Example 6.33 (Set Operations and Functions Acting on Sets)
Define f W R ! R by f.x/ D x2
C 2 for all x 2 R. It will be helpful to use the
graph shown in Figure 6.11.
2
4
6
8
10
12
–3 –2 –1 1 2 3
Figure 6.11: Graph for Example 6.33
We will use the following closed intervals:
A D Œ0; 3 B D Œ 2; 1 C D Œ2; 6 D D Œ0; 3.

1. Verify that f.A/ D Œ2; 11, f.B/ D Œ2; 6, f 1.C / D Œ 2; 2, and that
f 1
.D/ D Œ 1; 1.
2. (a) Explain why f.A B/ D Œ2; 3 and f.A/ f.B/ D Œ2; 6. So in this
case, f.A B/ f.A/ f.B/.
(b) Explain why f.A [ B/ D Œ2; 11 and f.A/ [ f.B/ D Œ2; 11. So in
this case, f.A [ B/ D f.A/ [ f.B/.
(c) Explain why f 1.C D/ D Œ 1; 1 and f 1.C / f 1.D/ D
Œ 1; 1. So in this case, f 1
.C D/ D f 1
.C / f 1
.D/.
(d) Explain why f 1
.C [ D/ D Œ 2; 2 and f 1
.C / [ f 1
.D/ D
Œ 2; 2. So in this case, f 1.C [ D/ D f 1.C / [ f 1.D/.
3. Recall that A D Œ0; 3. Notice f.A/ D Œ2; 11 is a subset of the codomain,
R. Explain why f 1
.f.A// D Œ 3; 3. Since f 1
.f.A// is a subset of the
domain, R, we see that in this case, A f 1.f.A//.
4. Recall that C D Œ2; 6. Notice that f 1
.C / D Œ 2; 2 is a subset of the
domain, R. Explain why f f 1.C /

D Œ2; 6. Since f f 1.C /

is a
subset of the codomain, R, we see that in this case f f 1.C /

D C .
The examples in Progress Check 6.32 and Example 6.33 were meant to illus-
trate general results about how functions act on sets. In particular, we investigated
how the action of a function on sets interacts with the set operations of intersec-
tion and union. We will now state the theorems that these examples were meant to
illustrate. Some of the proofs will be left as exercises.
Theorem 6.34. Let f W S ! T be a function and let A and B be subsets of S.
Then
1. f.A B/ f.A/ f.B/
2. f.A [ B/ D f.A/ [ f.B/
Proof. We will prove Part (1). The proof of Part (2) is Exercise (5).
Assume that f W S ! T is a function and let A and B be subsets of S. We
will prove that f.A B/ f.A/ f.B/ by proving that for all y 2 T , if
y 2 f.A B/, then y 2 f.A/ f.B/.
We assume that y 2 f.A B/. This means that there exists an x 2 A B
such that f.x/ D y. Since x 2 A B, we conclude that x 2 A and x 2 B.

Since x 2 A and f.x/ D y, we conclude that y 2 f.A/.
Since x 2 B and f.x/ D y, we conclude that y 2 f.B/.
Since y 2 f.A/ and y 2 f.B/, y 2 f.A/ f.B/. This proves that if y 2
f.A B/, then y 2 f.A/ f.B/. Hence f.A B/ f.A/ f.B/.
Theorem 6.35. Let f W S ! T be a function and let C and D be subsets of T .
Then
1. f 1.C D/ D f 1.C / f 1.D/
2. f 1.C [ D/ D f 1.C / [ f 1.D/
Assume that f WS ! T is a function and that C and D are subsets of T . We
will prove that f 1
.C [ D/ D f 1
.C / [ f 1
.D/ by proving that each set is a
subset of the other.
We start by letting x be an element of f 1
.C [ D/. This means that f.x/ is
an element of C [ D. Hence,
f.x/ 2 C or f.x/ 2 D:
In the case where f.x/ 2 C , we conclude that x 2 f 1.C /, and hence that
x 2 f 1.C / [ f 1.D/. In the case where f.x/ 2 D, we see that x 2 f 1.D/,
and hence that x 2 f 1
.C /[f 1
.D/. So in both cases, x 2 f 1
.C /[f 1
.D/,
and we have proved that f 1.C [ D/ f 1.C / [ f 1.D/.
We now let t 2 f 1
.C / [ f 1
.D/. This means that
t 2 f 1
.C / or t 2 f 1
.D/:
In the case where t 2 f 1
.C /, we conclude that f.t/ 2 C and hence that
f.t/ 2 C [ D. This means that t 2 f 1.C [ D/.
Similarly, when t 2 f 1
.D/, it follows that f.t/ 2 D and hence that
f.t/ 2 C [ D. This means that t 2 f 1.C [ D/.
These two cases prove that if t 2 f 1
.C / [ f 1
.D/, then t 2 f 1
.C [ D/.
Therefore, f 1.C / [ f 1.D/ f 1.C [ D/.
Since we have now proved that each of the two sets is a subset of the other set,
we can conclude that f 1
.C [ D/ D f 1
.C / [ f 1
.D/.

Theorem 6.36. Let f WS ! T be a function and let A be a subset of S and let C
be a subset of T . Then
1. A f 1
.f.A// 2. f f 1
.C /

C
To prove Part (1), we will prove that for all a 2 S, if a 2 A, then
a 2 f 1.f.A//. So let a 2 A. Then, by definition, f.a/ 2 f.A/. We know
that f.A/ T , and so f 1
.f.A// S. Notice that
f 1
.f.A// D fx 2 S j f.x/ 2 f.A/g :
Since f.a/ 2 f.A/, we use this to conclude that a 2 f 1
.f.A//. This proves
that if a 2 A, then a 2 f 1.f.A//, and hence that A f 1.f.A//.
Exercises 6.6
1. Let f WS ! T , let A and B be subsets of S, and let C and D be subsets of
T . For x 2 S and y 2 T , carefully explain what it means to say that
? (a) y 2 f.A B/
(b) y 2 f.A [ B/
(c) y 2 f.A/ f.B/
?
(d) y 2 f.A/ [ f.B/
(e) x 2 f 1.C D/
? (f) x 2 f 1.C [ D/
(g) x 2 f 1
.C / f 1
.D/
?
(h) x 2 f 1
.C / [ f 1
.D/
2. Let f WR ! R by f.x/ D 2x C 1. Let
A D Œ2; 5 B D Œ 1; 3 C D Œ 2; 3 D D Œ1; 4.
Find each of the following:
(a) f.A/
? (b) f 1.f.A//
(c) f 1.C /
? (d) f.f 1.C //
? (e) f.A B/
? (f) f.A/ f.B/
(g) f 1.C D/
(h) f 1.C / f 1.D/

3. Let g W N N ! N by g.m; n/ D 2m3n, let A D f1; 2; 3g, and let C D
f1; 4; 6; 9; 12; 16; 18g. Find
?
(a) g.A A/
?
(b) g 1
.C /
(c) g 1
.g.A A//
(d) g g 1
.C /

4. ?
(a) Let S D f1; 2; 3; 4g. Define F WS ! N by F.x/ D x2
for each x 2 S.
What is the range of the function F and what is F.S/? How do these
two sets compare?
Now let A and B be sets and let f WA ! B be an arbitrary function from A
to B.
(b) Explain why f.A/ D range.f /.
(c) Define a function gWA ! f.A/ by g.x/ D f.x/ for all x in A. Prove
that the function g is a surjection.
?
5. Prove Part (2) of Theorem 6.34.
Let f W S ! T be a function and let A and B be subsets of S. Then
f.A [ B/ D f.A/ [ f.B/.
?
Let f W S ! T be a function and let C and D be subsets of T . Then
f 1.C D/ D f 1.C / f 1.D/.
Let f WS ! T be a function and let C T . Then f f 1
.C /

C .
8. Let f WS ! T and let A and B be subsets of S. Prove or disprove each of
the following:
(a) If A B, then f.A/ f.B/.
(b) If f.A/ f.B/, then A B.
? 9. Let f WS ! T and let C and D be subsets of T . Prove or disprove each of
the following:
(a) If C D, then f 1.C / f 1.D/.
(b) If f 1.C / f 1.D/, then C D.

10. Prove or disprove:
If f WS ! T is a function and A and B are subsets of S, then
f.A/ f.B/ f.A B/.
Note: Part (1) of Theorem 6.34 states that f.A B/ f.A/ f.B/.
11. Let f WS ! T be a function, let A S, and let C T .
(a) Part (1) of Theorem 6.36 states that A f 1
.f.A//. Give an example
where f 1.f.A// 6 A.
(b) Part (2) of Theorem 6.36 states that f f 1.C /

C . Give an exam-
ple where C 6 f f 1
.C /

.
If f WS ! T is an injection and A S, then f 1
.f.A// D A.
If f WS ! T is a surjection and C T , then f f 1.C /

D C .
14. Let f WS ! T . Prove that f.A B/ D f.A/ f.B/ for all subsets A and
B of S if and only if f is an injection.
Function, page 284
Domain of a function, page 285
Codomain of a function, page 285
Image of x under f , page 285
preimage of y under f , page 285
Independent variable, page 285
Dependent variable, page 285
Range of a function, page 287
Image of a function, page 287
Equal functions, page 298
Sequence, page 301
Injection, page 310
One-to-one function, page 310
Surjection, page 311

Onto function, page 311
Bijection, page 312
One-to-one and onto, page 312
Composition of f and g,
page 325
Composite function, page 325
f followed by g, page 325
Inverse of a function, page 338
Image of a set under a function,
page 351
preimage of a set under a func-
tion, page 351
Important Theorems and Results about Functions
Theorem 6.20. Let A, B, and C be nonempty sets and let f W A ! B and
gWB ! C .
1. If f and g are both injections, then g ı f is an injection.
2. If f and g are both surjections, then g ı f is a surjection.
3. If f and g are both bijections, then g ı f is a bijection.
Theorem 6.21. Let A, B, and C be nonempty sets and let f W A ! B and
gWB ! C .
1. If g ı f WA ! C is an injection, then f WA ! B is an injection.
2. If g ı f WA ! C is a surjection, then gWB ! C is a surjection.
Theorem 6.22. Let A and B be nonempty sets and let f be a subset of AB
that satisfies the following two properties:
For every a 2 A, there exists b 2 B such that .a; b/ 2 f ; and
For every a 2 A and every b; c 2 B, if .a; b/ 2 f and .a; c/ 2 f ,
then b D c.
If we use f.a/ D b whenever .a; b/ 2 f , then f is a function from A to B.
Theorem 6.25. Let A and B be nonempty sets and let f W A ! B. The
inverse of f is a function from B to A if and only if f is a bijection.
Theorem 6.26. Let A and B be nonempty sets and let f W A ! B be a
bijection. Then f 1
WB ! A is a function, and for every a 2 A and b 2 B,
f.a/ D b if and only if f 1
.b/ D a:

Corollary 6.28. Let A and B be nonempty sets and let f W A ! B be a
bijection. Then
1. For every x in A, .f 1
ı f /.x/ D x.
2. For every y in B, .f ı f 1
/.y/ D y.
Theorem 6.29. Let f WA ! B and g WB ! C be bijections. Then g ı f is
a bijection and .g ı f / 1
D f 1
ı g 1
.
Theorem 6.34. Let f WS ! T be a function and let A and B be subsets of
S. Then
1. f.A B/ f.A/ f.B/
2. f.A [ B/ D f.A/ [ f.B/
Theorem 6.35. Let f WS ! T be a function and let C and D be subsets of
T . Then
1. f 1.C D/ D f 1.C / f 1.D/
2. f 1.C [ D/ D f 1.C / [ f 1.D/
Theorem 6.36. Let f WS ! T be a function and let A be a subset of S and
let C be a subset of T . Then
1. A f 1
.f.A// 2. f f 1
.C /

C

Chapter 7
Equivalence Relations
7.1 Relations
Preview Activity 1 (The United States of America)
Recall from Section 5.4 that the Cartesian product of two sets A and B, written
A B, is the set of all ordered pairs .a; b/, where a 2 A and b 2 B. That is,
A B D f.a; b/ j a 2 A and b 2 Bg.
Let A be the set of all states in the United States and let
R D f.x; y/ 2 A A j x and y have a land border in commong:
For example, since California and Oregon have a land border, we can say that
.California, Oregon/ 2 R and .Oregon, California/ 2 R. Also, since Califor-
nia and Michigan do not share a land border, (California, Michigan) … R and
.Michigan, California/ … R.
1. Use the roster method to specify the elements in each of the following sets:
(a) B D fy 2 A j.Michigan, y/ 2 Rg
(b) C D fx 2 A j.x; Michigan/ 2 Rg
(c) D D fy 2 A j.Wisconsin, y/ 2 Rg
2. Find two different examples of two ordered pairs, .x; y/ and .y; z/ such that
.x; y/ 2 R, .y; z/ 2 R, but .x; z/ 62 R, or explain why no such example
exists. Based on this, is the following conditional statement true or false?
For all x; y; z 2 A, if .x; y/ 2 R and .y; z/ 2 R, then .x; z/ 2 R.
362

7.1. Relations 363
3. Is the following conditional statement true or false? Explain.
For all x; y 2 A, if .x; y/ 2 R, then .y; x/ 2 R.
Preview Activity 2 (The Solution Set of an Equation with Two Variables)
In Section 2.3, we introduced the concept of the truth set of an open sentence with
one variable. This was defined to be the set of all elements in the universal set that
can be substituted for the variable to make the open sentence a true proposition.
Assume that x and y represent real numbers. Then the equation
4x2
C y2
D 16
is an open sentence with two variables. An element of the truth set of this open
sentence (also called a solution of the equation) is an ordered pair .a; b/ of real
numbers so that when a is substituted for x and b is substituted for y, the predicate
becomes a true statement (a true equation in this case). We can use set builder
notation to describe the truth set S of this equation with two variables as follows:
S D
˚
.x; y/ 2 R R j 4x2
C y2
D 16 :
When a set is a truth set of an open sentence that is an equation, we also call the
set the solution set of the equation.
1. List four different elements of the set S.
2. The graph of the equation 4x2 C y2 D 16 in the xy-coordinate plane is an
ellipse. Draw the graph and explain why this graph is a representation of the
truth set (solutions set) of the equation 4x2 C y2 D 16.
3. Describe each of the following sets as an interval of real numbers:
(a) A D
˚
x 2 R j there exists a y 2 R such that 4x2
C y2
D 16 .
(b) B D
˚
y 2 R j there exists an x 2 R such that 4x2
C y2
D 16 .
Introduction to Relations
In Section 6.1, we introduced the formal definition of a function from one set to
another set. The notion of a function can be thought of as one way of relating the
elements of one set with those of another set (or the same set). A function is a

364 Chapter 7. Equivalence Relations
special type of relation in the sense that each element of the first set, the domain,
is “related” to exactly one element of the second set, the codomain.
This idea of relating the elements of one set to those of another set using or-
dered pairs is not restricted to functions. For example, we may say that one integer,
a, is related to another integer, b, provided that a is congruent to b modulo 3. No-
tice that this relation of congruence modulo 3 provides a way of relating one integer
to another integer. However, in this case, an integer a is related to more than one
other integer. For example, since
5 5 .mod 3/; 5 2 .mod 3/; and 5 1 .mod 3/;
we can say that 5 is related to 3, 5 is related to 2, and 5 is related to 1. Notice that,
as with functions, each relation of the form a b .mod 3/ involves two integers
a and b and hence involves an ordered pair .a; b/, which is an element of Z Z.
Definition. Let A and B be sets. A relation R from the set A to the set B
is a subset of A B. That is, R is a collection of ordered pairs where the first
coordinate of each ordered pair is an element of A, and the second coordinate
of each ordered pair is an element of B.
A relation from the set A to the set A is called a relation on the set A. So a
relation on the set A is a subset of A A.
In Section 6.1, we defined the domain and range of a function. We make similar
definitions for a relation.
Definition. If R is a relation from the set A to the set B, then the subset of
A consisting of all the first coordinates of the ordered pairs in R is called the
domain of R. The subset of B consisting of all the second coordinates of the
ordered pairs in R is called the range of R.
We use the notation dom.R/ for the domain of R and range.R/ for the range
of R. So using set builder notation,
dom.R/ D fu 2 A j .u; y/ 2 R for at least one y 2 Bg
range.R/ D fv 2 B j .x; v/ 2 R for at least one x 2 Ag:

7.1. Relations 365
Example 7.1 (Domain and Range)
A relation was studied in each of the Preview Activities for this section. For Pre-
view Activity 2, the set S D
˚
.x; y/ 2 R R j 4x2 C y2 D 16 is a subset of
R R and, hence, S is a relation on R. In Problem (3) of Preview Activity 2, we
actually determined the domain and range of this relation.
dom.S/ D A D
˚
x 2 R j there exists a y 2 R such that 4x2
C y2
D 16
range.S/ D B D
˚
y 2 R j there exists an x 2 R such that 4x2
C y2
D 16
So from the results in Preview Activity 2, we can say that the domain of the relation
S is the closed interval Œ 2; 2 and the range of S is the closed interval Œ 4; 4.
Progress Check 7.2 (Examples of Relations)
1. Let T D
˚
.x; y/ 2 R R j x2 C y2 D 64 .
(a) Explain why T is a relation on R.
(b) Find all values of x such that .x; 4/ 2 T . Find all values of x such that
.x; 9/ 2 T.
(c) What is the domain of the relation T ? What is the range of T ?
(d) Since T is a relation on R, its elements can be graphed in the coordinate
plane. Describe the graph of the relation T.
2. From Preview Activity 1, A is the set of all states in the United States, and
R D f.x; y/ 2 A A j x and y have a border in commong:
(a) Explain why R is a relation on A.
(b) What is the domain of the relation R? What is the range of the relation
R?
(c) Are the following statements true or false? Justify your conclusions.
i. For all x; y 2 A, if .x; y/ 2 R, then .y; x/ 2 R.
ii. For all x; y; z 2 A, if .x; y/ 2 R and .y; z/ 2 R, then .x; z/ 2 R.
Some Standard Mathematical Relations
There are many different relations in mathematics. For example, two real numbers
can be considered to be related if one number is less than the other number. We

call this the “less than” relation on R. If x; y 2 R and x is less than y, we often
write x y. As a set of ordered pairs, this relation is R, where
R D f.x; y/ 2 R R j x yg:
With many mathematical relations, we do not write the relation as a set of ordered
pairs even though, technically, it is a set of ordered pairs. Table 7.1 describes some
standard mathematical relations.
Name Open Relation as a Set of
Sentence Ordered Pairs
The “less than” re-
lation on R
x y f.x; y/ 2 R R j x yg
The “equality” rela-
tion on R
x D y f.x; y/ 2 R R j x D yg
The “divides” rela-
tion on Z
m j n f.m; n/ 2 Z Z j m divides ng
The “subset” rela-
tion on P .U /
S T f.S; T / 2 P .U / P .U / j S T g
The “element of”
relation from U to
P .U /
x 2 S f.x; S/ 2 U P .U / j x 2 Sg
The “congruence
modulo n” relation
on Z
a b .mod n/ f.a; b/ 2 Z Z j a b .mod n/g
Table 7.1: Standard Mathematical Relations
Notation for Relations
The mathematical relations in Table 7.1 all used a relation symbol between the two
elements that form the ordered pair in AB. For this reason, we often do the same
thing for a general relation from the set A to the set B. So if R is a relation from A
to B, and x 2 A and y 2 B, we use the notation
x R y to mean .x; y/ 2 R; and
x 6R y to mean .x; y/ … R.
In some cases, we will even use a generic relation symbol for defining a new rela-
tion or speaking about relations in a general context. Perhaps the most commonly

7.1. Relations 367
used symbol is “”, read “tilde” or “squiggle” or “is related to.” When we do this,
we will write
x y means the same thing as .x; y/ 2 R; and
x œ y means the same thing as .x; y/ … R.
Progress Check 7.3 (The Divides Relation)
Whenever we have spoken about one integer dividing another integer, we have
worked with the “divides” relation on Z. In particular, we can write
D D f.m; n/ 2 Z Z j m divides ng:
In this case, we have a specific notation for “divides,” and we write
m j n if and only if .m; n/ 2 D.
1. What is the domain of the “divides” relation? What is the range of the “di-
vides” relation?
2. Are the following statements true or false? Explain.
(a) For every nonzero integer a, a j a.
(b) For all nonzero integers a and b, if a j b, then b j a.
(c) For all nonzero integers a, b, and c, if a j b and b j c, then a j c.
Functions as Relations
If we have a function f WA ! B, we can generate a set of ordered pairs f that is
a subset of A B as follows:
f D f.a; f.a// j a 2 Ag or f D f.a; b/ 2 A B j b D f.a/g :
This means that f is a relation from A to B. Since, dom.f / D A, we know that
(1) For every a 2 A, there exists a b 2 B such that .a; b/ 2 f .
When .a; b/ 2 f , we write b D f.a/. In addition, to be a function, each input can
produce only one output. In terms of ordered pairs, this means that there will never
be two ordered pairs .a; b/ and .a; c/ in the function f , where a 2 A, b; c 2 B,
and b ¤ c. We can formulate this as a conditional statement as follows:

(2) For every a 2 A and every b; c 2 B, if .a; b/ 2 f and .a; c/ 2 f , then
b D c.
This means that a function f from A to B is a relation from A to B that satisfies
conditions (1) and (2). (See Theorem 6.22 in Section 6.5.) Not every relation,
however, will be a function. For example, consider the relation T in Progress
Check 7.2.
T D
˚
.x; y/ 2 R R j x2
C y2
D 64
This relation fails condition (2) above since a counterexample comes from the facts
that .0; 8/ 2 T and .0; 8/ 2 T and 8 ¤ 8.
Progress Check 7.4 (A Set of Ordered Pairs)
Let F D
˚
.x; y/ 2 R R j y D x2
. The set F can then be considered to be
relation on R since it is a subset of R R.
1. List five different ordered pairs that are in the set F .
2. Use the roster method to specify the elements of each of the following the
sets:
(a) A D fx 2 R j .x; 4/ 2 F g
(b) B D fx 2 R j .x; 10/ 2 F g
(c) C D fy 2 R j .5; y/ 2 F g
(d) D D fy 2 R j . 3; y/ 2 F g
3. Since each real number x produces only one value of y for which y D x2,
the set F can be used to define a function from the set R to R. Draw a graph
of this function.
Visual Representations of Relations
In Progress Check 7.4, we were able to draw a graph of a relation as a way to visu-
alize the relation. In this case, the relation was a function from R to R. In addition,
in Progress Check 7.2, we were also able to use a graph to represent a relation.
In this case, the graph of the relation T D
˚
.x; y/ 2 R R j x2
C y2
D 64 is a
circle of radius 8 whose center is at the origin.
When R is a relation from a subset of the real numbers R to a subset of R,
we can often use a graph to provide a visual representation of the relation. This is
especially true if the relation is defined by an equation or even an inequality. For
example, if
R D
˚
.x; y/ 2 R R j y x2
;

7.1. Relations 369
then we can use the following graph as a way to visualize the points in the plane
that are also in this relation.
x
y
Figure 7.1: Graph of y x2
The points .x; y/ in the relation R are the points on the graph of y D x2
or are in
the shaded region. This because for these points, y x2. One of the shortcomings
of this type of graph is that the graph of the equation and the shaded region are
actually unbounded and so we can never show the entire graph of this relation.
However, it does allow us to see that the points in this relation are either on the
parabola defined by the equation y D x2 or are “inside” the parabola.
When the domain or range of a relation is infinite, we cannot provide a visu-
alization of the entire relation. However, if A is a (small) finite set, a relation R
on A can be specified by simply listing all the ordered pairs in R. For example, if
A D f1; 2; 3; 4g, then
R D f.1; 1/; .4; 4/; .1; 3/; .3; 2/; .1; 2/; .2; 1/g
is a relation on A. A convenient way to represent such a relation is to draw a point
in the plane for each of the elements of A and then for each .x; y/ 2 R (or x R y),
we draw an arrow starting at the point x and pointing to the point y. If .x; x/ 2 R
(or x R x), we draw a loop at the point x. The resulting diagram is called a directed
graph or a digraph. The diagram in Figure 7.2 is a digraph for the relation R.
In a directed graph, the points are called the vertices. So each element of A
corresponds to a vertex. The arrows, including the loops, are called the directed
edges of the directed graph. We will make use of these directed graphs in the next
section when we study equivalence relations.

3 4
2
1
Figure 7.2: Directed Graph for a Relation
Progress Check 7.5 (The Directed Graph of a Relation)
Let A D f1; 2; 3; 4; 5; 6g. Draw a directed graph for the following two relations on
the set A. For each relation, it may be helpful to arrange the vertices of A as shown
in Figure 7.3.
R D f.x; y/ 2 A A j x divides yg; T D f.x; y/ 2 A A j x C y is eveng:
1 2
3
4
5
6
Figure 7.3: Vertices for A
Exercises 7.1
?
1. Let A D fa; b; cg, B D fp; q; rg, and let R be the set of ordered pairs
defined by R D f.a; p/ ; .b; q/ ; .c; p/ ; .a; q/g.

7.1. Relations 371
(a) Use the roster method to list all the elements of A B. Explain why
A B can be considered to be a relation from A to B.
(b) Explain why R is a relation from A to B.
(c) What is the domain of R? What is the range of R?
? 2. Let A D fa; b; cg and let R D f.a; a/ ; .a; c/ ; .b; b/ ; .b; c/ ; .c; a/ ; .c; b/g
(so R is a relation on A). Are the following statements true or false? Explain.
(a) For each x 2 A, x R x.
(b) For every x; y 2 A, if x R y, then y R x.
(c) For every x; y; z 2 A, if x R y and y R z, then x R z.
(d) R is a function from A to A.
3. Let A be the set of all females citizens of the United States. Let D be the
relation on A defined by
D D f.x; y/ 2 A A j x is a daughter of yg:
That is, x D y means that x is a daughter of y.
? (a) Describe those elements of A that are in the domain of D.
?
(b) Describe those elements of A that are in the range of D.
(c) Is the relation D a function from A to A? Explain.
? 4. Let U be a nonempty set, and let R be the “subset relation” on P.U /. That
is,
R D f.S; T / 2 P.U / P.U / j S T g:
(a) Write the open sentence .S; T / 2 R using standard subset notation.
(b) What is the domain of this subset relation, R?
(c) What is the range of this subset relation, R?
(d) Is R a function from P.U / to P.U /? Explain.
5. Let U be a nonempty set, and let R be the “element of” relation from U to
P .U /. That is,
R D f.x; S/ 2 U P.U / j x 2 Sg:
(a) What is the domain of this “element of” relation, R?

(b) What is the range of this “element of” relation, R?
(c) Is R a function from U to P.U /? Explain.
?
6. Let S D
˚
.x; y/ 2 R R j x2
C y2
D 100 .
(a) Determine the set of all values of x such that .x; 6/ 2 S, and determine
the set of all values of x such that .x; 9/ 2 S.
(b) Determine the domain and range of the relation S and write each set
using set builder notation.
(c) Is the relation S a function from R to R? Explain.
(d) Since S is a relation on R, its elements can be graphed in the coordinate
plane. Describe the graph of the relation S. Is the graph consistent with
your answers in Exercises (6a) through (6c)? Explain.
7. Repeat Exercise( 6) using the relation on R defined by
S D
n
.x; y/ 2 R R j y D
p
100 x2
o
:
What is the connection between this relation and the relation in Exercise (6)?
8. Determine the domain and range of each of the following relations on R and
sketch the graph of each relation.
(a) R D
˚
.x; y/ 2 R R j x2
C y2
D 10
(b) S D
˚
.x; y/ 2 R R j y2 D x C 10
(c) T D f.x; y/ 2 R R j jxj C jyj D 10g
(d) R D
˚
.x; y/ 2 R R j x2
D y2
9. Let R be the relation on Z where for all a; b 2 Z, aR b if and only if
ja bj 2.
? (a) Use set builder notation to describe the relation R as a set of ordered
pairs.
? (b) Determine the domain and range of the relation R .
(c) Use the roster method to specify the set of all integers x such that x R 5
and the set of all integers x such that 5R x.
(d) If possible, find integers x and y such that x R 8, 8R y, but x 6R y.

7.1. Relations 373
(e) If a 2 Z, use the roster method to specify the set of all x 2 Z such that
x R a.
10. Let R D f.x; y/ 2 R R j x yg. This means that R is the “less than”
relation on R.
(a) What is the domain of the relation R?
(b) What is the range of the relation R?
(c) Is the relation R a function from R to R? Explain.
Note: Remember that a relation is a set. Consequently, we can talk about
one relation being a subset of another relation. Another thing to remember
is that the elements of a relation are ordered pairs.
11. The Inverse of a Relation. In Section 6.5, we introduced the inverse of a
function. If A and B are nonempty sets and if f W A ! B is a function,
then the inverse of f , denoted by f 1
, is defined as
f 1
D f.b; a/ 2 B A j f .a/ D bg
D f.b; a/ 2 B A j .a; b/ 2 f g:
Now that we know about relations, we see that f 1 is always a relation from
B to A. The concept of the inverse of a function is actually a special case of
the more general concept of the inverse of a relation, which we now define.
Definition. Let R be a relation from the set A to the set B. The inverse of R,
written R 1
and read “R inverse,” is the relation from B to A defined by
R 1
D f.y; x/ 2 B A j .x; y/ 2 Rg , or
R 1
D f.y; x/ 2 B A j x R yg:
That is, R 1 is the subset of B A consisting of all ordered pairs .y; x/ such
that x R y.
For example, let D be the “divides” relation on Z. See Progress Check 7.3.
So
D D f.m; n/ 2 Z Z j m divides ng:

This means that we can write m j n if and only if .m; n/ 2 D. So, in this
case,
D 1
D f.n; m/ 2 Z Z j .m; n/ 2 Dg
D f.n; m/ 2 Z Z j m divides ng:
Now, if we would like to focus on the first coordinate instead of the second
coordinate in D 1, we know that “m divides n” means the same thing as “n
is a multiple of m.” Hence,
D 1
D f.n; m/ 2 Z Z j n is a multiple of mg:
We can say that the inverse of the “divides” relation on Z is the “is a multiple
of” relation on Z.
Theorem 7.6, which follows, contains some elementary facts about inverse
relations.
Theorem 7.6. Let R be a relation from the set A to the set B. Then
The domain of R 1 is the range of R. That is, dom R 1

D range.R/.
The range of R 1 is the domain of R. That is, range R 1

D dom.R/.
The inverse of R 1
is R. That is, R 1
1
D R.
To prove the first part of Theorem 7.6, observe that the goal is to prove that
two sets are equal,
dom R 1

D range.R/:
One way to do this is to prove that each is a subset of the other. To prove
that dom R 1

range.R/, we can start by choosing an arbitrary ele-
ment of dom R 1

. So let y 2 dom R 1

. The goal now is to prove that
y 2 range.R/. What does it mean to say that y 2 dom R 1

? It means
that there exists an x 2 A such that
.y; x/ 2 R 1
:
Now what does it mean to say that .y; x/ 2 R 1
? It means that .x; y/ 2 R.
What does this tell us about y?
Complete the proof of the first part of Theorem 7.6. Then, complete the
proofs of the other two parts of Theorem 7.6.

7.2. Equivalence Relations 375
7.2 Equivalence Relations
Preview Activity 1 (Properties of Relations)
In previous mathematics courses, we have worked with the equality relation. For
example, let R be the relation on Z defined as follows: For all a; b 2 Z, aR b if and
only if a D b. We know this equality relation on Z has the following properties:
For each a 2 Z, a D b and so aR a.
For all a; b 2 Z, if a D b, then b D a. That is, if aR b, then b R a.
For all a; b; c 2 Z, if a D b and b D c, then a D c. That is, if aR b and
b R c, then aR c.
In mathematics, when something satisfies certain properties, we often ask if other
things satisfy the same properties. Before investigating this, we will give names to
these properties.
Definition. Let A be a nonempty set and let R be a relation on A.
The relation R is reflexive on A provided that for each x 2 A, x R x
or, equivalently, .x; x/ 2 R.
The relation R is symmetric provided that for every x; y 2 A, if x R y,
then y R x or, equivalently, for every x; y 2 A, if .x; y/ 2 R, then
.y; x/ 2 R.
The relation R is transitive provided that for every x; y; z 2 A, if
x R y and y R z, then x R z or, equivalently, for every x; y; z 2 A, if
.x; y/ 2 R and .y; z/ 2 R, then .x; z/ 2 R.
Before exploring examples, for each of these properties, it is a good idea to
understand what it means to say that a relation does not satisfy the property. So let
A be a nonempty set and let R be a relation on A.
1. Carefully explain what it means to say that the relation R is not reflexive on
the set A.
2. Carefully explain what it means to say that the relation R is not symmetric.
3. Carefully explain what it means to say that the relation R is not transitive.

To illustrate these properties, we let A D f1; 2; 3; 4g and define the relations R and
T on A as follows:
R D f.1; 1/; .2; 2/; .3; 3/; .4; 4/; .1; 3/; .3; 2/g
T D f.1; 1/; .1; 4/; .2; 4/; .4; 1/; .4; 2/g
4. Draw a directed graph for the relation R. Then explain why the relation R is
reflexive on A, is not symmetric, and is not transitive.
5. Draw a directed graph for the relation T . Is the relation T reflexive on A? Is
the relation T symmetric? Is the relation T transitive? Explain.
Preview Activity 2 (Review of Congruence Modulo n)
1. Let a; b 2 Z and let n 2 N. On page 92 of Section 3.1, we defined what
it means to say that a is congruent to b modulo n. Write this definition and
state two different conditions that are equivalent to the definition.
2. Explain why congruence modulo n is a relation on Z.
3. Carefully review Theorem 3.30 and the proofs given on page 148 of Sec-
tion 3.5. In terms of the properties of relations introduced in Preview Activ-
ity 1, what does this theorem say about the relation of congruence modulo n
on the integers?
4. Write a complete statement of Theorem 3.31 on page 150 and Corollary 3.32.
5. Write a proof of the symmetric property for congruence modulo n. That is,
prove the following:
Let a; b 2 Z and let n 2 N. If a b .mod n/, then b a .mod n/.
Directed Graphs and Properties of Relations
In Section 7.1, we used directed graphs, or digraphs, to represent relations on finite
sets. Three properties of relations were introduced in Preview Activity 1 and will
be repeated in the following descriptions of how these properties can be visualized
on a directed graph.
Let A be a nonempty set and let R be a relation on A.

The relation R is reflexive on A provided that for each x 2 A, x R x or,
equivalently, .x; x/ 2 R.
This means that if a reflexive relation is represented on a digraph, there
would have to be a loop at each vertex, as is shown in the following figure.
The relation R is symmetric provided that for every x; y 2 A, if x R y, then
y R x or, equivalently, for every x; y 2 A, if .x; y/ 2 R, then .y; x/ 2 R.
This means that if a symmetric relation is represented on a digraph, then
anytime there is a directed edge from one vertex to a second vertex, there
would be a directed edge from the second vertex to the first vertex, as is
shown in the following figure.
The relation R is transitive provided that for every x; y; z 2 A, if x R y and
y R z, then x R z or, equivalently, for every x; y; z 2 A, if .x; y/ 2 R and
.y; z/ 2 R, then .x; z/ 2 R. So if a transitive relation is represented by a
digraph, then anytime there is a directed edge from a vertex x to a vertex y
and a directed edge from y to a vertex z, there would be a directed edge from
x to z.
In addition, if a transitive relation is represented by a digraph, then anytime
there is a directed edge from a vertex x to a vertex y and a directed edge from
y to the vertex x, there would be loops at x and y. These two situations are
illustrated as follows:
x y
z
x y

Progress Check 7.7 (Properties of Relations)
Let A D fa; b; c; dg and let R be the following relation on A:
R D f.a; a/; .b; b/; .a; c/; .c; a/; .b; d/; .d; b/g:
Draw a directed graph for the relation R and then determine if the relation R is
reflexive on A, if the relation R is symmetric, and if the relation R is transitive.
Definition of an Equivalence Relation
In mathematics, as in real life, it is often convenient to think of two different things
as being essentially the same. For example, when you go to a store to buy a cold
soft drink, the cans of soft drinks in the cooler are often sorted by brand and type
of soft drink. The Coca Colas are grouped together, the Pepsi Colas are grouped
together, the Dr. Peppers are grouped together, and so on. When we choose a
particular can of one type of soft drink, we are assuming that all the cans are es-
sentially the same. Even though the specific cans of one type of soft drink are
physically different, it makes no difference which can we choose. In doing this, we
are saying that the cans of one type of soft drink are equivalent, and we are using
the mathematical notion of an equivalence relation.
An equivalence relation on a set is a relation with a certain combination of
properties that allow us to sort the elements of the set into certain classes. In this
section, we will focus on the properties that define an equivalence relation, and in
the next section, we will see how these properties allow us to sort or partition the
elements of the set into certain classes.
Definition. Let A be a nonempty set. A relation on the set A is an equiv-
alence relation provided that is reflexive, symmetric, and transitive. For
a; b 2 A, if is an equivalence relation on A and a b, we say that a is
equivalent to b.
Most of the examples we have studied so far have involved a relation on a
small finite set. For these examples, it was convenient to use a directed graph to
represent the relation. It is now time to look at some other type of examples, which
may prove to be more interesting. In these examples, keep in mind that there is a
subtle difference between the reflexive property and the other two properties. The
reflexive property states that some ordered pairs actually belong to the relation R,
or some elements of A are related. The reflexive property has a universal quantifier
and, hence, we must prove that for all x 2 A, x R x. Symmetry and transitivity, on

the other hand, are defined by conditional sentences. We often use a direct proof
for these properties, and so we start by assuming the hypothesis and then showing
that the conclusion must follow from the hypothesis.
Example 7.8 (A Relation that Is Not an Equivalence Relation)
Let M be the relation on Z defined as follows:
For a; b 2 Z, aM b if and only if a is a multiple of b.
So aM b if and only if there exists a k 2 Z such that a D bk.
The relation M is reflexive on Z since for each x 2 Z, x D x 1 and, hence,
x M x.
Notice that 4M 2, but 2 6M 4. So there exist integers x and y such that
x M y but y 6M x. Hence, the relation M is not symmetric.
Now assume that x M y and y M z. Then there exist integers p and q such
that
x D yp and y D zq:
Using the second equation to make a substitution in the first equation, we see
that x D z .pq/. Since pq 2 Z, we have shown that x is a multiple of z and
hence x M z. Therefore, M is a transitive relation.
The relation M is reflexive on Z and is transitive, but since M is not symmetric, it
is not an equivalence relation on Z.
Progress Check 7.9 (A Relation that Is an Equivalence Relation)
Define the relation on Q as follows: For all a; b 2 Q, a b if and only if
a b 2 Z. For example:

3
4

7
4
since
3
4
7
4
D 1 and 1 2 Z.

3
4
6
1
2
since
3
4
1
2
D
1
4
and
1
4
… Z.
To prove that is reflexive on Q, we note that for all q 2 Q, a a D 0. Since
0 2 Z, we conclude that a a. Now prove that the relation is symmetric and
transitive, and hence, that is an equivalence relation on Q.

Congruence Modulo n
One of the important equivalence relations we will study in detail is that of con-
gruence modulo n. We reviewed this relation in Preview Activity 2.
Theorem 3.30 on page 148 tells us that congruence modulo n is an equivalence
relation on Z. Recall that by the Division Algorithm, if a 2 Z, then there exist
unique integers q and r such that
a D nq C r and 0 r n:
Theorem 3.31 and Corollary 3.32 then tell us that a r .mod n/. That is, a is
congruent modulo n to its remainder r when it is divided by n. When we use the
term “remainder” in this context, we always mean the remainder r with 0 r n
that is guaranteed by the Division Algorithm. We can use this idea to prove the
following theorem.
Theorem 7.10. Let n 2 N and let a; b 2 Z. Then a b .mod n/ if and only if a
and b have the same remainder when divided by n.
Proof. Let n 2 N and let a; b 2 Z. We will first prove that if a and b have the
same remainder when divided by n, then a b .mod n/. So assume that a and b
have the same remainder when divided by n, and let r be this common remainder.
Then, by Theorem 3.31,
a r .mod n/ and b r .mod n/:
Since congruence modulo n is an equivalence relation, it is a symmetric relation.
Hence, since b r .mod n/, we can conclude that r b .mod n/. Combining
this with the fact that a r .mod n/, we now have
a r .mod n/ and r b .mod n/:
We can now use the transitive property to conclude that a b .mod n/. This
proves that if a and b have the same remainder when divided by n, then
a b .mod n/.
We will now prove that if a b .mod n/, then a and b have the same re-
mainder when divided by n. Assume that a b .mod n/, and let r be the least
nonnegative remainder when b is divided by n. Then 0 r n and, by Theo-
rem 3.31,
b r .mod n/:

Now, using the facts that a b .mod n/ and b r .mod n/, we can use the
transitive property to conclude that
a r .mod n/:
This means that there exists an integer q such that a r D nq or that
a D nq C r:
Since we already know that 0 r n, the last equation tells us that r is the
least nonnegative remainder when a is divided by n. Hence we have proven that if
a b .mod n/, then a and b have the same remainder when divided by n.
Examples of Other Equivalence Relations
1. The relation on Q from Progress Check 7.9 is an equivalence relation.
2. Let A be a nonempty set. The equality relation on A is an equivalence
relation. This relation is also called the identity relation on A and is denoted
by IA, where
IA D f.x; x/ j x 2 Ag:
3. Define the relation on R as follows:
For a; b 2 R, a b if and only if there exists an integer k such that
a b D 2k.
We will prove that the relation is an equivalence relation on R. The rela-
tion is reflexive on R since for each a 2 R, a a D 0 D 2 0 .
Now, let a; b 2 R and assume that a b. We will prove that b a. Since
a b, there exists an integer k such that
a b D 2k:
By multiplying both sides of this equation by 1, we obtain
. 1/.a b/ D . 1/.2k/
b a D 2. k/:
Since k 2 Z, the last equation proves that b a. Hence, we have proven
that if a b, then b a and, therefore, the relation is symmetric.

To prove transitivity, let a; b; c 2 R and assume that a b and b c. We
will prove that a c. Now, there exist integers k and n such that
a b D 2k and b c D 2n:
By adding the corresponding sides of these two equations, we see that
.a b/ C .b c/ D 2k C 2n
a c D 2.k C n/:
By the closure properties of the integers, k C n 2 Z. So this proves that
a c and, hence the relation is transitive.
We have now proven that is an equivalence relation on R. This equivalence
relation is important in trigonometry. If a b, then there exists an integer
k such that a b D 2k and, hence, a D b C k.2/. Since the sine and
cosine functions are periodic with a period of 2, we see that
sin a D sin.b C k.2// D sin b; and
cos a D cos.b C k.2// D cos b:
Therefore, when a b, each of the trigonometric functions have the same
value at a and b.
4. For an example from Euclidean geometry, we define a relation P on the set
L of all lines in the plane as follows:
For l1; l2 2 L, l1 P l2 if and only if l1 is parallel to l2 or l1 D l2.
We added the second condition to the definition of P to ensure that P is re-
flexive on L. Theorems from Euclidean geometry tell us that if l1 is parallel
to l2, then l2 is parallel to l1, and if l1 is parallel to l2 and l2 is parallel to l3,
then l1 is parallel to l3. (Drawing pictures will help visualize these proper-
ties.) This tells us that the relation P is reflexive, symmetric, and transitive
and, hence, an equivalence relation on L.
Progress Check 7.11 (Another Equivalence Relation)
Let U be a finite, nonempty set and let P .U / be the power set of U . Recall that
P .U / consists of all subsets of U . (See page 222.) Define the relation on P .U /
as follows:
For A; B 2 P .U /, A B if and only if card .A/ D card .B/.
For the definition of the cardinality of a finite set, see page 223. This relation states
that two subsets of U are equivalent provided that they have the same number of
elements. Prove that is an equivalence relation on

Exercises 7.2
? 1. Let A D fa; bg and let R D f.a; b/g. Is R an equivalence relation on A? If
not, is R reflexive, symmetric, or transitive? Justify all conclusions.
2. Let A D fa; b; cg. For each of the following, draw a directed graph that
represents a relation with the specified properties.
(a) A relation on A that is symmetric but not transitive
(b) A relation on A that is transitive but not symmetric
(c) A relation on A that is symmetric and transitive but not reflexive
on A
(d) A relation on A that is not reflexive on A, is not symmetric, and is not
transitive
(e) A relation on A, other than the identity relation, that is an equivalence
relation on A
? 3. Let A D f1; 2; 3; 4; 5g. The identity relation on A is
IA D f.1; 1/; .2; 2/; .3; 3/; .4; 4/; .5; 5/g :
Determine an equivalence relation on A that is different from IA or explain
why this is not possible.
?
4. Let R D f.x; y/ 2 R R j jxj C jyj D 4g. Then R is a relation on R. Is R
an equivalence relation on R? If not, is R reflexive, symmetric, or transitive?
Justify all conclusions.
5. A relation R is defined on Z as follows: For all a; b 2 Z, aR b if and only
if ja bj 3. Is R an equivalence relation on R? If not, is R reflexive,
symmetric, or transitive? Justify all conclusions.
?
6. Let f W R ! R be defined by f.x/ D x2
4 for each x 2 R. Define a
relation on R as follows:
For a; b 2 R; a b if and only if f.a/ D f.b/:
(a) Is the relation an equivalence relation on R? Justify your conclusion.
(b) Determine all real numbers in the set C D fx 2 R j x 5g.

7. Repeat Exercise (6) using the function f WR ! R that is defined by f.x/ D
x2
3x 7 for each x 2 R.
8. (a) Repeat Exercise (6a) using the function f W R ! R that is defined by
f.x/ D sinx for each x 2 R.
(b) Determine all real numbers in the set C D fx 2 R j x g.
9. Define the relation on Q as follows: For a; b 2 Q, a b if and only
if a b 2 Z. In Progress Check 7.9, we showed that the relation is an
equivalence relation on Q.
(a) List four different elements of the set C D

x 2 Q j x
5
7

.
(b) Use set builder notation (without using the symbol ) to specify the set
C .
(c) Use the roster method to specify the set C .
10. Let and be relations on Z defined as follows:
For a; b 2 Z, a b if and only if 2 divides a C b.
For a; b 2 Z, a b if and only if 3 divides a C b.
(a) Is an equivalence relation on Z? If not, is this relation reflexive,
symmetric, or transitive?
(b) Is an equivalence relation on Z? If not, is this relation reflexive,
11. Let U be a finite, nonempty set and let P.U / be the power set of U . That
is, P.U / is the set of all subsets of U . Define the relation on P.U / as
follows: For A; B 2 P.U /, A B if and only if A B D ;. That is, the
ordered pair .A; B/ is in the relation if and only if A and B are disjoint.
Is the relation an equivalence relation on P.U /? If not, is it reflexive,
symmetric, or transitive? Justify all conclusions.
12. Let U be a nonempty set and let P.U / be the power set of U . That is, P .U /
is the set of all subsets of U .
For A and B in P.U /, define A B to mean that there exists a bijection
f W A ! B. Prove that is an equivalence relation on P.U /.
Hint: Use results from Sections 6.4 and 6.5.

13. Let and be relations on Z defined as follows:
For a; b 2 Z, a b if and only if 2a C 3b 0 .mod 5/.
For a; b 2 Z, a b if and only if a C 3b 0 .mod 5/.
(a) Is an equivalence relation on Z? If not, is this relation reflexive,
(b) Is an equivalence relation on Z? If not, is this relation reflexive,
14. Let and be relations on R defined as follows:
For x; y 2 R, x y if and only if xy 0.
For x; y 2 R, x y if and only if xy 0.
(a) Is an equivalence relation on R? If not, is this relation reflexive,
(b) Is an equivalence relation on R? If not, is this relation reflexive,
15. Define the relation on R R as follows: For .a; b/ ; .c; d/ 2 R R,
.a; b/ .c; d/ if and only if a2
C b2
D c2
C d2
.
(a) Prove that is an equivalence relation on R R.
(b) List four different elements of the set
C D f.x; y/ 2 R R j .x; y/ .4; 3/g :
? (c) Give a geometric description of set C .
(a) Proposition. Let R be a relation on a set A. If R is symmetric and
transitive, then R is reflexive.
Proof. Let x; y 2 A. If x R y, then y R x since R is symmetric. Now,
x R y and y R x, and since R is transitive, we can conclude that x R x.
Therefore, R is reflexive.

(b) Proposition. Let be a relation on Z where for all a; b 2 Z, a b if
and only if .a C 2b/ 0 .mod 3/. The relation is an equivalence
relation on Z.
Proof. Assume a a. Then .a C 2a/ 0 .mod 3/ since
.3a/ 0 .mod 3/. Therefore, is reflexive on Z. In addition, if
a b, then .a C 2b/ 0 .mod 3/, and if we multiply both sides of
this congruence by 2, we get
2 .a C 2b/ 2 0 .mod 3/
.2a C 4b/ 0 .mod 3/
.2a C b/ 0 .mod 3/
.b C 2a/ 0 .mod 3/ :
This means that b a and hence, is symmetric.
We now assume that .a C 2b/ 0 .mod 3/ and .b C 2c/ 0 .mod 3/.
By adding the corresponding sides of these two congruences, we obtain
.a C 2b/ C .b C 2c/ 0 C 0 .mod 3/
.a C 3b C 2c/ 0 .mod 3/
.a C 2c/ 0 .mod 3/ :
Hence, the relation is transitive and we have proved that is an
equivalence relation on Z.
17. Other Types of Relations. In this section, we focused on the properties of
a relation that are part of the definition of an equivalence relation. However,
there are other properties of relations that are of importance. We will study
two of these properties in this activity.
A relation R on a set A is a circular relation provided that for all x, y, and
z in A, if x R y and y R z, then z R x.
(a) Carefully explain what it means to say that a relation R on a set A is
not circular.
(b) Let A D f1; 2; 3g. Draw a directed graph of a relation on A that is
circular and draw a directed graph of a relation on A that is not circular.

7.3. Equivalence Classes 387
(c) Let A D f1; 2; 3g. Draw a directed graph of a relation on A that is
circular and not transitive and draw a directed graph of a relation on A
that is transitive and not circular.
(d) Prove the following proposition:
A relation R on a set A is an equivalence relation if and only if it
is reflexive and circular.
A relation R on a set A is an antisymmetric relation provided that for all
x; y 2 A, if x R y and y R x, then x D y.
(e) Carefully explain what it means to say that a relation on a set A is not
antisymmetric.
(f) Let A D f1; 2; 3g. Draw a directed graph of a relation on A that is
antisymmetric and draw a directed graph of a relation on A that is not
antisymmetric.
(g) Are the following propositions true or false? Justify all conclusions.
If a relation R on a set A is both symmetric and antisymmetric,
then R is transitive.
If a relation R on a set A is both symmetric and antisymmetric,
then R is reflexive.
7.3 Equivalence Classes
Preview Activity 1 (Sets Associated with a Relation)
As was indicated in Section 7.2, an equivalence relation on a set A is a relation
with a certain combination of properties (reflexive, symmetric, and transitive) that
allow us to sort the elements of the set into certain classes. This is done by means
of certain subsets of A that are associated with the elements of the set A. This will
be illustrated with the following example. Let A D fa; b; c; d; eg, and let R be the
relation on the set A defined as follows:
aR a b R b c R c d R d eR e
aR b b R a b R e e R b
aR e eR a c R d d R c
For each y 2 A, define the subset RŒy of A as follows:
RŒy D fx 2 A j x R yg :

That is, RŒy consists of those elements in A such that x R y. For example, using
y D a, we see that aR a, b R a,and e R a, and so RŒa D fa; b; eg.
1. Determine RŒb, RŒc, RŒd, and RŒe.
2. Draw a directed graph for the relation R and explain why R is an equivalence
relation on A.
3. Which of the sets RŒa, RŒb, RŒc, RŒd, and RŒe are equal?
4. Which of the sets RŒa, RŒb, RŒc, RŒd, and RŒe are disjoint?
As we will see in this section, the relationships between these sets is typical for an
equivalence relation. The following example will show how different this can be
for a relation that is not an equivalence relation.
Let A D fa; b; c; d; eg, and let S be the relation on the set A defined as follows:
b S b c S c d S d e S e
aS b aS d b S c
cS d d S c
5. Draw a digraph that represents the relation S on A. Explain why S is not an
equivalence relation on A.
For each y 2 A, define the subset SŒy of A as follows:
SŒy D fx 2 A j x S yg D fx 2 A j .x; y/ 2 Sg :
For example, using y D b, we see that SŒb D fa; bg since .a; b/ 2 S and
.b; b/ 2 S. In addition, we see that SŒa D ; since there is no x 2 A such that
.x; a/ 2 S.
6. Determine SŒc, SŒd, and SŒe.
7. Which of the sets SŒa, SŒb, SŒc, SŒd, and SŒe are equal?
8. Which of the sets SŒb, SŒc, SŒd, and SŒe are disjoint?

Preview Activity 2 (Congruence Modulo 3)
An important equivalence relation that we have studied is congruence modulo n
on the integers. We can also define subsets of the integers based on congruence
modulo n. We will illustrate this with congruence modulo 3. For example, we can
define C Œ0 to be the set of all integers a that are congruent to 0 modulo 3. That is,
C Œ0 D fa 2 Z j a 0 .mod 3/g:
Since an integer a is congruent to 0 modulo 3 if an only if 3 divides a, we can use
the roster method to specify this set as follows:
C Œ0 D f; : : :; 9; 6; 3; 0; 3; 6; 9; : : :g :
1. Use the roster method to specify each of the following sets:
(a) The set C Œ1 of all integers a that are congruent to 1 modulo 3. That is,
(b) The set C Œ2 of all integers a that are congruent to 2 modulo 3. That is,
(c) The set C Œ3 of all integers a that are congruent to 3 modulo 3. That is,
2. Now consider the three sets, C Œ0, C Œ1, and C Œ2.
(a) Determine the intersection of any two of these sets. That is, determine
C Œ0 C Œ1, C Œ0 C Œ2, and C Œ1 C Œ2.
(b) Let n D 734. What is the remainder when n is divided by 3? Which of
the three sets, if any, contains n D 734?
(c) Repeat Part (2b) for n D 79 and for n D 79.
(d) Do you think that C Œ0 [ C Œ1 [ C Œ2 D Z? Explain.
(e) Is the set C Œ3 equal to one of the sets C Œ0; C Œ1, or C Œ2?
(f) We can also define C Œ4 D fa 2 Z j a 4 .mod 3/g: Is this set equal
to any of the previous sets we have studied in this part? Explain.

The Definition of an Equivalence Class
We have indicated that an equivalence relation on a set is a relation with a certain
combination of properties (reflexive, symmetric, and transitive) that allow us to
sort the elements of the set into certain classes. We saw this happen in the preview
activities. We can now illustrate specifically what this means. For example, in
Preview Activity 2, we used the equivalence relation of congruence modulo 3 on Z
to construct the following three sets:
C Œ0 D fa 2 Z j a 0 .mod 3/g;
C Œ1 D fa 2 Z j a 1 .mod 3/g; and
The main results that we want to use now are Theorem 3.31 and Corollary 3.32 on
page 150. This corollary tells us that for any a 2 Z, a is congruent to precisely one
of the integers 0, 1, or 2. Consequently, the integer a must be congruent to 0, 1, or
2, and it cannot be congruent to two of these numbers. Thus
1. For each a 2 Z, a 2 C Œ0, a 2 C Œ1, or a 2 C Œ2; and
2. C Œ0 C Œ1 D ;, C Œ0 C Œ2 D ; , and C Œ1 C Œ2 D ;.
This means that the relation of congruence modulo 3 sorts the integers into
three distinct sets, or classes, and that each pair of these sets have no elements in
common. So if we use a rectangle to represent Z, we can divide that rectangle
into three smaller rectangles, corresponding to C Œ0, C Œ1, and C Œ2, and we might
picture this situation as follows:
The Integers
C Œ0 consisting of
all integers with a
remainder of 0 when
divided by 3
C Œ1 consisting of
all integers with a
remainder of 1 when
divided by 3
C Œ2 consisting of
all integers with a
remainder of 2 when
divided by 3
Each integer is in exactly one of the three sets C Œ0, C Œ1, or C Œ2, and two
integers are congruent modulo 3 if and only if they are in the same set. We will see
that, in a similar manner, if n is any natural number, then the relation of congruence
modulo n can be used to sort the integers into n classes. We will also see that in
general, if we have an equivalence relation R on a set A, we can sort the elements
of the set A into classes in a similar manner.

Definition. Let be an equivalence relation on a nonempty set A. For each
a 2 A, the equivalence class of a determined by is the subset of A, denoted
by Œa, consisting of all the elements of A that are equivalent to a. That is,
Œa D fx 2 A j x ag:
We read Œa as “the equivalence class of a” or as “bracket a.”
Notes
1. We use the notation Œa when only one equivalence relation is being used.
If there is more than one equivalence relation, then we need to distinguish
between the equivalence classes for each relation. We often use something
like Œa, or if R is the name of the relation, we can use RŒa or ŒaR for the
equivalence class of a determined by R. In any case, always remember that
when we are working with any equivalence relation on a set A if a 2 A, then
the equivalence class Œa is a subset of A.
2. We know that each integer has an equivalence class for the equivalence rela-
tion of congruence modulo 3. But as we have seen, there are really only three
distinct equivalence classes. Using the notation from the definition, they are:
Œ0 D fa 2 Z j a 0 .mod 3/g ; Œ1 D fa 2 Z j a 1 .mod 3/g ; and
Œ2 D fa 2 Z j a 2 .mod 3/g :
Progress Check 7.12 (Equivalence Classes from Preview Activity 1)
Without using the terminology at that time, we actually determined the equivalence
classes of the equivalence relation R in Preview Activity 1. What are the distinct
equivalence classes for this equivalence relation?
Congruence Modulo n and Congruence Classes
In Preview Activity 2, we used the notation C Œk for the set of all integers that are
congruent to k modulo 3. We could have used a similar notation for equivalence
classes, and this would have been perfectly acceptable. However, the notation Œa
is probably the most common notation for the equivalence class of a. We will now
use this same notation when dealing with congruence modulo n when only one
congruence relation is under consideration.

Definition. Let n 2 N. Congruence modulo n is an equivalence relation on
Z. So for a 2 Z,
Œa D fx 2 Z j x a .mod n/g:
In this case, Œa is called the congruence class of a modulo n.
We have seen that congruence modulo 3 divides the integers into three distinct
congruence classes. Each congruence class consists of those integers with the same
remainder when divided by 3. In a similar manner, if we use congruence modulo
2, we simply divide the integers into two classes. One class will consist of all the
integers that have a remainder of 0 when divided by 2, and the other class will
consist of all the integers that have a remainder of 1 when divided by 2. That is,
congruence modulo 2 simply divides the integers into the even and odd integers.
Progress Check 7.13 (Congruence Modulo 4)
Determine all of the distinct congruence classes for the equivalence relation of
congruence modulo 4 on the integers. Specify each congruence class using the
roster method.
Properties of Equivalence Classes
As we have seen, in Preview Activity 1, the relation R was an equivalence relation.
For that preview activity, we used RŒy to denote the equivalence class of y 2 A,
and we observed that these equivalence classes were either equal or disjoint.
However, in Preview Activity 1, the relation S was not an equivalence relation,
and hence we do not use the term “equivalence class” for this relation. We should
note, however, that the sets SŒy were not equal and were not disjoint. This exhibits
one of the main distinctionsbetween equivalence relations and relations that are not
equivalence relations.
In Theorem 7.14, we will prove that if is an equivalence relation on the set A,
then we can “sort” the elements of A into distinct equivalence classes. The prop-
erties of equivalence classes that we will prove are as follows: (1) Every element
of A is in its own equivalence class; (2) two elements are equivalent if and only
if their equivalence classes are equal; and (3) two equivalence classes are either
identical or they are disjoint.
Theorem 7.14. Let A be a nonempty set and let be an equivalence relation on
the set A. Then,
1. For each a 2 A, a 2 Œa.

2. For each a; b 2 A, a b if and only if Œa D Œb.
3. For each a; b 2 A, Œa D Œb or Œa Œb D ;.
Proof. Let A be a nonempty set and assume that is an equivalence relation on
A. To prove the first part of the theorem, let a 2 A. Since is an equivalence
relation on A, it is reflexive on A. Thus, a a, and we can conclude that a 2 Œa.
The second part of this theorem is a biconditional statement. We will prove
it by proving two conditional statements. We will first prove that if a b, then
Œa D Œb. So let a; b 2 A and assume that a b. We will now prove that the two
sets Œa and Œb are equal. We will do this by proving that each is a subset of the
other.
First, assume that x 2 Œa. Then, by definition, x a. Since we have assumed
that a b, we can use the transitive property of to conclude that x b, and this
means that x 2 Œb. This proves that Œa Œb.
We now assume that y 2 Œb. This means that y b, and hence by the
symmetric property, that b y. Again, we are assuming that a b. So we have
a b and b y.
We use the transitiveproperty to conclude that a y and then, using the symmetric
property, we conclude that y a. This proves that y 2 Œa and, hence, that
Œb Œa. This means that we can conclude that if a b, then Œa D Œb.
We must now prove that if Œa D Œb, then a b. Let a; b 2 A and assume
that Œa D Œb. Using the first part of the theorem, we know that a 2 Œa and since
the two sets are equal, this tells us that a 2 Œb. Hence by the definition of Œb, we
conclude that a b. This completes the proof of the second part of the theorem.
For the third part of the theorem, let a; b 2 A. Since this part of the theorem is
a disjunction, we will consider two cases: Either
Œa Œb D ; or Œa Œb ¤ ;:
In the case where Œa Œb D ;, the first part of the disjunction is true, and
hence there is nothing to prove. So we assume that Œa Œb ¤ ; and will show
that Œa D Œb. Since Œa Œb ¤ ;, there is an element x in A such that
x 2 Œa Œb:
This means that x 2 Œa and x 2 Œb. Consequently, x a and x b, and so
we can use the first part of the theorem to conclude that Œx D Œa and Œx D Œb.
Hence, Œa D Œb, and we have proven that Œa D Œb or Œa Œb D ;.

Theorem 7.14 gives the primary properties of equivalence classes. Conse-
quences of these properties will be explored in the exercises. The following table
restates the properties in Theorem 7.14 and gives a verbal description of each one.
Formal Statement from
Theorem 7.14
Verbal Description
For each a 2 A, a 2 Œa. Every element of A is in its own
equivalence class.
For each a; b 2 A, a b if and
only if Œa D Œb.
Two elements of A are equivalent if
and only if their equivalence classes
are equal.
For each a; b 2 A, Œa D Œb or
Œa Œb D ;.
Any two equivalence classes are
either equal or they are disjoint.
This means that if two equivalence
classes are not disjoint then they
must be equal.
Progress Check 7.15 (Equivalence Classes)
4 for each x 2 R. Define a relation
on R as follows:
For a; b 2 R, a b if and only if f.a/ D f.b/.
In Exercise (6) of Section 7.2, we proved that is an equivalence relation on
R. Consequently, each real number has an equivalence class. For this equivalence
relation,
1. Determine the equivalence classes of 5, 5, 10, 10, , and .
2. Determine the equivalence class of 0.
3. If a 2 R, use the roster method to specify the elements of the equivalence
class Œa.
The results of Theorem 7.14 are consistent with all the equivalence relations
studied in the preview activities and in the progress checks. Since this theorem
applies to all equivalence relations, it applies to the relation of congruence modulo
n on the integers. Because of the importance of this equivalence relation, these
results for congruence modulo n are given in the following corollary.

Corollary 7.16. Let n 2 N. For each a 2 Z, let Œa represent the congruence class
of a modulo n.
1. For each a 2 Z, a 2 Œa.
2. For each a; b 2 Z, a b .mod n/ if and only if Œa D Œb.
3. For each a; b 2 Z, Œa D Œb or Œa Œb D ;.
For the equivalence relation of congruence modulo n, Theorem 3.31 and Corol-
lary 3.32 tell us that each integer is congruent to its remainder when divided by n,
and that each integer is congruent modulo n to precisely one of one of the integers
0; 1; 2; : : :; n 1. This means that each integer is in precisely one of the congru-
ence classes Œ0, Œ1, Œ2; : : :; Œn 1. Hence, Corollary 7.16 gives us the following
result.
Corollary 7.17. Let n 2 N. For each a 2 Z, let Œa represent the congruence class
of a modulo n.
1. Z D Œ0 [ Œ1 [ Œ2 [ [ Œn 1
2. For j; k 2 f0; 1; 2; : ::; n 1g, if j ¤ k, then Œj Œk D ;.
Partitions and Equivalence Relations
A partition of a set A is a collection of subsets of A that “breaks up” the set A into
disjoint subsets. Technically, each pair of distinct subsets in the collection must
be disjoint. We then say that the collection of subsets is pairwise disjoint. We
introduce the following formal definition.
Definition. Let A be a nonempty set, and let C be a collection of subsets of
A. The collection of subsets C is a partition of A provided that
1. For each V 2 C, V ¤ ;.
2. For each x 2 A, there exists a V 2 C such that x 2 V.
3. For every V; W 2 C, V D W or V W D ;.

There is a close relation between partitions and equivalence classes since the
equivalence classes of an equivalence relation form a partition of the underlying
set, as will be proven in Theorem 7.18. The proof of this theorem relies on the
results in Theorem 7.14.
Theorem 7.18. Let be an equivalence relation on the nonempty set A. Then the
collection C of all equivalence classes determined by is a partition of the set A.
Proof. Let be an equivalence relation on the nonempty set A, and let C be the
collection of all equivalence classes determined by . That is,
C D fŒa j a 2 Ag:
We will use Theorem 7.14 to prove that C is a partition of A.
Part (1) of Theorem 7.14 states that for each a 2 A, a 2 Œa. In terms of the
equivalence classes, this means that each equivalence class is nonempty since each
element of A is in its own equivalence class. Consequently, C, the collection of
all equivalence classes determined by , satisfies the first two conditions of the
definition of a partition.
We must now show that the collection C of all equivalence classes determined
by satisfies the third condition for being a partition. That is, we need to show
that any two equivalence classes are either equal or are disjoint. However, this is
exactly the result in Part (3) of Theorem 7.14.
Hence, we have proven that the collection C of all equivalence classes deter-
mined by is a partition of the set A.
Note: Theorem 7.18 has shown us that if is an equivalence relation on a nonempty
set A, then the collection of the equivalence classes determined by form a parti-
tion of the set A.
This process can be reversed. This means that given a partition C of a nonempty
set A, we can define an equivalence relation on A whose equivalence classes are
precisely the subsets of A that form the partition. This will be explored in Exer-
cise (12).
Exercises 7.3
?
1. Let A D fa; b; c; d; eg and let be the relation on A that is represented by
the directed graph in Figure 7.4.

d e
c
b
a
Figure 7.4: Directed Graph for the Relation in Exercise (1)
Prove that is an equivalence relation on the set A, and determine all of the
equivalence classes determined by this equivalence relation.
?
2. Let A D fa; b; c; d; e; f g, and assume that is an equivalence relation on
A. Also assume that it is known that
a b a 6 c e f
a d a 6 f e 6 c.
Draw a complete directed graph for the equivalence relation on the set
A, and then determine all of the equivalence classes for this equivalence
relation.
? 3. Let A D f0; 1; 2; 3; : : :; 999; 1000g. Define the relation R on A as follows:
For x; y 2 A, x R y if and only if x and y have the same number of
digits.
Prove that R is an equivalence relation on the set A and determine all of the
distinct equivalence classes determined by R.
4. Determine all of the congruence classes for the relation of congruence mod-
ulo 5 on the set of integers.
5. Let Z9 D f0; 1; 2; 3; 4; 5; 6; 7; 8g.
?
(a) Define the relation on Z9 as follows: For all a; b 2 Z9, a b if and
only if a2 b2 .mod 9/. Prove that is an equivalence relation on Z9
and determine all of the distinct equivalence classes of this equivalence
relation.

(b) Define the relation on Z9 as follows: For all a; b 2 Z9, a b if and
only if a3
b3
.mod 9/. Prove that is an equivalence relation on Z9
and determine all of the distinct equivalence classes of this equivalence
relation.
6. Define the relation on Q as follows: For a; b 2 Q, a b if and only if
a b 2 Z. In Progress Check 7.9 of Section 7.2, we showed that the relation
is an equivalence relation on Q. Also, see Exercise (9) in Section 7.2.
?
(a) Prove that

5
7

D

m C
5
7
ˇ
ˇ
ˇ
ˇ m 2 Z

.
(b) If a 2 Z, then what is the equivalence class of a?
(c) If a 2 Z, prove that there is a bijection from Œa to

5
7

.
7. Define the relation on R as follows:
For x; y 2 R, x y if and only if x y 2 Q.
(a) Prove that is an equivalence relation on R.
(b) List four different real numbers that are in the equivalence class of
p
2.
(c) If a 2 Q, what is the equivalence class of a?
(d) Prove that
hp
2
i
D
n
r C
p
2 j r 2 Q
o
.
(e) If a 2 Q, prove that there is a bijection from Œa to
hp
2
i
.
8. Define the relation on Z as follows: For a; b 2 Z, a b if and only if
2a C 3b 0 .mod 5/. The relation is an equivalence relation on Z. (See
Exercise (13) in Section 7.2). Determine all the distinct equivalence classes
for this equivalence relation.
9. Let A D Z .Z f0g/. That is, A D f.a; b/ 2 Z Z j b ¤ 0g. Define the
relation on A as follows:
For .a; b/ ; .c; d/ 2 A, .a; b/ .c; d/ if and only if ad D bc.
? (a) Prove that is an equivalence relation on A.
(b) Why was it necessary to include the restriction that b ¤ 0 in the defi-
nition of the set A?
?
(c) Determine an equation that gives a relation between a and b if .a; b/ 2
A and .a; b/ .2; 3/.

(d) Determine at least four different elements in Œ.2; 3/, the equivalence
class of .2; 3/.
(e) Use set builder notation to describe Œ.2; 3/, the equivalence class of
.2; 3/.
10. For .a; b/ ; .c; d/ 2 R R, define .a; b/ .c; d/ if and only if a2 C b2 D
c2
C d2
. In Exercise (15) of Section 7.2, we proved that is an equivalence
relation on R R.
(a) Determine the equivalence class of .0; 0/.
(b) Use set builder notation (and do not use the symbol ) to describe the
equivalence class of .2; 3/ and then give a geometric description of this
equivalence class.
(c) Give a geometric description of a typical equivalence class for this
equivalence relation.
(d) Let R D fx 2 R j x 0g. Prove that there is a one-to-one correspon-
dence (bijection) between R
and the set of all equivalence classes for
this equivalence relation.
11. Let A be a nonempty set and let be an equivalence relation on A. Prove
each of the following:
(a) For each a; b 2 A, a œ b if and only if Œa Œb D ;.
(b) For each a; b 2 A, if Œa ¤ Œb, then Œa Œb D ;.
(c) For each a; b 2 A, if Œa Œb ¤ ; then Œa D Œb.
12. A Partition Defines an Equivalence Relation. Let A D fa; b; c; d; eg and
let C D ffa; b; cg ; fd; egg.
(a) Explain why C is a partition of A.
Define a relation on A as follows: For x; y 2 A, x y if and only if there
exists a set U in C such that x 2 U and y 2 U.
(b) Prove that is an equivalence relation on the set A, and then deter-
mine all the equivalence classes for . How does the collection of all
equivalence classes compare to C?

What we did for the specific partition in Part (12b) can be done for any
partition of a set. So to generalize Part (12b), we let A be a nonempty set
and let C be a partition of A. We then define a relation on A as follows:
For x; y 2 A, x y if and only if there exists a set U in C such that
x 2 U and y 2 U.
(c) Prove that is an equivalence relation on the set A.
(d) Let a 2 A and let U 2 C such that a 2 U . Prove that Œa D U .
13. Equivalence Relations on a Set of Matrices. The following exercises re-
quire a knowledge of elementary linear algebra. We let Mn;n .R/ be the set
of all n by n matrices with real number entries.
(a) Define a relation on Mn;n.R/ as follows: For all A; B 2 Mn;n.R/,
A B if and only if there exists an invertible matrix P in Mn;n.R/
such that B D PAP 1. Is an equivalence relation on Mn;n.R/?
(b) Define a relation R on Mn;nR/ as follows: For all A; B 2 Mn;n.R/,
AR B if and only if det.A/ D det.B/. Is R an equivalence relation on
Mn;n.R/? Justify your conclusion.
(c) Let be an equivalence relation on R. Define a relation on Mn;n.R/
as follows: For all A; B 2 Mn;n.R/, A B if and only if det.A/
det.B/. Is an equivalence relation on Mn;n.R/? Justify your con-
clusion.
7.4 Modular Arithmetic
Preview Activity 1 (Congruence Modulo 6)
For this preview activity, we will only use the relation of congruence modulo 6 on
the set of integers.
1. Find five different integers a such that a 3 .mod 6/ and find five different
integers b such that b 4 .mod 6/. That is, find five different integers in
Œ3, the congruence class of 3 modulo 6 and five different integers in Œ4, the
congruence class of 4 modulo 6.
2. Calculate s D aCb using several values of a in Œ3 and several values of b in
Œ4 from Part (1). For each sum s that is calculated, find r so that 0 r 6
and s r .mod 6/. What do you observe?

7.4. Modular Arithmetic 401
3. Calculate p D a b using several values of a in Œ3 and several values of
b in Œ4 from Part (1). For each product p that is calculated, find r so that
0 r 6 and p r .mod 6/. What do you observe?
4. Calculate q D a2
using several values of a in Œ3 from Part (1). For each
product q that is calculated, find r so that 0 r 6 and q r .mod 6/.
Preview Activity 2 (The Remainder When Dividing by 9)
If a and b are integers with b 0, then from the Division Algorithm, we know
that there exist unique integers q and r such that
In this activity, we are interested in the remainder r. Notice that r D a bq. So,
given a and b, if we can calculate q, then we can calculate r.
We can use the “int” function on a calculator to calculate q. [The “int” function
is the “greatest integer function.” If x is a real number, then int.x/ is the greatest
integer that is less than or equal to x.]
So, in the context of the Division Algorithm, q D int
a
b

. Consequently,
r D a b int
a
b

:
If n is a positive integer, we will let s .n/ denote the sum of the digits of n. For
example, if n D 731, then
s.731/ D 7 C 3 C 1 D 11:
For each of the following values of n, calculate
The remainder when n is divided by 9, and
The value of s.n/ and the remainder when s.n/ is divided by 9.
1. n D 498
2. n D 7319
3. n D 4672
4. n D 9845
5. n D 51381
6. n D 305877

The Integers Modulo n
Let n 2 N. Since the relation of congruence modulo n is an equivalence relation
on Z, we can discuss its equivalence classes. Recall that in this situation, we refer
to the equivalence classes as congruence classes.
Definition. Let n 2 N. The set of congruence classes for the relation of
congruence modulo n on Z is the set of integers modulo n, or the set of
integers mod n. We will denote this set of congruence classes by Zn.
Corollary 7.17 tells us that
Z D Œ0 [ Œ1 [ Œ2 [ [ Œn 1:
In addition, we know that each integer is congruent to precisely one of the integers
0; 1; 2; : : :; n 1. This tells us that one way to represent Zn is
Zn D fŒ0; Œ1; Œ2; : : :Œn 1g :
Consequently, even though each integer has a congruence class, the set Zn has only
n distinct congruence classes.
The set of integers Z is more than a set. We can add and multiply integers.
That is, there are the arithmetic operations of addition and multiplication on the set
Z, and we know that Z is closed with respect to these two operations.
One of the basic problems dealt with in modern algebra is to determine if the
arithmetic operations on one set “transfer” to a related set. In this case, the related
set is Zn. For example, in the integers modulo 5, Z5, is it possible to add the
congruence classes Œ4 and Œ2 as follows?
Œ4 ˚ Œ2 D Œ4 C 2
D Œ6
D Œ1:
We have used the symbol ˚ to denote addition in Z5 so that we do not confuse
it with addition in Z. This looks simple enough, but there is a problem. The
congruence classes Œ4 and Œ2 are not numbers, they are infinite sets. We have to
make sure that we get the same answer no matter what element of Œ4 we use and
no matter what element of Œ2 we use. For example,
9 4 .mod 5/ and so Œ9 D Œ4: Also,
7 2 .mod 5/ and so Œ7 D Œ2:

Do we get the same result if we add Œ9 and Œ7 in the way we did when we added
Œ4 and Œ2? The following computation confirms that we do:
Œ9 ˚ Œ7 D Œ9 C 7
D Œ16
D Œ1:
This is one of the ideas that was explored in Preview Activity 1. The main differ-
ence is that in this preview activity, we used the relation of congruence, and here
we are using congruence classes. All of the examples in Preview Activity 1 should
have illustrated the properties of congruence modulo 6 in the following table. The
left side shows the properties in terms of the congruence relation and the right side
shows the properties in terms of the congruence classes.
If a 3 .mod 6/ and b 4 .mod 6/, then
.a C b/ .3 C 4/ .mod 6/;
.a b/ .3 4/ .mod 6/.
If Œa D Œ3 and Œb D Œ4 in
Z6, then
Œa C b D Œ3 C 4;
Œa b D Œ3 4.
These are illustrations of general properties that we have already proved in Theo-
rem 3.28. We repeat the statement of the theorem here because it is so important
for defining the operations of addition and multiplication in Zn.
Theorem 3.28 Let n be a natural number and let a; b; c; and d be integers. Then
1. If a b .mod n/ and c d .mod n/, then .a C c/ .b C d/ .mod n/.
2. If a b .mod n/ and c d .mod n/, then ac bd .mod n/.
3. If a b .mod n/ and m 2 N, then am
bm
.mod n/.
Since x y .mod n/ if and only if Œx D Œy, we can restate the result of this
Theorem 3.28 in terms of congruence classes in Zn.
Corollary 7.19. Let n be a natural number and let a; b; c and d be integers. Then,
in Zn,

1. If Œa D Œb and Œc D Œd, then Œa C c D Œb C d.
2. If Œa D Œb and Œc D Œd, then Œa c D Œb d.
3. If Œa D Œb and m 2 N, then Œam D Œbm.
Because of Corollary 7.19, we know that the following formal definition of ad-
dition and multiplication of congruence classes in Zn is independent of the choice
of the elements we choose from each class. We say that these definitions of addi-
tion and multiplication are well defined.
Definition. Let n 2 N. Addition and multiplication in Zn are defined as
follows: For Œa; Œc 2 Zn,
Œa ˚ Œc D Œa C c and Œa ˇ Œc D Œac:
The term modular arithmetic is used to refer to the operations of addition
and multiplication of congruence classes in the integers modulo n.
So if n 2 N, then we have an addition and multiplication defined on Zn, the
integers modulo n.
Always remember that for each of the equations in the definitions, the oper-
ations on the left, ˚ and ˇ, are the new operations that are being defined. The
operations on the right side of the equations .C and / are the known operations of
addition and multiplication in Z.
Since Zn is a finite set, it is possible to construct addition and multiplication
tables for Zn. In constructing these tables, we follow the convention that all sums
and products should be in the form Œr, where 0 r n. For example, in Z3, we
see that by the definition, Œ1 ˚ Œ2 D Œ3, but since 3 0 .mod 3/, we see that
Œ3 D Œ0 and so we write
Œ1 ˚ Œ2 D Œ3 D Œ0:
Similarly, by definition, Œ2 ˇ Œ2 D Œ4, and in Z3, Œ4 D Œ1. So we write
Œ2 ˇ Œ2 D Œ4 D Œ1:
The complete addition and multiplication tables for Z3 are
˚ Œ0 Œ1 Œ2 ˇ Œ0 Œ1 Œ2
Œ0 Œ0 Œ1 Œ2 Œ0 Œ0 Œ0 Œ0
Œ1 Œ1 Œ2 Œ0 Œ1 Œ0 Œ1 Œ2
Œ2 Œ2 Œ0 Œ1 Œ2 Œ0 Œ2 Œ1

Progress Check 7.20 (Modular Arithmetic in Z2, Z5, and Z6)
1. Construct addition and multiplication tables for Z2, the integers modulo 2.
2. Verify that the following addition and multiplication tables for Z5 are cor-
rect.
˚ Œ0 Œ1 Œ2 Œ3 Œ4 ˇ Œ0 Œ1 Œ2 Œ3 Œ4
Œ0 Œ0 Œ1 Œ2 Œ3 Œ4 Œ0 Œ0 Œ0 Œ0 Œ0 Œ0
Œ1 Œ1 Œ2 Œ3 Œ4 Œ0 Œ1 Œ0 Œ1 Œ2 Œ3 Œ4
Œ2 Œ2 Œ3 Œ4 Œ0 Œ1 Œ2 Œ0 Œ2 Œ4 Œ1 Œ3
Œ3 Œ3 Œ4 Œ0 Œ1 Œ2 Œ3 Œ0 Œ3 Œ1 Œ4 Œ2
Œ4 Œ4 Œ0 Œ1 Œ2 Œ3 Œ4 Œ0 Œ4 Œ3 Œ2 Œ1
3. Construct complete addition and multiplication tables for Z6.
4. In the integers, the following statement is true. We sometimes call this the
zero product property for the integers.
For all a; b 2 Z, if a b D 0, then a D 0 or b D 0.
Write the contrapositive of the conditional statement in this property.
(a) For all Œa; Œb 2 Z5, if Œa ˇ Œb D Œ0, then Œa D Œ0 or Œb D Œ0.
(b) For all Œa; Œb 2 Z6, if Œa ˇ Œb D Œ0, then Œa D Œ0 or Œb D Œ0.
Divisibility Tests
Congruence arithmetic can be used to proof certain divisibility tests. For example,
you may have learned that a natural number is divisible by 9 if the sum of its
digits is divisible by 9. As an easy example, note that the sum of the digits of
5823 is equal to 5 C 8 C 2 C 3 D 18, and we know that 18 is divisible by 9.
It can also be verified that 5823 is divisible by 9. (The quotient is 647.) We can
actually generalize this property by dealing with remainders when a natural number
is divided by 9.
Let n 2 N and let s.n/ denote the sum of the digits of n. For example, if
n D 7319, then s.7319/ D 7 C 3 C 1 C 9 D 20. In Preview Activity 2, we saw
that

7319 2 .mod 9/ and 20 2 .mod 9/.
In fact, for every example in Preview Activity 2, we saw that n and s.n/ were
congruent modulo 9 since they both had the same remainder when divided by 9.
The concepts of congruence and congruence classes can help prove that this is
always true.
We will use the case of n D 7319 to illustrate the general process. We must
use our standard place value system. By this, we mean that we will write 7319 as
follows:
7319 D 7 103

C 3 102

C 1 101

C 9 100

: (1)
The idea is to now use the definition of addition and multiplication in Z9 to convert
equation (1) to an equation in Z9. We do this as follows:
Œ7319 D Œ 7 103

C 3 102

C 1 101

C 9 100

D Œ7 103
˚ Œ3 102
˚ Œ1 101
˚ Œ9 100

D Œ7 ˇ Œ103

˚ Œ3 ˇ Œ102

˚ Œ1 ˇ Œ101

˚ .Œ9 ˇ Œ1/ : (2)
Since 103
1 .mod 9/, 102
1 .mod 9/ and 10 1 .mod 9/, we can conclude
that Œ103 D Œ1, Œ102 D Œ1 and Œ10 D Œ1. Hence, we can use these facts and
equation (2) to obtain
Œ7319 D Œ7 ˇ Œ103

˚ Œ3 ˇ Œ102

˚ .Œ1 ˇ Œ10/ ˚ .Œ9 ˇ Œ1/
D .Œ7 ˇ Œ1/ ˚ .Œ3 ˇ Œ1/ ˚ .Œ1 ˇ Œ1/ ˚ .Œ9 ˇ Œ1/
D Œ7 ˚ Œ3 ˚ Œ1 ˚ Œ9
D Œ7 C 3 C 1 C 9: (3)
Equation (3) tells us that 7319 has the same remainder when divided by 9 as the
sum of its digits. It is easy to check that the sum of the digits is 20 and hence has a
remainder of 2. This means that when 7319 is divided by 9, the remainder is 2.
To prove that any natural number has the same remainder when divided by 9
as the sum of its digits, it is helpful to introduce notation for the decimal represen-
tation of a natural number. The notation we will use is similar to the notation for
the number 7319 in equation (1).
In general, if n 2 N, and n D akak 1 a1a0 is the decimal representation of
n, then
n D

ak 10k

C

ak 1 10k 1

C C a1 101

C a0 100

:

This can also be written using summation notation as follows:
n D
k
X
j D0

aj 10j

:
Using congruence classes for congruence modulo 9, we have
Œn D Œ

ak 10k

C

ak 1 10k 1

C C a1 101

C a0 100

D Œak 10k
˚ Œak 1 10k 1
˚ ˚ Œa1 101
˚ Œa0 100

D

Œak ˇ Œ10k

˚

Œak 1 ˇ Œ10k 1

˚
˚ Œa1 ˇ Œ101

˚ Œa0 ˇ Œ100

: (4)
One last detail is needed. It is given in Proposition 7.21. The proof by mathemat-
ical induction is Exercise (6).
Proposition 7.21. If n is a nonnegative integer, then 10n
1 .mod 9/, and hence
for the equivalence relation of congruence modulo 9, Œ10n
D Œ1.
If we let s.n/ denote the sum of the digits of n, then
s.n/ D ak C ak 1 C C a1 C a0;
Now using equation (4) and Proposition 7.21, we obtain
Œn D .Œak ˇ Œ1/ ˚ .Œak 1 ˇ Œ1/ ˚ ˚ .Œa1 ˇ Œ1/ ˚ .Œa0 ˇ Œ1/
D Œak ˚ Œak 1 ˚ ˚ Œa1 ˚ Œa0
D Œak C ak 1 C C a1 C a0:
D Œs.n/:
This completes the proof of Theorem 7.22.
Theorem 7.22. Let n 2 N and let s.n/ denote the sum of the digits of n. Then
1. Œn D Œs.n/, using congruence classes modulo 9.
2. n s.n/ .mod 9/.
3. 9 j n if and only if 9 j s.n/.
Part (3) of Theorem 7.22 is called a divisibility test. If gives a necessary and
sufficient conditionfor a natural number to be divisibleby 9. Other divisibilitytests
will be explored in the exercises. Most of these divisibility tests can be proved in a
manner similar to the proof of the divisibility test for 9.

Exercises 7.4
1. ?
(a) Complete the addition and multiplication tables for Z4.
?
(b) Complete the addition and multiplication tables for Z7.
(c) Complete the addition and multiplication tables for Z8.
2. The set Zn contains n elements. One way to solve an equation in Zn is
to substitute each of these n elements in the equation to check which ones
are solutions. In Zn, when parentheses are not used, we follow the usual
order of operations, which means that multiplications are done first and then
additions. Solve each of the following equations:
?
(a) Œx2
D Œ1 in Z4
(b) Œx2 D Œ1 in Z8
(c) Œx4 D Œ1 in Z5
(d) Œx2 ˚ Œ3 ˇ Œx D Œ3 in Z6
?
(e) Œx2
˚ Œ1 D Œ0 in Z5
(f) Œ3 ˇ Œx ˚ Œ2 D Œ0 in Z5
? (g) Œ3 ˇ Œx ˚ Œ2 D Œ0 in Z6
(h) Œ3 ˇ Œx ˚ Œ2 D Œ0 in Z9
?
3. In each case, determine if the statement is true or false.
(a) For all Œa 2 Z6, if Œa ¤ Œ0, then there exists a Œb 2 Z6 such that
Œa ˇ Œb D Œ1.
(b) For all Œa 2 Z5, if Œa ¤ Œ0, then there exists a Œb 2 Z5 such that
Œa ˇ Œb D Œ1.
4. In each case, determine if the statement is true or false.
(a) For all Œa; Œb 2 Z6, if Œa ¤ Œ0 and Œb ¤ Œ0, then Œa ˇ Œb ¤ Œ0.
(b) For all Œa; Œb 2 Z5, if Œa ¤ Œ0 and Œb ¤ Œ0, then Œa ˇ Œb ¤ Œ0.
5. ? (a) Prove the following proposition:
For each Œa 2 Z5, if Œa ¤ Œ0, then Œa2
D Œ1 or Œa2
D Œ4.
(b) Does there exist an integer a such that a2 D 5; 158; 232; 468; 953; 153?
Use your work in Part (a) to justify your conclusion. Compare to Exer-
cise (11) in Section 3.5.
6. Use mathematical induction to prove Proposition 7.21.
If n is a nonnegative integer, then 10n
1 .mod 9/, and hence for the
equivalence relation of congruence modulo 9, Œ10n
D Œ1.

7. Use mathematical induction to prove that if n is a nonnegative integer, then
10n
1 .mod 3/. Hence, for congruence classes modulo 3, if n is a non-
negative integer, then Œ10n D Œ1.
8. Let n 2 N and let s.n/ denote the sum of the digits of n. So if we write
n D

ak 10k

C

ak 1 10k 1

C C a1 101

C a0 100

;
then s.n/ D ak C ak 1 C C a1 C a0. Use the result in Exercise (7) to
help prove each of the following:
(a) Œn D Œs.n/, using congruence classes modulo 3.
(b) n s.n/ .mod 3/.
(c) 3 j n if and only if 3 j s.n/.
9. Use mathematical induction to prove that if n is an integer and n 1, then
10n
0 .mod 5/. Hence, for congruence classes modulo 5, if n is an integer
and n 1, then Œ10n D Œ0.
10. Let n 2 N and assume
n D

ak 10k

C

ak 1 10k 1

C C a1 101

C a0 100

:
Use the result in Exercise (9) to help prove each of the following:
(a) Œn D Œa0, using congruence classes modulo 5.
(b) n a0 .mod 5/.
(c) 5 j n if and only if 5 j a0.
10n
0 .mod 4/. Hence, for congruence classes modulo 4, if n is an integer
and n 2, then Œ10n D Œ0.
n D

ak 10k

C

ak 1 10k 1

C C a1 101

C a0 100

:
(a) Œn D Œ10a1 C a0, using congruence classes modulo 4.
(b) n .10a1 C a0/ .mod 4/.

(c) 4 j n if and only if 4 j .10a1 C a0/.
10n 0 .mod 8/. Hence, for congruence classes modulo 8, if n is an integer
and n 3, then Œ10n
D Œ0.
n D

ak 10k

C

ak 1 10k 1

C C a1 101

C a0 100

:
Use the result in Exercise (13) to help develop a divisibility test for 8. Prove
that your divisibility test is correct.
15. Use mathematical induction to prove that if n is a nonnegative integer then
10n . 1/n
.mod 11/. Hence, for congruence classes modulo 11, if n is a
nonnegative integer, then Œ10n
D Œ. 1/n
.
n D

ak 10k

C

ak 1 10k 1

C C a1 101

C a0 100

:
(a) n
k
P
j D0
. 1/j
aj .mod 11/.
(b) Œn D Œ
k
P
j D0
. 1/j
aj , using congruence classes modulo 11.
(c) 11 divides n if and only if 11 divides
k
P
j D0
. 1/j
aj .
17. ?
(a) Prove the following proposition:
For all Œa; Œb 2 Z3. if Œa2 C Œb2 D Œ0, then Œa D 0 and
Œb D Œ0.
(b) Use Exercise (17a) to prove the following proposition:
Let a; b 2 Z. If a2 C b2

0 .mod 3/, then a 0 .mod 3/
and b 0 .mod 3/.
(c) Use Exercise (17b) to prove the following proposition:
For all a; b 2 Z, if 3 divides a2
C b2

, then 3 divides a and 3
divides b.

For each a 2 Z, if there exist integers b and c such that a D b4 C c4,
then the units digit of a must be 0, 1, 2, 5, 6, or 7.
Let n 2 Z. If n is odd, then 8 j n2
1

. Hint: What are the possible
values of n (mod 8)?
Let n 2 N. If n 7 .mod 8/, then n is not the sum of three squares.
That is, there do not exist natural numbers a, b, and c such that
n D a2
C b2
C c2
.
21. Using Congruence Modulo 4. The set Zn is a finite set, and hence one
way to prove things about Zn is to simply use the n elements in Zn as the n
cases for a proof using cases. For example, if n 2 Z, then in Z4, Œn D Œ0,
Œn D Œ1, Œn D Œ2, or Œn D Œ3.
(a) Prove that if n 2 Z, then in Z4, Œn2
D Œ0 or Œn2
D Œ1. Use this to
conclude that in Z4, Œn2 D Œ0 or Œn2 D Œ1.
(b) Translate the equations

n2

D Œ0 and

n2

D Œ1 in Z4 into congru-
ences modulo 4.
(c) Use a result in Exercise (12) to determine the value of r so that r 2 Z,
0 r 3, and
104 257 833 259 r .mod 4/:
That is, Œ104 257 833 259 D Œr in Z4.
(d) Is the natural number 104 257 833 259 a perfect square? Justify your
conclusion.

Relation from A to B, page 364
Relation on A, page 364
Domain of a relation, page 364
Range of a relation, page 364
Inverse of a relation, page 373
Reflexive relation, page 375
Symmetric relation, page 375
Transitive relation, page 375
Equivalence relation, page 378
Equivalence class, page 391
Congruence class, page 392
Partition of a set, page 395
Integers modulo n, page 402
Addition in Zn, page 404
Multiplication in Zn, page 404
Important Theorems and Results about Relations, Equivalence Rela-
tions, and Equivalence Classes
Theorem 7.6. Let R be a relation from the set A to the set B. Then
1. The domain of R 1
is the range of R. That is, dom.R 1
/ D range.R/.
2. The range of R 1
is the domain of R. That is, range.R 1
/ D dom.R/.
3. The inverse of R 1
is R. That is, R 1
1
D R.
Theorem 7.10. Let n 2 N and let a; b 2 Z. Then a b .mod n/ if and only
if a and b have the same remainder when divided by n.
Theorem 7.14. Let A be a nonempty set and let be an equivalence relation
on A.
1. For each a 2 A, a 2 Œa.
2. For each a; b 2 A, a b if and only if Œa D Œb.
3. For each a; b 2 A, Œa D Œb or Œa Œb D ;.
Corollary 7.16. Let n 2 N. For each a 2 Z, let Œa represent the congruence
class of a modulo n.
1. For each a 2 Z, a 2 Œa.

2. For each a; b 2 Z, a b .mod n/ if and only if Œa D Œb.
3. For each a; b 2 Z, Œa D Œb or Œa Œb D ;.
Corollary 7.17. Let n 2 N. For each a 2 Z, let Œa represent the congruence
class of a modulo n.
1. Z D Œ0 [ Œ1 [ Œ2 [ [ Œn 1
2. For j; k 2 f0; 1; 2; : ::; n 1g, if j ¤ k, then Œj Œk D ;.
Theorem 7.18. Let be an equivalence relation on the nonempty set A.
Then the collection C of all equivalence classes determined by is a parti-
tion of the set A.

Chapter 8
Topics in Number Theory
8.1 The Greatest Common Divisor
Preview Activity 1 (The Greatest Common Divisor)
1. Explain what it means to say that a nonzero integer m divides an integer n.
Recall that we use the notation m j n to indicate that the nonzero integer m
divides the integer n.
2. Let m and n be integers with m ¤ 0. Explain what it means to say that m
does not divide n.
Definition. Let a and b be integers, not both 0. A common divisor of a and
b is any nonzero integer that divides both a and b. The largest natural number
that divides both a and b is called the greatest common divisor of a and b.
The greatest common divisor of a and b is denoted by gcd .a; b/.
3. Use the roster method to list the elements of the set that contains all the
natural numbers that are divisors of 48.
4. Use the roster method to list the elements of the set that contains all the
natural numbers that are divisors of 84.
5. Determine the intersection of the two sets in Parts (3) and (4). This set
contains all the natural numbers that are common divisors of 48 and 84.
6. What is the greatest common divisor of 48 and 84?
414

8.1. The Greatest Common Divisor 415
7. Use the method suggested in Parts (3) through (6) to determine each of the
following: gcd.8; 12/, gcd.0; 5/, gcd.8; 27/, and gcd.14; 28/.
8. If a and b are integers, make a conjecture about how the common divisors
of a and b are related to the greatest common divisor of a and b.
Preview Activity 2 (The GCD and the Division Algorithm)
When we speak of the quotient and the remainder when we “divide an integer a
by the positive integer b,” we will always mean the quotient q and the remainder r
guaranteed by the Division Algorithm. (See Section 3.5, page 143.)
1. Each row in the following table contains values for the integers a and b.
In this table, the value of r is the remainder (from the Division Algorithm)
when a is divided by b. Complete each row in this table by determining
gcd.a; b/, r, and gcd.b; r/.
a b gcd.a; b/ Remainder r gcd.b; r/
44 12
75 21
50 33
2. Formulate a conjecture based on the results of the table in Part (1).
The System of Integers
Number theory is a study of the system of integers, which consists of the set of
integers, Z D f : : : ; 3; 2; 1; 0; 1; 2; 3; : : :g and the various properties of this
set under the usual operations of addition and multiplication and under the usual
ordering relation of “less than.” The properties of the integers in Table 8.1 will be
considered axioms in this text.
We will also assume the properties of the integers shown in Table 8.2. These
properties can be proven from the properties in Table 8.1. (However, we will not
do so here.)

416 Chapter 8. Topics in Number Theory
For all integers a, b, and c:
Closure Properties for Addition and
Multiplication
a C b 2 Z and
ab 2 Z
Commutative Properties for Addition
and Multiplication
a C b D b C a, and
ab D ba
Associative Properties for Addition
and Multiplication
.a C b/ C c D a C .b C c/ and
.ab/ c D a .bc/
Distributive Properties of Multiplica-
tion over Addition
a .b C c/ D ab C ac, and
.b C c/ a D ba C ca
Additive and Multiplicative Identity
Properties
a C 0 D 0 C a D a, and
a 1 D 1 a D a
Additive Inverse Property a C . a/ D . a/ C a D 0
Table 8.1: Axioms for the Integers
Zero Property of Multipli-
cation
If a 2 Z, then a 0 D 0 a D 0.
Cancellation Properties If a; b; c 2 Z and a C b D a C c, then b D c.
of Addition and Multipli-
cation
If a; b; c 2 Z, a ¤ 0 and ac D bc, then b D c.
Table 8.2: Properties of the Integers
We have already studied a good deal of number theory in this text in our dis-
cussion of proof methods. In particular, we have studied even and odd integers,
divisibility of integers, congruence, and the Division Algorithm. See the summary
for Chapter 3 on page 166 for a summary of results concerning even and odd in-
tegers as well as results concerning properties of divisors. We reviewed some of
these properties and the Division Algorithm in the Preview Activities.
The Greatest Common Divisor
One of the most important concepts in elementary number theory is that of the
greatest common divisor of two integers. The definition for the greatest common
divisor of two integers (not both zero) was given in Preview Activity 1.
1. If a; b 2 Z and a and b are not both 0, and if d 2 N, then d D gcd.a; b/
provided that it satisfies all of the following properties:

d j a and d j b. That is, d is a common divisor of a and b.
If k is a natural number such that k j a and k j b, then k d. That is,
any other common divisor of a and b is less than or equal to d.
2. Consequently, a natural number d is not the greatest common divisor of a
and b provided that it does not satisfy at least one of these properties. That
is, d is not equal to gcd.a; b/ provided that
d does not divide a or d does not divide b; or
There exists a natural number k such that k j a and k j b and k d.
This means that d is not the greatest common divisor of a and b provided
that it is not a common divisor of a and b or that there exists a common
divisor of a and b that is greater than d.
In the preview activities, we determined the greatest common divisors for several
pairs of integers. The process we used was to list all the divisors of both integers,
then list all the common divisors of both integers and, finally, from the list of all
common divisors, find the greatest (largest) common divisor. This method works
reasonably well for small integers but can get quite cumbersome if the integers are
large. Before we develop an efficient method for determining the greatest common
divisor of two integers, we need to establish some properties of greatest common
divisors.
One property was suggested in Preview Activity 1. If we look at the results
in Part (7) of that preview activity, we should observe that any common divisor of
a and b will divide gcd.a; b/. In fact, the primary goals of the remainder of this
section are
1. To find an efficient method for determining gcd.a; b/, where a and b are
integers.
2. To prove that the natural number gcd.a; b/ is the only natural number d that
satisfies the following properties:
d divides a and d divides b; and
if k is a natural number such that k j a and k j b, then k j d.
The second goal is only slightly different from the definition of the greatest com-
mon divisor. The only difference is in the second condition where k d is re-
placed by k j d.

We will first consider the case where a and b are integers with a ¤ 0 and
b 0. The proof of the result stated in the second goal contains a method
(called the Euclidean Algorithm) for determining the greatest common divisors
of the two integers a and b. The main idea of the method is to keep replacing the
pair of integers .a; b/ with another pair of integers .b; r/, where 0 r b and
gcd.b; r/ D gcd.a; b/. This idea was explored in Preview Activity 2. Lemma 8.1
is a conjecture that could have been formulated in Preview Activity 2.
Lemma 8.1. Let c and d be integers, not both equal to zero. If q and r are integers
such that c D d q C r, then gcd.c; d/ D gcd.d; r/.
Proof. Let c and d be integers, not both equal to zero. Assume that q and r are
integers such that c D d q C r. For ease of notation, we will let
m D gcd.c; d/ and n D gcd.d; r/:
Now, m divides c and m divides d. Consequently, there exist integers x and y such
that c D mx and d D my. Hence,
r D c d q
r D mx .my/q
r D m.x yq/:
But this means that m divides r. Since m divides d and m divides r, m is less than
or equal to gcd.d; r/. Thus, m n.
Using a similar argument, we see that n divides d and n divides r. Since
c D d q C r, we can prove that n divides c. Hence, n divides c and n divides d.
Thus, n gcd.c; d/ or n m. We now have m n and n m. Hence, m D n
and gcd.c; d/ D gcd.d; r/.
Progress Check 8.2 (Illustrations of Lemma 8.1)
We completed several examples illustrating Lemma 8.1 in Preview Activity 2. For
another example, let c D 56 and d D 12. The greatest common divisor of 56 and
12 is 4.
1. According to the Division Algorithm, what is the remainder r when 56 is
divided by 12?
2. What is the greatest common divisor of 12 and the remainder r?

The key to finding the greatest common divisor (in more complicated cases) is
to use the Division Algorithm again, this time with 12 and r. We now find integers
q2 and r2 such that
12 D r q2 C r2:
3. What is the greatest common divisor of r and r2?
The Euclidean Algorithm
The example in Progress Check 8.2 illustrates the main idea of the Euclidean Al-
gorithm for finding gcd.a; b/, which is explained in the proof of the following
theorem.
Theorem 8.3. Let a and b be integers with a ¤ 0 and b 0. Then gcd.a; b/ is
the only natural number d such that
(a) d divides a and d divides b, and
(b) if k is an integer that divides both a and b, then k divides d.
Proof. Let a and b be integers with a ¤ 0 and b 0, and let d D gcd.a; b/. By
the Division Algorithm, there exist integers q1 and r1 such that
a D b q1 C r1, and 0 r1 b: (1)
If r1 D 0, then equation (1) implies that b divides a. Hence, b D d D gcd.a; b/
and this number satisfies Conditions (a) and (b).
If r1 0, then by Lemma 8.1, gcd.a; b/ D gcd.b; r1/. We use the Division
Algorithm again to obtain integers q2 and r2 such that
b D r1 q2 C r2, and 0 r2 r1: (2)
If r2 D 0, then equation (2) implies that r1 divides b. This means that r1 D
gcd.b; r1/. But we have already seen that gcd.a; b/ D gcd.b; r1/. Hence, r1 D
gcd.a; b/. In addition, if k is an integer that divides both a and b, then, using
equation (1), we see that r1 D a b q1 and, hence k divides r1. This shows that
r1 D gcd.a; b/ satisfies Conditions (a) and (b).
If r2 0, then by Lemma 8.1, gcd.b; r1/ D gcd.r1; r2/. But we have already
seen that gcd.a; b/ D gcd.b; r1/. Hence, gcd.a; b/ D gcd.r1; r2/. We now con-
tinue to apply the Division Algorithm to produce a sequence of pairs of integers

(all of which have the same greatest common divisor). This is summarized in the
following table:
Original Equation from Inequality from New
Pair Division Algorithm Division Algorithm Pair
.a; b/ a D b q1 C r1 0 r1 b .b; r1/
.b; r1/ b D r1 q2 C r2 0 r2 r1 .r1; r2/
.r1; r2/ r1 D r2 q3 C r3 0 r3 r2 .r2; r3/
.r2; r3/ r2 D r3 q4 C r4 0 r4 r3 .r3; r4/
.r3; r4/ r3 D r4 q5 C r5 0 r5 r4 .r4; r5/
:
:
:
:
:
:
:
:
:
:
:
:
From the inequalities in the third column of this table, we have a strictly decreasing
sequence of nonnegative integers .b r1 r2 r3 r4 /. Consequently, a
term in this sequence must eventually be equal to zero. Let p be the smallest
natural number such that rpC1 D 0. This means that the last two rows in the
preceding table will be
Original Equation from Inequality from New
Pair Division Algorithm Division Algorithm Pair
rp 2; rp 1

rp 2 D rp 1 qp C rp 0 rp rp 1 rp 1; rp

rp 1; rp

rp 1 D rp qpC1 C 0
Remember that this table was constructed by repeated use of Lemma 8.1 and that
the greatest common divisor of each pair of integers produced equals gcd.a; b/.
Also, the last row in the table indicates that rp divides rp 1. This means that
gcd.rp 1; rp/ D rp and hence rp D gcd.a; b/.
This proves that rp D gcd.a; b/ satisfies Condition (a) of this theorem. Now
assume that k is an integer such that k divides a and k divides b. We proceed
through the table row by row. First, since r1 D a b q, we see that
k must divide r1.
The second row tells us that r2 D b r1 q2. Since k divides b and k divides r1,
we conclude that
k divides r2.

Continuing with each row, we see that k divides each of the remainders r1, r2, r3,
: : : ; rp. This means that rp D gcd.a; b/ satisfies Condition (b) of the theorem.
Progress Check 8.4 (Using the Euclidean Algorithm)
1. Use the Euclidean Algorithm to determine gcd.180; 126/. Notice that we
have deleted the third column (Inequality from Division Algorithm) from
the following table. It is not needed in the computations.
Original Equation from New
Pair Division Algorithm Pair
.180; 126/ 180 D 126 1 C 54 .126; 54/
.126; 54/ 126 D
Consequently, gcd.180; 126/ D .
2. Use the Euclidean Algorithm to determine gcd.4208; 288/.
.4208; 288/ 4208 D 288 14 C 176 .288; /
Consequently, gcd.4208; 288/ D .
Some Remarks about Theorem 8.3
Theorem 8.3 was proven with the assumptions that a; b 2 Z with a ¤ 0 and b 0.
A more general version of this theorem can be proven with a; b 2 Z and b ¤ 0.
This can be proven using Theorem 8.3 and the results in the following lemma.
Lemma 8.5. Let a; b 2 Z with b ¤ 0. Then
1. gcd.0; b/ D jbj.
2. If gcd.a; b/ D d, then gcd.a; b/ D d.
The proofs of these results are in Exercise (4). An application of this result is given
in the next example.

Example 8.6 (Using the Euclidean Algorithm)
Let a D 234 and b D 42. We will use the Euclidean Algorithm to determine
gcd.234; 42/.
Step Original Equation from New
1 .234; 42/ 234 D 42 5 C 24 .42; 24/
2 .42; 24/ 42 D 24 1 C 18 .24; 18/
3 .24; 18/ 24 D 18 1 C 6 .18; 6/
4 .18; 6/ 18 D 6 3
So gcd.234; 42/ D 6 and hence gcd.234; 42/ D 6.
Writing gcd.a; b/ in Terms of a and b
We will use Example 8.6 to illustrate another use of the Euclidean Algorithm. It
is possible to use the steps of the Euclidean Algorithm in reverse order to write
gcd.a; b/ in terms of a and b. We will use these steps in reverse order to find
integers m and n such that gcd.234; 42/ D 234m C 42n. The idea is to start
with the row with the last nonzero remainder and work backward as shown in the
following table:
Explanation Result
First, use the equation in Step 3 to 6 D 24 18 1
write 6 in terms of 24 and 18.
Use the equation in Step 2 to write 6 D 24 18 1
18 D 42 24 1. Substitute this into D 24 .42 24 1/
the preceding result and simplify. D 42 . 1/ C 24 2
We now have written 6 in terms of 6 D 42 . 1/ C 24 2
42 and 24. Use the equation in D 42 . 1/ C .234 42 5/ 2
Step 1 to write 24 D 234 42 5. D 234 2 C 42 . 11/
Substitute this into the preceding
result and simplify.
Hence, we can write
gcd.234; 42/ D 234 2 C 42 . 11/:
(Check this with a calculator.) In this case, we say that we have written
gcd.234; 42/ as a linear combination of 234 and 42. More generally, we have
the following definition.

Definition. Let a and b be integers. A linear combination of a and b is an
integer of the form ax C by, where x and y are integers.
Progress Check 8.7 (Writing the gcd as a Linear Combination)
Use the results from Progress Check 8.4 to
1. Write gcd.180; 126/ as a linear combination of 180 and 126.
2. Write gcd.4208; 288/ as a linear combination of 4208 and 288.
The previous example and progress check illustrate the following important
result in number theory, which will be used in the next section to help prove some
other significant results.
Theorem 8.8. Let a and b be integers, not both 0. Then gcd.a; b/ can be written
as a linear combination of a and b. That is, there exist integers u and v such that
gcd.a; b/ D au C bv.
We will not give a formal proof of this theorem. Hopefully, the examples
and activities provide evidence for its validity. The idea is to use the steps of the
Euclidean Algorithm in reverse order to write gcd.a; b/ as a linear combination of
a and b. For example, assume the completed table for the Euclidean Algorithm is
Step Original Equation from New
1 .a; b/ a D b q1 C r1 .b; r1/
2 .b; r1/ b D r1 q2 C r2 .r1; r2/
3 .r1; r2/ r1 D r2 q3 C r3 .r2; r3/
4 .r2; r3/ r2 D r3 q4 C 0 .r3; r4/
We can use Step 3 to write r3 D gcd.a; b/ as a linear combination of r1 and r2. We
can then solve the equation in Step 2 for r2 and use this to write r3 D gcd.a; b/ as
a linear combination of r1 and b. We can then use the equation in Step 1 to solve
for r1 and use this to write r3 D gcd.a; b/ as a linear combination of a and b.
In general, if we can write rp D gcd.a; b/ as a linear combination of a pair
in a given row, then we can use the equation in the preceding step to write rp D
gcd.a; b/ as a linear combination of the pair in this preceding row.
The notational details of this induction argument get quite involved. Many
mathematicians prefer to prove Theorem 8.8 using a property of the natural num-
bers called the Well-Ordering Principle. The Well-Ordering Principle for the

natural numbers states that any nonempty set of natural numbers must contain a
least element. It can be proven that the Well-Ordering Principle is equivalent to the
Principle of Mathematical Induction.
Exercises 8.1
1. Find each of the following greatest common divisors by listing all of the
positive common divisors of each pair of integers.
? (a) gcd.21; 28/
? (b) gcd. 21; 28/
? (c) gcd.58; 63/
? (d) gcd.0; 12/
(e) gcd.110; 215/
(f) gcd.110; 215/
2. ?
(a) Let a 2 Z and let k 2 Z with k ¤ 0. Prove that if k j a and k j .a C 1/,
then k j 1, and hence k D ˙1.
(b) Let a 2 Z. Find the greatest common divisor of the consecutive inte-
gers a and a C 1. That is, determine gcd.a; a C 1/.
3. (a) Let a 2 Z and let k 2 Z with k ¤ 0. Prove that if k j a and k j .a C 2/,
then k j 2.
(b) Let a 2 Z. What conclusions can be made about the greatest common
divisor of a and a C 2?
? 4. Let a; b 2 Z with b ¤ 0. Prove each of the following:
(a) gcd.0; b/ D jbj
(b) If gcd.a; b/ D d, then gcd.a; b/ D d. That is,
gcd .a; b/ D gcd.a; b/.
5. For each of the following pairs of integers, use the Euclidean Algorithm to
find gcd.a; b/ and to write gcd.a; b/ as a linear combination of a and b. That
is, find integers m and n such that d D am C bn.
?
(a) a D 36; b D 60
?
(b) a D 901; b D 935
(c) a D 72; b D 714
(d) a D 12628; b D 21361
?
(e) a D 901, b D 935
(f) a D 36, b D 60
6. ? (a) Find integers u and v such that 9u C 14v D 1 or explain why it is not
possible to do so. Then find integers x and y such that 9x C 14y D 10
or explain why it is not possible to do so.

(b) Find integers x and y such that 9x C 15y D 10 or explain why it is not
possible to do so.
(c) Find integers x and y such that 9x C 15y D 3162 or explain why it is
not possible to do so.
7. ?
(a) Notice that gcd.11; 17/ D 1. Find integers x and y such that
11x C 17y D 1.
? (b) Let m; n 2 Z. Write the sum
m
11
C
n
17
as a single fraction.
(c) Find two rational numbers with denominators of 11 and 17, respec-
tively , whose sum is equal to
10
187
. Hint: Write the rational numbers
in the form
m
11
and
n
17
, where m; n 2 Z. Then write
m
11
C
n
17
D
10
187
:
Use Exercises (7a) and (7b) to determine m and n.
(d) Find two rational numbers with denominators 17 and 21, respectively,
whose sum is equal to
326
357
(e) Find two rational numbers with denominators 9 and 15, respectively,
whose sum is equal to
10
225
8. Linear Combinations and the Greatest Common Divisor
(a) Determine the greatest common divisor of 20 and 12?
(b) Let d D gcd.20; 12/. Write d as a linear combination of 20 and 12.
(c) Generate at least six different linear combinations of 20 and 12. Are
these linear combinations of 20 and 12 multiples of gcd.20; 12/?
(d) Determine the greatest common divisor of 21 and 6 and then generate
at least six different linear combinations of 21 and 6. Are these linear
combinations of 21 and 6 multiples of gcd.21; 6/?
(e) The following proposition was first introduced in Exercise (18) on
page 243 in Section 5.2. Complete the proof of this proposition if you
have not already done so.

Proposition 5.16 Let a, b, and t be integers with t ¤ 0. If t divides a
and t divides b, then for all integers x and y, t divides (ax + by).
Proof. Let a, b, and t be integers with t ¤ 0, and assume that t divides
a and t divides b. We will prove that for all integers x and y, t divides
.ax C by/.
So let x 2 Z and let y 2 Z. Since t divides a, there exists an integer m
such that : : : :
(f) Now let a and b be integers, not both zero and let d D gcd.a; b/. The-
orem 8.8 states that d is a linear combination of a and b. In addition,
let S and T be the following sets:
S D fax C by j x; y 2 Zg and T D fkd j k 2 Zg:
That is, S is the set of all linear combinations of a and b, and T is the
set of all multiples of the greatest common divisor of a and b. Does the
set S equal the set T ? If not, is one of these sets a subset of the other
set? Justify your conclusions.
Note: In Parts (c) and (d), we were exploring special cases for these
two sets.
8.2 Prime Numbers and Prime Factorizations
Preview Activity 1 (Exploring Examples where a Divides b c)
1. Find at least three different examples of nonzero integers a, b, and c such
that a j .bc/ but a does not divide b and a does not divide c. In each case,
compute gcd.a; b/ and gcd.a; c/.
2. Find at least three different examples of nonzero integers a, b, and c such
that gcd.a; b/ D 1 and a j .bc/. In each example, is there any relation
between the integers a and c?
3. Formulate a conjecture based on your work in Parts (1) and (2).
Preview Activity 2 (Prime Factorizations)
Recall that a natural number p is a prime number provided that it is greater than
1 and the only natural numbers that divide p are 1 and p. A natural number other
than 1 that is not a prime number is a composite number. The number 1 is neither
prime nor composite. (See Exercise 13 from Section 2.4 on page 78.)

8.2. Prime Numbers and Prime Factorizations 427
1. Give examples of four natural numbers that are prime and four natural num-
bers that are composite.
Theorem 4.9 in Section 4.2 states that every natural number greater than 1 is
either a prime number or a product of prime numbers.
When a composite number is written as a product of prime numbers, we say
that we have obtained a prime factorization of that composite number. For exam-
ple, since 60 D 22 3 5, we say that 22 3 5 is a prime factorization of 60.
2. Write the number 40 as a product of prime numbers by first writing 40 D
2 20 and then factoring 20 into a product of primes. Next, write the number
40 as a product of prime numbers by first writing 40 D 58 and then factoring
8 into a product of primes.
3. In Part (2), we used two different methods to obtain a prime factorization
of 40. Did these methods produce the same prime factorization or different
prime factorizations? Explain.
4. Repeat Parts (2) and (3) with 150. First, start with 150 D 3 50, and then
start with 150 D 5 30.
Greatest Common Divisors and Linear Combinations
In Section 8.1, we introduced the concept of the greatest common divisor of two
integers. We showed how the Euclidean Algorithm can be used to find the greatest
common divisor of two integers, a and b, and also showed how to use the results
of the Euclidean Algorithm to write the greatest common divisor of a and b as a
linear combination of a and b.
In this section, we will use these results to help prove the so-called Fundamen-
tal Theorem of Arithmetic, which states that any natural number greater than 1 that
is not prime can be written as product of primes in “essentially” only one way. This
means that given two prime factorizations, the prime factors are exactly the same,
and the only difference may be in the order in which the prime factors are written.
We start with more results concerning greatest common divisors. We first prove
Proposition 5.16, which was part of Exercise (18) on page 243 in Section 5.2 and
Exercise (8) on page 425in Section 8.1.
Proposition 5.16 Let a, b, and t be integers with t ¤ 0. If t divides a and t divides
b, then for all integers x and y, t divides (ax + by).

Proof. Let a, b, and t be integers with t ¤ 0, and assume that t divides a and t
divides b. We will prove that for all integers x and y, t divides .ax C by/.
So let x 2 Z and let y 2 Z. Since t divides a, there exists an integer m such
that a D mt and since t divides b, there exists an integer n such that b D nt.
Using substitution and algebra, we then see that
ax C by D .mt/x C .nt/y
D t.mx C ny/
Since .mx C ny/ is an integer, the last equation proves that t divides ax C by and
this proves that for all integers x and y, t divides .ax C by/.
We now let a; b 2 Z, not both 0, and let d D gcd.a; b/. Theorem 8.8 states
that d can be written as a linear combination of a and b. Now, since d j a and
d j b, we can use the result of Proposition 5.16 to conclude that for all x; y 2 Z,
d j .ax C by/. This means that d divides every linear combination of a and b.
In addition, this means that d must be the smallest positive number that is a linear
combination of a and b. We summarize these results in Theorem 8.9.
Theorem 8.9. Let a; b 2 Z, not both 0.
1. The greatest common divisor, d, is a linear combination of a and b. That is,
there exist integers m and n such that d D am C bn.
2. The greatest common divisor, d, divides every linear combination of a and
b. That is, for all x; y 2 Z , d j .ax C by/.
3. The greatest common divisor, d, is the smallest positive number that is a
linear combination of a and b.
Relatively Prime Integers
In Preview Activity 1, we constructed several examples of integers a, b, and c such
that a j .bc/ but a does not divide b and a does not divide c. For each example,
we observed that gcd.a; b/ ¤ 1 and gcd.a; c/ ¤ 1.
We also constructed several examples where a j .bc/ and gcd.a; b/ D 1. In all
of these cases, we noted that a divides c. Integers whose greatest common divisor
is equal to 1 are given a special name.
Definition. Two nonzero integers a and b are relatively prime provided that
gcd.a; b/ D 1.

Progress Check 8.10 (Relatively Prime Integers)
1. Construct at least three different examples where p is a prime number, a 2
Z , and p j a. In each example, what is gcd.a; p/? Based on these examples,
formulate a conjecture about gcd.a; p/ when p j a.
2. Construct at least three different examples where p is a prime number, a 2
Z , and p does not divide a. In each example, what is gcd.a; p/? Based
on these examples, formulate a conjecture about gcd.a; p/ when p does not
divide a.
3. Give at least three different examples of integers a and b where a is not
prime, b is not prime, and gcd.a; b/ D 1, or explain why it is not possible to
construct such examples.
Theorem 8.11. Let a and b be nonzero integers, and let p be a prime number.
1. If a and b are relatively prime, then there exist integers m and n such that
am C bn D 1. That is, 1 can be written as linear combination of a and b.
2. If p j a, then gcd.a; p/ D p.
3. If p does not divide a, then gcd.a; p/ D 1.
Part (1) of Theorem 8.11 is actually a corollary of Theorem 8.9. Parts (2)
and (3) could have been the conjectures you formulated in Progress Check 8.10.
The proofs are included in Exercise (1).
Given nonzero integers a and b, we have seen that it is possible to use the
Euclidean Algorithmto write their greatest common divisor as a linear combination
of a and b. We have also seen that this can sometimes be a tedious, time-consuming
process, which is why people have programmed computers to do this. Fortunately,
in many proofs of number theory results, we do not actually have to construct this
linear combination since simply knowing that it exists can be useful in proving
results. This will be illustrated in the proof of Theorem 8.12, which is based on
work in Preview Activity 1.
Theorem 8.12. Let a, b, be nonzero integers and let c be an integer. If a and b are
relatively prime and a j .bc/, then a j c.

The explorations in Preview Activity 1 were related to this theorem. We will
first explore the forward-backward process for the proof. The goal is to prove that
a j c. A standard way to do this is to prove that there exists an integer q such that
c D aq: (1)
Since we are given that a j .bc/, there exists an integer k such that
bc D ak: (2)
It may seem tempting to divide both sides of equation (2) by b, but if we do so,
we run into problems with the fact that the integers are not closed under division.
Instead, we look at the other part of the hypothesis, which is that a and b are
relatively prime. This means that gcd.a; b/ D 1. How can we use this? This
means that a and b have no common factors except for 1. In light of equation (2),
it seems reasonable that any factor of a must also be a factor of c. But how do we
formalize this?
One conclusion that we can use is that since gcd.a; b/ D 1, by Theorem 8.11,
there exist integers m and n such that
am C bn D 1: (3)
We may consider solving equation (3) for b and substituting this into equa-
tion (2). The problem, again, is that in order to solve equation (3) for b, we need
to divide by n.
Before doing anything else, we should look at the goal in equation (1). We
need to introduce c into equation (3). One way to do this is to multiply both sides
of equation (3) by c. (This keeps us in the system of integers since the integers are
closed under multiplication.) This gives
.am C bn/ c D 1 c
acm C bcn D c: (4)
Notice that the left side of equation (4) contains a term, bcn, that contains bc. This
means that we can use equation (2) and substitute bc D ak in equation (4). After
doing this, we can factor the left side of the equation to prove that a j c.
Progress Check 8.13 (Completing the Proof of Theorem 8.12)
Write a complete proof of Theorem 8.12.

Corollary 8.14.
1. Let a; b 2 Z, and let p be a prime number. If p j .ab/, then p j a or p j b.
2. Let p be a prime number, let n 2 N, and let a1; a2; : : :; an 2 Z. If
p j .a1a2 an/, then there exists a natural number k with 1 k n
such that p j ak.
Part (1) of Corollary 8.14 is a corollary of Theorem 8.12. Part (2) is proved us-
ing mathematical induction. The basis step is the case where n D 1, and Part (1) is
the case where n D 2. The proofs of these two results are included in Exercises (2)
and (3).
Historical Note
Part (1) of Corollary 8.14 is known as Euclid’s Lemma. Most people associate
geometry with Euclid’s Elements, but these books also contain many basic results
in number theory. Many of the results that are contained in this section appeared
in Euclid’s Elements.
Prime Numbers and Prime Factorizations
We are now ready to prove the Fundamental Theorem of Arithmetic. The first part
of this theorem was proved in Theorem 4.9 in Section 4.2. This theorem states
that each natural number greater than 1 is either a prime number or is a product of
prime numbers. Before we state the Fundamental Theorem of Arithmetic, we will
discuss some notational conventions that will help us with the proof. We start with
an example.
We will use n D 120. Since 5 j 120, we can write 120 D 5 24. In addition,
we can factor 24 as 24 D 2 2 2 3. So we can write
120 D 5 24
D 5 .2 2 2 3/ :
This is a prime factorization of 120, but it is not the way we usually write this
factorization. Most often, we will write the prime number factors in ascending
order. So we write
120 D 2 2 2 3 5 or 120 D 23
3 5:

Now, let n 2 N. To write the prime factorization of n with the prime factors in
ascending order requires that if we write n D p1p2 pr , where p1; p2; pr are
prime numbers, we will have p1 p2 pr .
Theorem 8.15 (The Fundamental Theorem of Arithmetic).
1. Each natural number greater than 1 is either a prime number or is a product
of prime numbers.
2. Let n 2 N with n 1. Assume that
n D p1p2 pr and that n D q1q2 qs;
where p1; p2; pr and q1; q2; qs are primes with p1 p2 pr
and q1 q2 qs. Then r D s, and for each j from 1 to r, pj D qj .
Proof. The first part of this theorem was proved in Theorem 4.9. We will prove
the second part of the theorem by induction on n using the Second Principle of
Mathematical Induction. (See Section 4.2.) For each natural number n with n 1,
let P.n/ be
If n D p1p2 pr and n D q1q2 qs, where p1; p2; pr and q1; q2; qs
are primes with p1 p2 pr and q1 q2 qs, then r D s,
and for each j from 1 to r, pj D qj .
For the basis step, we notice that since 2 is a prime number, its only factoriza-
tion is 2 D 1 2. This means that the only equation of the form 2 D p1p2 pr ,
where p1; p2; pr are prime numbers, is the case where r D 1 and p1 D 2. This
proves that P.2/ is true.
For the inductive step, let k 2 N with k 2. We will assume that
P.2/; P.3/; : : :; P.k/ are true. The goal now is to prove that P.k C 1/ is true.
To prove this, we assume that .k C 1/ has two prime factorizations and then prove
that these prime factorizations are the same. So we assume that
k C 1 D p1p2 pr and that k C 1 D q1q2 qs, where p1; p2; pr and
q1; q2; qs are primes with p1 p2 pr and q1 q2 qs.
We must now prove that r D s, and for each j from 1 to r, pj D qj . We can
break our proof into two cases: (1) p1 q1; and (2) q1 p1. Since one of these
must be true, and since the proofs will be similar, we can assume, without loss of
generality, that p1 q1.

Since k C 1 D p1p2 pr , we know that p1 j .k C 1/, and hence we may
conclude that p1 j .q1q2 qs/. We now use Corollary 8.14 to conclude that there
exists a j with 1 j s such that p1 j qj . Since p1 and qj are primes, we
conclude that
p1 D qj :
We have also assumed that q1 qj for all j and, hence, we know that q1 p1.
However, we have also assumed that p1 q1. Hence,
p1 D q1:
We now use this and the fact that k C 1 D p1p2 pr D q1q2 qs to conclude
that
p2 pr D q2 qs:
The product in the previous equation is less that k C 1. Hence, we can apply our
induction hypothesis to these factorizations and conclude that r D s, and for each
j from 2 to r, pj D qj .
This completes the proof that if P.2/; P.3/; : : :; P.k/ are true, then P.k C 1/
is true. Hence, by the Second Principle of Mathematical Induction, we conclude
that P.n/ is true for all n 2 N with n 2. This completes the proof of the
theorem.
Note: : We often shorten the result of the Fundamental Theorem of Arithmetic by
simply saying that each natural number greater than one that is not a prime has a
unique factorization as a product of primes. This simply means that if n 2 N,
n 1, and n is not prime, then no matter how we choose to factor n into a product
of primes, we will always have the same prime factors. The only difference may
be in the order in which we write the prime factors.
Further Results and Conjectures about Prime Numbers
1. The Number of Prime Numbers
Prime numbers have fascinated mathematicians for centuries. For example,
we can easily start writing a list of prime numbers in ascending order. Fol-
lowing is a list of the prime numbers less than 100.
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71,
73, 79, 83, 89, 97

This list contains the first 25 prime numbers. Does this list ever stop? The
question was answered in Euclid’s Elements, and the result is stated in The-
orem 8.16. The proof of this theorem is considered to be one of the classical
proofs by contradiction.
Theorem 8.16. There are infinitely many prime numbers.
Proof. We will use a proof by contradiction. We assume that there are only
finitely many primes, and let
p1; p2; : : :; pm
be the list of all the primes. Let
M D p1p2 pm C 1: (1)
Notice that M ¤ 1. So M is either a prime number or, by the Fundamental
Theorem of Arithmetic, M is a product of prime numbers. In either case,
M has a factor that is a prime number. Since we have listed all the prime
numbers, this means that there exists a natural number j with 1 j m
such that pj j M. Now, we can rewrite equation (1) as follows:
1 D M p1p2 pm: (2)
We have proved pj j M, and since pj is one of the prime factors of p1p2 pm,
we can also conclude that pj j .p1p2 pm/. Since pj divides both of the
terms on the right side of equation (2), we can use this equation to conclude
that pj divides 1. This is a contradiction since a prime number is greater
than 1 and cannot divide 1. Hence, our assumption that there are only finitely
many primes is false, and so there must be infinitely many primes.
2. The Distribution of Prime Numbers
There are infinitely many primes, but when we write a list of the prime num-
bers, we can see some long sequences of consecutive natural numbers that
contain no prime numbers. For example, there are no prime numbers be-
tween 113 and 127. The following theorem shows that there exist arbitrarily
long sequences of consecutive natural numbers containing no prime num-
bers. A guided proof of this theorem is included in Exercise (15).
Theorem 8.17. For any natural number n, there exist at least n consecutive
natural numbers that are composite numbers.

There are many unanswered questions about prime numbers, two of which will
now be discussed.
3. The Twin Prime Conjecture
By looking at the list of the first 25 prime numbers, we see several cases
where consecutive prime numbers differ by 2. Examples are: 3 and 5; 11
and 13; 17 and 19; 29 and 31. Such pairs of prime numbers are said to be
twin primes. How many twin primes exist? The answer is not known. The
Twin Prime Conjecture states that there are infinitely many twin primes.
As of June 25, 2010, this is still a conjecture as it has not been proved or
disproved.
For some interesting information on prime numbers, visit the Web site The
Prime Pages (http://primes.utm.edu/), where there is a link to The Largest
Known Primes Web site. According to information at this site as of June 25,
2010, the largest known twin primes are
65516468355 2333333
1

and 65516468355 2333333
C 1

:
Each of these prime numbers contains 100355 digits.
4. Goldbach’s Conjecture
Given an even natural number, is it possible to write it as a sum of two prime
numbers? For example,
4 D 2 C 2 6 D 3 C 3 8 D 5 C 3
78 D 37 C 41 90 D 43 C 47 138 D 67 C 71
One of the most famous unsolved problems in mathematics is a conjecture
made by Christian Goldbach in a letter to Leonhard Euler in 1742. The
conjecture, now known as Goldbach’s Conjecture, is as follows:
Every even integer greater than 2 can be expressed as the sum of two
(not necessarily distinct) prime numbers.
As of June 25, 2010, it is not known if this conjecture is true or false, al-
though most mathematicians believe it to be true.
Exercises 8.2
?
1. Prove the second and third parts of Theorem 8.11.

(a) Let a be a nonzero integer, and let p be a prime number. If p j a, then
gcd.a; p/ D p.
(b) Let a be a nonzero integer, and let p be a prime number. If p does not
divide a, then gcd.a; p/ D 1.
?
2. Prove the first part of Corollary 8.14.
Let a; b 2 Z, and let p be a prime number. If p j .ab/, then p j a or p j b.
Hint: Consider two cases: (1) p j a; and (2) p does not divide a.
? 3. Use mathematical induction to prove the second part of Corollary 8.14.
Let p be a prime number, let n 2 N, and let a1; a2; : : :; an 2 Z. If
p j .a1a2 an/, then there exists a k 2 N with 1 k n such that
p j ak.
?
4. (a) Let a and b be nonzero integers. If there exist integers x and y such that
ax C by D 1, what conclusion can be made about gcd.a; b/? Explain.
(b) Let a and b be nonzero integers. If there exist integers x and y such that
ax C by D 2, what conclusion can be made about gcd.a; b/? Explain.
5. (a) Let a 2 Z. What is gcd.a; a C 1/? That is, what is the greatest com-
mon divisor of two consecutive integers? Justify your conclusion.
Hint: Exercise (4) might be helpful.
(b) Let a 2 Z. What conclusion can be made about gcd.a; a C 2/? That
is, what conclusion can be made about the greatest common divisor of
two integers that differ by 2? Justify your conclusion.
6. (a) Let a 2 Z. What conclusion can be made about gcd.a; a C 3/? That
(b) Let a 2 Z. What conclusion can be made about gcd.a; a C 4/? That
7. ? (a) Let a D 16 and b D 28. Determine the value of d D gcd.a; b/, and
then determine the value of gcd

a
d
;
b
d

.
?
(b) Repeat Exercise (7a) with a D 10 and b D 45.

(c) Let a; b 2 Z, not both equal to 0, and let d D gcd.a; b/. Explain why
a
d
and
b
d
are integers. Then prove that gcd

a
d
;
b
d

D 1. Hint: Start
by writing d as a linear combination of a and b.
This says that if you divide both a and b by their greatest common divisor,
the result will be two relatively prime integers.
8. Are the following propositions true or false? Justify your conclusions.
(a) For all integers a; b; and c, if a j c and b j c, then .ab/ j c.
(b) For all integers a; b; and c, if a j c, b j c and gcd.a; b/ D 1, then
.ab/ j c.
? 9. In Exercise (17) in Section 3.5, it was proved that if n is an odd integer, then
8 j n2
1

. (This result was also proved in Exercise (19) in Section 7.4.)
Now, prove the following proposition:
If n is an odd integer and 3 does not divide n, then 24 j n2 1

.
10. (a) Prove the following proposition:
For all a; b; c 2 Z, gcd.a; bc/ D 1 if and only if gcd.a; b/ D 1 and
gcd.a; c/ D 1.
(b) Use mathematical induction to prove the following proposition:
Let n 2 N and let a; b1; b2; : : :; bn 2 Z. If gcd.a; bi/ D 1 for all i 2 N
with 1 i n, then gcd.a; b1b2 bn/ D 1.
? 11. Is the following proposition true or false? Justify your conclusion.
For all integers a; b; and c, if gcd.a; b/ D 1 and c j .a C b/, then
gcd.a; c/ D 1 and gcd.b; c/ D 1.
If n 2 N, then gcd.5n C 2; 12n C 5/ D 1.
13. Let y 2 N. Use the Fundamental Theorem of Arithmetic to prove that
there exists an odd natural number x and a nonnegative integer k such that
y D 2k
x.
14. (a) Determine five different primes that are congruent to 3 modulo 4.
(b) Prove that there are infinitely many primes that are congruent to 3 mod-
ulo 4.
15. (a) Let n 2 N. Prove that 2 divides Œ.n C 1/Š C 2.

(b) Let n 2 N with n 2. Prove that 3 divides Œ.n C 1/Š C 3.
(c) Let n 2 N. Prove that for each k 2 N with 2 k .n C 1/, k divides
Œ.n C 1/Š C k.
(d) Use the result of Exercise (15c) to prove that for each n 2 N, there
exist at least n consecutive composite natural numbers.
16. The Twin Prime Conjecture states that there are infinitely many twin primes,
but it is not known if this conjecture is true or false. The answers to the
following questions, however, can be determined.
(a) How many pairs of primes p and q exist where q p D 3? That is,
how many pairs of primes are there that differ by 3? Prove that your
answer is correct. (One such pair is 2 and 5.)
(b) How many triplets of primes of the form p, p C2, and p C4 are there?
That is, how many triplets of primes exist where each prime is 2 more
than the preceding prime? Prove that your answer is correct. Notice
that one such triplet is 3, 5, and 7.
Hint: Try setting up cases using congruence modulo 3.
Let n 2 N. For each a 2 Z, if gcd.a; n/ D 1, then for every b 2 Z,
there exists an x 2 Z such that ax b .mod n/.
Hint: One way is to start by writing 1 as a linear combination of a and n.
For all natural numbers m and n, if m and n are twin primes other than
the pair 3 and 5, then 36 divides mnC1 and mnC1 is a perfect square.
Hint: Look at several examples of twin primes. What do you notice about
the number that is between the two twin primes? Set up cases based on this
observation.
19. Square Roots and Irrational Numbers. In Chapter 3, we proved that some
square roots (such as
p
2 and
p
3) are irrational numbers. In this activity, we
will use the Fundamental Theorem of Arithmetic to prove that if a natural
number is not a perfect square, then its square root is an irrational number.

8.3. Linear Diophantine Equations 439
(a) Let n be a natural number. Use the Fundamental Theorem of Arith-
metic to explain why if n is composite, then there exist prime numbers
p1; p2; : : :; pr and natural numbers ˛1; ˛2; : : :; ˛r such that
n D p˛1
1 p˛2
2 p˛r
r : (1)
Then, if we use r D 1 and ˛1 D 1 for a prime number, explain why we
can write any natural number in the form given in equation (1).
(b) A natural number b is a perfect square if and only if there exists a
natural number a such that b D a2. Explain why 36, 400, and 15876
are perfect squares. Then determine the prime factorization of these
perfect squares. What do you notice about these prime factorizations?
(c) Let n be a natural number written in the form given in equation (1) in
part (a). Prove that n is a perfect square if and only if for each natural
number k with 1 k r, ˛k is even.
(d) Prove that for all natural numbers n, if n is not a perfect square, then
p
n is an irrational number. Hint: Use a proof by contradiction.
8.3 Linear Diophantine Equations
Preview Activity 1 (Integer Solutions for Linear Equations in One Variable)
1. Does the linear equation 6x D 42 have a solutions that is an integer? Ex-
plain.
2. Does the linear equation 7x D 21 have a solution that is an integer? Ex-
plain.
3. Does the linear equation 4x D 9 have a solution that is an integer? Explain.
4. Does the linear equation 3x D 20 have a solution that is an integer? Ex-
plain.
5. Prove the following theorem:
Theorem 8.18. Let a; b 2 Z with a ¤ 0.
If a divides b, then the equation ax D b has exactly one solution that
is an integer.

If a does not divide b, then the equation ax D b has no solution that is
an integer.
Preview Activity 2 (Linear Equations in Two Variables)
1. Find integers x and y so that 2x C 6y D 25 or explain why it is not possible
to find such a pair of integers.
2. Find integers x and y so that 6x 9y D 100 or explain why it is not possible
to find such a pair of integers.
3. Notice that x D 2 and y D 1 is a solution of the equation 3x C 5y D 11,
and that x D 7 and y D 2 is also a solution of the equation 3x C 5y D 11.
(a) Find two pairs of integers x and y so that x 7 and 3x C 5y D 11.
(Try to keep the integer values of x as small as possible.)
(b) Find two pairs of integers x and y so that x 2 and 3x C 5y D 11.
(Try to keep the integer values of x as close to 2 as possible.)
(c) Determine formulas (one for x and one for y) that will generate pairs
of integers x and y so that 3x C 5y D 11.
Hint: The two formulas can be written in the form x D 2 C km and
y D 1 C kn, where k is an arbitrary integer and m and n are specific
integers.
4. Notice that x D 4 and y D 0 is a solution of the equation 4x C 6y D 16,
and that x D 7 and y D 2 is a solution of the equation 4x C 6y D 16.
(a) Find two pairs of integers x and y so that x 7 and 4x C 6y D 16.
(Try to keep the integer values of x as small as possible.)
(b) Find twopairs of integers x and y so that x 4 and 4x C 6y D 16.
(Try to keep the integer values of x as close to 4 as possible.)
(c) Determine formulas (one for x and one for y) that will generate pairs
of integers x and y so that 4x C 6y D 16.
Hint: The two formulas can be written in the form x D 4 C km and
y D 0 C kn, where k is an arbitrary integer and mand n are specific
integers.
In the two preview activities, we were interested only in integer solutions for
certain equations. In such instances, we give the equation a special name.

Definition. An equation whose solutions are required to be integers is called
a Diophantine equation.
Diophantine equations are named in honor of the Greek mathematician Dio-
phantus of Alexandria (circa 300 c.e.). Very little is known about Diophantus’ life
except that he probably lived in Alexandria in the early part of the fourth century
c.e. and was probably the first to use letters for unknown quantities in arith-
metic problems. His most famous work, Arithmetica, consists of approximately
130 problems and their solutions. Most of these problems involved solutions of
equations in various numbers of variables. It is interesting to note that Diophantus
did not restrict his solutions to the integers but recognized rational number solu-
tions as well. Today, however, the solutions for a so-called Diophantine equation
must be integers.
Definition. If a and b are integers with a ¤ 0, then the equation ax D b is a
linear Diophantine equation in one variable.
Theorem 8.18 in Preview Activity 1 provides us with results that allows us to
determine which linear diophantine equations in one variable have solutions and
which ones do not have a solution.
A linear Diophantine equation in two variables can be defined in a manner
similar to the definition for a linear Diophantine equation in one variable.
Definition. Let a, b, and c be integers with a ¤ 0 and b ¤ 0. The Diophan-
tine equation ax C by D c is called a linear Diophantine equation in two
variables.
The equations that were investigated in Preview Activity 2 were linear Dio-
phantine equations in two variables. The problem of determining all the solutions
of a linear Diophantine equation has been completely solved. Before stating the
general result, we will provide a few more examples.
Example 8.19 (A Linear Diophantine Equation in Two Variables)
The following example is similar to the examples studied in Preview Activity 2.
We can use substitution to verify that x D 2 and y D 1 is a solution of the
linear Diophantine equation
4x C 3y D 5:
The following table shows other solutions of this Diophantine equation.

x y x y
2 1 1 3
5 5 4 7
8 9 7 11
11 13 10 15
It would be nice to determine the pattern that these solutions exhibit. If we consider
the solution x D 2 and y D 1 to be the “starting point,” then we can see that
the other solutions are obtained by adding 3 to x and subtracting 4 from y in the
previous solution. So we can write these solutions to the equation as
x D 2 C 3k and y D 1 4k;
where k is an integer. We can use substitution and algebra to verify that these
expressions for x and y give solutions of this equation as follows:
4x C 3y D 4 .2 C 3k/ C 3 . 1 4k/
D .8 C 12k/ C . 3 12k/
D 5:
We should note that we have not yet proved that these solutions are all of the
solutions of the Diophantine equation 4x C 3y D 5. This will be done later.
If the general form for a linear Diophantine equation is ax C by D c, then for
this example, a D 4 and b D 3. Notice that for this equation, we started with one
solution and obtained other solutions by adding b D 3 to x and subtracting a D 4
from y in the previous solution. Also, notice that gcd.3; 4/ D 1.
Progress Check 8.20 (An Example of a Linear Diophantine Equation)
1. Verify that the following table shows some solutions of the linear Diophan-
tine equation 6x C 9y D 12.
x y x y
2 0 1 2
5 2 4 4
8 4 7 6
11 6 10 8
2. Follow the pattern in this table to determine formulas for x and y that will
generate integer solutions of the equation 6x C 9y D 12. Verify that the
formulas actually produce solutions for the equation 6x C 9y D 12.

Progress Check 8.21 (Revisiting Preview Activity 2)
Do the solutions for the linear Diophantine equations in Preview Activity 2 show
the same type of pattern as the solutions for the linear Diophantine equations in
Example 8.19 and Progress Check 8.20? Explain.
The solutions for the linear Diophantine equations in Preview Activity 2, Ex-
ample 8.19, and Progress Check 8.20 provide examples for the second part of The-
orem 8.22.
Theorem 8.22. Let a, b, and c be integers with a ¤ 0 and b ¤ 0, and let
d D gcd.a; b/.
1. If d does not divide c, then the linear Diophantine equation ax C by D c
has no solution.
2. If d divides c, then the linear Diophantine equation ax C by D c has in-
finitely many solutions. In addition, if .x0; y0/ is a particular solution of this
equation, then all the solutions of this equation can be written in the form
x D x0 C
b
d
k and y D y0
a
d
k;
for some integer k.
Proof. The proof of Part (1) is Exercise (1). For Part (2), we let a, b, and c be
integers with a ¤ 0 and b ¤ 0, and let d D gcd.a; b/. We also assume that d j c.
Since d D gcd.a; b/, Theorem 8.8 tells us that d is a linear combination of a and
b. So there exist integers s and t such that
d D as C bt: (1)
Since d j c, there exists an integer m such that c D dm. We can now multiply
both sides of equation (1) by m and obtain
dm D .as C bt/m
c D a.sm/ C b.tm/:
This means that x D sm, y D tm is a solution of ax C by D c, and we have
proved that the Diophantine equation ax C by D c has at least one solution.
Now let x D x0, y D y0 be any particular solution of ax C by D c, let k 2 Z,
and let
x D x0 C
b
d
k y D y0
a
d
k: (2)

We now verify that for each k 2 Z, the equations in (2) produce a solution of
ax C by D c.
ax C by D a

x0 C
b
d
k

C b

y0
a
d
k

D ax0 C
ab
d
k C by0
ab
d
k
D ax0 C by0
D c:
This proves that the Diophantine equation ax C by D c has infinitely many solu-
tions.
We now show that every solution of this equation can be written in the form
described in (2). So suppose that x and y are integers such that ax Cby D c. Then
.ax C by/ .ax0 C by0/ D c c D 0;
and this equation can be rewritten in the following form:
a.x x0/ D b.y0 y/: (3)
Dividing both sides of this equation by d, we obtain
a
d

.x x0/ D

b
d

.y0 y/ :
This implies that
a
d
divides

b
d

.y0 y/:
However, by Exercise (7) in Section 8.2, gcd

a
d
;
b
d

D 1, and so by Theo-
rem 8.12, we can conclude that
a
d
divides .y0 y/. This means that there exists
an integer k such that y0 y D
a
d
k, and solving for y gives
y D y0
a
d
k:
Substituting this value for y in equation (3) and solving for x yields
x D x0 C
b
d
k:
This proves that every solution of the Diophantine equation ax C by D c can be
written in the form prescribed in (2).

The proof of the following corollary to Theorem 8.22 is Exercise (2).
Corollary 8.23. Let a, b, and c be integers with a ¤ 0 and b ¤ 0. If a and b are
relatively prime, then the linear Diophantine equation ax C by D c has infinitely
many solutions. In addition, if x0, y0 is a particular solution of this equation, then
all the solutions of the equation are given by
x D x0 C bk y D y0 ak
where k 2 Z.
Progress Check 8.24 (Linear Diophantine Equations)
1. Use the Euclidean Algorithm to verify that gcd.63; 336/ D 21. What con-
clusion can be made about linear Diophantine equation 63x C 336y D 40
using Theorem 8.22? If this Diophantine equation has solutions, write for-
mulas that will generate the solutions.
2. Use the Euclidean Algorithm to verify that gcd.144; 225/ D 9. What con-
clusion can be made about linear Diophantine equation 144x C 225y D 27
using Theorem 8.22? If this Diophantine equation has solutions, write for-
mulas that will generate the solutions.
Exercises 8.3
1. Prove Part (1) of Theorem 8.22:
Let a, b, and c be integers with a ¤ 0 and b ¤ 0, and let d D gcd.a; b/. If
d does not divide c, then the linear Diophantine equation ax C by D c has
no solution.
2. Prove Corollary 8.23.
Let a, b, and c be integers with a ¤ 0 and b ¤ 0. If a and b are relatively
prime, then the linear Diophantineequation axCby D c has infinitely many
solutions. In addition, if .x0; y0/ is a particular solutionof this equation, then
all the solutions of the equation are given by
x D x0 C bk y D y0 ak;
where k 2 Z.
3. Determine all solutions of the following linear Diophantine equations.

? (a) 9x C 14y D 1
? (b) 18x C 22y D 4
? (c) 48x 18y D 15
? (d) 12x C 9y D 6
(e) 200x C 49y D 10
(f) 200x C 54y D 21
(g) 10x 7y D 31
(h) 12x C 18y D 6
?
4. A certain rare artifact is supposed to weigh exactly 25 grams. Suppose that
you have an accurate balance scale and 500 each of 27 gram weights and 50
gram weights. Explain how to use Theorem 8.22 to devise a plan to check
the weight of this artifact.
Hint: Notice that gcd.50; 27/ D 1. Start by writing 1 as a linear combination
of 50 and 27.
?
5. On the night of a certain banquet, a caterer offered the choice of two dinners,
a steak dinner for $25 and a vegetarian dinner for $16. At the end of the
evening, the caterer presented the host with a bill (before tax and tips) for
$1461. What is the minimum number of people who could have attended the
banquet? What is the maximum number of people who could have attended
the banquet?
6. The goal of this exercise is to determine all (integer) solutions of the linear
Diophantine equation in three variables 12x1 C 9x2 C 16x3 D 20.
? (a) First, notice that gcd.12; 9/ D 3. Determine formulas that will gener-
ate all solutions for the linear Diophantine equation
3y C 16x3 D 20.
?
(b) Explain why the solutions (for x1 and x2) of the Diophantine equation
12x1 C 9x2 D 3y can be used to generate solutions for
12x1 C 9x2 C 16x3 D 20.
? (c) Use the general value for y from Exercise (6a) to determine the solu-
tions of 12x1 C 9x2 D 3y.
(d) Use the results from Exercises (6a) and (6c) to determine formulas that
will generate all solutions for the Diophantine equation
12x1 C 9x2 C 16x3 D 20.
Note: These formulas will involve two arbitrary integer parameters.
Substitutespecific values for these integers and then check the resulting
solution in the original equation. Repeat this at least three times.
(e) Check the general solution for 12x1 C 9x2 C 16x3 D 20 from Exer-
cise (6d).

7. Use the method suggested in Exercise (6) to determine formulas that will
generate all solutions of the Diophantine equation 8x1 C 4x2 6x3 D 6.
Check the general solution.
8. Explain why the Diophantine equation 24x1 18x2 C 60x3 D 21 has no
solution.
9. The purpose of this exercise will be to prove that the nonlinear Diophantine
equation 3x2
y2
D 2 has no solution.
(a) Explain why if there is a solution of the Diophantine equation 3x2
y2
D 2, then that solution must also be a solution of the congruence
3x2
y2
2 .mod 3/.
(b) If there is a solution to the congruence 3x2 y2 2 .mod 3/,
explain why there then must be an integer y such that y2
2 .mod 3/.
(c) Use a proof by contradiction to prove that the Diophantine equation
3x2 y2 D 2 has no solution.
10. Use the method suggested in Exercise (9) to prove that the Diophantineequa-
tion 7x2 C 2 D y3 has no solution.
11. Linear Congruences in One Variable. Let n be a natural number and let
a; b 2 Z with a ¤ 0. A congruence of the form ax b .mod n/ is called a
linear congruence in one variable. This is called a linear congruence since
the variable x occurs to the first power.
A solution of a linear congruence in one variable is defined similarly to
the solution of an equation. A solution is an integer that makes the result-
ing congruence true when the integer is substituted for the variable x. For
example,
The integer x D 3 is a solution for the congruence 2x 1 .mod 5/
since 2 3 1 .mod 5/ is a true congruence.
The integer x D 7 is not a solution for the congruence 3x 1 .mod 6/
since 3 7 1 .mod 6/ is a not a true congruence.
(a) Verify that x D 2 and x D 5 are the only solutions the linear congru-
ence 4x 2 .mod 6/ with 0 x 6.

(b) Show that the linear congruence 4x 3 .mod 6/ has no solutions with
0 x 6.
(c) Determine all solutions of the linear congruence 3x 7 .mod 8/ with
0 x 8.
The following parts of this activity show that we can use the results of Theo-
rem 8.22 to help find all solutions of the linear congruence 6x 4 .mod 8/.
(d) Verify that x D 2 and x D 6 are the only solutions for the linear
congruence 6x 4 .mod 8/ with 0 x 8.
(e) Use the definition of “congruence” to rewrite the congruence
6x 4 .mod 8/ in terms of “divides.”
(f) Use the definition of “divides” to rewrite the result in part (11e) in the
form of an equation. (An existential quantifier must be used.)
(g) Use the results of parts (11d) and (11f) to write an equation that will
generate all the solutions of the linear congruence 6x 4 .mod 8/.
Hint: Use Theorem 8.22. This can be used to generate solutions for x
and the variable introduced in part (11f). In this case, we are interested
only in the solutions for x.
Now let n be a natural number and let a; c 2 Z with a ¤ 0. A general linear
congruence of the form ax c .mod n/ can be handled in the same way
that we handled in 6x 4 .mod 8/.
(h) Use the definition of “congruence” to rewrite ax c .mod n/ in terms
of “divides.”
(i) Use the definition of “divides” to rewrite the result in part (11h) in the
form of an equation. (An existential quantifier must be used.)
(j) Let d D gcd.a; n/. State and prove a theorem about the solutions of
the linear congruence ax c .mod n/ in the case where d does not
divide c.
Hint: Use Theorem 8.22.
(k) Let d D gcd.a; n/. State and prove a theorem about the solutions of
the linear congruence ax c .mod n/ in the case where d divides c.

Greatest common divisor of two integers, page 414
Linear combination of two integers, page 423
Prime number, page 426
Composite number, page 426
Prime factorization, page 427
Relatively prime integers, page 428
Diophantine equation, page 441
Linear Diophantine equation in two variables, page 441
Important Theorems and Results about Relations, Equivalence Rela-
tions, and Equivalence Classes
Theorem 8.3. Let a and b be integers with a ¤ 0 and b 0. Then gcd.a; b/
is the only natural number d such that
(a) d divides a,
(b) d divides b, and
(c) if k is an integer that divides both a and b, then k divides d.
Theorem 8.8. Let a and b be integers, not both 0. Then gcd.a; b/ can be
written as a linear combination of a and b. That is, there exist integers u
and v such that gcd.a; b/ D au C bv.
Theorem 8.9.
1. The greatest common divisor, d, is a linear combination of a and b.
That is, there exist integers m and n such that d D am C bn.
2. The greatest common divisor, d, divides every linear combination of a
and b. That is, for all x; y 2 Z , d j .ax C by/.
3. The greatest common divisor, d, is the smallest positive number that is
a linear combination of a and b.

Theorem 8.11. Let a and b be nonzero integers, and let p be a prime num-
ber.
1. If a and b are relatively prime, then there exist integers m and n such
that am C bn D 1. That is, 1 can be written as linear combination of
a and b.
2. If p j a, then gcd.a; p/ D p.
3. If p does not divide a, then gcd.a; p/ D 1.
Theorem 8.12 Let a, b, and c be integers. If a and b are relatively prime
and a j .bc/, then a j c.
Corollary 8.14
1. Let a; b 2 Z, and let p be a prime number. If p j .ab/, then p j a or
p j b.
2. Let p be a prime number, let n 2 N, and let a1; a2; : : :; an 2 Z. If
p j .a1a2 an/, then there exists a natural number k with 1 k n
such that p j ak.
Theorem 8.15, The Fundamental Theorem of Arithmetic
1. Each natural number greater than 1 is either a prime number or is a
product of prime numbers.
2. Let n 2 N with n 1. Assume that
n D p1p2 pr and that n D q1q2 qs;
where p1; p2; pr and q1; q2; qs are primes with p1 p2
pr and q1 q2 qs. Then r D s, and for each j
from 1 to r, pj D qj .
Theorem 8.16. There are infinitely many prime numbers.
Theorem 8.22. Let a, b, and c be integers with a ¤ 0 and b ¤ 0, and let
d D gcd.a; b/.
1. If d does not divide c, then the linear Diophantineequation axCby D
c has no solution.

2. If d divides c, then the linear Diophantine equation ax C by D c
has infinitely many solutions. In addition, if .x0; y0/ is a particular
solution of this equation, then all the solutionsof the equation are given
by
x D x0 C
b
d
k and y D y0
a
d
k;
where k 2 Z.
Corollary 8.23. Let a, b, and c be integers with a ¤ 0 and b ¤ 0. If a and
b are relatively prime, then the linear Diophantine equation ax C by D c
has infinitely many solutions. In addition, if x0, y0 is a particular solution
of this equation, then all the solutions of the equation are given by
x D x0 C bk and y D y0 ak;
where k 2 Z.

Chapter 9
Finite and Infinite Sets
9.1 Finite Sets
Preview Activity 1 (Equivalent Sets, Part 1)
1. Let A and B be sets and let f be a function from A to B. .f W A ! B/.
Carefully complete each of the following using appropriate quantifiers: (If
necessary, review the material in Section 6.3.)
(a) The function f is an injection provided that : : : :
(b) The function f is not an injection provided that : : : :
(c) The function f is a surjection provided that : : : :
(d) The function f is not a surjection provided that : : : :
(e) The function f is a bijection provided that : : ::
Definition. Let A and B be sets. The set A is equivalent to the set B provided
that there exists a bijection from the set A onto the set B. In this case, we write
A B.
When A B, we also say that the set A is in one-to-one correspondence
with the set B and that the set A has the same cardinality as the set B.
Note: When A is not equivalent to B, we write A 6 B.
2. For each of the following, use the definition of equivalent sets to determine
if the first set is equivalent to the second set.
452

9.1. Finite Sets 453
(a) A D f1; 2; 3g and B D fa; b; cg
(b) C D f1; 2g and B D fa; b; cg
(c) X D f1; 2; 3; : : :; 10g and Y D f57; 58; 59; : : :; 66g
3. Let DC be the set of all odd natural numbers. Prove that the function
f W N ! DC
defined by f .x/ D 2x 1, for all x 2 N, is a bijection
and hence that N DC.
4. Let RC
be the set of all positive real numbers. Prove that the function
g W R ! RC
defined by g.x/ D ex
, for all x 2 R is a bijection and hence,
that R RC.
Preview Activity 2 (Equivalent Sets, Part 2)
1. Review Theorem 6.20 in Section 6.4, Theorem 6.26 in Section 6.5, and Ex-
ercise (9) in Section 6.5.
2. Prove each part of the following theorem.
Theorem 9.1. Let A, B, and C be sets.
(a) For each set A, A A.
(b) For all sets A and B, if A B, then B A.
(c) For all sets A, B, and C , if A B and B C , then A C .
Equivalent Sets
In Preview Activity 1, we introduced the concept of equivalent sets. The motiva-
tion for this definition was to have a formal method for determining whether or not
two sets “have the same number of elements.” This idea was described in terms of
a one-to-one correspondence (a bijection) from one set onto the other set. This idea
may seem simple for finite sets, but as we will see, this idea has surprising conse-
quences when we deal with infinite sets. (We will soon provide precise definitions
for finite and infinite sets.)
Technical Note: The three properties we proved in Theorem 9.1 in Preview Ac-
tivity 2 are very similar to the concepts of reflexive, symmetric, and transitive re-
lations. However, we do not consider equivalence of sets to be an equivalence

454 Chapter 9. Finite and Infinite Sets
relation on a set U since an equivalence relation requires an underlying (universal)
set U . In this case, our elements would be the sets A, B, and C , and these would
then have to subsets of some universal set W (elements of the power set of W ). For
equivalence of sets, we are not requiring that the sets A, B, and C be subsets of the
same universal set. So we do not use the term relation in regards to the equivalence
of sets. However, if A and B are sets and A B, then we often say that A and B
are equivalent sets.
Progress Check 9.2 (Examples of Equivalent Sets)
We will use the definition of equivalent sets from in Preview Activity 1 in all parts
of this progress check. It is no longer sufficient to say that two sets are equivalent
by simply saying that the two sets have the same number of elements.
1. Let A D f1; 2; 3; : : :; 99; 100g and let B D f351; 352; 353; : : :; 449; 450g.
Define f WA ! B by f.x/ D x C 350, for each x in A. Prove that f is a
bijection from the set A to the set B and hence, A B.
2. Let E be the set of all even integers and let D be the set of all odd integers.
Prove that E D by proving that F W E ! D, where F .x/ D x C 1, for
all x 2 E, is a bijection.
3. Let .0; 1/ be the open interval of real numbers between 0 and 1. Similarly, if
b 2 R with b 0, let .0; b/ be the open interval of real numbers between 0
and b.
Prove that the function f W.0; 1/ ! .0; b/ by f .x/ D bx, for all x 2 .0; 1/,
is a bijection and hence .0; 1/ .0; b/.
In Part (3) of Progress Check 9.2, notice that if b 1, then .0; 1/ is a proper
subset of .0; b/ and .0; 1/ .0; b/.
Also, in Part (3) of Preview Activity 1, we proved that the set D of all odd
natural numbers is equivalent to N, and we know that D is a proper subset of N.
These results may seem a bit strange, but they are logical consequences of
the definition of equivalent sets. Although we have not defined the terms yet, we
will see that one thing that will distinguish an infinite set from a finite set is that
an infinite set can be equivalent to one of its proper subsets, whereas a finite set
cannot be equivalent to one of its proper subsets.

Finite Sets
In Section 5.1, we defined the cardinality of a finite set A, denoted by card .A/,
to be the number of elements in the set A. Now that we know about functions and
bijections, we can define this concept more formally and more rigorously. First,
for each k 2 N, we define Nk to be the set of all natural numbers between 1 and k,
inclusive. That is,
Nk D f1; 2; : : :; kg:
We will use the concept of equivalent sets introduced in Preview Activity 1 to
define a finite set.
Definition. A set A is a finite set provided that A D ; or there exists a natural
number k such that A Nk. A set is an infinite set provided that it is not a
finite set.
If A Nk, we say that the set A has cardinality k (or cardinal number k),
and we write card .A/ D k.
In addition, we say that the empty set has cardinality 0 (or cardinal number
0), and we write card .;/ D 0.
Notice that by this definition, the empty set is a finite set. In addition, for each
k 2 N, the identity function on Nk is a bijection and hence, by definition, the set
Nk is a finite set with cardinality k.
Theorem 9.3. Any set equivalent to a finite nonempty set A is a finite set and has
the same cardinality as A.
Proof. Suppose that A is a finite nonempty set, B is a set, and A B. Since A
is a finite set, there exists a k 2 N such that A Nk. We also have assumed that
A B and so by part (b) of Theorem 9.1 (in Preview Activity 2), we can conclude
that B A. Since A Nk, we can use part (c) of Theorem 9.1 to conclude that
B Nk. Thus, B is finite and has the same cardinality as A.
It may seem that we have done a lot of work to prove an “obvious” result in
Theorem 9.3. The same may be true of the remaining results in this section, which
give further results about finite sets. One of the goals is to make sure that the con-
cept of cardinality for a finite set corresponds to our intuitive notion of the number
of elements in the set. Another important goal is to lay the groundwork for a more
rigorous and mathematical treatment of infinite sets than we have encountered be-
fore. Along the way, we will see the mathematical distinction between finite and
infinite sets.

The following two lemmas will be used to prove the theorem that states that
every subset of a finite set is finite.
Lemma 9.4. If A is a finite set and x … A, then A [ fxg is a finite set and
card.A [ fxg/ D card.A/ C 1.
Proof. Let A be a finite set and assume card.A/ D k, where k D 0 or k 2 N.
Assume x … A.
If A D ;, then card.A/ D 0 and A [ fxg D fxg, which is equivalent to N1.
Thus, A [ fxg is finite with cardinality 1, which equals card.A/ C 1.
If A ¤ ;, then A Nk, for some k 2 N. This means that card.A/ D k, and
there exists a bijection f W A ! Nk. We will now use this bijection to define a
function gWA [ fxg ! NkC1 and then prove that the function g is a bijection. We
define gWA [ fxg ! NkC1 as follows: For each t 2 A [ fxg,
g .t/ D
(
f .t/ if t 2 A
k C 1 if t D x:
To prove that g is an injection, we let x1; x2 2 A [ fxg and assume x1 ¤ x2.
If x1; x2 2 A, then since f is a bijection, f.x1/ ¤ f.x2/, and this implies
that g.x1/ ¤ g.x2/.
If x1 D x, then since x2 ¤ x1, we conclude that x2 ¤ x and hence x2 2 A.
So g.x1/ D k C 1, and since f.x2/ 2 Nk and g.x2/ D f.x2/, we can
conclude that g.x1/ ¤ g.x2/.
This proves that the function g is an injection. The proof that g is a surjection is
Exercise (1). Since g is a bijection, we conclude that A [ fxg NkC1, and
card.A [ fxg/ D k C 1:
Since card .A/ D k, we have proved that card.A [ fxg/ D card.A/ C 1:
Lemma 9.5. For each natural number m, if A Nm, then A is a finite set and
card .A/ m.
Proof. We will use a proof using induction on m. For each m 2 N, let P.m/ be,
“If A Nm, then A is finite and card.A/ m.”

We first prove that P.1/ is true. If A N1, then A D ; or A D f1g, both
of which are finite and have cardinality less than or equal to the cardinality of N1.
This proves that P.1/ is true.
For the inductive step, let k 2 N and assume that P.k/ is true. That is, assume
that if B Nk, then B is a finite set and card.B/ k. We need to prove that
P.k C 1/ is true.
So assume that A is a subset of NkC1. Then A fk C 1g is a subset of Nk.
Since P.k/ is true, A fk C 1g is a finite set and
card.A fk C 1g/ k:
There are two cases to consider: Either k C 1 2 A or k C 1 62 A.
If k C 1 62 A, then A D A fk C 1g. Hence, A is finite and
card.A/ k k C 1:
If k C 1 2 A, then A D .A fk C 1g/ [ fk C 1g. Hence, by Lemma 9.4, A is
a finite set and
card.A/ D card.A fk C 1g/ C 1:
Since card.A fk C 1g/ k, we can conclude that card.A/ k C 1.
This means that we have proved the inductive step. Hence, by mathematical
induction, for each m 2 N, if A Nm, then A is finite and card.A/ m.
The preceding two lemmas were proved to aid in the proof of the following
theorem.
Theorem 9.6. If S is a finite set and A is a subset of S, then A is a finite set and
card.A/ card.S/.
Proof. Let S be a finite set and assume that A is a subset of S. If A D ;, then A
is a finite set and card.A/ card.S/. So we assume that A ¤ ;.
Since S is finite, there exists a bijection f W S ! Nk for some k 2 N. In this
case, card.S/ D k. We need to show that A is equivalent to a finite set. To do this,
we define gWA ! f.A/ by
g.x/ D f.x/ for each x 2 A.

Since f is an injection, we conclude that g is an injection. Now let y 2 f.A/.
Then there exists an a 2 A such that f.a/ D y. But by the definition of g, this
means that g.a/ D y, and hence g is a surjection. This proves that g is a bijection.
Hence, we have proved that A f.A/. Since f.A/ is a subset of Nk, we use
Lemma 9.5 to conclude that f.A/ is finite and card.f.A// k. In addition, by
Theorem 9.3, A is a finite set and card.A/ D card.f.A//. This proves that A is a
finite set and card.A/ card.S/.
Lemma 9.4 implies that adding one element to a finite set increases its cardi-
nality by 1. It is also true that removing one element from a finite nonempty set
reduces the cardinality by 1. The proof of Corollary 9.7 is Exercise (4).
Corollary 9.7. If A is a finite set and x 2 A, then A fxg is a finite set and
card.A fxg/ D card.A/ 1.
The next corollary will be used in the next section to provide a mathematical
distinction between finite and infinite sets.
Corollary 9.8. A finite set is not equivalent to any of its proper subsets.
Proof. Let B be a finite set and assume that A is a proper subset of B. Since A is
a proper subset of B, there exists an element x in B A. This means that A is a
subset of B fxg. Hence, by Theorem 9.6,
card.A/ card.B fxg/ :
Also, by Corollary 9.7
card.B fxg/ D card.B/ 1:
Hence, we may conclude that card.A/ card.B/ 1 and that
card.A/ card.B/:
Theorem 9.3 implies that B 6 A. This proves that a finite set is not equivalent to
any of its proper subsets.

The Pigeonhole Principle
The last property of finite sets that we will consider in this section is often called
the Pigeonhole Principle. The “pigeonhole” version of this property says, “If m
pigeons go into r pigeonholes and m r, then at least one pigeonhole has more
than one pigeon.”
In this situation, we can think of the set of pigeons as being equivalent to a
set P with cardinality m and the set of pigeonholes as being equivalent to a set
H with cardinality r. We can then define a function f W P ! H that maps each
pigeon to its pigeonhole. The Pigeonhole Principle states that this function is not
an injection. (It is not one-to-one since there are at least two pigeons “mapped” to
the same pigeonhole.)
Theorem 9.9 (The Pigeonhole Principle). Let A and B be finitesets. If card.A/
card.B/, then any function f WA ! B is not an injection.
Proof. Let A and B be finite sets. We will prove the contrapositive of the theorem,
which is, if there exists a function f WA ! B that is an injection, then card.A/
card.B/.
So assume that f W A ! B is an injection. As in Theorem 9.6, we define a
function gWA ! f.A/ by
g.x/ D f.x/ for each x 2 A.
As we saw in Theorem 9.6, the function g is a bijection. But then A f.A/ and
f.A/ B. Hence,
card.A/ D card.f.A// and card.f.A// card.B/.
Hence, card.A/ card.B/, and this proves the contrapositive. Hence, if card.A/
card.B/, then any function f WA ! B is not an injection.
The Pigeonhole Principle has many applications in the branch of mathematics
called “combinatorics.” Some of these will be explored in the exercises.
Exercises 9.1
1. Prove that the function g W A [ fxg ! NkC1 in Lemma 9.4 is a surjection.

? 2. Let A be a subset of some universal set U . Prove that if x 2 U, then
A fxg A.
? 3. Let EC be the set of all even natural numbers. Prove that N EC.
? 4. Prove Corollary 9.7.
If A is a finite set and x 2 A, then A fxg is a finite set and
card.A fxg/ D card.A/ 1.
Hint: One approach is to use the fact that A D .A fxg/ [ fxg.
5. Let A and B be sets. Prove that
? (a) If A is a finite set, then A B is a finite set.
? (b) If A [ B is a finite set, then A and B are finite sets.
(c) If A B is an infinite set, then A is an infinite set.
(d) If A is an infinite set or B is an infinite set, then A [ B is an infinite
set.
6. There are over 7 million people living in New York City. It is also known
that the maximum number of hairs on a human head is less than 200,000.
Use the Pigeonhole Principle to prove that there are at least two people in
the city of New York with the same number of hairs on their heads.
7. Prove the following propositions:
?
(a) If A, B, C , and D are sets with A B and C D, then
A C B D.
(b) If A, B, C , and D are sets with A B and C D and if A and C
are disjoint and B and D are disjoint, then A [ C B [ D.
Hint: Since A B and C D, there exist bijections f W A ! B and
g W C ! D. To prove that A C B D, prove that h W A C ! B D
is a bijection, where h.a; c/ D .f.a/; g.c//, for all .a; c/ 2 A C .
8. Let A D fa; b; cg.
?
(a) Construct a function f W N5 ! A such that f is a surjection.
(b) Use the function f to construct a function g W A ! N5 so that
f ı g D IA, where IA is the identity function on the set A. Is the
function g an injection? Explain.

9. This exercise is a generalization of Exercise (8). Let m be a natural num-
ber, let A be a set, and assume that f W Nm ! A is a surjection. Define
g W A ! Nm as follows:
For each x 2 A, g.x/ D j , where j is the least natural number in
f 1.fxg/.
Prove that f ı g D IA, where IA is the identity function on the set A and
prove that g is an injection.
10. Let B be a finite, nonempty set and assume that f W B ! A is a surjection.
Prove that there exists a function h W A ! B such that f ı h D IA and h is
an injection.
Hint: Since B is finite, there exists a natural number m such that Nm B.
This means there exists a bijection k W Nm ! B. Now let h D k ı g, where
g is the function constructed in Exercise (9).
11. Using the Pigeonhole Principle. For this activity, we will consider subsets
of N30 that contain eight elements.
(a) One such set is A D f3; 5; 11; 17; 21; 24; 26; 29g. Notice that
f3; 21; 24; 26g A and 3 C 21 C 24 C 26 D 74
f3; 5; 11; 26; 29g A and 3 C 5 C 11 C 26 C 29 D 74:
Use this information to find two disjoint subsets of A whose elements
have the same sum.
(b) Let B D f3; 6; 9; 12; 15; 18; 21; 24g. Find two disjoint subsets of B
whose elements have the same sum. Note: By convention, if T D fag,
where a 2 N, then the sum of the elements in T is equal to a.
(c) Now let C be any subset of N30 that contains eight elements.
i. How many subsets does C have?
ii. The sum of the elements of the empty set is 0. What is the maxi-
mum sum for any subset of N30 that contains eight elements.? Let
M be this maximum sum.
iii. Now define a function f W P.C / ! NM so that for each X 2
P.C /, f.X/ is equal to the sum of the elements in X.
Use the Pigeonhole Principle to prove that there exist two subsets
of C whose elements have the same sum.

(d) If the two subsets in part (11(c)iii) are not disjoint, use the idea pre-
sented in part (11a) to prove that there exist two disjoint subsets of C
whose elements have the same sum.
(e) Let S be a subset of N99 that contains 10 elements. Use the Pigeon-
hole Principle to prove that there exist two disjoint subsets of S whose
elements have the same sum.
9.2 Countable Sets
Preview Activity 1 (Introduction to Infinite Sets)
In Section 9.1, we defined a finite set to be the empty set or a set A such that
A Nk for some natural number k. We also defined an infinite set to be a set that
is not finite, but the question now is, “How do we know if a set is infinite?” One
way to determine if a set is an infinite set is to use Corollary 9.8, which states that
a finite set is not equivalent to any of its subsets. We can write this as a conditional
statement as follows:
If A is a finite set, then A is not equivalent to any of its proper subsets.
or more formally as
For each set A, if A is a finite set, then for each proper subset B of A, A 6 B.
1. Write the contrapositive of the preceding conditional statement. Then ex-
plain how this statement can be used to determine if a set is infinite.
2. Let DC
be the set of all odd natural numbers. In Preview Activity 1 from
Section 9.1, we proved that N DC
.
(a) Use this to explain carefully why N is an infinite set.
(b) Is DC a finite set or an infinite set? Explain carefully how you know.
3. Let b be a positive real number. Let .0; 1/ and .0; b/ be the open intervals
from 0 to 1 and 0 to b, respectively. In Part (3) of Progress Check 9.2 (on
page 454), we proved that .0; 1/ .0; b/.
(a) Use a value for b where 0 b 1 to explain why .0; 1/ is an infinite
set.
(b) Use a value for b where b 1 to explain why .0; b/ is an infinite set.

9.2. Countable Sets 463
Preview Activity 2 (A Function from N to Z)
In this preview activity, we will define and explore a function f W N ! Z. We will
start by defining f.n/ for the first few natural numbers n.
f.1/ D 0
f.2/ D 1 f.3/ D 1
f.4/ D 2 f.5/ D 2
f.6/ D 3 f.7/ D 3
Notice that if we list the outputs of f in the order f.1/; f.2/; f.3/; : : :, we create
the following list of integers: 0; 1; 1; 2; 2; 3; 3; : : :. We can also illustrate the
outputs of this function with the following diagram:
1 2 3 4 5 6 7 8 9 10
# # # # # # # # # #
0 1 1 2 2 3 3 4 4 5
Figure 9.1: A Function from N to Z
1. If the pattern suggested by the function values we have defined continues,
what are f.11/ and f.12/? What is f.n/ for n from 13 to 16?
2. If the pattern of outputs continues, does the function f appear to be an in-
jection? Does f appear to be a surjection? (Formal proofs are not required.)
We will now attempt to determine a formula for f.n/, where n 2 N. We will
actually determine two formulas: one for when n is even and one for when n is
odd.
3. Look at the pattern of the values of f.n/ when n is even. What appears to
be a formula for f.n/ when n is even?
4. Look at the pattern of the values of f.n/ when n is odd. What appears to be
a formula for f.n/ when n is odd?
5. Use the work in Part (3) and Part (4) to complete the following: Define
f WN ! Z, where
f.n/ D
8̂

:̂
‹‹ if n is even
‹‹ if n is odd:

6. Use the formula in Part (5) to
(a) Calculate f.1/ through f.10/. Are these results consistent with the
pattern exhibited at the beginning of this preview activity?
(b) Calculate f.1000/ and f.1001/.
(c) Determine the value of n so that f.n/ D 1000.
In this section, we will describe several infinite sets and define the cardinal
number for so-called countable sets. Most of our examples will be subsets of some
of our standard numbers systems such as N, Z, and Q.
Infinite Sets
In Preview Activity 1, we saw how to use Corollary 9.8 to prove that a set is infinite.
This corollary implies that if A is a finite set, then A is not equivalent to any of its
proper subsets. By writing the contrapositive of this conditional statement, we can
restate Corollary 9.8 in the following form:
Corollary 9.8 If a set A is equivalent to one of its proper subsets, then A is infinite.
In Preview Activity 1, we used Corollary 9.8 to prove that
The set of natural numbers, N, is an infinite set.
The open interval .0; 1/ is an infinite set.
Although Corollary 9.8 provides one way to prove that a set is infinite, it is some-
times more convenient to use a proof by contradiction to prove that a set is infinite.
The idea is to use results from Section 9.1 about finite sets to help obtain a contra-
diction. This is illustrated in the next theorem.
Theorem 9.10. Let A and B be sets.
1. If A is infinite and A B, then B is infinite.
Proof. We will prove part (1). The proof of part (2) is exercise (3) on page 473.

To prove part (1), we use a proof by contradiction and assume that A is an
infinite set, A B, and B is not infinite. That is, B is a finite set. Since A B
and B is finite, Theorem 9.3 on page 455 implies that A is a finite set. This is a
contradiction to the assumption that A is infinite. We have therefore proved that if
A is infinite and A B, then B is infinite.
Progress Check 9.11 (Examples of Infinite Sets)
1. In Preview Activity 1, we used Corollary 9.8 to prove that N is an infinite
set. Now use this and Theorem 9.10 to explain why our standard number
systems .Z; Q; and R/ are infinite sets. Also, explain why the set of all
positive rational numbers, QC
, and the set of all positive real numbers, RC
,
are infinite sets.
2. Let DC be the set of all odd natural numbers. In Part (2) of Preview Activ-
ity 1, we proved that DC
N. Use Theorem 9.10 to explain why DC
is an
infinite set.
3. Prove that the set EC
of all even natural numbers is an infinite set.
Countably Infinite Sets
In Section 9.1, we used the set Nk as the standard set with cardinality k in the sense
that a set is finite if and only if it is equivalent to Nk. In a similar manner, we will
use some infinite sets as standard sets for certain infinite cardinal numbers. The
first set we will use is N.
We will formally define what it means to say the elements of a set can be
“counted” using the natural numbers. The elements of a finite set can be “counted”
by defining a bijection (one-to-one correspondence) between the set and Nk for
some natural number k. We will be able to “count” the elements of an infinite set
if we can define a one-to-one correspondence between the set and N.
Definition. The cardinality of N is denoted by @0. The symbol @ is the first
letter of the Hebrew alphabet, aleph. The subscript 0 is often read as “naught”
(or sometimes as “zero” or “null”). So we write
card.N/ D @0
and say that the cardinality of N is “aleph naught.”

Definition. A set A is countably infinite provided that A N. In this case,
we write
card.A/ D @0:
A set that is countably infinite is sometimes called a denumerable set. A set
is countable provided that it is finite or countably infinite. An infinite set that
is not countably infinite is called an uncountable set.
Progress Check 9.12 (Examples of Countably Infinite Sets)
1. In Preview Activity 1 from Section 9.1, we proved that N DC
, where DC
is the set of all odd natural numbers. Explain why card.DC/ D @0.
2. Use a result from Progress Check 9.11 to explain why card.EC
/ D @0.
3. At this point, if we wish to prove a set S is countably infinite, we must find
a bijection between the set S and some set that is known to be countably
infinite.
Let S be the set of all natural numbers that are perfect squares. Define a
function
f WS ! N
that can be used to prove that S N and, hence, that card.S/ D @0.
The fact that the set of integers is a countably infinite set is important enough to
be called a theorem. The function we will use to establish that N Z was explored
in Preview Activity 2.
Theorem 9.13. The set Z of integers is countably infinite, and so card.Z/ D @0.
Proof. To prove that N Z, we will use the following function: f W N ! Z,
where
f.n/ D
8̂
ˆ
ˆ

ˆ
ˆ
:̂
n
2
if n is even
1 n
2
if n is odd:
From our work in Preview Activity 2, it appears that if n is an even natural number,
then f.n/ 0, and if n is an odd natural number, then f.n/ 0. So it seems
reasonable to use cases to prove that f is a surjection and that f is an injection.
To prove that f is a surjection, we let y 2 Z.

If y 0, then 2y 2 N, and
f.2y/ D
2y
2
D y:
If y 0, then 2y 0 and 1 2y is an odd natural number. Hence,
f.1 2y/ D
1 .1 2y/
2
D
2y
2
D y:
These two cases prove that if y 2 Z, then there exists an n 2 N such that
f.n/ D y. Hence, f is a surjection.
To prove that f is an injection, we let m; n 2 N and assume that f.m/ D f.n/.
First note that if one of m and n is odd and the other is even, then one of f.m/ and
f.n/ is positive and the other is less than or equal to 0. So if f.m/ D f.n/, then
both m and n must be even or both m and n must be odd.
If both m and n are even, then
f.m/ D f.n/ implies that
m
2
D
n
2
and hence that m D n.
If both m and n are odd, then
f.m/ D f.n/ implies that
1 m
2
D
1 n
2
.
From this, we conclude that 1 m = 1 n and hence that m D n. This
proves that if f.m/ D f.n/, then m D n and hence that f is an injection.
Since f is both a surjection and an injection, we see that f is a bijection and,
therefore, N Z. Hence, Z is countably infinite and card.Z/ D @0.
The result in Theorem 9.13 can seem a bit surprising. It exhibits one of the
distinctions between finite and infinite sets. If we add elements to a finite set, we
will increase its size in the sense that the new set will have a greater cardinality
than the old set. However, with infinite sets, we can add elements and the new
set may still have the same cardinality as the original set. For example, there is a
one-to-one correspondence between the elements of the sets N and Z. We say that
these sets have the same cardinality.
Following is a summary of some of the main examples dealing with the cardi-
nality of sets that we have explored.

The sets Nk, where k 2 N, are examples of sets that are countable and finite.
The sets N, Z, the set of all odd natural numbers, and the set of all even nat-
ural numbers are examples of sets that are countable and countably infinite.
We have not yet proved that any set is uncountable.
The Set of Positive Rational Numbers
If we expect to find an uncountable set in our usual number systems, the rational
numbers might be the place to start looking. One of the main differences between
the set of rational numbers and the integers is that given any integer m, there is a
next integer, namely m C 1. This is not true for the set of rational numbers. We
know that Q is closed under division (by nonzero rational numbers) and we will
see that this property implies that given any two rational numbers, we can also find
a rational number between them. In fact, between any two rational numbers, we
can find infinitely many rational numbers. It is this property that may lead us to
believe that there are “more” rational numbers than there are integers.
The basic idea will be to “go half way” between two rational numbers. For
example, if we use a D
1
3
and b D
1
2
, we can use
a C b
2
D
1
2

1
3
C
1
2

D
5
12
as a rational number between a and b. We can then repeat this process to find a
rational number between
5
12
and
1
2
.
So we will now let a and b be any two rational numbers with a b and let
c1 D
a C b
2
. We then see that
c1 a D
a C b
2
a b c1 D b
a C b
2
D
a C b
2
2a
2
D
2b
2
a C b
2
D
b a
2
D
b a
2
Since b a, we see that b a 0 and so the previous equations show that
c1 a 0 and b c1 0. We can then conclude that a c1 b.

We can now repeat this process by using c2 D
c1 C b
2
and proving that c1
c2 b. In fact, for each natural number, we can define
ckC1 D
ck C b
2
and obtain the result that a c1 c2 cn b and this proves
that the set fck j k 2 Ng is a countably infinite set where each element is a rational
number between a and b. (A formal proof can be completed using mathematical
induction. See Exercise ().
This result is true no matter how close together a and b are. For example,
we can now conclude that there are infinitely many rational numbers between 0
and
1
10000
. This might suggest that the set Q of rational numbers is uncountable.
Surprisingly, this is not the case. We start with a proof that the set of positive
rational numbers is countable.
Theorem 9.14. The set of positive rational numbers is countably infinite.
Proof. We can write all the positive rational numbers in a two-dimensional array
as shown in Figure 9.2. The top row in Figure 9.2 represents the numerator of the
rational number, and the left column represents the denominator. We follow the
arrows in Figure 9.2 to define f W N ! QC
. The idea is to start in the upper left
corner of the table and move to successive diagonals as follows:
We start with all fractions in which the sum of the numerator and denomina-
tor is 2

only
1
1

. So f.1/ D
1
1
.
We next use those fractions in which the sum of the numerator and denomi-
nator is 3. So f.2/ D
2
1
and f.3/ D
1
2
.
We next use those fractions in which the sum of the numerator and denom-
inator is 4. So f.4/ D
1
3
, f.5/ D
3
1
. We skipped
2
2
since
2
2
D
1
1
. In this
way, we will ensure that the function f is a one-to-one function.
We now continue with successive diagonals omitting fractions that are not in low-
est terms. This process guarantees that the function f will be an injection and a
surjection. Therefore, N QC and card.QC/ D @0.
Note: For another proof of Theorem 9.14, see exercise (14) on page 475.

1 2 3 4 5
1
2
3
4
5
1
1
2
1
3
1
4
1
5
1
1
2
2
2
3
2
4
2
5
2
1
3
2
3
3
3
4
3
5
3
1
4
2
4
3
4
4
4
5
4
1
5
2
5
3
5
4
5
5
5
Figure 9.2: Counting the Positive Rational Numbers
Since QC
is countable, it seems reasonable to expect that Q is countable. We will
explore this soon. On the other hand, at this point, it may also seem reasonable to
ask,
“Are there any uncountable sets?”
The answer to this question is yes, but we will wait until the next section to prove
that certain sets are uncountable. We still have a few more issues to deal with
concerning countable sets.
Countably Infinite Sets
Theorem 9.15. If A is a countably infinite set, then A [ fxg is a countably infinite
set.
Proof. Let A be a countably infinite set. Then there exists a bijection f WN ! A.
Since x is either in A or not in A, we can consider two cases.

If x 2 A, then A [ fxg D A and A [ fxg is countably infinite.
If x … A, define gWN ! A [ fxg by
g.n/ D
(
x if n D 1
f.n 1/ if n 1.
The proof that the function g is a bijection is Exercise (4). Since g is a bijection,
we have proved that A [ fxg N and hence, A [ fxg is countably infinite.
Theorem 9.16. If A is a countably infinite set and B is a finite set, then A [ B is
a countably infinite set.
Proof. Exercise (5) on page 473.
Theorem 9.16 says that if we add a finite number of elements to a countably in-
finite set, the resulting set is still countably infinite. In other words, the cardinality
of the new set is the same as the cardinality of the original set. Finite sets behave
very differently in the sense that if we add elements to a finite set, we will change
the cardinality. What may even be more surprising is the result in Theorem 9.17
that states that the union of two countably infinite (disjoint) sets is countably infi-
nite. The proof of this result is similar to the proof that the integers are countably
infinite (Theorem 9.13). In fact, if A D fa1; a2; a3; : : :g and B D fb1; b2; b3; : : :g,
then we can use the following diagram to help define a bijection from N to A [ B.
1 2 3 4 5 6 7 8 9 10
# # # # # # # # # #
a1 b1 a2 b2 a3 b3 a4 b4 a5 b5
Figure 9.3: A Function from N to A [ B
Theorem 9.17. If A and B are disjoint countably infinite sets, then A [ B is a
countably infinite set.
Proof. Let A and B be countably infinite sets and let f W N ! A and g W N ! B
be bijections. Define hWN ! A [ B by
h.n/ D
8̂
ˆ
ˆ

ˆ
ˆ
:̂
f

n C 1
2

if n is odd
g
n
2

if n is even:

It is left as Exercise (6) on page 474 to prove that the function h is a bijection.
Since we can write the set of rational numbers Q as the union of the set of non-
negative rational numbers and the set of rational numbers, we can use the results in
Theorem 9.14, Theorem 9.15, and Theorem 9.17 to prove the following theorem.
Theorem 9.18. The set Q of all rational numbers is countably infinite.
In Section 9.1, we proved that any subset of a finite set is finite (Theorem 9.6).
A similar result should be expected for countable sets. We first prove that every
subset of N is countable. For an infinite subset B of N, the idea of the proof is to
define a function gWN ! B by removing the elements from B from smallest to the
next smallest to the next smallest, and so on. We do this by defining the function g
recursively as follows:
Let g.1/ be the smallest natural number in B.
Remove g.1/ from B and let g.2/ be the smallest natural number in
B fg.1/g.
Remove g.2/ and let g.3/ be the smallest natural number in
B fg.1/; g.2/g.
We continue this process. The formal recursive definition of g W N ! B is
included in the proof of Theorem 9.19.
Theorem 9.19. Every subset of the natural numbers is countable.
Proof. Let B be a subset of N. If B is finite, then B is countable. So we next
assume that B is infinite. We will next give a recursive definition of a function
gWN ! B and then prove that g is a bijection.
Let g.1/ be the smallest natural number in B.
For each n 2 N, the set B fg.1/; g.2/; : : :; g.n/g is not empty since B is in-
finite. Define g.n C 1/ to be the smallest natural number in
B fg.1/; g.2/; : : :; g.n/g.
The proof that the function g is a bijection is Exercise (11) on page 474.

Corollary 9.20. Every subset of a countable set is countable.
Exercises 9.2
? 1. State whether each of the following is true or false.
(a) If a set A is countably infinite, then A is infinite.
(b) If a set A is countably infinite, then A is countable.
(c) If a set A is uncountable, then A is not countably infinite.
(d) If A Nk for some k 2 N, then A is not countable.
2. Prove that each of the following sets is countably infinite.
(a) The set F C
of all natural numbers that are multiples of 5
(b) The set F of all integers that are multiples of 5
(c)

1
2k
ˇ
ˇ
ˇ
ˇ k 2 N

(d) fn 2 Z j n 10g
? (e) N f4; 5; 6g
? (f) fm 2 Z j m 2 .mod 3/g
3. Prove part (2) of Theorem 9.10.
Let A and B be sets. If A is infinite and A B, then B is infinite.
4. Complete the proof of Theorem 9.15 by proving the following:
Let A be a countably infinite set and x … A. If f WN ! A is a bijection, then
g is a bijection, where gWN ! A [ fxg by
g.n/ D
(
x if n D 1
f.n 1/ if n 1:
?
5. Prove Theorem 9.16.
If A is a countably infinite set and B is a finite set, then A [ B is a countably
infinite set.
Hint: Let card.B/ D n and use a proof by induction on n. Theorem 9.15 is
the basis step.

? 6. Complete the proof of Theorem 9.17 by proving the following:
Let A and B be disjoint countably infinite sets and let f W N ! A and
gWN ! B be bijections. Define hWN ! A [ B by
h.n/ D
8̂
ˆ
ˆ

ˆ
ˆ
:̂
f

n C 1
2

if n is odd
g
n
2

if n is even:
Then the function h is a bijection.
? 7. Prove Theorem 9.18.
The set Q of all rational numbers is countable.
Hint: Use Theorem 9.15 and Theorem 9.17.
?
8. Prove that if A is countably infinite and B is finite, then A B is countably
infinite.
9. Define f WN N ! N as follows: For each .m; n/ 2 N N,
f.m; n/ D 2m 1
.2n 1/:
(a) Prove that f is an injection. Hint: If f.m; n/ D f.s; t/, there are three
cases to consider: m s, m s, and m D s. Use laws of exponents
to prove that the first two cases lead to a contradiction.
(b) Prove that f is a surjection. Hint: You may use the fact that if y 2 N,
then y D 2k
x, where x is an odd natural number and k is a non-
negative integer. This is actually a consequence of the Fundamental
Theorem of Arithmetic, Theorem 8.15. [See Exercise (13) in Sec-
tion 8.2.]
(c) Prove that N N N and hence that card .N N/ D @0.
10. Use Exercise (9) to prove that if A and B are countably infinite sets, then
A B is a countably infinite set.
11. Complete the proof of Theorem 9.19 by proving that the function g defined
in the proof is a bijection from N to B.
Hint: To prove that g is an injection, it might be easier to prove that for all
r; s 2 N, if r ¤ s, then g.r/ ¤ g.s/. To do this, we may assume that r s
since one of the two numbers must be less than the other. Then notice that
g.r/ 2 fg.1/; g.2/; : : :; g.s 1/g.

To prove that g is a surjection, let b 2 B and notice that for some k 2 N,
there will be k natural numbers in B that are less than b.
12. Prove Corollary 9.20, which states that every subset of a countable set is
countable.
Hint: Let S be a countable set and assume that A S. There are two cases:
A is finite or A is infinite. If A is infinite, let f W S ! N be a bijection and
define gWA ! f.A/ by g.x/ D f.x/, for each x 2 A.
13. Use Corollary 9.20 to prove that the set of all rational numbers between 0
and 1 is countably infinite.
14. Another Proof that QC
Is Countable. For this activity, it may be help-
ful to use the Fundamental Theorem of Arithmetic (see Theorem 8.15 on
page 432). Let QC
be the set of positive rational numbers. Every posi-
tive rational number has a unique representation as a fraction
m
n
, where m
and n are relatively prime natural numbers. We will now define a function
f WQC ! N as follows:
If x 2 QC and x D
m
n
, where m; n 2 N, n ¤ 1 and gcd.m; n/ D 1, we
write
m D p˛1
1 p˛2
2 p˛r
r ; and
n D q
ˇ1
1 q
ˇ2
2 qˇs
s ;
where p1, p2, ..., pr are distinct prime numbers, q1, q2, ..., qs are distinct
prime numbers, and ˛1, ˛2, ..., ˛r and ˇ1, ˇ2, ..., ˇs are natural numbers.
We also write 1 D 20
when m D 1. We then define
f.x/ D p2˛1
1 p2˛2
2 p2˛r
r q
2ˇ1 1
1 q
2ˇ2 1
2 q2ˇs 1
s :
If x D
m
1
, then we define f.x/ D p2˛1
1 p2˛2
2 p2˛r
r D m2.
(a) Determine f

2
3

, f

5
6

, f.6/, f.
12
25
/, f

375
392

, and f

23
113
3 54

.
(b) If possible, find x 2 QC such that f.x/ D 100.
(c) If possible, find x 2 QC
such that f.x/ D 12.

(d) If possible, find x 2 QC such that f.x/ D 28 35 13 172.
(e) Prove that the function f an injection.
(f) Prove that the function f a surjection.
(g) What has been proved?
9.3 Uncountable Sets
Preview Activity 1 (The Game of Dodge Ball)
(From The Heart of Mathematics: An Invitation to Effective Thinking by Edward
B. Burger and Michael Starbird, Key Publishing Company, c 2000 by Edward B.
Burger and Michael Starbird.)
Dodge Ball is a game for two players. It is played on a game board such as
the one shown in Figure 9.4. Player One has a 6 by 6 array to complete and Player
Two has a 1 by 6 row to complete. Each player has six turns as described next.
Player One begins by filling in the first horizontal row of his or her table with
a sequence of six X’s and O’s, one in each square in the first row.
Then Player Two places either an X or an O in the first box of his or her row.
At this point, Player One has completed the first row and Player Two has
filled in the first box of his or her row with one letter.
The game continues with Player One completing a row with six letters (X’s
and O’s), one in each box of the next row followed by Player Two writing
one letter (an X or an O) in the next box of his or her row. The game is
completed when Player One has completed all six rows and Player Two has
completed all six boxes in his or her row.
Winning the Game
Player One wins if any horizontal row in the 6 by 6 array is identical to the
row that Player Two created. (Player One matches Player Two.)
Player Two wins if Player Two’s row of six letters is different than each of
the six rows produced by Player One. (Player Two “dodges” Player One.)

9.3. Uncountable Sets 477
Player One’s Array
1
2
3
4
5
6
Player Two’s Row
1 2 3 4 5 6
Figure 9.4: Game Board for Dodge Ball
There is a winning strategy for one of the two players. This means that there is
plan by which one of the two players will always win. Which player has a winning
strategy? Carefully describe this winning strategy.
Applying the Winning Strategy to Lists of Real Numbers
Following is a list of real numbers between 0 and 1. Each real number is written as
a decimal number.
a1 D 0:1234567890 a6 D 0:0103492222
a2 D 0:3216400000 a7 D 0:0011223344
a3 D 0:4321593333 a8 D 0:7077700022
a4 D 0:9120930092 a9 D 0:2100000000
a5 D 0:0000234102 a10 D 0:9870008943

Use a method similar to the winning strategy in Cantor’s dodge ball to write a real
number (in decimal form) between 0 and 1 that is not in this list of 10 numbers.
1. Do you think your method could be used for any list of 10 real numbers
between 0 and 1 if the goal is to write a real number between 0 and 1 that is
not in the list?
2. Do you think this method could be extended to a list of 20 different real
numbers? To a list of 50 different real numbers?
3. Do you think this method could be extended to a list consisting of countably
infinite list of real numbers?
Preview Activity 2 (Functions from a Set to Its Power Set)
Let A be a set. In Section 5.1, we defined the power set P.A/ of A to be the set of
all subsets of A. This means that
X 2 P.A/ if and only if X A.
Theorem 5.5 in Section 5.1 states that if a set A has n elements, then A has 2n
subsets or that P.A/ has 2n elements. Using our current notation for cardinality,
this means that
if card.A/ D n, then card.P.A// D 2n.
(The proof of this theorem was Exercise (17) on page 229.)
We are now going to define and explore some functions from a set A to its
power set P.A/. This means that the input of the function will be an element of A
and the output of the function will be a subset of A.
1. Let A D f1; 2; 3; 4g. Define f WA ! P.A/ by
f.1/ D f1; 2; 3g
f.2/ D f1; 3; 4g
f.3/ D f1; 4g
f.4/ D f2; 4g.
(a) Is 1 2 f.1/? Is 2 2 f.2/? Is 3 2 f.3/? Is 4 2 f.4/?
(b) Determine S D fx 2 A j x … f.x/g.
(c) Notice that S 2 P.A/. Does there exist an element t in A such that
f.t/ D S? That is, is S 2 range.f /?

2. Let A D f1; 2; 3; 4g. Define f WA ! P.A/ by
f.x/ D A fxg for each x 2 A.
(a) Determine f.1/. Is 1 2 f.1/?
(b) Determine f.2/. Is 2 2 f.2/?
(c) Determine f.3/. Is 3 2 f.3/?
(d) Determine f.4/. Is 4 2 f.4/?
(e) Determine S D fx 2 A j x … f.x/g.
(f) Notice that S 2 P.A/. Does there exist an element t in A such that
3. Define f WN ! P.N/ by
f.n/ D N
˚
n2
; n2
2n , for each n 2 N.
(a) Determine f.1/, f.2/, f.3/, and f.4/. In each of these cases, deter-
mine if k 2 f.k/.
(b) Prove that if n 3, then n 2 f.n/. Hint: Prove that if n 3, then
n2 n and n2 2n n.
(c) Determine S D fx 2 N j x … f.x/g.
(d) Notice that S 2 P.N/. Does there exist an element t in N such that
We have seen examples of sets that are countably infinite, but we have not
yet seen an example of an infinite set that is uncountable. We will do so in this
section. The first example of an uncountable set will be the open interval of real
numbers .0; 1/. The proof that this interval is uncountable uses a method similar
to the winning strategy for Player Two in the game of Dodge Ball from Preview
Activity 1. Before considering the proof, we need to state an important results
about decimal expressions for real numbers.
Decimal Expressions for Real Numbers
In its decimal form, any real number a in the interval .0; 1/ can be written as
a D 0:a1a2a3a4 : : : ; where each ai is an integer with 0 ai 9. For example,
5
12
D 0:416666 : : ::

We often abbreviate this as
5
12
D 0:416 to indicate that the 6 is repeated. We can
also repeat a block of digits. For example,
5
26
D 0:19230769 to indicate that the
block 230769 repeats. That is,
5
26
D 0:19230769230769230769 : : ::
There is only one situation in which a real number can be represented as a decimal
in more than one way. A decimal that ends with an infinite string of 9’s is equal to
one that ends with an infinite string of 0’s. For example, 0:3199999 : : : represents
the same real number as 0:3200000 : : :: Geometric series can be used to prove that
a decimal that ends with an infinite string of 9’s is equal to one that ends with an
infinite string of 0’s, but we will not do so here.
Definition. A decimal representation of a real number a is in normalized
form provided that there is no natural number k such that for all natural num-
bers n with n k, an D 9. That is, the decimal representation of a is in
normalized form if and only if it does not end with an infinite string of 9’s.
One reason the normalized form is important is the following theorem (which will
not be proved here).
Theorem 9.21. Two decimal numbers in normalized form are equal if and only if
they have identical digits in each decimal position.
Uncountable Subsets of R
In the proof that follows, we will use only the normalized form for the decimal
representation of a real number in the interval .0; 1/.
Theorem 9.22. The open interval .0; 1/ is an uncountable set.
Proof. Since the interval .0; 1/ contains the infinite subset

1
2
;
1
3
;
1
4
; : : :

, we can
use Theorem 9.10, to conclude that .0; 1/ is an infinite set. So .0; 1/ is either count-
ably infinite or uncountable. We will prove that .0; 1/ is uncountable by proving
that any injection from .0; 1/ to N cannot be a surjection, and hence, there is no
bijection between .0; 1/ and N.

So suppose that the function f W N ! .0; 1/ is an injection. We will show
that f cannot be a surjection by showing that there exists an element in .0; 1/ that
cannot be in the range of f . Writing the images of the elements of N in normalized
form, we can write
f.1/ D 0:a11a12a13a14a15 : : :
f.2/ D 0:a21a22a23a24a25 : : :
f.3/ D 0:a31a32a33a34a35 : : :
f.4/ D 0:a41a42a43a44a45 : : :
f.5/ D 0:a51a52a53a54a55 : : :
:
:
:
f.n/ D 0:an1an2an3an4an5 : : :
:
:
:
Notice the use of the double subscripts. The number aij is the j th digit to the right
of the decimal point in the normalized decimal representation of f.i/.
We will now construct a real number b D 0:b1b2b3b4b5 : : : in .0; 1/ and in
normalized form that is not in this list.
Note: The idea is to start in the upper left corner and move down the diagonal in
a manner similar to the winning strategy for Player Two in the game in Preview
Activity 1. At each step, we choose a digit that is not equal to the diagonal digit.
Start with a11 in f.1/. We want to choose b1 so that b1 ¤ 0, b1 ¤ a11, and
b1 ¤ 9. (To ensure that we end up with a decimal that is in normalized form, we
make sure that each digit is not equal to 9.) We then repeat this process with a22,
a33, a44, a55, and so on. So we let b be the real number b D 0:b1b2b3b4b5 : : : ;
where for each k 2 N
bk D
(
3 if akk ¤ 3
5 if akk D 3.
(The choice of 3 and 5 is arbitrary. Other choices of distinct digits will also work.)
Now for each n 2 N, b ¤ f.n/ since b and f.n/ are in normalized form
and b and f.n/ differ in the nth decimal place. This proves that any function
from N to .0; 1/ cannot be surjection and hence, there is no bijection from N to
.0; 1/. Therefore, .0; 1/ is not countably infinite and hence must be an uncountable
set.

Progress Check 9.23 (Dodge Ball and Cantor’s Diagonal Argument)
The proof of Theorem 9.22 is often referred to as Cantor’s diagonal argument.
It is named after the mathematician Georg Cantor, who first published the proof
in 1874. Explain the connection between the winning strategy for Player Two in
Dodge Ball (see Preview Activity 1) and the proof of Theorem 9.22 using Cantor’s
diagonal argument.
The open interval .0; 1/ is our first example of an uncountable set. The cardinal
number of .0; 1/ is defined to be c, which stands for the cardinal number of
the continuum. So the two infinite cardinal numbers we have seen are @0 for
countably infinite sets and c.
Definition. A set A is said to have cardinality c provided that A is equivalent
to .0; 1/. In this case, we write card.A/ D c and say that the cardinal number
of A is c.
The proof of Theorem 9.24 is included in Progress Check 9.25.
Theorem 9.24. Let a and b be real numbers with a b. The open interval .a; b/
is uncountable and has cardinality c.
Progress Check 9.25 (Proof of Theorem 9.24)
1. In Part (3) of Progress Check 9.2, we proved that if b 2 R and b 0, then
the open interval .0; 1/ is equivalent to the open interval .0; b/. Now let a
and b be real numbers with a b. Find a function
f W.0; 1/ ! .a; b/
that is a bijection and conclude that .0; 1/ .a; b/.
Hint: Find a linear function that passes through the points .0; a/ and .1; b/.
Use this to define the function f . Make sure you prove that this function f
is a bijection.
2. Let a; b; c; d be real numbers with a b and c d. Prove that
.a; b/ .c; d/.
Theorem 9.26. The set of real numbers R is uncountable and has cardinality c.
Proof. Let f W

2
;

2

! R be defined by f.x/ D tan x, for each x 2 R. The
function f is a bijection and, hence,

2
;

2

R. So by Theorem 9.24, R is
uncountable and has cardinality c.

Cantor’s Theorem
We have now seen two different infinite cardinal numbers, @0 and c. It can seem
surprising that there is more than one infinite cardinal number. A reasonable ques-
tion at this point is, “Are there any other infinite cardinal numbers?” The astonish-
ing answer is that there are, and in fact, there are infinitely many different infinite
cardinal numbers. The basis for this fact is the following theorem, which states that
a set is not equivalent to its power set. The proof is due to Georg Cantor (1845–
1918), and the idea for this proof was explored in Preview Activity 2. The basic
idea of the proof is to prove that any function from a set A to its power set cannot
be a surjection.
Theorem 9.27 (Cantor’s Theorem). For every set A, A and P.A/ do not have
the same cardinality.
Proof. Let A be a set. If A D ;, then P.A/ D f;g, which has cardinality 1.
Therefore, ; and P.;/ do not have the same cardinality.
Now suppose that A ¤ ;, and let f WA ! P.A/. We will show that f cannot
be a surjection, and hence there is no bijection from A to P.A/. This will prove
that A is not equivalent to P.A/. Define
S D fx 2 A j x … f.x/g :
Assume that there exists a t in A such that f.t/ D S. Now, either t 2 S or t … S.
If t 2 S, then t 2 fx 2 A j x … f.x/g. By the definition of S, this means
that t … f.t/. However, f.t/ D S and so we conclude that t … S. But now
we have t 2 S and t … S. This is a contradiction.
If t … S, then t … fx 2 A j x … f.x/g. By the definition of S, this means
that t 2 f.t/. However, f.t/ D S and so we conclude that t 2 S. But now
we have t … S and t 2 S. This is a contradiction.
So in both cases we have arrived at a contradiction. This means that there does not
exist a t in A such that f.t/ D S. Therefore, any function from A to P.A/ is not
a surjection and hence not a bijection. Hence, A and P.A/ do not have the same
cardinality.
Corollary 9.28. P.N/ is an infinite set that is not countably infinite.

Proof. Since P.N/ contains the infinite subset ff1g ; f2g ; f3g: : : g, we can use The-
orem 9.10, to conclude that P.N/ is an infinite set. By Cantor’s Theorem (Theo-
rem 9.27), N and P.N/ do not have the same cardinality. Therefore, P.N/ is not
countable and hence is an uncountable set.
Some Final Comments about Uncountable Sets
1. We have now seen that any open interval of real numbers is uncountable and
has cardinality c. In addition, R is uncountable and has cardinality c. Now,
Corollary 9.28 tells us that P.N/ is uncountable. A question that can be
asked is,
“Does P.N/ have the same cardinality as R?”
The answer is yes, although we are not in a position to prove it yet. A
proof of this fact uses the following theorem, which is known as the Cantor-
Schröder-Bernstein Theorem.
Theorem 9.29 (Cantor-Schröder-Bernstein). Let A and B be sets. If there exist
injections f WA ! B and gWB ! A, then A B.
In the statement of this theorem, notice that it is not required that the function g
be the inverse of the function f . We will not prove the Cantor-Schröder-Bernstein
Theorem here. The following items will show some uses of this important theorem.
2. The Cantor-Schröder-Bernstein Theorem can also be used to prove that the
closed interval Œ0; 1 is equivalent to the open interval .0; 1/. See Exercise (6)
on page 486.
3. Another question that was posed earlier is,
“Are there other infinite cardinal numbers other than @0 and c?”
Again, the answer is yes, and the basis for this is Cantor’s Theorem (Theo-
rem 9.27). We can start with card.N/ D @0. We then define the following
infinite cardinal numbers:
card.P.N// D ˛1.
card.P.P.N/// D ˛2.
card.P.P.P.N//// D ˛3.
:
:
:

Cantor’s Theorem tells us that these are all different cardinal numbers, and
so we are just using the lowercase Greek letter ˛ (alpha) to help give names
to these cardinal numbers. In fact, although we will not define it here, there
is a way to “order” these cardinal numbers in such a way that
@0 ˛1 ˛2 ˛3 :
Keep in mind, however, that even though these are different cardinal num-
bers, Cantor’s Theorem does not tell us that these are the only cardinal num-
bers.
4. In Comment (1), we indicated that P.N/ and R have the same cardinality.
Combining this with the notation in Comment (3), this means that
˛1 D c:
However, this does not necessarily mean that c is the “next largest” cardinal
number after @0. A reasonable question is, “Is there an infinite set with
cardinality between @0 and c?” Rewording this in terms of the real number
line, the question is, “On the real number line, is there an infinite set of
points that is not equivalent to the entire line and also not equivalent to the
set of natural numbers?” This question was asked by Cantor, but he was
unable to find any such set. He conjectured that no such set exists. That
is, he conjectured that c is really the next cardinal number after @0. This
conjecture has come to be known as the Continuum Hypothesis. Stated
somewhat more formally, the Continuum Hypothesis is
There is no set X such that @0 card.X/ c.
The question of whether the Continuum Hypothesis is true or false is one of
the most famous problems in modern mathematics.
Through the combined work of Kurt Gödel in the 1930s and Paul Cohen in
1963, it has been proved that the Continuum Hypothesis cannot be proved or
disproved from the standard axioms of set theory. This means that either the
Continuum Hypothesis or its negation can be added to the standard axioms
of set theory without creating a contradiction.

Exercises 9.3
1. Use an appropriate bijection to prove that each of the following sets has
cardinality c.
? (a) .0; 1/
? (b) .a; 1/, for any a 2 R
(c) R f0g
(d) R fag, for any a 2 R
?
2. Is the set of irrational numbers countable or uncountable? Prove that your
answer is correct.
?
3. Prove that if A is uncountable and A B, then B is uncountable.
? 4. Do two uncountable sets always have the same cardinality? Justify your
conclusion.
5. Let C be the set of all infinite sequences, each of whose entries is the digit 0
or the digit 1. For example,
.1; 0; 1; 0; 1; 0; 1; 0; : : :/ 2 C I
.0; 1; 0; 1; 1; 0; 1; 1; 1; 0; 1; 1; 1; 1; : : :/ 2 C I
.2; 1; 0; 1; 1; 0; 1; 1; 1; 0; 1; 1; 1; 1; : : :/ … C:
Is the set C a countable set or an uncountable set? Justify your conclusion.
6. The goal of this exercise is to use the Cantor-Schröder-Bernstein Theorem
to prove that the cardinality of the closed interval Œ0; 1 is c.
(a) Find an injection f W.0; 1/ ! Œ0; 1.
(b) Find an injection hWŒ0; 1 ! . 1; 2/.
(c) Use the fact that . 1; 2/ .0; 1/ to prove that there exists an injection
g W Œ0; 1 ! .0; 1/. (It is only necessary to prove that the injection g
exists. It is not necessary to determine a specific formula for g.x/.)
Note: Instead of doing Part (b) as stated, another approach is to find an
injection k WŒ0; 1 ! .0; 1/. Then, it is possible to skip Part (c) and go
directly to Part (d).
(d) Use the Cantor-Schröder-Bernstein Theorem to conclude that
Œ0; 1 .0; 1/ and hence that the cardinality of Œ0; 1 is c.

7. In Exercise (6), we proved that the closed interval Œ0; 1 is uncountable and
has cardinality c. Now let a; b 2 R with a b. Prove that Œa; b Œ0; 1
and hence that Œa; b is uncountable and has cardinality c.
8. Is the set of all finite subsets of N countable or uncountable? Let F be the
set of all finite subsets of N. Determine the cardinality of the set F .
Consider defining a function f W F ! N that produces the following.
If A D f1; 2; 6g, then f.A/ D 213256.
If B D f3; 6g, then f.B/ D 23
36
.
If C D fm1; m2; m3; m4g with m1 m2 m3 m4, then f.C / D
2m13m25m37m4.
It might be helpful to use the Fundamental Theorem of Arithmeticon page 432
and to denote the set of all primes as P D fp1; p2; p3; p4; : : :g with p1
p2 p3 p4 . Using the sets A, B, and C defined above, we would
then write
f.A/ D p1
1p2
2p6
3; f.B/ D p3
1p6
2; and f.C / D pm1
1 pm2
2 pm3
3 pm4
4 :
9. In Exercise (2), we showed that the set of irrational numbers is uncountable.
However, we still do not know the cardinality of the set of irrational numbers.
Notice that we can use Qc
to stand for the set of irrational numbers.
(a) Construct a function f WQc ! R that is an injection.
We know that any real number a can be represented in decimal form as
follows:
a D A:a1a2a3a4 an ;
where A is an integer and the decimal part .0:a1a2a3a4 / is in normalized
form. (See page 480.) We also know that the real number a is an irrational
number if and only a has an infinite non-repeating decimal expansion. We
now associate with a the real number
A:a10a211a3000a41111a500000a6111111 : (1)
Notice that to construct the real number in (1), we started with the decimal
expansion of a, inserted a 0 to the right of the first digit after the decimal
point, inserted two 1’s to the right of the second digit to the right of the
decimal point, inserted three 0’s to the right of the third digit to the right of
the decimal point, and so on.

(b) Explain why the real number in (1) is an irrational number.
(c) Use these ideas to construct a function gWR ! Qc that is an injection.
(d) What can we now conclude by using the Cantor-Schröder-Bernstein
Theorem?
10. Let J be the unit open interval. That is, J D fx 2 R j 0 x 1g and let
S D f.x; y/ 2 R R j 0 x 1 and 0 y 1g. We call S the unit open
square. We will now define a function f from S to J. Let .a; b/ 2 S and
write the decimal expansions of a and b in normalized form as
a D 0:a1a2a3a4 an
b D 0:b1b2b3b4 bn :
We then define f.a; b/ D 0:a1b1a2b2a3b3a4b4 anbn .
(a) Determine the values of f.0:3; 0:625/, f

1
3
;
1
4

, and f

1
6
;
5
6

.
(b) If possible, find .x; y/ 2 S such that f.x; y/ D 0:2345.
(c) If possible, find .x; y/ 2 S such that f.x; y/ D
1
3
.
(d) If possible, find .x; y/ 2 S such that f.x; y/ D
1
2
.
(e) Explain why the function f WS ! J is an injection but is not a surjec-
tion.
(f) Use the Cantor-Schröder-Bernstein Theorem to prove that the cardi-
nality of the unit open square S is equal to c. If this result seems
surprising, you are in good company. In a letter written in 1877 to
the mathematician Richard Dedekind describing this result that he had
discovered, Georg Cantor wrote, “I see it but I do not believe it.”
11. The Closed Interval Œ0; 1. In Exercise (6), the Cantor-Schröder-Bernstein
Theorem was used to prove that the closed interval Œ0; 1 has cardinality c.
This may seem a bit unsatisfactory since we have not proved the Cantor-
Schröder-Bernstein Theorem. In this activity, we will prove that card.Œ0; 1/ D
c by using appropriate bijections.

(a) Let f WŒ0; 1 ! Œ0; 1/ by
f.x/ D
8

:
1
n C 1
if x D
1
n
for some n 2 N
x otherwise:
i. Determine f.0/, f.1/, f

1
2

, f

1
3

, f

1
4

, and f

1
5

.
ii. Sketch a graph of the function f . Hint: Start with the graph of
y D x for 0 x 1. Remove the point .1; 1/ and replace it with
the point

1;
1
2

. Next, remove the point

1
2
;
1
2

and replace it
with the point

1
2
;
1
3

. Continue this process of removing points
on the graph of y D x and replacing them with the points deter-
mined from the information in Part (11(a)i). Stop after repeating
this four or five times so that pattern of this process becomes ap-
parent.
iii. Explain why the function f is a bijection.
iv. Prove that Œ0; 1 Œ0; 1/.
(b) Let gWŒ0; 1/ ! .0; 1/ by
g.x/ D
8̂
ˆ
ˆ

ˆ
ˆ
:̂
1
2
if x D 0
1
n C 1
if x D
1
n
for some n 2 N
x otherwise:
i. Follow the procedure suggested in Part (11a) to sketch a graph of
g.
ii. Explain why the function g is a bijection.
iii. Prove that Œ0; 1/ .0; 1/.
(c) Prove that Œ0; 1 and Œ0; 1/ are both uncountable and have cardinality c.

Equivalent sets, page 452
Sets with the same cardinality,
page 452
Finite set, page 455
Infinite set, page 455
Cardinality of a finite set,
page 455
Cardinality of N, page 465
@0, page 465
Countably infinite set, page 466
Denumerable set, page 466
Uncountable set, page 466
Important Theorems and Results about Finite and Infinite Sets
Theorem 9.3. Any set equivalent to a finite nonempty set A is a finite set and
has the same cardinality as A.
Theorem 9.6. If S is a finite set and A is a subset of S, then A is finite and
card.A/ card.S/.
Corollary 9.8. A finite set is not equivalent to any of its proper subsets.
Theorem 9.9 [The Pigeonhole Principle]. Let A and B be finite sets. If
card.A/ card.B/, then any function f WA ! B is not an injection.
Theorem 9.10. Let A and B be sets.
Theorem 9.13. The set Z of integers is countably infinite, and so card.Z/ D
@0.
Theorem 9.14. The set of positive rational numbers is countably infinite.
Theorem 9.16. If A is a countably infinite set and B is a finite set, then
A [ B is a countably infinite set.
Theorem 9.17. If A and B are disjoint countably infinite sets, then A [ B is
a countably infinite set.

Theorem 9.18. The set Q of all rational numbers is countably infinite.
Theorem 9.19. Every subset of the natural numbers is countable.
Corollary 9.20. Every subset of a countable set is countable.
Theorem 9.22. The open interval .0; 1/ is an uncountable set.
Theorem 9.24. Let a and b be real numbers with a b. The open interval
.a; b/ is uncountable and has cardinality c.
Theorem 9.26. The set of real numbers R is uncountableand has cardinality
c.
Theorem 9.27 [Cantor’s Theorem]. For every set A, A and P.A/ do not
have the same cardinality.
Corollary 9.28. P.N/ is an infinite set that is not countably infinite.
Theorem 9.29 [Cantor-Schröder-Bernstein]. Let A and B be sets. If there
exist injections f1 W A ! B and f2 W B ! A, then A B.

Appendix A
Guidelines for Writing
Mathematical Proofs
One of the most important forms of mathematical writing is writing mathematical
proofs. The writing of mathematical proofs is an acquired skill and takes a lot
of practice. Throughout the textbook, we have introduced various guidelines for
writing proofs. These guidelines are in Sections 1.1, 1.2, 3.1, 3.2, 3.3, and 4.1.
Following is a summary of all the writing guidelines introduced in the text.
This summary contains some standard conventions that are usually followed when
writing a mathematical proof.
1. Know your audience. Every writer should have a clear idea of the intended
audience for a piece of writing. In that way, the writer can give the right
amount of information at the proper level of sophistication to communicate
effectively. This is especially true for mathematical writing. For example, if
a mathematician is writing a solution to a textbook problem for a solutions
manual for instructors, the writing would be brief with many details omitted.
However, if the writing was for a students’ solution manual, more details
would be included.
2. Begin with a carefully worded statement of the theorem or result to be
proven. The statement should be a simple declarative statement of the prob-
lem. Do not simply rewrite the problem as stated in the textbook or given on
a handout. Problems often begin with phrases such as “Show that” or “Prove
that.” This should be reworded as a simple declarative statement of the the-
orem. Then skip a line and write “Proof” in italics or boldface font (when
492

Appendix A. Writing Guidelines 493
using a word processor). Begin the proof on the same line. Make sure that
all paragraphs can be easily identified. Skipping a line between paragraphs
or indenting each paragraph can accomplish this.
As an example, an exercise in a text might read, “Prove that if x is an odd
integer, then x2
is an odd integer.” This could be started as follows:
Theorem. If x is an odd integer, then x2
is an odd integer.
Proof: We assume that x is an odd integer ....
3. Begin the proof with a statement of your assumptions. Follow the state-
ment of your assumptions with a statement of what you will prove.
Proof. We assume that x and y are odd integers and will prove that x y is
an odd integer.
4. Use the pronoun “we.” If a pronoun is used in a proof, the usual convention
is to use “we” instead of “I.” The idea is to stress that you and the reader
are doing the mathematics together. It will help encourage the reader to
continue working through the mathematics. Notice that we started the proof
of Theorem 1.8 with “We assume that : : ::”
5. Use italics for variables when using a word processor. When using a
word processor to write mathematics, the word processor needs to be capa-
ble of producing the appropriate mathematical symbols and equations. The
mathematics that is written with a word processor should look like typeset
mathematics. This means that variables need to be italicized, boldface is
used for vectors, and regular font is used for mathematical terms such as the
names of the trigonometric functions and logarithmic functions.
For example, we do not write sin x or sin x. The proper way to typeset this
is sinx.
6. Do not use for multiplication or ˆ for exponents. Leave this type of
notation for writing computer code. The use of this notation makes it difficult
for humans to read. In addition, avoid using = for division when using a
complex fraction.
For example, it is very difficult to read x3
3x2
C 1=2

= .2x=3 7/; the
fraction
x3 3x2 C
1
2
2x
3
7
is much easier to read.

494 Appendix A. Writing Guidelines
7. Use complete sentences and proper paragraph structure. Good grammar
is an important part of any writing. Therefore, conform to the accepted rules
of grammar. Pay careful attention to the structure of sentences. Write proofs
using complete sentences but avoid run-on sentences. Also, do not forget
punctuation, and always use a spell checker when using a word processor.
8. Keep the reader informed. Sometimes a theorem is proven by proving the
contrapositive or by using a proof by contradiction. If either proof method
is used, this should be indicated within the first few lines of the proof. This
also applies if the result is going to be proven using mathematical induction.
Examples:
We will prove this result by proving the contrapositive of the statement.
We will prove this statement using a proof by contradiction.
We will assume to the contrary that : : ::
We will use mathematical induction to prove this result.
In addition, make sure the reader knows the status of every assertion that
is made. That is, make sure it is clearly stated whether an assertion is an
assumption of the theorem, a previously proven result, a well-known result,
or something from the reader’s mathematical background.
9. Display important equations and mathematical expressions. Equations
and manipulations are often an integral part of the exposition. Do not write
equations, algebraic manipulations, or formulas in one column with reasons
given in another column (as is often done in geometry texts). Important equa-
tions and manipulations should be displayed. This means that they should
be centered with blank lines before and after the equation or manipulations,
and if one side of an equation does not change, it should not be repeated. For
example,
x y D .2m C 1/ .2n C 1/
D 4mn C 2m C 2n C 1
D 2 .2mn C m C n/ C 1:
Since m and n are integers, we conclude that : : : :

Appendix A. Writing Guidelines 495
10. Equation numbering guidelines. If it is necessary to refer to an equation
later in a proof, that equation should be centered and displayed, and it should
be given a number. The number for the equation should be written in paren-
theses on the same line as the equation at the right-hand margin.
Example:
Since x is an odd integer, there exists an integer n such that
x D 2n C 1: (1)
Later in the proof, there may be a line such as
Then, using the result in equation (1), we obtain : : : :
Please note that we should only number those equations we will be
referring to later in the proof. Also, note that the word “equation” is not
capitalized when we are referring to an equation by number. Although it
may be appropriate to use a capital “E,” the usual convention in
mathematics is not to capitalize.
11. Do not use a mathematical symbol at the beginning of a sentence.
For example, we should not write, “Let n be an integer. n is an odd integer
provided that : : : :” Many people find this hard to read and often have to re-
read it to understand it. It would be better to write, “An integer n is an odd
integer provided that : : ::”
12. Use English and minimize the use of cumbersome notation. Do not use
the special symbols for quantifiers 8 (for all), 9 (there exists), Ö (such that),
or ) (therefore) in formal mathematical writing. It is often easier to write,
and usually easier to read, if the English words are used instead of the sym-
bols. For example, why make the reader interpret
.8x 2 R/ .9y 2 R/ .x C y D 0/
or more succinctly (if appropriate)

496 Appendix A. Writing Guidelines
13. Tell the reader when the proof has been completed. Perhaps the best way
to do this is to say outright that, “This completes the proof.” Although it
may seem repetitive, a good alternative is to finish a proof with a sentence
that states precisely what has been proven. In any case, it is usually good
practice to use some “end of proof symbol” such as .
14. Keep it simple. It is often difficult to understand a mathematical argument
no matter how well it is written. Do not let your writing help make it more
difficult for the reader. Use simple, declarative sentences and short para-
graphs, each with a simple point.
15. Write a first draft of your proof and then revise it. Remember that a proof
is written so that readers are able to read and understand the reasoning in the
proof. Be clear and concise. Include details but do not ramble. Do not be
satisfied with the first draft of a proof. Read it over and refine it. Just like
any worthwhile activity, learning to write mathematics well takes practice
and hard work. This can be frustrating. Everyone can be sure that there will
be some proofs that are difficult to construct, but remember that proofs are a
very important part of mathematics. So work hard and have fun.

Appendix B
Answers for the Progress Checks
Section 1.1
Progress Check 1.2
1. This proposition is false. A counterexample is a D 2 and b D 1. For these
values, .a C b/2
D 9 and a2
C b2
D 5.
2. This proposition is true, as we can see by using x D 3 and y D 7. We could
also use x D 2 and y D 9. There are many other possible choices for x
and y.
3. This proposition appears to be true. Anytime we use an example where x is
an even integer, the number x2 is an even integer. However, we cannot claim
that this is true based on examples since we cannot list all of the examples
where x is an even integer.
4. This proposition appears to be true. Anytime we use an example where x
and y are both integers, the number x y is an odd integer. However, we
cannot claim that this is true based on examples since we cannot list all of
the examples where both x and y are odd integers.
Progress Check 1.4
1. (a) This does not mean the conditional statement is false since when x D
3, the hypothesis is false, and the only time a conditional statement is
false is when the hypothesis is true and the conclusion is false.
497

498 Appendix B. Answers for Progress Checks
(b) This does not mean the conditional statment is true since we have not
checked all positive real numbers, only the one where x D 4.
(c) All examples should indicate that the conditional statement is true.
2. The number n2 n C 41

will be a prime number for all examples of n that
are less than 41. However, when n D 41, we get
n2
n C 41 D 412
41 C 41
n2
n C 41 D 412
So in the case where n D 41, the hypothesis is true (41 is a positive inte-
ger) and the conclusion is false 412 is not prime

. Therefore, 41 is a coun-
terexample that shows the conditional statement is false. There are other
counterexamples (such as n D 42, n D 45, and n D 50), but only one
counterexample is needed to prove that the statement is false.
Progress Check 1.5
1. We can conclude that this function is continuous at 0.
2. We can make no conclusion about this function from the theorem.
3. We can make no conclusion about this function from the theorem.
4. We can conclude that this function is not differentiable at 0.
Progress Check 1.7
1. The set of rational numbers is closed under addition since
a
b
C
c
d
D
ad C bc
bd
.
2. The set of integers is not closed under division. For example,
2
3
is not an
integer.
3. The set of rational numbers is closed under subtraction since
a
b
c
d
D
ad bc
bd
.

Appendix B. Answers for Progress Checks 499
Section 1.2
Progress Check 1.10
All examples should indicate the proposition is true. Following is a proof.
Proof. We assume that m is an odd integer and will prove that 3m2
C 4m C 6

.
Since m is an odd integer, there exists an integer k such that m D 2k C 1. Substi-
tuting this into the expression 3m2
C 4m C 6

and using algebra, we obtain
3m2
C 4m C 6 D 3 .2k C 1/2
C 4 .2k C 1/ C 6
D 12k2
C 12k C 3

C .8k C 4/ C 6
D 12k2
C 20k C 13
D 12k2
C 20k C 12 C 1
D 2 6k2
C 10k C 6

C 1
By the closure properties of the integers, 6k2 C 10k C 6

is an integer, and hence,
the last equation shows that 3m2
C 4m C 6 is an odd integer. This proves that if m
is an odd integer, then 3m2 C 4m C 6

is an odd integer.
Progress Check 1.11
Proof. We let m be a real number and assume that m, m C 1, and m C 2 are the
lengths of the three sides of a right triangle. We will use the Pythagorean Theorem
to prove that m D 3. Since the hypotenuse is the longest of the three sides, the
Pythagorean Theorem implies that m2
C .m C 1/2
D .m C 2/2
. We will now use
algebra to rewrite both sides of this equation as follows:
m2
C m2
C 2m C 1

D m2
C 4m C 4
2m2
C 2m C 1 D m2
C 4m C 4
The last equation is a quadratic equation. To solve for m, we rewrite the equation
in standard form and then factor the left side. This gives
m2
2m 3 D 0
.m 3/.m C 1/ D 0

The two solutions of this equation are m D 3 and m D 1. However, since m is
the length of a side of a right triangle, m must be positive and we conclude that
m D 3. This proves that if m, m C 1, and m C 2 are the lengths of the three sides
of a right triangle, then m D 3.
Section 2.1
Progress Check 2.1
1. Whenever a quadrilateral is a square, it is a rectangle, or a quadrilateral is a
rectangle whenever it is a square.
2. A quadrilateral is a square only if it is a rectangle.
3. Being a rectangle is necessary for a quadrilateral to be a square.
4. Being a square is sufficient for a quadrilateral to be a rectangle.
Progress Check 2.2
P Q P ^ :Q : .P ^ Q/ :P ^ :Q :P _ :Q
T T F F F F
T F T T F T
F T F T F T
F F F T T T
Statements (2) and (4) have the same truth table.
Progress Check 2.4
P :P P _ :P P ^ :P
T F T F
F T T F

P Q P _ Q P ! .P _ Q/
T T T T
T F T T
F T T T
F F F T
P ! .P _ Q/ is a tautology:
Section 2.2
Progress Check 2.7
1. Starting with the suggested equivalency, we obtain
.P ^ :Q/ ! R : .P ^ :Q/ _ R
.:P _ : .:Q// _ R
:P _ .Q _ R/
P ! .Q _ R/
2. For this, let P be, “3 is a factor of a b,” let Q be, “ is a factor of a,” and let
R be, “3 is a factor of b.” Then the stated proposition is written in the form
P ! .Q _ R/. Since this is logically equivalent to .P ^ :Q/ ! R, if we
prove that
if 3 is a factor of a b and 3 is not a factor of a, then 3 is a factor of b,
then we have proven the original proposition.
Section 2.3
Progress Check 2.9
1. 10 2 A, 22 2 A, 13 … A, 3 … A, 0 2 A, 12 … A
2. A D B, A B, B A, A C , A D, B C , B D

Progress Check 2.11
1. (a) Two values of x for which P.x/ is false are x D 3 and x D 4.
(c) The set of all x for which P.x/ is true is the set f 2; 1; 0; 1; 2g.
2. (a) Two examples for which R.x; y; z/ is false are: x D 1, y D 1, z D 1
and x D 3, y D 1, z D 5.
(b) Two examples for which R.x; y; z/ is true are: x D 3, y D 4, z D 5
and x D 5, y D 12, z D 13.
Progress Check 2.13
1. The truth set is the set of all real numbers whose square is less than or equal
to 9. The truth set is
˚
x 2 Rj x2 9 D fx 2 Rj 3 x 3g.
2. The truth set is the set of all integers whose square is less than or equal to 9.
The truth set is f 3; 2; 1; 0; 1; 2; 3g.
3. The truth sets in Parts (1) and (2) equal are not equal. One purpose of this
progress check is to show that the truth set of a predicate depends on the
predicate and on the universal set.
Progress Check 2.15
A D f4n 3 j n 2 Ng D fx 2 N j x D 4n 3 for some natural number ng :
B D f 2n j n is a nonnegative integerg :
C D
p
2
2m 1
j m 2 N

D
np
2
n
j n is an odd natural number
o
:
D D f3n
j n is a nonnegative integerg :
Section 2.4
Progress Check 2.18
1. For each real number a, a C 0 D a.

.9a 2 R/.a C 0 ¤ a/.
There exists a real number a such that a C 0 ¤ a.
2. For each real number x, sin.2x/ D 2.sinx/.cos x/.
.9x 2 R/.sin.2x/ ¤ 2.sin x/.cos x//.
There exists a real number x such that sin.2x/ ¤ 2 .sinx/ .cos x/.
3. For each real number x, tan2
x C 1 D sec2
x.
.9x 2 R/ tan2
x C 1 ¤ sec2
x

.
There exists a real number x such that tan2 x C 1 ¤ sec2 x.
4. There exists a rational number x such that x2
3x 7 D 0.
.8x 2 Q/ x2 3x 7 ¤ 0

.
For each rational number x, x2
3x 7 ¤ 0.
5. There exists a real number x such that x2 C 1 D 0.
.8x 2 R/ x2
C 1 ¤ 0

.
For each real number x, x2 C 1 ¤ 0.
Progress Check 2.19
1. A counterexample is n D 4 since 42 C 4 C 1 D 21, and 21 is not prime.
2. A counterexample is x D
1
4
since
1
4
is positive and 2

1
4
2
D
1
8
and
1
8

1
4
.
Progress Check 2.20
1. An integer n is a multiple of 3 provided that .9k 2 Z/ .n D 3k/.
4. An integer n is not a multiple of 3 provided that .8k 2 Z/ .n ¤ 3k/.
5. An integer n is not a multiple of 3 provided that for every integer k, n ¤ 3k.
Progress Check 2.21
.9x 2 Z/ .9y 2 Z/ .x C y ¤ 0/.
There exist integers x and y such that x C y ¤ 0.

Section 3.1
Progress Check 3.2
2. For each example in Part (1), the integer a divides the sum b C c.
3. Conjecture: For all integers a, b, and c with a ¤ 0, if a divides b and a
divides c, then a divides b C c.
4. A Know-show table for a proof of the conjecture in Part (3).
Step Know Reason
P a j b and a j c Hypothesis
P1 .9s 2 Z/ .b D a s/
.9t 2 Z/ .c D a t/
Definition of “divides”
P 2 b C c D as C at Substituting for b and c
P 3 b C c D a .s C t/ Distributive property
Q1 s C t is an integer Z is closed under addition
Q aj .b C c/ Definition of “divides”
Step Show Reason
Progress Check 3.3
A counterexample for this statement will be values of a and b for which 5 divides a
or 5 divides b, and 5 does not divide 5aCb. One counterexample for the statement
is a D 5 and b D 1. For these values, the hyothesis is true since 5 divides a and
the conclusion is false since 5a C b D 26 and 5 does not divide 26.
Progress Check 3.4
1. Some integers that are congruent to 5 modulo 8 are 11; 3; 5; 13, and 21.
2. fx 2 Zj x 5 .mod 8/g D f: : : ; 19; 11; 3; 5; 13; 21; 29; :: :g.
3. For example, 3 C 5 D 2, 11 C 29 D 18, 13 C 21 D 34.
4. If we subtract 2 from any of the sums obtained in Part (3), the result will be
a multiple of 8. This means that the sum is congruent to 2 modulo 8. For
example, 2 2 D 0, 18 2 D 16, 34 2 D 32.

Progress Check 3.6
1. To prove that 8 divides .a C b 2/, we can prove that there exists an integer
q such that .a C b 2 D 8q.
2. Since 8 divides .a 5/ and .b 5/, there exist integers k and m such that
a 5 8k and b 5 D 8m.
3. a D 5 C 8k and b D 5 C 8m.
4. a C b 2 D .5 C 8k/ C 5 C 8m/ 2 D 8 C 8k C 8m D 8.1 C k C m/.
5. Proof. Let a and b be integers and assume that a 5 .mod 8/ and b
5 .mod 8/. We will prove that .a C b/ 2 .mod 8/. Since 8 divides
.a 5/ and .b 5/, there exist integers k and m such that a 5 D 8k and
b 5 D 8m. We then see that
a C b 2 D .5 C 8k/ C .5 C 8m/ 2
D 8 C 8k C 8m
D 8.1 C k C m/
By the closure properties of the integers, .1 C k C m/ is an integer and so
the last equation proves that 8 divides .a C b 2 and hence, .a C b/
2 .mod 8/. This proves that if a 5 .mod 8/ and b 5 .mod 8/, then
.a C b/ 2 .mod 8/.
Section 3.2
Progress Check 3.8
1. For all real numbers a and b, if ab D 0, then a D 0 or b D 0.
2. For all real numbers a and b, if ab D 0 and a ¤ 0, b D 0.
3. This gives
1
a
.ab/ D
1
a
0:

We now use the associative property on the left side of this equation and
simplify both sides of the equation to obtain

1
a
a

b D 0
1 b D 0
b D 0
Therefore, b D 0 and this completes the proof of a statement that is logically
equivalent to the contrapositive. Hence, we have proved the proposition.
Section 3.3
Progress Check 3.15
1. There exists a real number x such that x is irrational and 3
p
x is rational.
2. There exists a real number x such that

x C
p
2

is rational and

x C
p
2

is rational.
3. There exist integers a and b such that 5 divides ab and 5 does not divide a
and 5 does not divide b.
4. There exist real numbers a and b such that a 0 and b 0 and
2
a
C
2
b
D
4
a C b
.
Progress Check 3.16
1. Some integers that are congruent to 2 modulo 4 are 6; 2; 2; 6; 10, and
some integers that are congruent to 3 modulo 6 are: 9; 3; 3; 9; 15. There
are no integers that are in both of the lists.
2. For this proposition, it is reasonable to try a proof by contradiction since the
conclusion is stated as a negation.
3. Proof. We will use a proof by contradiction. Let n 2 Z and assume that
n 2 .mod 4/ and that n 3 .mod 6/. Since n 2 .mod 4/, we know
that 4 divides n 2. Hence, there exists an integer k such that
n 2 D 4k: (1)

We can also use the assumption that n 3 .mod 6/ to conclude that 6
divides n 3 and that there exists an integer m such that
n 3 D 6m: (2)
If we now solve equations (1) and (2) for n and set the two expressions equal
to each other, we obtain
4k C 2 D 6m C 3:
However, this equation can be rewritten as
2 .2k C 1/ D 2 .3m C 1/ C 1:
Since 2k C 1 is an integer and 3m C 1 is an integer, this last equation is
a contradiction since the left side is an even integer and the right side is
an odd integer. Hence, we have proven that if n 2 .mod 4/, then n 6
3 .mod 6/.
Progress Check 3.18
1. x2
C y2
D .2m C 1/2
C .2n C 1/2
D 2 2m2
C 2m C 2n2
C 2n C 1

.
2. Using algebra to rewrite the last equation, we obtain
4m2
C 4m C 4n2
C 4n C 2 D 4k2
:
If we divide both sides of this equation by 2, we see that 2m2 C2mC2n2 C
2n C 1 D 2k2
or
2 m2
C m C n2
C n

C 1 D 2k2
:
However, the left side of the last equation is an odd integer and the right side
is an even integer. This is a contradiction, and so we have proved that for
all integers x and y, if x and y are odd integers, then there does not exist an
integer z such that x2
C y2
D z2
.

Section 3.4
Progress Check 3.21
Proposition. For each integer n, n2 5n C 7 is an odd integer.
Proof. Let n be an integer. We will prove that n2 5n C 7 is an odd integer by
examining the case where n is even and the case where n is odd.
In the case where n is even, there exists an integer m such that n D 2m. So in this
case,
n2
5n C 7 D .2m/2
5 .2m/ C 7
D 4m2
10m C 6 C 1
D 2 2m2
5m C 3

C 1:
Since 2m2
5m C 3

is an integer, the last equation shows that if n is even, then
n2 5n C 7 is odd.
In the case where n is odd, there exists an integer m such that n D 2m C 1. So
in this case,
n2
5n C 7 D .2m C 1/2
5 .2m C 1/ C 7
D 4m2
14m C 3
D 2 2m2
7m C 1

C 1:
Since 2m2 7m C 1

is an integer, the last equation shows that if n is odd, then
n2
5n C 7 is odd. Hence, by using these two cases, we have shown that for each
integer n, n2 5n C 7 is an odd integer.
Progress Check 3.24
1. j4:3j D 4:3 and j j D j
2. (a) t D 12 or t D 12.
(b) t C 3 D 5 or t C 3 D 5. So t D 2 or t D 8.
(c) t 4 D
1
5
or t 4 D
1
5
. So t D
21
5
or t D
19
5
.
(d) 3t 4 D 8 or 3t 4 D 8. So t D 4 or t D
4
3
.

Section 3.5
Progress Check 3.26
1. (a) The possible remainders are 0, 1, 2, and 3.
(b) The possible remainders are 0, 1, 2, 3, 4, 5, 6, 7, and 8.
(a) 17 D 5 3 C 2
(b) 17 D . 6/ 3 C 1
(c) 73 D 10 7 C 3
(d) 73 D . 11/ 7 C 4
(e) 436 D 16 27 C 4
(f) 539 D 4 110 C 99
Progress Check 3.29
Proof. Let n be a natural number and let a; b; c; and d be integers. We assume that
a b .mod n/ and c d .mod n/ and will prove that .a C c/ .b C d/ .mod n/.
Since a b .mod n/ and c d .mod n/, n divides .a b/ and .c d/ and so
there exist integers k and q such that a b D nk and c d D nq. We can then
write a D b C nk and c D d C nq and obtain
a C c D .b C nk/ C .d C nq/
D .b C d/ C n.k C q/
By subtracting .b C d/ from both sides of the last equation, we see that
.a C c/ .b C d/ D n.k C q/:
Since .k C q/ is an integer, this proves that n divides .aC c/ .b C d/, and hence,
we can conclude that .a C c/ .b C d/ .mod n/.
Progress Check 3.34
a2
22
.mod 5/ or a2
4 .mod 5/ :
This proves that if a 2 .mod 5/, then a2 4 .mod 5/.
a2
32
.mod 5/ or a2
9 .mod 5/ :

We also know that 9 4 .mod 5/. So we have a2 9 .mod 5/ and 9
4 .mod 5/, and we can now use the transitive property of congruence (Theo-
rem 3.30) to conclude that a2 4 .mod 5/. This proves that if a 3 .mod 5/,
then a2
4 .mod 5/.
Section 4.1
Progress Check 4.1
1. It is not possible to tell if 1 2 T and 5 2 T .
2. True.
3. True. The contrapositive is, “If 2 2 T , then 5 2 T ,” which is true.
4. True.
5. False. If k 2 T , then k C 1 2 T .
6. True, since “k … T or k C 1 2 T ” is logically equivalent to “If k 2 T , then
k C 1 2 T .”
7. It is not possible to tell if this is true. It is the converse of the conditional
statement, “For each integer k, if k 2 T , then k C 1 2 T .”
8. True. This is the contrapositive of the conditional statement, “For each inte-
ger k, if k 2 T , then k C 1 2 T .”
Progress Check 4.3
Proof. Let P.n/ be the predicate, “1 C 2 C 3 C C n D
n .n C 1/
2
.” For the
basis step, notice that the equation 1 D
1 .1 C 1/
2
shows that P.1/ is true. Now
let k be a natural number and assume that P.k/ is true. That is, assume that
1 C 2 C 3 C C k D
k .k C 1/
2
: (1)
We now need to prove that P.k C 1/ is true or that
1 C 2 C 3 C C k C .k C 1/ D
.k C 1/ .k C 2/
2
: (2)

By adding .k C 1/ to both sides of equation (1), we see that
1 C 2 C 3 C C k C .k C 1/ D
k .k C 1/
2
C .k C 1/
D
k .k C 1/ C 2 .k C 1/
2
D
k2 C 3k C 2
2
D
.k C 1/ .k C 2/
2
:
By comparing the last equation to equation (2), we see that we have proved that if
P.k/ is true, then P.k C 1/ is true, and the inductive step has been established.
Hence, by the Principle of Mathematical Induction, we have proved that for each
integer n, 1 C 2 C 3 C C n D
n .n C 1/
2
.
Progress Check 4.5
For the inductive step, let k be a natural number and assume that P.k/ is true. That
is, assume that 5k
1 .mod 4/.
1. To prove that P.k C 1/ is true, we must prove 5kC1 1 .mod 4/.
2. Since 5kC1
D 5 5k
, we multiply both sides of the congruence 5k

1 .mod 4/ by 5 and obtain
5 5k
5 1 .mod 4/ or 5kC1
5 .mod 4/ :
3. Since 5kC1
5 .mod 4/ and we know that 5 1 .mod 4/, we can use
the transitive property of congruence to obtain 5kC1 1 .mod 4/. This
proves that if P.k/ is true, then P.k C 1/ is true, and hence, by the Principle
of Mathematical Induction, we have proved that for each natural number n,
5n 1 .mod 4/.
Section 4.2
Progress Check 4.8
1. For each natural number n, if n 3, then 3n
1 C 2n
.

2. For each natural number n, if n 6, then 2n .n C 1/2.
3. For each natural number n, if n 6, then

1 C
1
n
n
2:5.
Progress Check 4.10
Construct the following table and use it to answer the first two questions. The table
shows that P .3/, P .5/, and P .6/ are true. We can also see that P.2/, P.4/, and
P.7/ are false. It also appears that if n 2 N and n 8, then P .n/ is true.
x 0 1 2 3 4 0 1 2 0 1 1
y 0 0 0 0 0 1 1 1 2 2 3
3x C 5y 0 3 6 9 12 5 8 11 10 13 18
The following proposition provides answers for Problems (3) and (4).
Proposition 4.11. For all natural numbers n with n 8, there exist non-negative
Proof. (by mathematical induction) Let Z D fx 2 Z j x 0g, and for each natu-
ral number n, let P .n/ be, “there exist x; y 2 Z
such that n D 3x C 5y.”
Basis Step: Using the table above, we see that P .8/, P .9/, and P .10/ are true.
Inductive Step: Let k 2 N with k 10. Assume that P .8/, P .9/, ..., P .k/ are
true. Now, notice that
k C 1 D 3 C .k 2/ :
Since k 10, we can conclude that k 2 8 and hence P .k 2/ is true.
Therefore, there exist non-negative integers u and v such that k 2 D .3u C 5v/.
Using this equation, we see that
k C 1 D 3 C .3u C 5v/
D 3 .1 C u/ C 5v:
Hence, we can conclude that P .k C 1/ is true. This proves that if P .8/, P .9/,
..., P .k/are true, then P .k C 1/ is true. Hence, by the Second Principle of Math-
ematical Induction, for all natural numbers n with n 8, there exist nonnegative

Section 4.3
Progress Check 4.12
Proof. We will use a proof by induction. For each natural number n, we let P.n/
be,
f3n is an even natural number.
Since f3 D 2, we see that P.1/ is true and this proves the basis step.
For the inductive step, we let k be a natural number and assume that P.k/ is true.
That is, assume that f3k is an even natural number. This means that there exists an
integer m such that
f3k D 2m: (1)
We need to prove that P.k C 1/ is true or that f3.kC1/ is even. Notice that
3.k C 1/ D 3k C 3 and, hence, f3.kC1/ D f3kC3. We can now use the recur-
sion formula for the Fibonacci numbers to conclude that
f3kC3 D f3kC2 C f3kC1:
Using the recursion formula again, we get f3kC2 D f3kC1 C f3k. Putting this all
together, we see that
f3.kC1/ D f3kC3
D f3kC2 C f3kC1
D .f3kC1 C f3k/ C f3kC1
D 2f3kC1 C f3k: (2)
We now substitute the expression for f3k in equation (1) into equation (2). This
gives
f3.kC1/ D 2f3kC1 C 2m
f3.kC1/ D 2 .f3kC1 C m/
This preceding equation shows that f3.kC1/ is even. Hence it has been proved that
if P .k/ is true, then P .k C 1/ is true and the inductive step has been established.
By the Principle of Mathematical Induction, this proves that for each natural num-
ber n, the Fibonacci number f3n is an even natural number.

Section 5.1
Progress Check 5.3
A 6; 6¤ B ; ; ; ¤ A
5 2 B f5g ; ; ¤ B
A ¤ C f1; 2g ¤ C
f1; 2g ; ; ¤ A f4; 2; 1g ; D A
6 … A B ¤ ;
Progress Check 5.4
1.
A B
C
U
1
2
3
4
5
6
7
8
Using the standard Venn diagram for
three sets shown above:
(a) For the set .A B/ C , region
5 is shaded.
(b) For the set .A B/ [ C , the
regions 2, 4, 5, 6, 7 are shaded.
(c) For the set .Ac [B, the regions
2, 3, 5, 6, 7, 8 are shaded.
(d) For the set .Ac .B [ C /, the
regions 3, 6, 7 are shaded.
2.
A
B
C
3.
A
B C
Section 5.2
Progress Check 5.8
A D fx 2 Z j x is a multiple of 9g and B D fx 2 Z j x is a multiple of 3g.
1. The set A is a subset of B. To prove this, we let x 2 A. Then there exists an
integer m such that x D 9m, which can be written as
x D 3 .3m/:

Since 3m 2 Z, the last equation proves that x is a multiple of 3 and so x 2 B.
Therefore, A B.
2. The set A is not equal to the set B. We note that 3 2 B but 3 … A. Therefore,
B 6 A and, hence, A ¤ B.
Progress Check 5.9
Step Know Reason
P A B Hypothesis
P1 Let x 2 Bc. Choose an arbitrary element
of Bc
.
P 2 If x 2 A, then x 2 B. Definition of “subset”
P 3 If x … B, then x … A. Contrapositive
P 4 If x 2 Bc
, then x 2 Ac
. Step P 3 and definition of
“complement”
Q2 The element x is in Ac
. Steps P1 and P 4
Q1 Every element of Bc is an
element of Ac
.
The choose-an-element
method with Steps P1 and
Q2.
Q Bc
Ac
Definition of “subset”
Progress Check 5.12
Proof. Let A and B be subsets of some universal set. We will prove that A B D
A Bc by proving that each set is a subset of the other set. We will first prove
that A B A Bc
. Let x 2 A B. We then know that x 2 A and x … B.
However, x … B implies that x 2 Bc. Hence, x 2 A and x 2 Bc, which means
that x 2 A Bc
. This proves that A B A Bc
.
To prove that A Bc
A B, we let y 2 A Bc
. This means that y 2 A
and y 2 Bc, and hence, y 2 A and y … B. Therefore, y 2 A B and this proves
that A Bc
A B. Since we have proved that each set is a subset of the other
set, we have proved that A B D A Bc.

Progress Check 5.15
Proof. Let A D fx 2 Z j x 3 .mod 12/g and B D fy 2 Z j y 2 .mod 8/g.
We will use a proof by contradiction to prove that A B D ;. So we assume that
A B ¤ ; and let x 2 A B. We can then conclude that x 3 .mod 12/ and
that x 2 .mod 8/. This means that there exist integers m and n such that
x D 3 C 12m and x D 2 C 8n:
By equating these two expressions for x, we obtain 3 C 12m D 2 C 8n, and this
equation can be rewritten as 1 D 8n 12m. This is a contradiction since 1 is
an odd integer and 8n 12m is an even integer. We have therefore proved that
A B D ;.
Section 5.3
Progress Check 5.19
1. In our standard configuration for a Venn diagram with three sets, regions 1,
2, 4, 5, and 6 are the shaded regions for both A [ .B C / and .A [ B/
.A [ C/.
2. Based on the Venn diagrams in Part (1), it appears that A [ .B C / D
.A [ B/ .A [ C/.
Progress Check 5.21
1. Using our standard configuration for a Venn diagram with three sets, re-
gions 1, 2, and 3 are the regions that are shaded for both .A [ B/ C and
.A C/ [ .B C /.
2.
.A [ B/ C D .A [ B/ C c
(Theorem 5.20)
D C c
.A [ B/ (Commutative Property)
D C c
A

[ C c
B

Distributive Property
D A C c

[ B C c

(Commutative Property)
D .A C / [ .B C / (Theorem 5.20)

Section 5.4
Progress Check 5.23
1. Let A D f1; 2; 3g, T D f1; 2g, B D fa; bg, and C D fa; cg.
(a) A B D f.1; a/ ; .1; b/ ; .2; a/ ; .2; b/ ; .3; a/ ; .3; b/g
(b) T B D f.1; a/ ; .1; b/ ; .2; a/ ; .2; b/g
(c) A C D f.1; a/ ; .1; c/ ; .2; a/ ; .2; c/ ; .3; a/ ; .3; c/g
(d) A .B C/ D f.1; a/ ; .2; a/ ; .3; a/g
(e) .A B/ .A C / D f.1; a/ ; .2; a/ ; .3; a/g
(f) A .B [ C/ D
f.1; a/ ; .1; b/ ; .1; c/ ; .2; a/ ; .2; b/ ; .2; c/ ; .3; a/ ; .3; b/ ; .3; c/g
(g) .A B/ [ .A C / D
f.1; a/ ; .1; b/ ; .1; c/ ; .2; a/ ; .2; b/ ; .2; c/ ; .3; a/ ; .3; b/ ; .3; c/g
(h) A .B C/ D f.1; b/ ; .2; b/ ; .3; b/g
(i) .A B/ .A C / D f.1; b/ ; .2; b/ ; .3; b/g
(j) B A D f.a1/ ; .b; 1/ ; .a; 2/ ; .b; 2/ ; .a; 3/ ; .b; 3/g
2. T B A B
A .B C / D .A B/ .A C /
A .B [ C/ D .A B/ [ .A C /
A .B C/ D .A B/ .A C /
Progress Check 5.24
1. (a) A B D f.x; y/ 2 R R j 0 x 2 and 2 y 4g
(b) T B D f.x; y/ 2 R R j 1 x 2 and 2 y 4g
(c) A C D f.x; y/ 2 R R j 0 x 2 and 3 y 5g
(d) A .B C/ D f.x; y/ 2 R R j 0 x 2 and 3 y 4g
(e) .A B/ .A C / D f.x; y/ 2 R R j 0 x 2 and 3 y 4g
(f) A .B [ C/ D f.x; y/ 2 R R j 0 x 2 and 2 y 5g
(g) .A B/ [ .A C / D f.x; y/ 2 R R j 0 x 2 and 2 y 5g
(h) A .B C/ D f.x; y/ 2 R R j 0 x 2 and 2 y 3g
(i) .A B/ .A C / D f.x; y/ 2 R R j 0 x 2 and 2 y 3g
(j) B A D f.x; y/ 2 R R j 2 x 4 and 0 y 2g

2. T B A B
A .B C / D .A B/ .A C /
A .B [ C / D .A B/ [ .A C /
A .B C / D .A B/ .A C /
Section 5.5
Progress Check 5.26
1.
6
S
j D1
Aj D f1; 2; 3; 4; 5; 6; 9; 16; 25; 36g
2.
6
T
j D1
Aj D f1g
3.
6
S
j D3
Aj D f3; 4; 5; 6; 9; 16; 25; 36g
4.
6
T
j D3
Aj D f1g
5.
1
S
j D1
Aj D N
6.
1
T
j D1
Aj D f1g
Progress Check 5.27
1. A1 D f7; 14g; A2 D f10; 12g; A3 D f10; 12g; A4 D f8; 14g.
2. The statement is false. For example, 2 ¤ 3 and A2 D A3.
3. The statement is false. For example, 1 ¤ 1 and B1 D B 1.
Progress Check 5.29
1. Since
S
˛2RC
A˛ D . 1; 1/,
S
˛2RC
A˛
!c
D . 1; 1.
2.
T
˛2RC
Ac
˛ D . 1; 1.
3. Since
T
˛2RC
A˛ D . 1; 0,
T
˛2RC
A˛
!c
D . 1; 1 [ .0; 1/.

4.
S
˛2RC
Ac
˛ D . 1; 1 [ .0; 1/.
Progress Check 5.32
All three families of sets .A; B; and C/ are disjoint families of sets. One the family
A is a pairwise disjoint family of sets.
Section 6.1
Progress Check 6.1
1. f. 3/ D 24
f.
p
8/ D 8 5
p
8
2. g.2/ D 6, g. 2/ D 14
3. f 1; 6g
4. f 1; 6g
5.
(
5 C
p
33
2
;
5
p
33
2
)
6. ;
Progress Check 6.2
1. (a) The domain of the function f is the set of all people.
(b) A codomain for the function f is the set of all days in a leap year.
(c) This means that the range of the function f is equal to its codomain.
2. (a) The domain of the function s is the set of natural numbers.
(b) A codomain for the function s is the set of natural numbers.
(c) This means that the range of s is not equal to the set of natural numbers.
Progress Check 6.3
1. f. 1/ 3 and f.2/ 2:5.
2. Values of x for which f.x/ D 2 are approximately 2:8; 1:9; 0:3; 1:2, and
3.5.
3. The range of f appears to be the closed interval Œ 3:2; 3:2 or
fy 2 R j 3:2 y3:2g.

Progress Check 6.4
Only the arrow diagram in Figure (a) can be used to represent a function from A to
B. The range of this function is the set fa; bg.
Section 6.2
Progress Check 6.5
1. f.0/ D 0, f.1/ D 1, f.2/ D 1, f.3/ D 1, f.4/ D 1.
2. g.0/ D 0, g.1/ D 1, g.2/ D 2, g.3/ D 3, g.4/ D 4.
Progress Check 6.6
IZ5 ¤ f and IZ5 D g.
Progress Check 6.7
1. 3.5 2. 4.02
3.
C
p
2
4
4. The process of finding the average of a finite set of real numbers can be
thought of as a function from F.R/ to R. So the domain is F.R/, the
codomain is R, and we can define a function avg W F.R/ ! R as follows: If
A 2 F.R/ and A D fa1; a2; : : :; ang, then avg.A/ D
a1 C a2 C an
n
.
Progress Check 6.8
1. The sixth term is
1
18
and the tenth term is
1
30
.
2. The sixth term is
1
36
and the tenth term is
1
100
.
3. The sixth term is 1 and the tenth term is 1.

Progress Check 6.9
1. g.0; 3/ D 3; g.3; 2/ D 11; g. 3; 2/ D 11; g.7; 1/ D 50.
2.
˚
.m; n/ 2 Z Z j n D m2
3.
˚
.m; n/ 2 Z Z j n D m2
5
Section 6.3
Progress Check 6.10
The functions k, F , and s are injections. The functions f and h are not injections.
Progress Check 6.11
The functions f and s are surjections. The functions k and F are not surjections.
Progress Check 6.15
The function f is an injection but not a surjection. To see that it is an injection, let
a; b 2 R and assume that f.a/ D f.b/. This implies that e a
D e b
. Now use
the natural logarithm function to prove that a D b. Since e x 0 for each real
number x, there is no x 2 R such that f.x/ D 1. So f is not a surjection.
The function g is an injection and is a surjection. The proof that g is an injection is
basically the same as the proof that f is an injection. To prove that g is a surjection,
let b 2 RC. To construct the real number a such that g.a/ D b, solve the equation
e a
D b for a. The solution is a D lnb. It can then be verified that g.a/ D b.
Progress Check 6.16
1. There are several ordered pairs .a; b/ 2 R R such that g.a; b/ D 2. For
example, g.0; 2/ D 2, g. 1; 4/ D 2, and g.2; 2/ D 2.
2. For each z 2 R, g.0; z/ D z.

3. Part (1) implies that the function g is not an injection. Part (2) implies that
the function g is a surjection since for each z 2 R, .0; z/ is in the domain of
g and g.0; z/ D z.
Section 6.4
Progress Check 6.17
The arrow diagram for g ı f W A ! B should show the following:
.g ı f /.a/ D g.f.a// .g ı f /.b/ D g.f.b//
D g.2/ D 1 D g.3/ D 2
.g ı f /.c/ D g.f.c// .g ı f /.d/ D g.f.d//
D g.1/ D 3 D g.2/ D 1
The arrow diagram for g ı g W B ! B should show the following:
.g ı g/.1/ D g.g.1// .g ı g/.2/ D g.g.2//
D g.3/ D 2 D g.1/ D 3
.g ı g/.3/ D g.g.3//
D g.2/ D 1
Progress Check 6.18
1. F D g ı f , where f W R ! R by f.x/ D x2 C 3, and g W R ! R by
g.x/ D x3
.
2. G D h ı f , where f W R ! RC
by f.x/ D x2
C 3, and h W RC
! R by
h.x/ D ln x.
3. f D g ı k, where k W R ! R by k.x/ D x2 3 and g W R ! R by
g.x/ D jxj.
4. g D h ı f , where f W R ! R by f.x/ D
2x 3
x2 C 1
and h W R ! R by
h.x/ D cos x.

Progress Check 6.19
For the examples that are constructed:
1. g ı f should be an injection.
2. g ı f should be a surjection.
3. g ı f should be a bijection.
Section 6.5
Progress Check 6.23
Neither set can be used to define a function.
1. The set F does not satisfy the first condition of Theorem 6.22.
2. The set G does not satisfy the second condition of Theorem 6.22.
Progress Check 6.24
2. f 1 D f.r; a/; .p; b/; .q; c/g
g 1 D f.p; a/; .q; b/; .p; c/g
h 1 D f.p; a/; .q; b/; .r; c/; .q; d/g
3. (a) f 1 is a function from C to A.
(b) g 1 is not a function from C to A since .p; a/ 2 g 1and .p; c/ 2 g 1.
(c) h 1 is not a function from C to B since .q; b/ 2 h 1 and .q; d/ 2 h 1.
5. In order for the inverse of a function F W S ! T to be a function from T to
S, the function F must be a bijection.

Section 6.6
Progress Check 6.30
1. f.A/ D fs; tg 2. f.B/ D ff.x/ j x 2 Bg D fsg
3. f 1.C / D fx 2 S j f.x/ 2 C g D fa; b; c; dg
4. f 1
.D/ D fx 2 S j f.x/ 2 Dg D fa; dg
Progress Check 6.32
1. f.0/ D 2
f.1/ D 3
f.2/ D 6
f.3/ D 3
f.4/ D 2
f.5/ D 3
f.6/ D 6
f.7/ D 3
2. (a) f.A/ D f2; 3; 6g
f.B/ D f2; 3; 6g
f 1.C / D f0; 1; 3; 4; 5; 7g
f 1
.D/ D f1; 3; 5; 7g
3. (a) f.A/ f.B/ D f2g f.A/ f.B/ D f2; 3; 6g
So in this case, f.A B/ f.A/ f.B/.
(b) f.A/ [ f.B/ D f2; 3; 6g f.A [ B/ D f2; 3; 6g
So in this case, f.A [ B/ D f.A/ [ f.B/.
(c) f 1.C / f 1.D/ D f 1.C D/ D f1; 3; 5; 7g. So in this case,
f 1
.C D/ D f 1
.C / f 1
.D/.
(d) f 1.C / [ f 1.D/ D f 1.C [ D/ D f0; 1; 3; 4; 5; 7g. So in this
case, f 1
.C [ D/ D f 1
.C / [ f 1
.D/.
4. f.A/ D f2; 3; 6g. Hence, f 1.f.A// D f0; 1; 2; 3; 4; 5; 6; 7g. So in this
case, A f 1
.f.A//.
5. f 1.C / D f0; 1; 3; 4; 5; 7g. So f f 1.C /

D f2; 3g. So in this case,
f f 1
.C /

C .

Section 7.1
Progress Check 7.2
1. (a) T is a relation on R since S is a subset of R R.
(b) Solve the equation x2 C 42 D 64. This gives x D ˙
p
48.
Solve the equation x2
C 92
D 64. There are no real number solutions.
So there does not exist an x 2 R such that .x; 9/ 2 S.
(c) dom.T / D fx 2 Rj 8 x 8g range.T / D fy 2 Rj 8 y 8g
(d) The graph is a circle of radius 8 whose center is at the origin.
2. (a) R is a relation on A since R is a subset of A A.
(b) If we assume that each state except Hawaii has a land border in com-
mon with itself, then the domain and range of R are the set of all states
except Hawaii. If we do not make this assumption, then the domain
and range are the set of all states except Hawaii and Alaska.
(c) The first statement is true. If x has a land border with y, then y has
a land border with x. The second statement is false. Following is a
counterexample: .Michigan,Indiana/ 2 R, .Indiana,Illinois/ 2 R, but
.Michigan,Illinois/ … R.
Progress Check 7.3
1. The domain of the divides relation is the set of all nonzero integers. The
range of the divides relation is the set of all integers.
2. (a) This statement is true since for each a 2 Z, a D a 1.
(b) This statement is false: For example, 2 divides 4 but 4 does not divide
2.
(c) This statement is true by Theorem 3.1 on page 88.
Progress Check 7.4
1. Each element in the set F is an ordered pair of the form .x; y/ where y D x2.
2.

(a) A D f 2; 2g
(b) B D f
p
10;
p
10g
(c) C D f25g
(d) D D f9g
3. The graph of y D x2
is a parabola with vertex at the origin that is concave
up.
Progress Check 7.5
The directed graph for Part (a) is on the left and the directed graph for Part (b) is
on the right.
1 2
3
4
5
6
1 2
3
4
5
6
Section 7.2
Progress Check 7.7
The relation R:
Is not reflexive since .c; c/ … R and .d; d/ … R.
Is symmetric.
Is not transitive. For example, .c; a/ 2 R, .a; c/ 2 R, but .c; c/ … R.
Progress Check 7.9
Proof that the relation is symmetric: Let a; b 2 Q and assume that a b.
This means that a b 2 Z. Therefore, .a b/ 2 Z and this means that
b a 2 Z, and hence, b a.

Proof that the relation is transitive: Let a; b; c 2 Q and assume that a b
and b c. This means that a b 2 Z and that b c 2 Z. Therefore,
..a b/ C .b c// 2 Z and this means that a c 2 Z, and hence, a c.
Progress Check 7.11
The relation is reflexive on P .U / since for all A 2 P .U /, card .A/ D card .A/.
The relation is symmetric since for all A; B 2 P .U /, if card .A/ D card .B/,
then using the fact that equality on Z is symmetric, we conclude that card .B/ D
card .A/. That is, if A has the same number of elements as B, then B has the same
number of elements as A.
The relation is transitive since for all A; B; C 2 P .U /, if card .A/ D card .B/
and card .B/ D card .C /, then using the fact that equality on Z is transitive, we
conclude that card .A/ D card .C /. That is, if A and B have the same number of
elements and B and C have the same number of elements, then A and C have the
same number of elements.
Therefore, the relation is an equivalence relation on P .U /.
Section 7.3
Progress Check 7.12
The distinct equivalence classes for the relation R are: fa; b; eg and fc; dg.
Progress Check 7.13
The distinct congruence classes for congruence modulo 4 are
Œ0 D f : : : ; 12; 8; 4; 0; 4; 8; 12; : ::g Œ1 D f : : :; 11; 7; 3; 1; 5; 9; 13; : : :g
Œ2 D f : : : ; 10; 6; 2; 2; 6; 10; 14; :: :g Œ3 D f : : :; 9; 5; 1; 3; 7; 11; 15; : : :g
Progress Check 7.15
1.

Œ5 D Œ 5 D f 5; 5g
Œ10 D Œ 10 D f 10; 10g
Œ D Œ D f ; g
2. Œ0 D f0g 3. Œa D f a; ag
Section 7.4
Progress Check 7.20
1.
˚ Œ0 Œ1 ˇ Œ0 Œ1
Œ0 Œ0 Œ1 Œ0 Œ0 Œ0
Œ1 Œ1 Œ0 Œ1 Œ0 Œ1
3.
˚ Œ0 Œ1 Œ2 Œ3 Œ4 Œ5 ˇ Œ0 Œ1 Œ2 Œ3 Œ4 Œ5
Œ0 Œ0 Œ1 Œ2 Œ3 Œ4 Œ5 Œ0 Œ0 Œ0 Œ0 Œ0 Œ0 Œ0
Œ1 Œ1 Œ2 Œ3 Œ4 Œ5 Œ0 Œ1 Œ0 Œ1 Œ2 Œ3 Œ4 Œ5
Œ2 Œ2 Œ3 Œ4 Œ5 Œ0 Œ1 Œ2 Œ0 Œ2 Œ4 Œ0 Œ2 Œ4
Œ3 Œ3 Œ4 Œ5 Œ0 Œ1 Œ2 Œ3 Œ0 Œ3 Œ0 Œ3 Œ0 Œ3
Œ4 Œ4 Œ5 Œ0 Œ1 Œ2 Œ3 Œ4 Œ0 Œ4 Œ2 Œ0 Œ4 Œ2
Œ5 Œ5 Œ0 Œ1 Œ2 Œ3 Œ4 Œ5 Œ0 Œ5 Œ4 Œ3 Œ2 Œ1
4. For all a; b 2 Z, if a ¤ 0 and b ¤ 0, then ab ¤ 0.
5. The statement in (a) is true and the statement in (b) is false. For example, in
Z6, Œ2 ˇ Œ3 D Œ0.
Section 8.1
Progress Check 8.2
1. The remainder is 8.
2. gcd.12; 8/ D 4
3. 12 D 8 1 C 4 and gcd.r; r2/ D
gcd.8; 4/ D 4

Progress Check 8.4
1.
.180; 126/ 180 D 126 1 C 54 .126; 54/
.126; 54/ 126 D 54 2 C 18 .54; 18/
.54; 18/ 54 D 18 3 C 0
Consequently, gcd.180; 126/ D 18.
2.
.4208; 288/ 4208 D 288 14 C 176 .288; 176/
.288; 176/ 288 D 176 1 C 112 .176; 112/
.176; 112/ 176 D 112 1 C 64 .112; 64/
.112; 64/ 112 D 64 1 C 48 .64; 48/
.64; 48/ 64 D 48 1 C 16 .48; 16/
.48; 16/ 48 D 16 3 C 0
Consequently, gcd.4208; 288/ D 16.
Progress Check 8.7
1. From Progress Check 8.4, gcd.180; 126/ D 18.
18 D 126 54 2
D 126 .180 126/ 2
D 126 3 C 180 . 2/
So gcd.180; 126/ D 18, and 18 D 126 3 C 180 . 2/.
2. From Progress Check 8.4, gcd.4208; 288/ D 16.
16 D 64 48
D 64 .112 64/ D 64 2 112
D .176 112/ 2 112 D 176 2 112 3
D 176 2 .288 176/ 3 D 176 5 288 3
D .4208 288 14/ 5 288 3
D 4208 5 C 288 . 73/
So gcd.4208; 288/ D 16, and 16 D 4208 5 C 288 . 73/.

Section 8.2
Progress Check 8.10
1. If a; p 2 Z, p is prime, and p divides a, then gcd.a; p/ D p.
2. If a; p 2 Z, p is prime, and p does not divide a, then gcd.a; p/ D 1.
3. Three examples are gcd.4; 9/ D 1, gcd.15; 16/ D 1, gcd.8; 25/ D 1.
Progress Check 8.13
Theorem 8.12. Let a, b, and c be integers. If a and b are relatively prime and
a j .bc/, then a j c.
Proof. Let a, b, and c be integers. Assume that a and b are relatively prime and
a j .bc/. We will prove that a divides c.
Since a divides bc, there exists an integer k such that
bc D ak: (1)
In addition,we are assuming that a and b are relatively prime and hence gcd.a; b/ D
1. So by Theorem 8.9, there exist integers m and n such that
am C bn D 1: (2)
We now multiply both sides of equation (2) by c. This gives
.am C bn/c D 1 c
acm C bcn D c (3)
We can now use equation (1) to substitute bc D ak in equation (3) and obtain
acm C akn D c:
If we now factor the left side of this last equation, we see that a.cm C kn/ D c.
Since .cm C kn/ is an integer, this proves that a divides c. Hence, we have proven
that if a and b are relatively prime and a j .bc/, then a j c.

Section 8.3
Progress Check 8.20
2. x D 2 C 3k and y D 0 2k, where k can be any integer. Again, this
does not prove that these are the only solutions.
Progress Check 8.21
One of the Diophantine equations in Preview Activity 2 was 3x C 5y D 11. We
were able to write the solutions of this Diophantine equation in the form
where k is an integer. Notice that x D 2 and y D 1 is a solution of this equation. If
we consider this equation to be in the form ax C by D c, then we see that a D 3,
b D 5, and c D 11. Solutions for this equation can be written in the form
x D 2 C bk and y D 1 ak;
where k is an integer.
The other equation was 4x C 6y D 16. So in this case, a D 4, b D 6, and
c D 16. Also notice that d D gcd.4; 6/ D 2. We note that x D 4 and y D 0 is one
solution of this Diophantine equation and solutions can be written in the form
where k is an integer. Using the values of a, b, and d given above, we see that the
solutions can be written in the form
x D 2 C
b
d
k and y D 0
a
d
;
where k is an integer.
Progress Check 8.24
1. Since 21 does not divide 40, Theorem 8.22 tells us that the Diophantine
equation 63xC336y D 40 has no solutions. Remember that this means there
is no ordered pair of integers .x; y/ such that 63x C 336y D 40. However,
if we allow x and y to be real numbers, then there are real number solutions.

In fact, we can graph the straight line whose equation is 63x C336y D 40 in
the Cartesian plane. From the fact that there is no pair of integers x; y such
that 63x C 336y D 40, we can conclude that there is no point on the graph
of this line in which both coordinates are integers.
2. To write formulas that will generate all the solutions, we first need to find
one solution for 144x C 225y D 27. This can sometimes be done by trial
and error, but there is a systematic way to find a solution. The first step is
to use the Euclidean Algorithm in reverse to write gcd.144; 225/ as a linear
combination of 144 and 225. See Section 8.1 to review how to do this. The
result from using the Euclidean Algorithm in reverse for this situation is
144 11 C 225 . 7/ D 9:
If we multiply both sides of this equation by 3, we obtain
144 33 C 225 . 21/ D 27:
This means that x0 D 33; y0 D 21 is a solution of the linear Diophantine
equation 144x C 225y D 27. We can now use Theorem 8.22 to conclude
that all solutions of this Diophantine equation can be written in the form
x D 33 C
225
9
k y D 21
144
9
k;
where k 2 Z. Simplifying, we see that all solutions can be written in the
form
x D 33 C 25k y D 21 16k;
where k 2 Z.
We can check this general solution as follows: Let k 2 Z. Then
144x C 225y D 144.33 C 25k/ C 225. 21 16k/
D .4752 C 3600k/ C . 4725 3600k/
D 27:
Section 9.1
Progress Check 9.2
1. We first prove that f W A ! B is an injection. So let x; y 2 A and assume
that f.x/ D f.y/. Then x C 350 D y C 350 and we can conclude that

x D y. Hence, f is an injection. To prove that f is a surjection, let b 2 B.
Then 351 b 450 and hence, 1 b 350 100 and so b 350 2 A.
In addition, f.b 350/ D .b 350/ C 350 D b. This proves that f is a
surjection. Hence, the function f is a bijection, and so, A B.
2. If x and t are even integers and F.x/ D F.t/, then x C 1 D t C 1 and,
hence, x D t. Therefore, F is an injection.
To prove that F is a surjection, let y 2 D. This means that y is an odd
integer and, hence, y 1 is an even integer. In addition,
F.y 1/ D .y 1/ C 1 D y:
Therefore, F is a surjection and hence, F is a bijection. We conclude that
E D.
3. Let x; t 2 .0; 1/ and assume that f.x/ D f.t/. Then bx D bt and, hence,
x D t. Therefore, f is an injection.
To prove that f is a surjection, let y 2 .0; b/. Since 0 y b, we conclude
that 0
y
b
1 and that
f
y
b

D b
y
b

D y:
Therefore, f is a surjection and hence f is a bijection. Thus, .0; 1/ .0; b/.
Section 9.2
Progress Check 9.11
1. The set of natural numbers N is a subset of Z, Q, and R. Since N is an
infinite set, we can use Part (2) of Theorem 9.10 to conclude that Z, Q, and
R are infinite sets.
2. Use Part (1) of Theorem 9.10.
3. Prove that EC
N and use Part (1) of Theorem 9.10.

Progress Check 9.12
1. Use the definition of a countably infinite set.
2. Since EC N, we can conclude that card.EC/ D @0.
3. One function that can be used is f WS ! N defined by f.m/ D
p
m for all
m 2 S.
Section 9.3
Progress Check 9.23
Player Two has a winning strategy. On the kth turn, whatever symbol Player One
puts in the kth position of the kth row, Player Two must put the other symbol in the
kth position of his or her row. This guarantees that the row of symbols produced
by Player Two will be different that any of the rows produced by Player One.
This is the same idea used in Cantor’s Diagonal Argument. Once we have a
“list” of real numbers in normalized form, we create a real number that is not in
the list by making sure that its kth decimal place is different than the kth decimal
place for the kth number in the list. The one complication is that we must make
sure that our new real number does not have a decimal expression that ends in all
9’s. This was done by using only 3’s and 5’s.
Progress Check 9.25
1. Proof. In order to find a bijection f W.0; 1/ ! .a; b/, we will use the linear
function through the points .0; a/ and .1; b/. The slope is .b a// and the
y-intercept is .0; a/. So define f W.0; 1/ ! .a; b/ by
f.x/ D .b a/x C a, for each x 2 .0; 1/.
Now, if x; t 2 .0; 1/ and f.x/ D f.t/, then
.b a/x C a D .b a/t C a:
This implies that .b a/x D .b a/t, and since b a ¤ 0, we can conclude
that x D t. Therefore, f is an injection.

To prove that f is a surjection, we let y 2 .a; b/. If x D
y a
b a
, then
f.x/ D f
y a
b a

D .b a/
y a
b a

C a
D .y a/ C a
D y
This proves that f is a surjection. Hence, f is a bijection and .0; 1/
.a; b/. Therefore, .a; b/ is uncountable and has cardinality c.
2. Now, if a; b; c; d are real numbers with a b and c d, then we know that
.a; b/ .0; 1/ and .c; d/ .0; 1/.
Since is an equivalence relation,we can conclude that .a; b/ .c; d/.

Appendix C
Answers and Hints for Selected
Exercises
Section 1.1
1. Sentences (a), (c),(e), (f), (j) and (k) are statements. Sentence (h) is a state-
ment if we are assuming that n is a prime number means that n is an integer.
2.
Hypothesis Conclusion
a. n is a prime number. n2
has three positive di-
visors.
b. a is an irrational num-
ber and b is an irrational
number.
ab is an irrational num-
ber.
c. p is a prime number. p D 2 or p is an odd
number.
d. p is a prime number and
p ¤ 2.
p is an odd number.
e. p ¤ 2 and p is an even
number
p is not prime.
3. Statements (a), (c), and (d) are true.
4. (a) True when a ¤ 3. (b) True when a D 3.
6. (a) This function has a maximum value when x D
5
16
.
536

Appendix C. Answers for Exercises 537
(b) The function h has a maximum value when x D
9
2
.
(c) No conclusion can be made about this function from this theorem.
9. (a) The set of natural numbers is not closed under division.
(b) The set of rational numbers is not closed under division since division
by zero is not defined.
(c) The set of nonzero rational numbers is closed under division.
(d) The set of positive rational numbers is closed under division.
(e) The set of positive real numbers is not closed under subtraction.
(f) The set of negative rational numbers is not closed under division.
(g) The set of negative integers is closed under addition.
Section 1.2
1. (a)
Step Know Reason
P m is an even integer. Hypothesis
P1 There exists an integers k
such that m D 2k.
Definition of an even inte-
ger
P 2 m C 1 D 2k C 1 Algebra
Q1 There exists an integer q
such that mC1 D 2q C1.
Substitution of k D q
Q m C 1 is an odd integer. Definition of an odd inte-
ger
2. (c) We assume that x and y are odd integers and will prove that x C y is
an even integer. Since x and y are odd, there exist integers m and n
such that x D 2m C 1 and y D 2n C 1. Then
x C y D .2m C 1/ C .2n C 1/
D 2m C 2n C 2
D 2 .m C n C 1/ :
Since the integers are closed under addition, .m C n C 1/ is an integer,
and hence the last equation shows that x C y is even. Therefore, we
have proven that if x and y are odd integers, then x C y is an even
integer.

538 Appendix C. Answers for Exercises
3. (a)
Step Know Reason
P m is an even integer and n
is an integer.
Hypothesis
P1 There exists an integer k
m D 2k.
Definition of an even inte-
ger.
P 2 m n D .2k/ n Substitution
P 3 m n D 2 .kn/ Algebra
P 4 .kn/ is an integer Closure properties of the
integers
Q1 There exists an integer q
such that m n D 2q
q D kn.
Q m n is an even integer. Definition of an even inte-
ger.
(b) Use Part (a) to prove this.
4. (a) We assume that m is an even integer and will prove that 5m C 7 is an
odd integer. Since m is an even integer, there exists an integer k such
that m D 2k. Using substitution and algebra, we see that
5m C 7 D 5.2k/ C 7
D 10k C 6 C 1
D 2.5k C 3/ C 1
By the closure properties of the integers, we conclude that 5k C 3 is an
integer, and so the last equation proves that 5m C 7 is an odd integer.
Another proof. By Part (a) of Exercise 2, 5m is an even integer.
Hence, by Part (b) of Exercise 2, 5m C 7 is an even integer.
5. (b) We assume that m is an odd integer and will prove that 3m2
C7mC12
is an even integer. Since m is odd, there exists an integer k such that
m D 2k C 1. Hence,
3m2
C 2m C 3 D 3 .2k C 1/2
C 7 .2k C 1/ C 12
D 12k2
C 26k C 22
D 2 6k2
C 13k C 11

By the closure properties of the integers, 6k2 C 13k C 11

is an inte-
ger. Hence, this proves that if m is odd, then 3m2
C7mC12 is an even
integer.

6. (a) Prove that their difference is equal to zero or prove that they are not
zero and their quotient is equal to 1.
(d) Prove that two of the sides have the same length. Prove that the triangle
has two congruent angles. Prove that an altitude of the triangle is a
perpendicular bisector of a side of the triangle.
9. (a) Some examples of type 1 integers are 5, 2, 1, 4, 7, 10.
(c) All examples should indicate the proposition is true.
10. (a) Let a and b be integers and assume that a and b are both type 1 integers.
Then, there exist integers m and n such that a D 3m C 1 and b D
3n C 1. Now show that
a C b D 3 .m C n/ C 2:
The closure properties of the integers imply that m C n is an integer.
Therefore, the last equation tells us that aCb is a type 2 integer. Hence,
we have proved that if a and b are both type 1 integers, then a C b is a
type 2 integer.
Section 2.1
1. The statement was true. When the hypothesis is false, the conditional state-
ment is true.
2. (a) P is false. (b) P ^ Q is false. (c) P _ Q is false.
4. (c) Statement P is true but since we do not know if R is true or false, we
cannot tell if P ^ R is true or false.
5. Statements (a) and (d) have the same truth table. Statements (b) and (c) have
the same truth tables.
P Q P ! Q Q ! P
T T T T
T F F T
F T F F
F F T T

7.
P Q R P ^ .Q _ R/ .P ^ Q/ _ .P ^ R/
T T T T T
T T F T T
T F T T T
T F F F F
F T T F F
F T F F F
F F T F F
F F F F F
The two statements have the same truth table.
9. (c) The integer x is even only if x2
is even.
(d) The integer x2 is even is necessary for x to be even.
11. (a) :Q _ .P ! Q/ is a tautology.
(b) Q ^ .P ^ :Q/ is a contradiction.
(c) .Q ^ P / ^ .P ! :Q/ is a contradiction.
(d) :Q ! .P ^ :P / is neither a tautology nor a contradiction.
Section 2.2
1. (a) Converse: If a2
D 25, then a D 5. Contrapositive: If a2
¤ 25, then
a ¤ 5.
(b) Converse: If Laura is playing golf, then it is not raining. Contraposi-
tive: If Laura is not playing golf, then it is raining.
(c) Converse: If a4 ¤ b4, then a ¤ b. Contrapositive: If a4 D b4, then
a D b.
(d) Converse: If 3a is an odd integer, then a is an odd integer. Contraposi-
tive: If 3a is an even integer, then a is an even integer.
2. (a) Disjunction: a ¤ 5 or a2
D 25. Negation: a D 5 and a2
¤ 25.
(b) Disjunction: It is raining or Laura is playing golf.
Negation: It is not raining and Laura is not playing golf.
(c) Disjunction: a D b or a4
¤ b4
. Negation: a ¤ b and a4
D b4
.

(d) Disjunction: a is an even integer or 3a is an odd integer.
Negation: a is an odd integer and 3a is an even integer.
3. (a) We will not win the first game or we will not win the second game.
(b) They will not lose the first game and they will not lose the second game.
(c) You mow the lawn and I will not pay you $20.
(d) We do not win the first game and we will play a second game.
(e) I will not wash the car and I will not mow the lawn.
7. (a) In this case, it may be better to work with the right side first.
.P ! R/ _ .Q ! R/ .:P _ R/ _ .:Q _ R/
.:P _ :Q/ _ .R _ R/
.:P _ :Q/ _ R
: .P ^ Q/ _ R
.P ^ Q/ ! R:
(b) In this case, we start with the left side.
ŒP ! .Q ^ R/ :P _ .Q ^ R/
.:P ^ Q/ _ .:P ^ R/
.P ! Q/ ^ .P ! R/
10. Statements (c) and (d) are logically equivalent to the given conditional state-
ment. Statement (f) is the negation of the given conditional statement.
11. (d) This is the contrapositive of the given statement and hence, it is logi-
cally equivalent to the given statement.
Section 2.3
1. (a) The set of all real number solutions of the equation 2x2 C 3x 2 D 0,
which is

1
2
; 2

.
(b) The set of all integer solutions of the equation 2x2
C3x 2 D 0, which
is f 2g.

(c) The set of all integers whose square is less than 25, which is
f 4; 3; 2; 1; 0; 1; 2; 3; 4g.
(d) The set of all natural numbers whose square is less than 25, which is
f1; 2; 3; 4g
(e) The set of all rational numbers that are 2 units from 2.5 on the number
line, which is f 0:5; 4:5g.
(f) The set of all integers that are less than or equal to 2.5 units from 2 on
the number line, which is f0; 1; 2; 3; 4g.
2. Possible answers for (b):
A D
˚
n2
j n 2 N B D f n
j n is a nonnegative integerg
C D f6n C 3 j is a nonnegative integerg D D f4n j n 2 Z and 0 n 25g
3. The sets in (b) and (c) are equal to the given set.
4. (a) f 3g (b) f 8; 8g
5. (a) fx 2 Z j x 5g (c) fx 2 Q j x 0g (e)
˚
x 2 R j x2
10
Section 2.4
1. (a) There exists a rational number x such that x2 3x 7 D 0. This
statement is false since the solutions of the equation are x D
3 ˙
p
37
2
,
which are irrational numbers.
(b) There exists a real number x such that x2
C 1 D 0. This statement is
false since the only solutions of the equation are i and i, which are
not real numbers.
(c) There exists a natural number m such that m2 1. This statement is
false because if m is a natural number, then m 1 and hence, m2
1.
2. (a) m D 1 is a counterexample. The negation is: There exists a natural
number m such that m2 is not even or there exists a natural number m
such that m2
is odd.
(b) x D 0 is a counterexample. The negation is: There exists a real number
x such that x2
0.

(f) x D

2
is a counterexample. The negation is: There exists a real number
x such that tan2
x C 1 ¤ sec2
x.
3. (a) There exists a rational number x such that x
p
2.
The negation is .8x 2 Q/

x
p
2

, which is, For each rational num-
ber x, x
p
2.
(c) For each integer x, x is even or x is odd.
The negation is .9x 2 Z/ .x is odd and x is even/, which is, There ex-
ists an integer x such that x is odd and x is even.
(e) For each integer x, if x2
is odd, then x is odd.
The negation is .9x 2 Z/ x2 is odd and x is even

, which is, There
exists an integer x such that x2
is odd and x is even.
(h) There exists a real number x such that cos .2x/ D 2 .cos x/.
The negation is .8x 2 R/ .cos .2x/ ¤ 2 .cos x//, which is, For each
real number x, cos .2x/ ¤ 2 .cos x/.
4. (a) There exist integers m and n such that m n.
(e) There exists an integer n such that for each integer m, m2 n.
5. (a) .8m/ .8n/ .m n/.
For all integers m and n , m n.
(e) .8n/ .9m/ m2 n

.
For each integer n, there exists an integer m such that m2
n.
6. (a) It is not a statement since x is an unquantified variable.
(b) It is a true statement.
(c) It is a false statement.
(d) It is a true statement.
(e) f 20; 10; 5; 4; 2; 1; 1; 2; 4; 5; 10; 20g
10. (a) A function f with domain R is strictly increasing provided that
.8x; y 2 R/ Œ.x y/ ! .f.x/ f.y//.

Section 3.1
1. (a) Since a j b and a j c, there exist integers m and n such that b D am
and c D an. Hence,
b c D am an
D a.m n/
Since m n is an integer (by the closure properties of the integers), the
last equation implies that a divides b c.
(b) What do you need to do in order to prove that n3
is odd? Notice that if
n is an odd integer, then there exists an integer k such that n D 2k C 1.
Remember that to prove that n3
is an odd integer, you need to prove
that there exists an integer q such that n3 D 2q C 1.
This can also be approached as follows: If n is odd, then by Theo-
rem 1.8, n2
is odd. Now use the fact that n3
D n n2
.
(c) If 4 divides .a 1/, then there exists an integer k such that a 1 D 4k
and so a D 4k C 1. Use algebra to rewrite a2
1

D .4k C 1/2
1.
2. (a) The natural number n D 6 is a counterexample since n2
1 D 35 and
3 does not divide 35.
(d) The integer a D 3 is a counterexample since a2
1 D 8 and a 1 D 2.
Since 4 divides 8 and 4 does not divide 2, this is an example where the
hypothesis of the conditional statement is true and the conclusion is
false.
3. (b) This statement is false. One counterexample is a D 3 and b D 2 since
this is an example where the hypothesis is true and the conclusion is
false.
(d) This statement is false. One counterexample is n D 5. Since n2
4 D
21 and n 2 D 3, this is an example where the hypothesis of the
conditional statement is true and the conclusion is false.
(e) Make sure you first try some examples. How do you prove that an
integer is an odd integer?
(f) The following algebra may be useful.
4 .2m C 1/2
C 7 .2m C 1/ C 6 D 6m2
C 30m C 17:
(g) This statement is false. One counterexample is a D 7, b D 1, and
d D 2. Why is this a counterexample?

4. (a) If xy D 1, then x and y are both divisors of 1, and the only divisors of
1 are 1 and 1.
(b) Part (a) is useful in proving this.
5. Another hint: .4n C 3/ 2 .2n C 1/ D 1.
8. Assuming a and b are both congruent to 2 modulo 3, there exist integers m
and n such that a D 3m C 2 and b D 3n C 2.
For part (a), show that
a C b 1 D 3 .m C n C 1/ :
We can then conclude that 3 divides .aCb/ 1 and this proves that a C b
1 .mod 3/.
For part (b), show that
a b 1 D 3.3mn C 2m C 2n C 1/:
We can then conclude that 3 divides a b 1 and this proves that a b
1 .mod 3/.
11. (a) Let n 2 N. If a is an integer, then a a D 0 and n divides 0. Therefore,
a a .mod n/.
(b) Let n 2 N, let a; b 2 Z and assume that a b .mod n/. We will prove
that b a .mod n/. Since a b .mod n/, n divides .a b/ and so
there exists an integer k such that a b D nk. From this, we can show
that b a D n. k/ and so n divides .b a/. Hence, if a b .mod n/,
then b a .mod n/.
12. The assumptions mean that n j .a b/ and that n j .c d/. Use these
divisibility relations to obtain an expression that is equal to a and to obtain
an expression that is equal to c. Then use algebra to rewrite the resulting
expressions for a C c and a c.
Section 3.2
1. (a) Let n be an even integer. Since n is even, there exists an integer k such
that n D 2k. Now use this to prove that n3 must be even.
(b) Prove the contrapositive.

(c) Explain why Parts (a) and (b) prove this.
(d) Explain why Parts (a) and (b) prove this.
2. (a) The contrapositive is, For all integers a and b, if ab 0.mod6/, then
a 0.mod6/ or b 0.mod6/.
3. (a) The contrapositive is: For all positive real numbers a and b, if a D b,
then
p
ab D
a C b
2
.
(b) The statement is true. If a D b, then
a C b
2
D
2a
2
D a, and
p
ab D
p
a2 D a. This proves the contrapositive.
4. (a) True. If a 2 .mod 5/, then there exists an integer k such that a 2 D
5k. Then,
a2
4 D .2 C 5k/2
4 D 20k C 25k2
:
This means that a2 4 D 5 4k C 5k2

, and hence, a2 4 .mod 5/.
(b) False. A counterexample is a D 3 since 32 4 .mod 5/ and 3 6
2 .mod 5/.
(c) False. Part (b) shows this is false.
6. (a) For each integer a, if a 3 .mod 7/, then .a2 C 5a/ 3 .mod 7/,
and for each integer a, if .a2
C5a/ 3 .mod 7/, then a 3 .mod 7/.
(b) For each integer a, if a 3 .mod 7/, then .a2 C 5a/ 3 .mod 7/
is true. To prove this, if a 3 .mod 7/, then there exists an integer k
such that a D 3 C 7k. We can then prove that
a2
C 5a

3 D 21 C 77k C 49k2
D 7.3 C 11k C 7k2
/:
This shows that .a2 C 5a/ 3 .mod 7/.
For each integer a,if .a2 C 5a/ 3 .mod 7/, then a 3 .mod 7/ is
false. A counterexample is a D 6. When a D 6, a2
C 5a D 66 and
66 3 .mod 7/ and 6 6 3 .mod 7/.
(c) Since one of the two conditional statements in Part (b) is false, the
given proposition is false.
8. Prove both of the conditional statements: (1) If the area of the right triangle
is c2
=4, then the right triangle is an isosceles triangle. (2) If the right triangle
is an isosceles triange, then the area of the right triangle is c2
=4.

9. The statement is true. It is easier to prove the contrapositive, which is:
For each positive real number x, if
p
x is rational, then x is rational.
Let x be a positive real number. If there exist positive integers m and n such
that
p
x D
m
n
, then x D
m2
n2
.
10. Remember that there are two conditional statements associated with this bi-
conditional statement. Be willing to consider the contrapositive of one of
these conditional statements.
15. Define an appropriate function and use the Intermediate Value Theorem.
17. (b) Since 4 divides a, there exist an integer n such that a D 4n. Using
this, we see that b3 D 16n2. This means that b3 is even and hence by
Exercise (1), b is even. So there exists an integer m such that b D 2m.
Use this to prove that m3 must be even and hence by Exercise (1), m is
even.
18. It may be necessary to factor a sum cubes. Recall that
u3
C v3
D .u C v/.u2
uv C v2
/:
Section 3.3
1. (a) P _ C
2. (a) This statement is true. Use a proof by contradiction. So assume that
there exist integers a and b such that a is even, b is odd, and 4 divides
a2 C b2. So there exist integers m and n such that
a D 2m and a2
C b2
D 4n:
Substitute a D 2m into the second equation and use algebra to rewrite
in the form b2
D 4.n m2
/. This means that b2
is even and hence,
that b is even. This is a contradiction to the assumption that b is odd.
(b) This statement is true. Use a proof by contradiction. So assume that
there exist integers a and b such that a is even, b is odd, and 6 divides
a2
C b2
. So there exist integers m and n such that
a D 2m and a2
C b2
D 6n:

Substitute a D 2m into the second equation and use algebra to rewrite
in the form b2
D 2.3n 2m2
/. This means that b2
is even and hence,
that b is even. This is a contr
(d) This statement is true. Use a direct proof. Let a and b be integers and
assume they are odd. So there exist integers m and n such that
a D 2m C 1 and b D 2n C 1:
We then see that
a2
C 3b2
D 4m2
C 4m C 1 C 12n2
C 12n C 3
D 4 m2
C m C 3n2
C 3n C 1

:
This shows that 4 divides a2 C 3b2.
3. (a) We would assume that there exists a positive real number r such that
r2
D 18 and r is a rational number.
(b) Do not attempt to mimic the proof that the square root of 2 is irrational
(Theorem 3.20). You should still use the definition of a rational number
but then use the fact that
p
18 D
p
9 2 D
p
9
p
2 D 3
p
2. So, if we
assume that r D
p
18 D 3
p
2 is rational, then
r
3
D
p
18
3
is rational
since the rational numbers are closed under division. Hence,
p
2 is
rational and this is a contradiction to Theorem 3.20.
5. (a) Use a proof by contradiction. So, we assume that there exist real num-
bers x and y such that x is rational, y is irrational, and xCy is rational.
Since the rational numbers are closed under addition, this implies that
.x C y/ x is a rational number. Since .x C y/ x D y, we conclude
that y is a rational number and this contradicts the assumption that y is
irrational.
(b) Use a proof by contradiction. So, we assume that there exist nonzero
real numbers x and y such that x is rational, y is irrational, and xy
is rational. Since the rational numbers are closed under division by
nonzero rational numbers, this implies that
xy
x
is a rational number.
Since
xy
x
D y, we conclude that y is a rational number and this con-
tradicts the assumption that y is irrational.
6. (a) This statement is false. A counterexample is x D
p
2.

(b) This statement is true since the contrapositive is true. The contraposi-
tive is:
For any real number x, if
p
x is rational, then x is rational.
If there exist integers a and b with b ¤ 0 such that
p
x D
a
b
, then
x2
D
a2
b2
and hence, x2
is rational.
11. Recall that log2.32/ is the real number a such that 2a D 32. That is, a D
log2.32/ means that 2a
D 32. If we assume that a is rational, then there
exist integers m and n, with n ¤ 0, such that a D
m
n
.
12. Hint: The only factors of 7 are 1; 1; 7; and 7.
13. (a) What happens if you expand Œsin./ C cos./2? Don’t forget your
trigonometric identities.
14. Hint: Three consecutive natural numbers can be represented by n, nC1, and
nC2, where n 2 N, or three consecutive natural numbers can be represented
by m 1, m, and m C 1, where m 2 N.
Section 3.4
1. Use the fact that n2 C n D n.n C 1/.
2. Do not use the quadratic formula. Try a proof by contradiction. If there
exists a solution of the equation x2
C x u D 0 that is an integer, then we
can conclude that there exists an integer n such that n2 C n u D 0. Then,
u D n .n C 1/ :
From Exercise (1), we know that n.nC 1/ is even and hence, u is even. This
contradicts the assumption that u is odd.
3. If n is an odd integer, then there exists an integer m such that n D 2m C 1.
Use two cases: (1) m is even; (2) m is odd. If m is even, then there exists an
integer k such that m D 2k and this means that n D 2.2k/C1 or n D 4kC1.
If m is odd, then there exists an integer k such that m D 2k C 1. Then
n D 2.2k C 1/ C 1 or n D 4k C 3.
4. If a 2 Z and a2 D a, then a.a 1/ D 0. Since the product is equal to zero,
at least one of the factors must be zero. In the first case, a D 0. In the second
case, a 1 D 0 or a D 1.

5. (c) For all integers a, b, and d with d ¤ 0, if d divides the product ab,
then d divides a or d divides b.
6. (a) The statement, for all integers m and n, if 4 divides m2
C n2
1

,
then m and n are consecutive integers, is false. A counterexample is
m D 2 and n D 5.
The statement, for all integers m and n, if m and n are consecutive
integers, then 4 divides m2 C n2 1

, is true. To prove this, let n D
m C 1. Then
m2
C n2
1 D 2m2
C 2m D 2m.m C 1/:
We have proven the m.m C 1/ is even. (See Exercise (1).) So this can
be used to prove that 4 divides m2 C n2 1

.
8. Try a proof by contradiction with two cases: a is even or a is odd.
10. (a) One way is to use three cases: (i) x 0; (ii) x D 0; and x 0. For
the first case, x 0 and j xj D . x/ D x D jxj.
11. (a) For each real number x, jxj a if and only if x a or x a.
Section 3.5
2. (a) The first case is when n 0 .mod 3/. We can then use Theorem 3.28
to conclude that n3
03
.mod 3/ or that n3
0 .mod 3/. So in this
case, n3 n .mod 3/.
For the second case, n 1 .mod 3/. We can then use Theorem 3.28
to conclude that n3
13
.mod 3/ or that n3
1 .mod 3/. So in this
case, n3 n .mod 3/.
The last case is when n 2 .mod 3/. We then get n3
23
.mod 3/
or n3
8 .mod 3/. Since 8 2 .mod 3/, we can use the transitive
property to conclude that n3 2 .mod 3/, and so n3 n .mod 3/.
Since we have proved it in all three cases, we conclude that for each
integer n, n3 n .mod 3/.
(b) Since n3 n .mod 3/, we use the definition of congruence to con-
clude that 3 divides n3
n

.
3. Let n 2 N. For a; b 2 Z, you need to prove that if a b .mod n/, then
b a .mod n/. So let a; b 2 Z and assume that a b .mod n/. So

n j .a b/ and there exists an integer k such that a b D nk. Then,
b a D n. k/ and b a .mod n/.
4. (a) The contrapositive is: For each integer a, if 3 does not divide a, then 3
divides a2.
(b) To prove the contrapositive, let a 2 Z and assume that 3 does not
divide a. So using the Division Algorithm, we can consider two cases:
(1) There exists a unique integer q such that a D 3q C 1. (2) There
exists a unique integer q such that a D 3q C 2.
For the first case, show that a2
D 3 3q2
C 2q

C 1. For the second
case, show that a2 D 3 3q2 C 4q C 1

C 1. Since the Division Algo-
rithm states that the remainder is unique, this shows that in both cases,
the remainder is 1 and so 3 does not divide a2.
5. (a) a 0 .mod n/ if and only if n j .a 0/.
(b) Let a 2 Z. Corollary 3.32 tell us that if a 6 0 .mod3/, then
a 1 .mod 3/ or a 2 .mod 3/.
(c) Part (b) tells us we can use a proof by cases using the following two
cases: (1) a 1 .mod 3/; (2) a 2 .mod 3/.
So, if a 1 .mod 3/, then by Theorem 3.28, a a 1 1 .mod 3/,
and hence, a2 1 .mod 3/.
If a 2 .mod 3/, then by Theorem 3.28, a a 2 2 .mod 3/,
and hence, a2 4 .mod 3/. Since 4 1 .mod 3/, this implies that
a2
1 .mod 3/.
6. The contrapositive is: Let a and b be integers. If a 6 0 .mod 3/ and
b 6 0 .mod 3/, then ab 6 0 .mod 3/.
Using Exercise 5(b), we can use the following four cases:
(1) a 1 .mod 3/ and b 1 .mod 3/;
(2) a 1 .mod 3/ and b 2 .mod 3/;
(3) a 2 .mod 3/ and b 1 .mod 3/;
(4) a 2 .mod 3/ and b 2 .mod 3/.
In all four cases, we use Theorem 3.28 to conclude that ab 6 0 .mod 3/.
For example, for the third case, we see that ab 2 1 .mod 3/. That is,
ab 2 .mod 3/.
7. (a) This follows from Excercise (5) and the fact that 3 j k if and only if
k 0 .mod 3/.
(b) This follows directly from Part (a) using a D b.

8. (a) Use a proof similar to the proof of Theorem 3.20. The result of Exer-
cise (7) may be helpful.
9. The result in Part (c) of Exercise (5) may be helpful in a proof by contradic-
tion.
10. (b) Factor n3 n.
(c) Consider using cases based on congruence modulo 6.
Section 4.1
1. The sets in Parts (a) and (b) are inductive.
2. A finite nonempty set is not inductive (why?) but the empty set is inductive
(why?).
3. (a) For each n 2 N, let P.n/ be, 2 C 5 C 8 C C .3n 1/ D
n.3n C 1/
2
.
When we use n D 1, the summation on the left side of the equation is
2, and the right side is
1.3 1 C 1/
2
D 2. Therefore, P.1/ is true. For
the inductive step, let k 2 N and assume that P .k/ is true. Then,
2 C 5 C 8 C cdots C .3k 1/ D
k .3k C 1/
2
:
We now add 3 .k C 1/ 1 to both sides of this equation. This gives
2 C 5 C 8 C C .3k 1/
C3 .k C 1/ 1 D
k .3k C 1/
2
C .3 .k C 1/ 1/
D
k .3k C 1/
2
C .3k C 2/
If we now combine the terms on the right side of the equation into a

single fraction, we obtain
2 C 5 C C .3k 1/ C 3 .k C 1/ 1 D
k .3k C 1/ C 6k C 4
2
D
3k2 C 7k C 4
2
D
.k C 1/ .3k C 4/
2
D
.k C 1/ .3 .k C 1/ C 1/
2
This proves that if P .k/ is true, then P .k C 1/ is true.
6. The conjecture is that for each n 2 N, 1 C 3 C 5 C C .2n 1/ D n2
. The
key to the inductive step is that
1 C 3 C 5 C C .2k 1/
C.2.k C 1/ 1/ D Œ1 C 3 C 5 C C .2k 1/ C .2.k C 1/ 1
D k2
C .2k C 1/
D .k C 1/2
7. (e) For each natural number n, 4n
1 .mod 3/.
(f) For each natural number n, let P.n/ be, “4n
1 .mod 3/.” Since
41
1 .mod 3/, we see that P.1/ is true. Now let k 2 N and assume
that P.k/ is true. That is, assume that
4k
1 .mod 3/ :
Multiplying both sides of this congruence by 4 gives
4kC1
4 .mod 3/ :
However, 4 1 .mod 3/ and so by using the transitive property of
congruence, we see that 4kC1
1 .mod 3/. This proves that if P.k/
is true, then P.k C 1/ is true.
8. (a) The key to the inductive step is that if 4k D 1 C 3m, then 4k 4 D
4.1 C 3m/, which implies that
4kC1
1 D 3.1 C 4m/:

13. Let k be a natural number. If ak bk .mod n/, then since we are also
assuming that a b .mod n/, we can use Part (2) of Theorem 3 to conclude
that a ak b bk .mod n/.
14. Three consecutive natural numbers may be represent by n, n C 1, and n C 2,
where n is a natural number. For the inductive step, think before you try to
do a lot of algebra. You should be able to complete a proof of the inductive
step by expanding the cube of only one expression.
Section 4.2
1. (a) Let P .n/ be, “3n 1 C 2n.” P .2/ is true since 32 D 9, 1 C 22 D 5,
and 9 5.
For the inductive step, we assume that P .k/ is true and so
3k
1 C 2k
: (1)
To prove that P.k C 1/ is true, we must prove that 3kC1 1 C 2kC1.
Multiplying both sides inequality (1) by 3 gives
3kC1
3 C 3 2k
:
Now, since 3 1 and 3 2k 2kC1, we see that 3 C 3 2k 1 C 2kC1
and hence, 3kC1
1 C 2kC1
. Thus, if P .k/ is true, then P .k C 1/ is
true. This proves the inductive step.
2. If n 5, then n2 2n. To prove this, we let P.n/ be n2 2n. For the
basis step, when n D 5, n2
D 25, 2n
D 32, and 25 32. For the inductive
step, we assume that k 5 and that P.k/ is true or that k2 2k. With these
assumptions, we need to prove that P.kC1/ is true or that .kC1/2 2kC1.
We first note that
.k C 1/2
D k2
C 2k C 1 2k
C 2k C 1: (1)
Since k 5, we see that 5k k2
and so 2k C 3k k2
. However, 3k 1
and so 2k C 1 2k C 3k k2. Combinining this with inequality (1),
we obtain .k C 1/2
2k
C k2
. Using the assumption that P.k/ is true

k2 2k

, we obtain
.k C 1/2
2k
C 2k
D 2 2k
.k C 1/2
2kC1
This proves that if P.k/ is true, then P.k C 1/ is true.

5. Let P.n/ be the predicate, “8n j .4n/Š.” Verify that P.0/; P.1/; P.2/, and
P.3/ are true. For the inductive step, the following fact about factorials may
be useful:
Œ4.k C 1/Š D .4k C 4/Š
D .4k C 4/.4k C 3/.4k C 2/.4k C 1/.4k/Š:
8. Let P.n/ be, “The natural number n can be written as a sum of natural
numbers, each of which is a 2 or a 3.” Verify that P.4/; P.5/; P.6/, and
P.7/ are true.
To use the Second Principle of Mathematical Induction, assume that k 2
N; k 5 and that P.4/; P.5/; : : :; P.k/ are true. Then notice that
k C 1 D .k 1/ C 2:
Since k 1 4, we have assumed that P.k 1/ is true. This means that
.k 1/ can be written as a sum of natural numbers, each of which is a 2 or a
3. Since k C 1 D .k 1/ C 2, we can conclude that .k C 1/ can be written
as a sum of natural numbers, each of which is a 2 or a 3. This completes the
proof of the inductive step.
12. Let P.n/ be, “Any set with n elements has
n.n 1/
2
2-element subsets.”
P.1/ is true since any set with only one element has no 2-element subsets.
Let k 2 N and assume that P.k/ is true. This means that any set with k
elements has
k.k 1/
2
2-element subsets. Let A be a set with kC1 elements,
and let x 2 A. Now use the inductive hypothesis on the set A fxg, and
determine how the 2-element subsets of A are related to the set A fxg.
16. (a) Use Theorem 4.9.
(b) Assume k ¤ q and consider two cases: (i) k q; (ii) k q.
Section 4.3
1. Let P.n/ be an D nŠ. Since a0 D 1 and 0Š D 1, we see that P.0/ is true.
For the inductive step, we assume that k 2 N [ 0 and that P.k/ is true or

that ak D kŠ.
akC1 D .k C 1/ ak
D .k C 1/ kŠ
D .k C 1/Š:
This proves the inductive step that if P.k/ is true, then P.k C 1/ is true.
2. (a) Let P.n/ be, “f4n is a multiple of 3.” Since f4 D 3, P.1/ is true. If
P.k/ is true, then there exists an integer m such that f4k D 3m. We
now need to prove that P.k C 1/ is true or that f4.kC1/ is a multiple of
3. We use the following:
f4.kC1/ D f4kC4
D f4kC3 C f4kC2
D .f4kC2 C f4kC1/ C .f4kC1 C f4k/
D f4kC2 C 2f4kC1 C f4k
D .f4kC1 C f4k/ C 2f4kC1 C f4k
D 3f4kC1 C 2f4k
We now use the assumption that f4k D 3m and the last equation to ob-
tain f4.kC1/ D 3f4kC1C23m and hence, f4.kC1/ D 3 .f4kC1 C 2m/.
Therefore, f4.kC1/ is a multiple of 3 and this completes the proof of
the inductive step.
(c) Let P.n/ be, “f1 C f2 C C fn 1 D fnC1 1.” Since f1 D f3 1,
P.2/ is true. For k 2, if P.k/ is true, then f1 C f2 C C fk 1 D
fkC1 1. Then
.f1 C f2 C C fk 1/ C fk D .fkC1 1/ C fk
D .fkC1 C fk/ 1
D fkC2 1:
(f) Let P.n/ be, “f 2
1 C f 2
2 C C f 2
n D fnfnC1.” For the basis step, we
notice that f 2
1 D 1 and f1 f2 D 1 and hence, P.1/ is true. For the
inductive step, we need to prove that if P.k/ is true, then P.k C 1/ is
true. That is, we need to prove that if f 2
1 C f 2
2 C C f 2
k
D fkfkC1,

then f 2
1 C f 2
2 C C f 2
k
C f 2
kC1
D fkC1fkC2. To do this, we can
use
f 2
1 C f 2
2 C C f 2
k

C f 2
kC1 D fkfkC1 C f 2
kC1
f 2
1 C f 2
2 C C f 2
k C f 2
kC1 D fkC1 .fk C fkC1/
D fkC1fkC2:
6. For the inductive step, if ak D a rk 1, then
akC1 D r ak
D r

a rk 1

D a rk
:
8. For the inductive step, use the assumption that Sk D a
1 rk
1 r
!
and the
recursive definiton to write SkC1 D a C r Sk.
9. (a) a2 D 7, a3 D 12, a4 D 17, a5 D 22, a6 D 27.
(b) One possibility is: For each n 2 N, an D 2 C 5.n 1/.
12. (a) a2 D
p
6, a3 D
pp
6 C 5 2:729, a4 2:780, a5 2:789,
a6 2:791
(b) Let P.n/ be, “an 3.” Since a1 D 1, P.1/ is true. For k 2 N, if P.k/
is true, then ak 3. Now
akC1 D
p
5 C ak:
Since ak 3, this implies that akC1
p
8 and hence, akC1 3.
13. (a) a3 D 7, a4 D 15, a5 D 31, a6 D 63
(b) Think in terms of powers of 2.
14. (a) a3 D
3
2
, a4 D
7
4
, a5 D
37
24
, a6 D
451
336

16. (b) a2 D 5
a3 D 23
a4 D 119
a5 D 719
a6 D 5039
a7 D 40319
a8 D 362879
a9 D 3628799
a10 D 39916799
18. (a) Let P.n/ be, “Ln D 2fnC1 fn.” First, verify that P.1/ and P.2/
are true. Now let k be a natural number with k 2 and assume that
P.1/, P.2/, ..., P.k/ are all true. Since P.k/ and P.k 1/ are both
assumed to be true, we can use them to help prove that P.k C 1/ must
then be true as follows:
LkC1 D Lk C Lk 1
D .2fkC1 fk/ C .2fk fk 1/
D 2 .fkC1 C fk/ .fk C fk 1/
D 2fkC2 fkC1:
Section 5.1
1. (a) A D B (c) C ¤ D (e) A 6 D
(b) A B (d) C D
2. In both cases, the two sets have preceisely the same elements.
3. A ; ; ¤ B ; ; ; ¤ A
5 2 C f5g ; ; ¤ C
A ; ; ¤ C f1; 2g ; ; ¤ B
f1; 2g ¤ A f3; 2; 1g ; ; ¤ D
4 … B D ¤ ;
card .A/ D card .D/ card .A/ ¤ card .B/
A 2 P .A/ A 2 P .B/
4. N Z N Q N R
Z Q Z R Q R
5. (a) The set fa; bg is a not a subset of fa; c; d; eg since b 2 fa; bg and
b … fa; c; d; eg.
(b) f 2; 0; 2g D
˚
x 2 Z j x is even and x2
5 since both sets have pre-
cisely the same elements.

(c) ; f1g since the following statement is true: For every x 2 U , if
x 2 ;, then x 2 f1g.
(d) The statement is false. The set fag is an element of P .A/.
6. (a) x … A B if and only if x … A or x … B.
7. (a) A B D f5; 7g
(b) A [ B D f1; 3; 4; 5; 6; 7; 9g
(c) .A [ B/c
D f2; 8; 10g
(d) Ac
Bc
D f2; 8; 10g
(e) .A [ B/ C D f3; 6; 9g
(f) A C D f3; 6g
(g) B C D f9g
(h) .A C / [ .B C/ D f3; 6; 9g
(i) B D D ;
(j) .B D/c
D U
(k) A D D f3; 5; 7g
(l) B D D f1; 5; 7; 9g
(m) .A D/ [ .B D/ D f1; 3; 5; 7; 9g
(n) .A [ B/ D D f1; 3; 5; 7; 9g
9. (b) There exists an x 2 U such that x 2 .P Q/ and x … .R S/. This
can be written as, There exists an x 2 U such that x 2 P , x … Q, and
x … R or x … S.
10. (a) The given statement is a conditional statement. We can rewrite the
subset relations in terms of conditional sentences: A B means, “For
all x 2 U , if x 2 A, then x 2 B,” and Bc Ac means, “For all
x 2 U , if x 2 Bc
, then x 2 Ac
.”
Section 5.2
1. (a) The set A is a subset of B. To prove this, we let x 2 A. Then 2
x 2. Since x 2, we conclude that x 2 B and hence, we have
proved that A is a subset of B.
(b) The set B is not a subset of A. There are many examples of a real
number that is in B but not in A. For example, 3 is in A, but 3 is
not in B.
3. (a) A D f: : : 9; 1; 7; 15; 23; : : :g and B D f: : : 9; 5; 1; 3; 7; 11; 15; : : :g.
(b) To prove that A B, let x 2 A. Then, x 7 .mod 8/ and so,
8 j .x 7/. This means that there exists an integer m such that x 7 D
8m. By adding 4 to both sides of this equation, we see that x 3 D

8mC4, or x 3 D 4 .2m C 1/. From this, we conclude that 4 j .x 3/
and that x 3 .mod 4/. Hence, x 2 B.
(c) B 6 A. For example, 3 2 B and 3 … A.
5. (a) We will prove that A D B. Notice that if x 2 A, then there exists an
integer m such that x 2 D 3m. We can use this equation to see that
2x 4 D 6m and so 6 divides .2x 4. Therefore, x 2 B and hence,
A B.
Conversely, if y 2 B, then there exists an integer m such that 2y 4 D
6m. Hence, y 2 D 3m, which implies that y 2 .mod 3/ and
y 2 A. Therefore, B A.
(c) A B D ;. To prove this, we will use a proof by contradiction and
assume that A B ¤ ;. So there exists an x in A B. We can then
conclude that there exist integers m and n such that x 1 D 5m and
x y D 10n. So x D 5m C 1 and x D 10n C 7. We then see that
5m C 1 D 10n C 7
5.m 2n/ D 6
The last equation implies that 5 divides 6, and this is a contradiction.
7. (a) Let x 2 AB. Then, x 2 A and x 2 B. This proves that if x 2 AB,
then x 2 A, and hence, A B A.
(b) Let x 2 A. Then, the statement “x 2 A or x 2 B” is true. Hence,
A A [ B.
(e) By Theorem 5.1, ; A ;. By Part (a), A ; ;. Therefore,
A ; D ;.
12. (a) Let x 2 A C . Then x 2 A and x 2 C . Since we are assum-
ing that A B, we see that x 2 B and x 2 C . This proves that
A C B C .
15. (a) “If A B, then A Bc
D ;” is Proposition 5.14. To prove the
other conditional statement, start with, “Let x 2 A.” Then use the
assumption that A Bc
D ; to prove that x must be in B.
(b) To prove “If A B, then A [ B D B,” first note that if x 2 B, then
x 2 A [ B and, hence, B A [ B. Now let x 2 A [ B and note that
since A B, if x 2 A, then x 2 B. Use this to argue that under the
assumption that A B, A [ B B.
To prove “If A [ B D B, then A B,” start with, Let x 2 A and use
this asusmption to prove that x must be an element of B.

Section 5.3
1. (a) Let x 2 .Ac/c. Then x … Ac, which means x 2 A. Hence, .Ac/c A.
Now let y 2 A. Then, y … Ac
and hence, y 2 .Ac
/c
. Therefore,
A .Ac/c
.
(c) Let x 2 U . Then x … ; and so x 2 ;c
. Therefore, U ;c
. Also,
since every set we deal with is a subset of the universal set, ;c U .
2. To prove that A .B [ C / .A B/ [ .A C /, we let x 2 A .B [ C /.
Then x 2 A and x 2 B [ C . So we will use two cases: (1) x 2 B; (2)
x 2 C .
In Case (1), x 2 A B and, hence, x 2 .A B/ [ .A C /. In Case (2),
x 2 AC and, hence, x 2 .AB/[.AC /. This proves that A.B[C /
.A B/ [ .A C /.
To prove that .A B/ [ .A C / A .B [ C /, let y 2 .A B/ [
.A C /. Then, y 2 A B or y 2 A C . If y 2 A B, then y 2 A
and y 2 B. Therefore, y 2 A and y 2 B [ C . So, we may conclude that
y 2 A.B [ C /. In a similar manner, we can prove that if y 2 AC , then
y 2 A .B [ C /. This proves that .A B/ [ .A C / A .B [ C /,
and hence that A .B [ C / D .A B/ [ .A C /.
4. (a) A .B [ C / D .A B/ .A C /.
(c) Using the algebra of sets, we obtain
.A B/ .A C / D .A Bc
/ .A C c
/
D .A A/ .Bc
C c
/
D A .B [ C /c
D A .B [ C /:
6. (a) Using the algebra of sets, we see that
.A C / .B C / D .A C c
/ .B C c
/
D .A B/ .C c
C c
/
D .A B/ C c
D .A B/ C:
9. (a) Use a proof by contradiction. Assume the sets are not disjoint and let
x 2 A .B A/. Then x 2 A and x 2 B A, which implies that
x … A.

Section 5.4
1. (a) A B D f.1; a/ ; .1; b/ ; .1; c/ ; .1; d/ ; .2; a/ ; .2; b/ ; .2; c/ ; .2; d/g.
(b) B A D f.a; 1/ ; .b; 1/ ; .c; 1/ ; .d; 1/ ; .a; 2/ ; .b; 2/ ; .c; 2/ ; .d; 2/g.
(c) A C D f.1; 1/ ; .1; a/ ; .1; b/ ; .2; 1/; .2; a/ ; .2; b/g.
(d) A2
D f.1; 1/; .1; 2/ ; .2; 1/; .2; 2/g.
(e) A .B C / D f.1; a/ ; .1; b/ ; .2; a/ ; .2; b/g.
(f) .A B/ .A C / D f.1; a/ ; .1; b/ ; .2; a/ ; .2; b/g.
(g) A ; D ;.
(h) B f2g D f.a; 2/ ; .b; 2/ ; .c; 2/ ; .d; 2/g.
3. Let u 2 A .B C/. Then, there exists x 2 A and there exists y 2 B C
such that u D .x; y/. Since y 2 B C , we know that y 2 B and y 2 C .
So, we have:
u D .x; y/ 2 A B and u D .x; y/ 2 A C .
Hence, u 2 .A B/.A C / and this proves that A.B C/ .A B/
.A C /.
Now let v 2 .A B/ .A C /. Then, v 2 A B and v 2 A C . So,
there exists an s in A and a t in B such that v D .s; t/. But, since v is
also in A C , we see that t must also be in C . Thus, t 2 B C and so,
v 2 A .B C /. This proves that .A B/ .A C / A .B C /.
4. Let u 2 .A [ B/ C . Then, there exists x 2 A [ B and there exists y 2 C
such that u D .x; y/. Since x 2 A [ B, we know that x 2 A or x 2 B. In
the case where x 2 A, we see that u 2 A C , and in the case where x 2 B,
we see that u 2 B C . Hence, u 2 .A C / [ .B C /. This proves that
.A [ B/ C .A C / [ .B C /.
We still need to prove that .A C / [ .B C / A .B [ C /.
Section 5.5
1. (a) f3; 4g (d) f3; 4; 5; 6; 7; 8; 9; 10g

2. (a) f5; 6; 7; : : :g
(c) ;
(d) f1; 2; 3; 4g
(f) ;
3. (a) fx 2 R j 100 x 100g (b) fx 2 R j 1 x 1g
4. (a) We let ˇ 2 ƒ and let x 2 Aˇ . Then x 2 A˛, for at least one ˛ 2 ƒ
and, hence, x 2
S
˛2ƒ
A˛. This proves that Aˇ
S
˛2ƒ
A˛.
5. (a) We first let x 2 B

S
˛2ƒ
A˛

. Then x 2 B and x 2
S
˛2ƒ
A˛.
This means that there exists an ˛ 2 ƒ such that x 2 A˛. Hence,
x 2 B A˛, which implies that x 2
S
˛2ƒ
.B A˛/. This proves that
B

S
˛2ƒ
A˛

S
˛2ƒ
.B A˛/.
We now let y 2
S
˛2ƒ
.B A˛/. So there exists an ˛ 2 ƒ such that
y 2 B A˛. Then y 2 B and y 2 A˛, which implies that y 2 B
and y 2
S
˛2ƒ
A˛. Therefore, y 2 B

S
˛2ƒ
A˛

, and this proves that
S
˛2ƒ
.B A˛/ B

S
˛2ƒ
A˛

.
8. (a) Let x 2 B. For each ˛ 2 ƒ, B A˛ and, hence, x 2 A˛. This means
that for each ˛ 2 ƒ, x 2 A˛ and, hence, x 2
T
˛2ƒ
A˛. Therefore,
B
T
˛2ƒ
A˛.
12. (a) We first rewrite the set difference and then use a distributive law.
[
˛2ƒ
A˛
!
B D
[
˛2ƒ
A˛
!
Bc
D
[
˛2ƒ
A˛ Bc

D
[
˛2ƒ
.A˛ B/

Section 6.1
1. (a) f. 3/ D 15, f. 1/ D 3, f.1/ D 1, f.3/ D 3.
(b) The set of preimages of 0 is f0; 2g. The set of preimages of 4 is
(
2
p
20
2
;
2 C
p
20
2
)
. (Use the quadratic formula.)
(d) range.f / D fy 2 R j y 1g
3. (a) f . 7/ D 10, f . 3/ D 6, f .3/ D 0, f .7/ D 4.
(b) The set of preimages of 5 is f 2g. There set of preimages of 4 is f 1g.
(c) range.f / D Z. Notice that for all y 2 Z (codomain), f.3 y/ D y
and .3 y/ 2 Z (domain).
4. (b) The set of preimages of 5 is f2g. There set of preimages of 4 is ;.
(c) The range of the function f is the set of all odd integers.
(d) The graph of the function f consists of an infinite set of discrete points.
5. (b) dom.F / D

x 2 R j x
1
2

, range.F / D R
(d) dom.g/ D fx 2 R j x ¤ 2 and x ¤ 2g,
range.g/ D fy 2 R j y 0g [ fy 2 R j y 1g
6. (a) d.1/ D 1, d.2/ D 2, d.3/ D 2, d.4/ D 3, d.5/ D 2, d.6/ D 4,
d.7/ D 2, d.8/ D 4, d.9/ D 3, d.10/ D 4, d.11/ D 2, d.12/ D 6.
(b) There is no natural number n other than 1 such that d.n/ D 1 since
every natural number greater than one has at least two divisors. The set
of preimages of 1 is f1g.
(c) The only natural numbers n such that d.n/ D 2 are the prime num-
bers. The set of preimages of the natural number 2 is the set of prime
numbers.
(d) The statement is false. A counterexample is m D 2 and n D 3 since
d.2/ D 2 and d.3/ D 2.
(e) d 20

D 1, d 21

D 2, d 22

D 3, d 23

D 4, d 24

D 5,
d 25

D 6, and d 26

D 7.
(f) For each nonnegative integer n, the divisors of 2n
are 20
; 21
; 22
; : : :2n 1
,
and 2n
. This is a list of n C 1 natural numbers and so d.2n
/ D n C 1.

(g) The statement is true. To prove this, let n be a natural number. Then
2n 1
2 N and d 2n 1

D .n 1/ C 1 D n.
7. (a) The domain of S is N. The power set of N, P.N/, can be the codomain.
The rule for determining outputs is that for each n 2 N, S.n/ is the set
of all distinct natural number factors of n.
(b) For example, S.8/ D f1; 2; 4; 8g, S.15/ D f1; 3; 5; 15g.
(c) For example, S.2/ D f1; 2g, S.3/ D f1; 3g, S.31/ D f1; 31g.
Section 6.2
1. (a) f.0/ D 4, f.1/ D 0; f.2/ D 3, f.3/ D 3, f.4/ D 0
(b) g.0/ D 4, g.1/ D 0; g.2/ D 3, g.3/ D 3, g.4/ D 0
(c) The two functions are equal.
3. (a) f.2/ D 9, f. 2/ D 9, f.3/ D 14, f.
p
2 D 7
(b) g.0/ D 5, g.2/ D 9, g. 2/ D 9, g.3/ D 14, g.
p
2/ D 7
(c) The function f is not equal to the function g since they do not have the
same domain.
(d) The function h is equal to the function f since if x ¤ 0, then
x2
C 5x
x
D
x2 C 5.
4. (a) hani, where an D
1
n2
for each n 2 N. The domain is N, and Q can be
the codomain.
(d) hani, where an D cos..n/ for each n 2 N. The domain is N, and
f 1; 1g can be the codomain. This sequence is equal to the sequence
in Part (c).
5. (a) p1.1; x/ D 1, p1.1; y/ D 1, p1.1; z/ D 1, p1.2; x/ D 2, p1.2; y/ D
2, p1.2; z/ D 2
(c) range..p1/ D A, range..p2/ D B
6. Start of the inductive step: Let P.n/ be “A convex polygon with n sides has
n.n 3/
2
diagonals.” Let k 2 D and assume that P..k/ is true, that is, a
convex polygon with k sides has
k.k 3/
2
diagonals. Now let Q be convex

polygon with .k C 1/ sides. Let v be one of the .k C 1/ vertices of Q and
let u and w be the two vertices adjacent to v. By drawing the line segment
from u to w and omitting the vertex v, we form a convex polygon with k
sides. Now complete the inductive step.
7. (a) f. 3; 4/ D 9, f. 2; 7/ D 23
(b) range.f / D f.m; n/ 2 Z Z j m D 4 3ng
8. (a) g.3; 5/ D .6; 2/, g. 1; 4/ D . 2; 5/.
(c) The set of preimages of .8; 3/ is f.4; 7/g.
9. (a) det

3 5
4 1

D 17; det

1 0
0 7

D 7; and det

3 2
5 0

D 10
Section 6.3
2. (a) Notice that f.0/ D 4, f.1/ D 0, f.2/ D 3, f.3/ D 3, and f.4/ D 0.
So the function f is not an injection and is not a surjection.
(c) Notice that F.0/ D 4, F.1/ D 0, F.2/ D 2, F.3/ D 1, and F.4/ D 3.
So the function F is an injection and is a surjection.
3. (a) The function f is an injection. To prove this, let x1; x2 2 Z and assume
that f.x1/ D f.x2/. Then,
3x1 C 1 D 3x2 C 1
3x1 D 3x2
x1 D x2:
Hence, f is an injection. Now, for each x 2 Z, 3x C 1 1 .mod 3/,
and hence f.x/ 1 .mod 3/. This means that there is no integer x
such that f.x/ D 0. Therefore, f is not a surjection.
(b) The proof that F is an injection is similar to the proof in Part (a) that
f is an injection. To prove that F is a surjection, let y 2 Q. Then,
y 1
3
2 Q and F

y 1
3

D y and hence, F is a surjection.
(h) Since h.1/ D h.4/, the function h is not an injection. Using calculus,
we can see that the function h has a maximum when x D 2 and a

minimum when x D 2, and so for each x 2 R, h. 2/ h.x/ h.2/
or
1
2
h.x/
1
2
:
This can be used to prove that h is not a surjection.
We can also prove that there is no x 2 R such that h.x/ D 1 using a
proof by contradiction. If such an x were to exist, then
2x
x2 C 4
D 1 or
2x D x2 C 4. Hence, x2 2x C 4 D 0. We can then use the quadratic
formula to prove that x is not a real number. Hence, there is no real
number x such that h.x/ D 1 and so h is not a surjection.
4. (a) Let F W R ! R be defined by F.x/ D 5x C 3 for all x 2 R. Let
x1; x2 2 R and assume that F.x1/ D F.x2/. Then 5x1 C3 D 5x2 C3.
Show that this implies that x1 D x2 and, hence, F is an injection.
Now let y 2 R. Then
y 3
5
2 R. Prove that F

y 3
5

D y. Thus,
F is a surjection and hence F is a bijection.
(b) The proof that G is an injection is similar to the proof in Part (a) that F
is an injection. Notice that for each x 2 Z, G.x/ 3 .mod 5/. Now
explain why G is not a surjection.
7. The birthday function is not an injection since there are two different people
with the same birthday. The birthday function is a surjection since for each
day of the year, there is a person that was born on that day.
9. (a) The function f is an injection and a surjection. To prove that f is an
injection, we assume that .a; b/ 2 R R, .c; d/ 2 R R, and that
f.a; b/ D f.c; d/. This means that
.2a; a C b/ D .2c; c C d/:
Since this equation is an equality of ordered pairs, we see that
2a D 2c, and
a C b D c C d:
The first equation implies that a D c. Substituting this into the second
equation shows that b D d. Hence,
.a; b/ D .c; d/;

and we have shown that if f.a; b/ D f.c; d/, then .a; b/ D .c; d/.
Therefore, f is an injection.
Now, to determine if f is a surjection, we let .r; s/ 2 R R. To find
an ordered pair .a; b/ 2 R R such that f.a; b/ D .r; s/, we need
.2a; a C b/ D .r; s/:
That is, we need
2a D r, and
a C b D s:
Solving this system for a and b yields
a D
r
2
and b D
2s r
2
:
Since r; s 2 R, we can conclude that a 2 R and b 2 R and hence that
.a; b/ 2 R R. So,
f.a; b/ D f

r
2
;
2s r
2

D

2
r
2

;
r
2
C
2s r
2

D .r; s/:
This proves that for all .r; s/ 2 R R, there exists .a; b/ 2 R R such
that f.a; b/ D .r; s/. Hence, the function f is a surjection. Since f is
both an injection and a surjection, it is a bijection.
(b) The proof that the function g is an injection is similar to the proof that
f is an injection in Part (a). Now use the fact that the first coordinate
of g.x; y/ is an even integer to explain why the fucntion g is not a
surjection.
Section 6.4
2. .g ı h/ W R ! R by .g ı h/.x/ D g.h.x// D g x3

D 3x3 C 2.
.h ı g/ W R ! R by .h ı g/.x/ D h.g.x// D h.3x C 2/ D .3x C 2/3
.
This shows that h ı g ¤ g ı h or that composition of functions is not com-
mutative.

3. (a) F .x/ D .g ı f / .x/, where f .x/ D ex and g .x/ D cos x.
(b) G .x/ D .g ı f / .x/ where f .x/ D cos x and g .x/ D ex.
(c) H .x/ D .g ı f / .x/, f .x/ D sinx, g .x/ D 1
x .
(d) K .x/ D .g ı f / .x/, f .x/ D e x2
, g .x/ D cos x.
4. (a) For each x 2 A, .f ı IA/.x/ D f.IA.x// D f.x/. Therefore,
f ı IA D f .
5. (a) Œ.h ı g/ ı f .x/ D 3
p
sin.x2/; Œh ı .g ı f / .x/ D 3
p
sin.x2/. This
proves that .h ı g/ ı f D h ı .g ı f / for these particular functions.
6. Start of a proof: Let A, B, and C be nonempty sets and let f WA ! B and
g W B ! C . Assume that f and g are both injections. Let x; y 2 A and
assume that .g ı f /.x/ D .g ı f /.y/.
7. (a) f WR ! R by f.x/ D x, g W R ! R by g.x/ D x2
. Then
g ı f W R ! R by .g ı f /.x/ D x2. The function f is a surjection,
but g ı f is not a surjection.
(b) f W R ! R by f .x/ D x, g W R ! R by g .x/ D x2. Then
g ı f W R ! R by .g ı f /.x/ D x2
. The function f is an injection,
but g ı f is not an injection.
(f) By Part (1) of Theorem 6.21, this is not possible since if g ı f is an
injection, then f is an injection.
Section 6.5
2. (b) f 1
D f.c; a/; .b; b/; .d; c/; .a; d/g
(d) For each x 2 S, f 1 ı f

.x/ D x D .f ı f 1/.x/. This illustrates
Corollary 6.28.
3. (a) This is a use of Corollary 6.28 since the cube root function and the
cubing function are inverse functions of each other and consequently,
the composition of the cubing function with the cube root function is
the identity function.
(b) This is a use of Corollary 6.28 since the natural logarithm function and
the exponential function with base e are inverse functions of each other
and consequently, the composition of the natural logarithm function
with the exponential function with base e is the identity function.

(c) They are similar because they both use the concept of an inverse func-
tion to “undo” one side of the equation.
4. Using the notation from Corollary 6.28, if y D f.x/ and x D f 1
.y/, then
f ı f 1

.y/ D f.f 1
.y//
D f.x/
D y
6. (a) Let x; y 2 A and assume that f.x/ D f.y/. Apply g to both sides of
this equation to prove that .gıf /.x/ D .gıf /.y/. Since gıf D IA,
this implies that x D y and hence that f is an injection.
(b) Start by assuming that f ı g D IB , and then let y 2 B. You need to
prove there exists an x 2 A such that f.x/ D y.
7. (a) f W R ! R is defined by f .x/ D e x2
. Since this function is not an
injection, the inverse of f is not a function.
(b) g W R
! .0; 1 is defined by g .x/ D e x2
. In this case, g is a
bijection and hence, the inverse of g is a function.
To see that g is an injection, assume that x; y 2 R and that e x2
D
e y2
. Then, x2 D y2 and since x; y 0, we see that x D y. To see
that g is a surjection, let y 2 .0; 1. Then, ln y 0 and ln y 0,
and g
p
ln y

D y.
Section 6.6
1. (a) There exists an x 2 A B such that f.x/ D y.
(d) There exists an a 2 A such that f.a/ D y or there exists a b 2 B such
that f.b/ D y.
(f) f.x/ 2 C [ D
(h) f.x/ 2 C or f.x/ 2 D
2. (b) f 1
.f.A// D Œ2; 5.
(d) f.f 1.C // D Œ 2; 3
(e) f.A B/ D Œ 5; 3
(f) f.A/ f.B/ D Œ 5; 3
3. (a) g.A A/ D f6; 12; 18; 24; 36; 54; 72; 108; 216g

(b) g 1.C / D f.1; 1/; .2; 1/; .1; 2/g
4. (a) range.F / D F.S/ D f1; 4; 9; 16g
5. To prove f .A [ B/ f .A/[f .B/, let y 2 f .A [ B/. Then there exists
an x 2 A[B such that f .x/ D y. Since x 2 AcupB, x 2 A or x 2 B. We
first note that if x 2 A, then y D f .x/ is in f .A/. In addition, if x 2 B,
then y D f .x/ is in f .B/. In both cases, y D f .x/ 2 f .A/ [ f .B/ and
hence, f .A [ B/ f .A/ [ f .B/.
Now let y 2 f .A/ [ f .B/. If y 2 f .A/, then there exists an x 2 A such
that y D f .x/. Since A A[B, this implies that y D f .x/ 2 f .A [ B/.
In a similar manner, we can prove that if y 2 f .B/, then y 2 f .A [ B/.
Therefore, f .A/ [ f .B/ f .A [ B/.
6. To prove that f 1
.C D/ f 1
.C / f 1
.D/, let x 2 f 1
.C D/.
Then f.x/ 2 C D. How do we prove that x 2 f 1.C / f 1.D/?
9. Statement (a) is true and Statement (b) is false.
Section 7.1
1. (a) The set A B contains nine ordered pairs. The set A B is a relation
from A to B since A B is a subset of A B.
(b) The set R is a relation from A to B since R A B.
(c) dom.R/ D A, range.R/ D fp; qg
2. (a) The statement is false since .c; c/ … R, which can be written as c6R d.
(b) The statement is true since whenever .x; y/ 2 R, .y; x/ is also in R.
That is, whenever x R y, y R x.
(c) The statement is false since .a; c/ 2 R, .c; b/ 2 R, but .a; b/ … R.
That is, aR c, cR b, but a6R b
(d) The statement is false since .a; a/ 2 R and .a; c/ 2 R.
3. (a) The domain of D consists of the female citizens of the United States
whose mother is a female citizen of the United States.
(b) The range of D consists of those female citizens of the United States
who have a daughter that is a female citizen of the United States.
4. (a) .S; T / 2 R means that S T .

(b) The domain of the subset relation is P.U /.
(c) The range of the subset relation is P.U /.
(d) The relation R is not a function from P.U / to P.U / since any proper
subset of U is a subset of more than one subset of U .
6. (a) fx 2 R j .x; 6/ 2 Sg D f 8; 8g.
fx 2 R j .x; 9/ 2 Sg D
n p
19;
p
19
o
.
(b) The domain of the relation S is the closed interval Œ 10; 10.
The range of the relation S is the closed interval Œ 10; 10.
(c) The relation S is not a function from R to R.
(d) The graph of the relation S is the circle of radius 10 whose center is at
the origin.
9. (a) R D f.a; b/ 2 Z Z jja bj 2g
(b) dom.R/ D Z and range.R/ D Z
Section 7.2
1. The relation R is not reflexive on A and is not symmetric. However, it is
transitive since the conditional statement “For all x; y; z 2 A, if x R y and
y R z, then x R z” is a true conditional statement since the hypothesis will
always be false.
3. There are many possible equivalence relations on this set. Perhaps one of the
easier ways to determine one is to first decide what elements will be equiv-
alent. For example, suppose we say that we want 1 and 2 to be equivalent
(and of course, all elements will be equivalent to themselves. So if we use
the symbol for the equivalence relation, then we need 1 2 and 2 1.
Using set notation, we can write this equivalence relation as
f.1; 1/; .2; 2/; .3; 3/; .4; 4/; .5; 5/; .1; 2/; .2; 1/g:
4. The relation R is not reflexive on A. For example, .4; 4/ … R. The relation R
is symmetric. If .a; b/ 2 R, then jajCjbj D 4. Therefore, jbjCjaj D 4, and
hence, .b; a/ 2 R. The relation R is not transitive. For example, .4; 0/ 2 R,
.0; 4/ 2 R, and .4; 4/ … R. The relation R is not an equivalence relation.

6. (a) The relation is an equivalence relation.
For a 2 R, a a since f .a/ D f .a/. So, is reflexive.
For a; b 2 R, if a b, then f .a/ D f .b/. So, f .b/ D f .a/.
Hence, b a and is symmetric.
For a; b; c 2 R, if a b and b c, then f .a/ D f .b/ and f .b/ D
f .c/. So, f .a/ D f .c/. Hence, a c and is transitive.
(b) C D f 5; 5g
10. (a) The relation is an equivalence relation on Z. It is reflexive since
for each integer a, a C a D 2a and hence, 2 divides a C a. Now let
a; b 2 Z and assume that 2 divides a C b. Since a C b D b C a,
2 divides b C a and hence, is symmetric. Finally, let a; b; c 2 Z
and assume that a b and b c. Since 2 divides a C b, a and b
must both be odd or both be even. In the case that a and b are both
odd, then b c implies that c must be odd. Hence, a C c is even and
a c. A similar proof shows that if a and b are both even, then a c.
Therefore, is transitive.
15. (c) The set C is a circle of radius 5 with center at the origin.
Section 7.3
1. Use the directed graph to examine all the cases necessary to prove that
is reflexive, symmetric, and transitive. The distinct equivalence classes are:
Œa D Œb D fa; bg; Œc D fcg; Œd D Œe D fd; eg
2. The equivalence class are
Œa D Œb D Œd D fa; b; dg; Œc D fcg; Œe D Œf D fe; f g:
Following is the directed graph for this equivalence relation.
a b
f
e d
c

3. Let x 2 A. Since x has the same number of digits as itself, the relation R is
reflexive. Now let x; y; z 2 A. If x R y, then x and y have the same number
of digits. Hence, y and x have the same number of digits and y R x, and so
R is symmetric.
If x R y and y R z, then x and y have the same number of digits and y and
z have the same number of digits. Hence, x and z have the same number of
digits, and so x R z. Therefore, R is transitive.
The equivalence classes are: f0; 1; 2; : : :; 9g, f10; 11; 12; : : :; 99g,
f100; 101; 102; : : :; 999g, f1000g.
4. The congruence classes for the relation of congruence modulo 5 on the set
of integers are
Œ0 D f5n j n 2 Zg
Œ1 D f5n C 1 j n 2 Zg
Œ4 D f5n C 4 j n 2 Zg.
5. (a) Let a; b; c 2 Z9. Since a2 a2 .mod 9/, we see that a a and
is reflexive. Let a; b; c 2 Z9. If a b, then a2
b2
.mod 9/ and
hence, by the symmetric property of congruence, b2 a2 .mod 9/.
This proves that is symmetric. Finally, if a b and b c, then
a2 b2 .mod 9/ and b2 c2 .mod 9/. By the transitive property
of congruence, we conclude that a2
c2
.mod 9/ and hence, a c.
This proves that is transitive. The distinct equivalence classes are
f0; 3; 6g, f1; 8g, f2; 7g, and f4; 5g.
6. (a) Let x 2

5
7

. Then x
5
7
2 Z, which means that there is an in-
teger m such that x
5
7
D m, or x D
5
7
C m. This proves that
x 2

m C
5
7
ˇ
ˇ
ˇ
ˇ m 2 Z

and, hence, that

5
7

m C
5
7
ˇ
ˇ
ˇ
ˇ m 2 Z

. We
still need to prove that

m C
5
7
ˇ
ˇ
ˇ
ˇ m 2 Z

5
7

.
9. (a) To prove the relation is symmetric, note that if .a; b/ .c; d/, then
ad D bc. This implies that cb D da and, hence, .c; d/ .a; b/.
(c) 3a D 2b

Section 7.4
1. (a)
˚ Œ0 Œ1 Œ2 Œ3 ˇ Œ0 Œ1 Œ2 Œ3
Œ0 Œ0 Œ1 Œ2 Œ3 Œ0 Œ0 Œ0 Œ0 Œ0
Œ1 Œ1 Œ2 Œ3 Œ0 Œ1 Œ0 Œ1 Œ2 Œ3
Œ2 Œ2 Œ3 Œ0 Œ1 Œ2 Œ0 Œ2 Œ0 Œ2
Œ3 Œ3 Œ0 Œ1 Œ2 Œ3 Œ0 Œ3 Œ2 Œ1
(b)
˚ Œ0 Œ1 Œ2 Œ3 Œ4 Œ5 Œ6
Œ0 Œ0 Œ1 Œ2 Œ3 Œ4 Œ5 Œ6
Œ1 Œ1 Œ2 Œ3 Œ4 Œ5 Œ6 Œ0
Œ2 Œ2 Œ3 Œ4 Œ5 Œ6 Œ0 Œ1
Œ3 Œ3 Œ4 Œ5 Œ6 Œ0 Œ1 Œ2
Œ4 Œ4 Œ5 Œ6 Œ0 Œ1 Œ2 Œ3
Œ5 Œ5 Œ6 Œ0 Œ1 Œ2 Œ3 Œ4
Œ6 Œ6 Œ0 Œ1 Œ2 Œ3 Œ4 Œ5
ˇ Œ0 Œ1 Œ2 Œ3 Œ4 Œ5 Œ6
Œ0 Œ0 Œ0 Œ0 Œ0 Œ0 Œ0 Œ0
Œ1 Œ1 Œ2 Œ3 Œ4 Œ5 Œ6 Œ0
Œ2 Œ0 Œ2 Œ4 Œ6 Œ1 Œ3 Œ5
Œ3 Œ0 Œ3 Œ6 Œ2 Œ5 Œ1 Œ4
Œ4 Œ0 Œ4 Œ1 Œ5 Œ2 Œ6 Œ3
Œ5 Œ0 Œ5 Œ3 Œ1 Œ6 Œ4 Œ2
Œ6 Œ0 Œ6 Œ5 Œ4 Œ3 Œ2 Œ1
2. (a) Œx D Œ1 or Œx D Œ3 (e) Œx D Œ2 or Œx D Œ3
(g) The equation has no solution.
3. (a) The statement is false. By using the multiplication table for Z6, we see
that a counterexample is Œa D Œ2.
(b) The statement is true. By using the multiplication table for Z5, we see
that:
Œ1 ˇ Œ1 D Œ1.
Œ2 ˇ Œ3 D Œ1.
Œ3 ˇ Œ2 D Œ1.
Œ4 ˇ Œ4 D Œ1.
5. (a) The proof consists of the following computations:

Œ11 D Œ1
Œ21
D Œ4
Œ31 D Œ9 D Œ4
Œ41
D Œ16 D Œ1.
17. (a) Prove the contrapositive by calculating Œa2
C Œb2
for all nonzero Œa
and Œb in Z3.
Section 8.1
1. (a) The set of positive common divisors of 21 and 28 is f1; 7g. So
gcd.21; 28/ D 7
(b) The set of positive common divisors of 21 and 28 is f1; 7g. So
gcd. 21; 28/ D 7
(c) The set of positive common divisors of 58 and 63 is f1g. So
gcd.58; 63/ D 1
(d) The set of positive common divisors of 0 and 12 is f1; 2; 3; 4; 6; 12g.
So gcd.0; 12/ D 12
2. (a) Hint: Prove that k j Œ.a C 1/ a.
4. (a) jbj is the largest natural number that divides 0 and b.
(b) The integers b and b have the same divisors. Therefore, gcd.a; b/ D
gcd.a; b/.
5. (a) gcd.36; 60/ D 12 12 D 36 2 C 60 . 1/
(b) gcd.901; 935/ D 17 17 D 901 27 C 935 . 26/
(e) gcd.901; 935/ D 17 17 D 901 27 C . 935/ .26/
6. (a) One possibility is u D 3 and v D 2. In this case, 9u C 14v D 1. We
then multiply both sides of this equation by 10 to obtain
9 . 30/ C 14 20 D 10:
So we can use x D 30 and y D 20.
7. (a) 11 . 3/ C 17 2 D 1 (b)
m
11
C
n
17
D
17m C 11n
187

Section 8.2
1. For both parts, use the fact that the only natural number divisors of a prime
number p are 1 and p.
2. Use cases: (1) p divides a; (2) p does not divide a. In the first case, the
conclusion is automatically true. For the second case, use the fact that
gcd.p; a/ D 1 and so we can use Theorem 8.12 to conclude that p divides
b. Another option is to write the number 1 as a linear combination of a and
p and then multiply both sides of the equation by b.
3. A hint for the inductive step: Write p j .a1a2 am/amC1. Then look at
two cases: (1) p j amC1; (2) p does not divide amC1.
4. (a) gcd.a; b/ D 1. Why?
(b) gcd.a; b/ D 1 or gcd.a; b/ D 2. Why?
7. (a) gcd .16; 28/ D 4. Also,
16
4
D 4,
28
4
D 7, and gcd .4; 7/ D 1.
(b) gcd .10; 45/ D 5. Also,
10
5
D 2,
45
5
D 9, and gcd .2; 9/ D 1.
9. Part (b) of Exercise (8) can be helpful.
11. The statement is true. Start of a proof: If gcd.a; b/ D 1 and c j .a C b/,
then there exist integers x and y such that ax C by D 1 and there exists an
integer m such that a C b D cm.

Section 8.3
3. (a) x D 3 C 14k, y D 2 9k
(b) x D 1 C 11k, y D 1 C 9k
(c) No solution
(d) x D 2 C 3k, y D 2 4k
4. There are several possible solutions to this problem, each of which can be
generated from the solutions of the Diophantine equation
27x C 50y D 25.
5. This problem can be solved by finding all solutions of a linear Diophantine
equation 25x C16y D 1461, where both x and y are positive. The mininum
number of people attending the banquet is 66.
6. (a) y D 12 C 16k; x3 D 1 3k
(b) If 3y D 12x1 C9x2 and 3y C16x3 D 20, we can substitute for 3y and
obtain 12x1 C 9x2 C 16x3 D 20.
(c) Rewrite the equation 12x1 C 9x2 D 3y as 4x1 C 3x2 D y. A general
solution for this linear Diophantine equation is
x1 D y C 3n
x2 D y 4n:
Section 9.1
2. One way to do this is to prove that the following function is a bijection:
f WA fxg ! A by f.a; x/ D a, for all .a; x/ 2 A fxg.
3. One way to prove that N EC is to find a bijection from N to EC. One
possibility is f W N ! EC
by f .n/ D 2n for all n 2 N. (We must prove
that this is a bijection.)
4. Notice that A D .A fxg/[fxg. Use Theorem 9.6 to conclude that A fxg
is finite. Then use Lemma 9.4.
5. (a) Since A B A, if A is finite, then Theorem 9.6 implies that A B
is finite.
(b) The sets A and B are subsets of A [ B. So if A [ B is finite, then A
and B are finite.

7. (a) Remember that two ordered pairs are equal if and only if their corre-
sponding coordinates are equal. So if .a1; c1/ ; .a2; c2/ 2 A C and
h .a1; c1/ D h .a2; c2/, then .f .a1/ ; g .c1// D .f .a2/ ; g .c2//. We
can then conclude that f .a1/ D f .a2/ and g .c1/ D g .c2/. Since f
and g are both injections, this means that a1 D a2 and c1 D c2 and
therefore, .a1; c1/ D .a2; c2/. This proves that f is an injection.
Now let .b; d/ 2 B D. Since f and g are surjections, there exists
a 2 A and c 2 C such that f .a/ D b and g .c/ D d. Therefore,
h .a; c/ D .b; d/. This proves that f is a surjection.
8. (a) If we define the function f by f.1/ D a, f.2/ D b, f.3/ D c,
f.4/ D a, and f.5/ D b, then we can use g.a/ D 1, g.b/ D 2, and
g.3/ D c. The function g is an injection.
Section 9.2
1. All except Part (d) are true.
2. (a) Prove that the function f W N ! F C defined by f .n/ D 5n for all
n 2 N is a bijection.
(e) One way is to define f WN ! N f4; 5; 6g by
f.n/ D
(
n if n D 1, n D 2, or n D 3
f.n C 3/ if n 4:
and then prove that the function f is a bijection.
It is also possible to use Corollary 9.20 to conclude that N f4; 5; 6g is
countable, but it must also be proved that N f4; 5; 6g cannot be finite.
To do this, assume that N f4; 5; 6g is finite and then prove that N is
finite, which is a contradiction.
(f) Let A D fm 2 Z j m 2 .mod 3/g D f3k C 2 j k 2 Zg. Prove that
the function f W Z ! A is a bijection, where f .x/ D 3x C 2 for all
x 2 Z. This proves that Z A and hence, N A.
5. For each n 2 N, let P.n/ be “If card.B/ D n, then A [ B is a countably
infinite set.”
Note that if card.B/ D k C 1 and x 2 B, then card.B fxg/ D k. Apply
the inductive assumption to B fxg.

6. Let m; n 2 N and assume that h .n/ D h .m/. Then since A and B are
disjoint, either h .n/ and h .m/ are both in A or are both in B. If they are
both in A, then both m and n are odd and
f

n C 1
2

D h .n/ D h .m/ D f

m C 1
2

:
Since f is an injection, this implies that
n C 1
2
D
m C 1
2
and hence that
n D m. Similary, if both h .n/ and h .m/ are in B, then m and n are even
and g
n
2

D g
m
2

, and since g is an injection,
n
2
D
m
2
and n D m.
Therefore, h is an injection.
Now let y 2 A [ B. There are only two cases to consider: y 2 A or y 2 B.
If y 2 A, then since f is a surjection, there exists an m 2 N such that
f .m/ D y. Let n D 2m 1. Then n is an odd natural number, m D
n C 1
2
,
and
h .n/ D f

n C 1
2

D f .m/ D y:
Now assume y 2 B and use the fact that g is a surjection to help prove that
there exists a natural number n such that h.n/ D y.
We can then conclude that h is a surjection.
7. By Theorem 9.14, the set QC of positive rational numbers is countably infi-
nite. So by Theorem 9.15, QC
[ f0g is countably infinite. Now prove that
the set Q of all negative rational numbers is countably infinite and then use
Theorem 9.17 to prove that Q is countably infinite.
8. Since A B A, the set A B is countable. Now assume A B is finite
and show that this leads to a contradiction.
Section 9.3
1. (a) One such bijection is f W .0; 1/ ! R by f.x/ D ln x for all x 2
.0; 1/
(b) One such bijection is g W .0; 1/ ! .a; 1/ by g.x/ D x C a for all
x 2 .0; 1/. The function g is a bijection and so .0; 1/ .a; 1/.
Then use Part (a).

2. Use a proof by contradiction. Let H be the set of irrational numbers and
assume that H is countable. Then R D Q[ H and Q and H are disjoint. Use
Theorem 9.17, to obtain a contradiction.
3. By Corollary 9.20, every subset of a countable set is countable. So if B is
countable, then A is countable.
4. By Cantor’s Theorem (Theorem 9.27), R and P .R/ do not have the same
cardinality.

Appendix D
List of Symbols
Symbol Meaning Page
! conditional statement 5, 33
R set of real numbers 10
Q set of rational numbers 10
Z set of integers 10
N set of natural numbers 54
y 2 A y is an element of A 55
z … A z is not an element of A 55
f j g set builder notation 58
8 universal quantifier 63
9 existential quanitifer 63
; the empty set 60
^ conjunction 33
_ disjunction 33
: negation 33
$ biconditional statement 39
logically equivalent 43
m j n m divides n 82
a b .mod n/ a is congruent to b modulo n 92
jxj absolute value of x 135
A D B A equals B (set equality) 55
A B A is a subset of B 55
A 6 B A is not a subset of B 219
A B A is a proper subset of B 218
582

Appendix D. List of Symbols 583
Symbol Meaning Page
P.A/ power set of A 222
jAj cardinality of a finite set A 223
A B intersection of A and B 216
A [ B union of A and B 216
Ac
complement of A 216
A B set difference of A and B 216
A B Cartesian product of A and B 256
.a; b/ ordered pair 256
R R Cartesian plane 258
R2 Cartesian plane 258
S
X2C
X union of a family of sets 265
T
X2C
X intersection of a family of sets 265
n
S
j D1
Aj union of a finite family of sets 267
n
T
j D1
Aj intersection of a finite family of sets 267
1
S
j D1
Bj union of an infinite family of sets 267
1
T
j D1
Bj intersection of an infinite family of sets 267
fA˛ j ˛ 2 ƒg indexed family of sets 268
S
˛2ƒ
A˛ union of an indexed family of sets 269
T
˛2ƒ
A˛ union of an indexed family of sets 269
nŠ n factorial 188
f1; f2; f3; : : : Fibonacci numbers 202
s.n/ sum of the divisors of n 283
f W A ! B function from A to B 285
dom.f / domain of the function f 285
codom.f / codmain of the function f 285
f.x/ image of x under f 285
range.f / range of the function f 287
d.n/ number of divisors of n 292
IA identity function on the set A 298
p1; p2 projection functions 304
det.A/ determinant of A 305
AT transpose of A 305
det W M2;2 ! R determinant function 323

584 Appendix D. List of Symbols
Symbol Meaning Page
g ı f W A ! C composition of functions f and g 325
f 1 the inverse of the function f 338
Sin the restricted sine function 349
Sin 1
the inverse sine function 349
dom.R/ domain of the relation R 364
range.R/ range of the relation R 364
x R y x is related to y 366
x 6R y x is not related to y 366
x y x is related to y 367
x œ y x is not related to y 367
R 1
the inverse of the relation R 373
Œa equivalence class of a 391
Œa congruence class of a 392
Zn the integers modulo n 402
Œa ˚ Œc addition in Zn 404
Œa ˇ Œc multiplication in Zn 404
gcd.a; b/ greatest common divisor of a and b 414
f.A/ image of A under the function f 351
f 1.C / pre-image of C under the funciton f 351
A B A is equivalent to B 452
A and B have the same cardinality
Nk Nk D f1; 2; : : :; kg 455
card.A/ D k cardinality of A is k 455
@0 cardinality of N 465
c cardinal number of the continuum 482

Index
absolute value, 135–141
additive identity, 254
additive inverse, 254
aleph, 465
algebra of sets, 245, 278
antisymmetric relation, 387
arcsine function, 349
arithmetic sequence, 208
arrow diagram, 290
associative laws
for real numbers, 254
for sets, 246, 278
average
of a finite set of numbers, 300
axiom, 85
basis step, 173, 191, 194
biconditional statement, 39, 103
forms of, 39
proof of, 107
bijection, 312
Binet’s formula, 208
birthday function, 283, 284, 319
Cantor’s diagonal argument, 482
Cantor’s Theorem, 483
Cantor, Georg, 483
Cantor-Schröder-Bernstein Theorem, 484
cardinal number, 455
cardinality, 223, 452, 455, 465, 482
@0, 465
c, 482
finite set, 223, 455
natural numbers, 465
Cartesian plane, 258
Cartesian product, 256, 362
cases, proof using, 132, 160
Caucy sequence, 78
chain rule, 327
choose an element method, 279
choose-an-element method, 231–235, 238
circular relation, 386
closed interval, 228
closed ray, 228
closed under addition, 10, 62
closed under multiplication, 10, 63
closed under subtraction, 10, 63
closure properties, 10–11
codomain, 281, 285
Cohen, Paul, 485
common divisor, 414
commutative laws
for sets, 246, 278
commutative operation, 77
complement of a set, 216
complex numbers, 224
composite function, 325
composite number, 78, 189, 426
composition of functions, 325
inner function, 325
outer function, 325
compound interest, 211
compound statement, 33
585

586 Index
conditional, 33
conditional statement, 5–10, 36
conclusion, 5, 37
contrapositive of, 44
converse of, 44
forms of, 36
hypothesis, 5, 37
logical equivalencies, 45
negation, 46
truth table, 6
congruence, 92–94, 147–149, 380
Division Algorithm, 150
reflexive property, 98, 149, 167
symmetric property, 98, 149, 167
transitive property, 98, 149, 167
congruence class, 392
congruent modulo n, 92
conjecture, 3, 86
conjunction, 33, 36
connective, 33
consecutive integers, 137
constant, 54
construction method, 88
constructive proof, 109–110, 161
continuous, 77
continuum, 482
Continuum Hypothesis, 485
contradiction, 40, 116, 159
contrapositive, 44, 104, 159
converse, 44
coordinates, 256
corollary, 86
countable set, 466
countably infinite set, 466
countably infinite sets
subsets of, 472–473
union of, 471
counterexample, 3, 66, 69, 90–91
De Moivre’s Theorem, 199
de Moivre, Abraham, 199
De Morgan’s Laws
for indexed family of sets, 271
for sets, 248, 278
for statements, 45
decimal expression
for a real number, 479
normalized form, 480
decomposing functions, 327
definition, 15, 86
by recursion, 200
denumerable set, 466
dependent variable, 285
derivative, 295
determinant, 305, 323
diagonal, 295
difference of two sets, 216
digraph, 369
Diophantine equation, 441
linear in one variable, 441
linear in two variables, 441–445
Diophantus of Alexandria, 441
direct proof, 16, 24–26, 104, 158
directed edge, 369
directed graph, 369
directed edge, 369
vertex, 369
disjoint
pairwise, 395
disjoint sets, 236
disjunction, 33, 36
distributive laws
for indexed family of sets, 271
for sets, 246, 278
for statements, 48, 49
divides, 82, 367
divisibility test, 407
for 11, 410
for 3, 409

Index 587
for 4, 410
for 5, 409
for 9, 407
Division Algorithm, 143–145
congruence, 150
using cases, 146–147
divisor, 82
properties, 167
Dodge Ball, 476, 482
domain
of a function, 281, 285
of a relation, 364
domino theory, 189
element-chasing proof, 238
empty set, 60
properties, 246, 278
equal functions, 298
equal sets, 55
equality relation, 381
equation numbers, 89
equivalence class, 391
properties of, 392
equivalence relation, 378
equivalent sets, 452, 455
Euclid’s Elements, 431
Euclid’s Lemma, 431
Euclidean Algorithm, 419
even integer, 15, 82
properties, 31, 166
exclusive or, 34
existence theorem, 110, 111, 161
existential quantifier, 63
Extended Principle of Mathematical In-
duction, 190, 213
factor, 82
factorial, 188, 201
family of sets, 264
indexed, 268
intersection, 265
union, 265
Fermat’s Last Theorem, 164
Fermat, Pierre, 164
Fibonacci numbers, 202
Fibonacci Quarterly, 204
finite set, 455, 462
properties of, 455–459
function, 281, 284
as set of ordered pairs, 336
bijective, 312
codomain, 281, 285
composite, 325
composition, 325
domain, 281, 285
equality, 298
injective, 310
inverse of, 338, 373
invertible, 341
of two variables, 302
one-to-one, 310
onto, 311
piecewise defined, 322
projection, 304
range, 287
real, 288, 292
surjective, 311
Fundamental Theorem
of Arithmetic, 432
of Calculus, 306
future value, 212
Gödel, Kurt, 485
geometric sequence, 206
geometric series, 206
Goldbach’s Conjecture, 163, 435
greatest common divisor, 414
half-open interval, 228
Hemachandra, Acharya, 203

588 Index
idempotent laws for sets, 246, 278
identity function, 298, 319, 331, 345
identity relation, 381
image
of a set, 351
of a union, 355, 361
of an element, 285
of an intersection, 355, 361
implication, 33
inclusive or, 34
increasing, strictly, 76
independent variable, 285
indexed family of sets, 268
De Morgan’s Laws, 271
distributive laws, 271
intersection, 269
union, 269
indexing set, 268
inductive assumption, 174
inductive hypothesis, 174
inductive set, 171
inductive step, 173, 191, 194
infinite set, 455, 462
initial condition, 200
injection, 310
inner function, 325
integers, 10, 54, 224
consecutive, 137
system, 415
integers modulo n, 402
addition, 404
multiplication, 404
Intermediate Value Theorem, 111
intersection
of a family of sets, 265
of an indexed family of sets, 269
of two sets, 216
interval, 228
Cartesian product, 259
closed, 228
half-open, 228
open, 228
inverse image of a set, 351
inverse of a function, 338, 373
inverse of a relation, 373
inverse sine function, 349
invertible function, 341
irrational numbers, 10, 113, 122, 224
know-show table, 18–21
backward question, 19
forward question, 19
know-show-table, 86
Kuratowski, Kazimierz, 263
Law of Trichotomy, 228
least upper bound, 79
lemma, 86
Leonardo of Pisa, 203
linear combination, 423
linear congruence, 447
linear Diophantine equations, 441–445
logical operator, 33
logically equivalent, 43
Lucas numbers, 210
magic square, 129
mapping, 284
mathematical induction
basis step, 173, 191, 194
Extended Principle, 190, 193, 213,
214
inductive step, 173, 191, 194
Principle, 173, 213
matrix, 305
determinant, 305
transpose, 305
modular arithmetic, 404
modus ponens, 42
multiple, 82
multiplicative identity, 254

Index 589
multiplicative inverse, 254
natural numbers, 10, 54, 223
necessary condition, 37
negation, 33, 36
of a conditional statement, 46, 69
of a quantified statement, 67–69
neighborhood, 78
normalized form
of a decimal expression, 480
number of divisors function, 292, 293,
319
octagon, 295
odd integer, 15
properties, 31, 166
one-to-one correspondence, 452
one-to-one function, 310
only if, 37
onto function, 311
open interval, 228
open ray, 228
open sentence, 56
ordered pair, 256, 263
ordered triple, 263
ordinary annuity, 212
outer function, 325
pairwise disjoint, 272, 395
partition, 395
pentagon, 295
perfect square, 70, 439
piecewise defined function, 322
Pigeonhole Principle, 459, 461
polygon, 294
diagonal, 295
regular, 295
power set, 222, 230, 478
cardinality, 230
predicate, 56
preimage
of a set, 351
of a union, 356, 361
of an element, 285
of an intersection, 356, 361
prime factorization, 427
prime number, 78, 189, 426
prime numbers
distribution of, 434
twin, 435
Principle of Mathematical Induction, 173,
213
projection function, 304, 320
proof, 22, 85
biconditional statement, 107
by contradiction, 116–118, 159
constructive, 110, 161
contrapositive, 104, 159
direct, 16, 24–26, 104, 158
element-chasing, 238
non-constructive, 111, 161
using cases, 132, 160
proper subset, 218
proposition, 1
propositional function, 56
Pythagorean Theorem, 26
Pythagorean triple, 29, 102, 128, 163,
164
quadratic equation, 29, 98
quadratic formula, 29
quadrilateral, 295
quantifier, 63, 70
existential, 63
universal, 63
quotient, 144
range, 287
of a relation, 364
rational numbers, 10, 54, 113, 122
ray

590 Index
closed, 228
open, 228
real function, 288, 292
real number system, 10
real numbers, 54, 224
recurrence relation, 200
recursive definition, 200
reflexive, 375, 377
regular polygon, 295
relation, 364
antisymmetric, 387
circular, 386
divides, 367
domain, 364
equality, 381
equivalence, 378
identity, 381
inverse of, 373
range, 364
reflexive on, 375, 377
symmetric, 375, 377
transitive, 375, 377
relative complement, 216
relatively prime integers, 428
remainder, 144
ring, 77
roster method, 53
Second Principleof Mathematical Induc-
tion, 193, 214
sequence, 200, 301
arithmetic, 208
geometric, 206
set
complement, 216
difference, 216
equality, 55
intersection, 216
power, 222
proving equality, 235
relative complement, 216
roster method, 53
union, 216
set builder notation, 58
set equality, 55
set of divisors function, 293
solution set, 58, 363
statement, 1, 3
biconditional, 103
compound, 33
conditional, 5–10
subset, 55
proper, 218
sufficient condition, 37
sum of divisors function, 284, 319
surjection, 311
syllogism, 42
symmetric, 375, 377
tautology, 40, 116
theorem, 86
transitive, 375, 377
transpose, 305, 323
triangle, 295
Triangle Inequality, 137
truth set, 58, 254, 363
truth table, 6
Twin Prime Conjecture, 435
twin primes, 163, 435
uncountable set, 466
undefined term, 85
union
of a family of sets, 265
of an indexed family of sets, 269
of two sets, 216
unique factorization, 433
universal quantifier, 63
universal set, 54
properties, 246, 278

Index 591
upper bound, 79
variable, 54
dependent, 285
independent, 285
Venn diagram, 217
vertex, 369
Wallis cosine formulas, 183
Wallis sine formulas, 183
Well Ordering Principle, 423
Wiles, Andrew, 283
writing guidelines, 22–24, 73, 89, 94–
95, 105, 108, 119, 176, 492–
496
zero divisor, 77

[Sundstrom_Ted.]_Mathematical_Reasoning_Writing - Copy.pdf

More Related Content

Similar to [Sundstrom_Ted.]_Mathematical_Reasoning_Writing - Copy.pdf (20)

More from Sahat Hutajulu (20)

Recently uploaded (20)

[Sundstrom_Ted.]_Mathematical_Reasoning_Writing - Copy.pdf