Bridge to Abstract Mathematics

Bridge to

AbstractMathematics

Ralph W. Oberste-Vorthand

Aristides Mouzakitis

1

Copyright c©2003 by Ralph W. Oberste-Vorth and Aristides Mouzakitis

i

Table of Contents

Contents i

Some Notes on Notation iii

Preface iv

PART I. THE AXIOMATIC METHOD 1

Chapter 1. Introduction 2Section 1. The History of Numbers 2Section 2. The Algebra of Numbers 2Section 3. The Axiomatic Method 3Section 4. Parallel Mathematical Universes 5

Chapter 2. Statements in Mathematics 8Section 1. Mathematical Statements 8Section 2. Mathematical Connectives 8Section 3. Symbolic Logic 8Section 4. Compound Statements in English 8Section 5. Predicates and Quantifiers 11

Supplemental Exercises 74

Chapter 3. Proofs in Mathematics 8Section 1. What is Mathematics? 12Section 2. Direct Proof 13Section 3. Contraposition and Proof by Contradiction 15Section 4. Proof by Induction 16Section 5. Examples and Counterexamples 22


How to THINK about mathematics: A Summary 24How to COMMUNICATE mathematics: A Summary 24

PART II. SET THEORY 25

Chapter 4. Basic Set Operations 26Section 1. Introduction 26Section 2. Subsets 27Section 3. Intersections and Unions 28Section 4. Differences and Complements 31

ii

Section 5. Power Sets 32Section 6. Russell’s Paradox 33


Chapter 5. Functions 34Section 1. Functions as Rules 34Section 2. Cartesian Products, Relations and Functions 34Section 3. Injective, Surjective and Bijective Functions 38Section 4. Compositions of Functions 40Section 5. Inverse Functions and Inverse Images of Functions 42Section 6. Another Approach to Compositions 44


Chapter 6. Relations on a Set 56Section 1. Properties of Relations 56Section 2. Order Relations 57Section 3. Equivalence Relations 61


Chapter 7. Cardinality of Sets 46Section 1. Introduction 46Section 2. Finite Sets 46Section 3. Infinite Sets 49Section 4. Countable Sets 51Section 5. Uncountable Sets 52


PART III. ALGEBRA OF NUMBER SYSTEMS 64

Chapter 8. Algebra of Number Systems 65Section 1. Primary Properties of Number Systems 65Section 2. Secondary Properties Involving Addition and Multiplication 66Section 3. Secondary Properties Involving Order 66Section 4. Isomorphisms and Embeddings 68


Chapter 9. Archimedean Ordered Fields 65Section 1. The Main Goal and Archimedean Principle 69


Chapter 10. The Natural Numbers 71Section 1. Introduction 71Section 2. Zero, the Natural Numbers and Addition 71Section 3. Multiplication 74


Summary of the Properties of Z+ 74

Chapter 11. The Integers 75Section 1. Introduction: Integers as Equivalence Classes 75

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Section 2. A Total Ordering of the Integers 75Section 3. Addition of Integers 75Section 4. Multiplication of Integers 75Section 5. Embedding the Natural Numbers in the Integers 78

Supplemental Exercises 74Summary of the Properties of Z 74

Chapter 12. The Rational Numbers 79Section 1. Introduction 79Section 2. A Total Ordering of the Rationals 75Section 3. Addition of Rationals 75Section 4. Multiplication of Rationals 75Section 5. An Ordered Field Containing the Integers 79


Summary of the Properties of Z+ 74

Chapter 13. The Real Numbers 82Section 1. Dedekind Cuts 82Section 2. Order and Addition of Real Numbers 82Section 3. Multiplication of Real Numbers 84Section 4. Embedding the Rationals in the Reals 85Section 5. Uniqueness of the set of Real Numbers 85


Chapter 14. The Complex Numbers 89Section 1. Introduction 89Section 2. Algebra of Complex Numbers 89Section 3. Order on the Complex Field 89Section 4. Embedding the Real Numbers in the Complex Numbers 90


Chapter 15. The Real Numbers According to Cantor 92Section 1. Convergence of Sequences of Rational Numbers 92Section 2. Cauchy Sequences of Rational Numbers 93Section 3. Cantor’s Set of Real Numbers 95Section 4. The Isomorphism from Cantor’s to Dedekind’s Reals 97


PART IV. HINTS 98

Chapter 16. Hints for the Exercises 99Hints for Chapter 1 99Hints for Chapter 3 99Hints for Chapter 4 99Hints for Chapter 5 100Hints for Chapter 7 100Hints for Chapter 6 102Hints for Chapter 8 103Hints for Chapter 10 104

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Hints for Chapter 11 104Hints for Chapter 12 105Hints for Chapter 13 106Hints for Chapter 14 107Hints for Chapter 15 108

Postscript and Selected References 109

Index 110

v

Some Notes on Notation

We will use the following sets of numbers:

N denotes the set of natural numbers: { 1, 2, 3, 4, 5, . . . }Zn denotes the set of the first n integers: { 1, 2, 3, . . . , n }∗Z denotes the set of integers

Z+ denotes the set of nonnegative integers: { 0, 1, 2, 3, 4, . . . }Z∗ denotes the set of nonzero integers

Q denotes the set of rational numbers

Q+ denotes the set of nonnegative rational numbers

Q ∗+ denotes the set of positive rational numbers

R denotes the set of real numbers

R+ denotes the set of nonnegative real numbers

R ∗+ denotes the set of positive real numbers

C denotes the set of complex numbers

(a, b) denotes the open interval: { x ∈ R | a < x < b }[a, b] denotes the closed interval: { x ∈ R | a ≤ x ≤ b }(a, b] denotes the interval: { x ∈ R | a < x ≤ b }[a, b) denotes the interval: { x ∈ R | a ≤ x < b }

Beware of our use of the following notation for subsets:

A ⊂ B A is a subset of B (A may equal B)†

A � B A is a proper subset of B (A �= B)

*Most authors use Zn = { 0, 1, 2, . . . , n − 1 }. This is especially true in number theoryand abstract algebra, where 0 and n are really the same. The choice here is more natural inour discussion of cardinality in Chapter 7.

†We never use the notation A ⊆ B. Many authors use A ⊆ B for subsets and A ⊂ Bfor proper subsets.

˜περ ¶δει δε›ξαι.

vi Notation

We use the following notation for relations and functions:

f : X → Y f is a function with domain X and codomain Y

f∣∣

S restriction of f to a subset S of its original domain

The symbol � will denote the end of a proof; purists may read it as “Q.E.D.”abbreviating the Latin quod erat demonstrandum, which translates the an-cient Greek of Euclid:

Similarly, we use the symbol to mark the end of an example or remark.

vii

Preface

’Mathematicians,’ [Uncle Petros] continued, ’find the same enjoyment in their studies

that chess players find in chess. In fact, the psychological make-up of the true mathematician

is closer to that of the poet or the musical composer, in other words of someone concerned with

the creation of Beauty and the search for Harmony and Perfection. He is the polar opposite

of the practical man, the engineer, the politician or the ...’ – he paused for a moment seeking

something even more abhorred in his scale of values – ’... indeed, the businessman.’

— Apostolos Doxiadis, Uncle Petros & Goldbach’s Conjecture (2000)

A Historical Perspective.We, the authors of this text, were undergraduates together at the end

of the 1970’s. We learned how to construct proofs essentially by osmosis.Also, we learned most of the material contained in this text in much the sameway. That is, we learned the art of proof mostly by imitating the experts,after listening to and watching their presentations, reading their textbooksand thinking these things through. For us, the process of learning howto prove theorems began in the calculus sequence and continued throughlinear algebra and beyond.

Undergraduate mathematics curricula in the U.S. have undergonesome changes since that time. The most visible change may be the endproduct of Calculus Reform. With an emphasis on the integration of thenumerical, analytical, and geometrical aspects of the subject as well asthe integration of computing technology, some topics have been removedor minimized in the curricula. We saw proofs of most everything in ourcalculus courses, even if we were not generally held responsible for themon the examinations. This is not always the case today. In fact, at manyinstitutions a variety of calculus sequences exists. Some of the calculustextbooks used in some of those courses do not give any proofs.

Calculus Reform is certainly not the only factor for the changing cur-ricula. The growing number of high school graduates who continue theireducations at post-secondary institutions has also put some pressure on thecalculus curricula by requiring increasingly extensive reviews of algebraand trigonometry. Also, the fluid nature of the liberal arts education oftenleads a future major in mathematics to take a calculus course designed forother majors.

viii Preface

As a result of these fundamental changes in calculus curricula, studentshave been enrolling in linear algebra courses with weaker backgrounds inconstructing and understanding proofs. Recently, many institutions haveinserted new (or, sometimes, recycled) courses into their curricula to ad-dress this problem. These courses go by many names. The choice ofmathematical content is not universal, but the prime objective is univer-sal. That objective is to prepare students to deal with proofs in their latercourses. At the University of South Florida, the faculty gave this course thesomewhat pretentious title “Bridge to Abstract Mathematics.” (Amazingly,the title was more contentious than the course itself!)

An interesting aside lies in the fact that the usual content for thesenew “Bridge” courses is set theory and related topics. Courses in set theorywere a standard part of the major earlier in the twentieth century. Betweenthe 1950’s and 1970’s most of these types of courses were squeezed out ofthe requirements for the mathematics major by a combination of courses inlinear algebra and abstract algebra as well as a relaxation of requirementsin favor of elective courses; many of the elective courses were new to theundergraduate curriculum.

We are curious how our own mathematical development would havebeen altered had we taken such a course during our own mathematicaltraining. We certainly hope that you—today’s majors—find this type ofcourse useful.

About This Text and How to Use It.This text has a dual purpose. As suggested above, the goal is both to

present appropriate mathematical content and to guide students on how todo proofs. While the content is, we hope, interesting, the process of trans-forming the thought processes from a passive computational orientation toan active creative orientation is at least as important.

This text consists of four parts. Part I is an introduction (with a fewhistorical comments). The second and third chapters should be impossibleto absorb on a cursory reading. You may want to read it carefully andrefer to it as you read the remaining parts. (If you wish to try to ignore itcompletely, we suggest you at least look at Section 5 of Chapter 2 on thelanguage of mathematics.)

Parts II and III present the mathematical content. Part II presentstopics from set theory and Part III presents the constructions of variousnumber systems in the context of some of their properties. The definitionsare presented with examples and comments. The proofs of propositions andtheorems are almost all left as exercises. Yes, you can learn mathematics—how to prove theorems—passively, by watching, listening, and reading.However, we strongly believe that you will be better off learning actively,that is, by supplying the proofs yourself. You have a number of resources

Preface ix

available to you. The text takes small steps forward, with some proofsprovided where they involve a larger leap forward. Of course, you can askthe experts. Lastly, do not overlook this resource: your fellow students. Itis difficult to do mathematics without discussing it with someone, even anon-expert.

We should mention here that the chapters on the constructions of thenatural numbers and the real numbers, Chapters 10 and 13, respectively, aremuch more difficult than the other chapters of Part III. They can be skippedas long as their properties are understood for use in the other chapters.

Part IV constitutes another resource for completing proofs. It containshints and comments for many of the exercises. Some of the hints will tell youhow to start a proof. Other hints may only be helpful once you have alreadythought about how to approach the proof. In any case, we recommend thatyou attempt each exercise without consulting the hints. If you get stuck,then consult the hints (or a fellow student or an expert). Once you havecompleted an exercise, read the hint and tell someone about your solution.

By the way, proper etiquette is to challenge everything your fellowstudents tell you about their proofs. Passive listeners are useless in thepursuit of knowledge. It is not rude to call someone an idiot (at leastwith respect to a faulty part of their proof), but you should be ready forsomeone else to call you an idiot (with respect to your own faulty proofs).Perhaps this is a bit overblown: calling a nonsensical statement made by afriend idiotic will hopefully be taken (and given) constructively, while suchbehavior with a total stranger could be quite dangerous. We are merelysupporting a ruthless attack on knowledge through free discourse and notan attack on your neighbors!

Acknowledgments.We would like to thank the Fall 2000, Spring 2001 and Summer 2001

classes of MGF 3301 taught by the first author at the University of SouthFlorida for “testing” drafts of this text. Their corrections and suggestionshave been invaluable. In particular, we would like to thank Paul Ander-son, Ray Burrus, Christie Burton, Leon Calleja, Thuc Cao, Nathan Chau,Teresa Chung, Jason Copenhaver, Mindy Eason, Adam Francis, AlynneFrewin, Russell Gerbers, Bridget Giroux, Erika Johnson, Kristy Kazemfar,Sarah Lahlou-Amine, Christopher Ledwith, Carson McCoy, Rose Nestor,Cheryl Ng, Ryan Parrish, Michelle Richardson, Patrick Robbins, EmilyRoberts, Cheryl Scilex, Anthony Upchurch, Adrianne Waltz, Jeanne Waserand Aimee Yates. We would also like to thank professors Edwin Clark andBoris Shekhtman for using drafts of this text at the University of SouthFlorida in 2001 to 2003.

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PART I

THE AXIOMATIC METHOD

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Chapter 1

Introduction

1. The History of Numbers.The seemingly easy concept of number is actually an abstraction which

came quite late in the history of our intellectual evolution. The ancientshepherd had no method of counting the sheep in his herd. He used themore primitive concept of a one-to-one correspondence instead, perhapsby notching a bone as many times as the number of his sheep.

The counting numbers or natural numbers 1, 2, 3, 4, . . . were the firstto appear on the scene. Pythagoras, a mystic who lived in the sixth centuryB.C. and who is famous for the theorem which bears his name, consideredthe natural numbers to be the very stuff out of which the universe is made.

The positive rational numbers, which come up when we divide a wholething into smaller pieces, were invented second. They constitute a naturalextension of the natural numbers. A number is positive rational if and onlyif it can be expressed as a quotient of natural numbers. Such quotients arecalled fractions.

Pythagoras and his disciples knew about fractions. However, thediscovery of numbers that are not fractions shook the foundations of thePythagorean school and caused the first big crisis in mathematics. Eudoxosdevised an ingenious theory of irrational quantities, which anticipated therigorous treatment given by G. Cantor and R. Dedekind in their construc-tions of the real number system almost two and one half millennia later.

The number zero was introduced by the Hindus in the ninth centuryA.D., but some historians conjecture that it was known to the Babyloni-ans. The negative and complex numbers go hand in hand because of thedefinition of imaginary numbers as the square roots of negative numbers.Negative and complex numbers were recognized at this time as well. How-ever, the existence of negative and complex numbers was viewed as absurdfor another millennium, despite their increased formal use.

Economic developments—for example, an international banking sys-tem spreading out of Italy in the fourteenth century just as Europe was

Section 2. The Algebra of Numbers 3

moving from the Middle Ages to the Renaissance—provided favorable cir-cumstances for wider acceptance of negative numbers. The use of negativenumbers was perhaps influenced by Fibonacci in the thirteenth century; henoted that a negative number could be regarded as a loss whereas a positivenumber could represent a gain.

The universal acceptance of the complex numbers and, hence, thenegative numbers, did not occur until the 1830’s. The early part of thenineteenth century saw developments in algebra, particularly the extensionsof the naturals to the integers and rationals and of the reals to the complexnumbers. The notion of equivalence classes stood at the heart of thesedevelopments, though this language was not developed until the twentiethcentury.

The real numbers, at least the positive real numbers, were certainlyaccepted since the time that Eudoxus developed the irrationals numbers.However, the axiomatic construction of the real numbers did not occuruntil the nineteenth century. Many mathematicians had attempted to putthe reals on a solid foundation; this was finally achieved by R. Dedekindand G. Cantor 1870’s.

This brings us full circle back to the naturals. The easiest of thenumbers were the most difficult, requiring the very axioms of set theorywhich were taking form at the beginning of the twentieth century. Today,we give B. Russell and A. Whitehead the credit for formalizing what wasknown and used for ages.

2. The Algebra of Numbers.Apart from any social conditions which may have played a role in

the development of number systems, there also exist intrinsic reasons fortheir development that peculiar to the inner dynamics of mathematical ac-tivity. The step from natural numbers to rational numbers derives easilyfrom human activity where an apple and yet another make two apples andwhere two people sharing an apple make for one half of an apple for each.Irrational numbers are at a deeper level; in the time of Eudoxus, it centeredon geometrical issues such as the lengths of diagonals of a rectangle whosesides had lengths expressed by natural numbers. Negative numbers areunderstandable, as by Fibonacci above, if not as easily realizable. Wheredo complex numbers come from?

Pause for a moment to reflect on the interesting names given to theexpanding sets of numbers. They surely reflect the prejudices of the namersor, at the very least, the harsh criticism their introduction produced! Con-sider the positive versus the negative, the rational versus the irrational andthe real versus the imaginary.

There is a purely algebraic way of explaining the need for most of thesenumbers and it is this need which first pointed to the complex numbers.

4 Chapter 1. Introduction

We will give successive elementary equations in terms of successive sets ofknown numbers whose solutions are not in those sets, but in successivelyexpanded sets.

Consider the equations

m + x = n,

for natural numbers m and n. It is true that exactly one of the followingholds: n > m or n = m or n < m. If n > m, then the equation has asolution in N, which is called n − m. If n = m, then the existence of 0 issuggested. Finally, if n < m, then the negative integers are needed for theequation to have a solution. The equations

qn = m,

for integers m and n, with n �= 0, lead us to the rational numbers. Polyno-mial equations with integral coefficients such as

an xn + an−1xn−1 + an−2xn−2 + · · · + a1x + a0 = 0

lead us to irrational numbers such as the solutions of x2 − 2 = 0 andcomplex numbers such as the solutions of x2 + 1 = 0.

The set that is missing from the above discussion is the set of realnumbers. The set of all real solutions of polynomial equations with integercoefficients is called the set of algebraic numbers. Some transcendentalnumbers (real numbers, such as π , which are not algebraic) were knownbefore the real numbers were rigorously defined. The actual set of realnumbers—not just the rationals and some algebraic irrationals—was notfully explained until the nineteenth century. One has to view the realnumbers as some sort of completion of the rational numbers.

Consider the rational numbers. All can be written as repeating deci-mals such as

11 = 1.0 . . . , 1

2 = 0.50 . . . , and 13 = 0.3 . . . .

We can now imagine all other possible decimals, where the digits need notform any pattern, and call this the set of real numbers. Making this preciseis not as easy as it may appear!

3. The Axiomatic Method.Mathematics is highly respected in our culture. To most people it

inspires awe or fear, while to a few it is a source of joy and admiration, anartistic game of the highest quality, the very language in which the book ofnature is written.

Section 3. The Axiomatic Method 5

Whence mathematics derives its power? Is there a royal road to it?Is it accessible to all mortals? It certainly requires hard work, discipline,concentration, self-respect and self-confidence. Armed with these qualities,you have to plunge into this book and be actively engaged in decipheringits content. That means that you have to ask yourself questions and, bytrying to answer them, ask more questions; you have to try to discover thereal path that the intellect has followed to derive a proof, a path that isnow hidden either for aesthetic purposes or for the sake of precision andsimplicity.

Mathematics started as an empirical discipline in Mesopotamia andEgypt to serve everyday practical needs. Rules were invented to measuredistances, areas, and volumes, thus constituting an embryonic form of thebody of knowledge called geometry.

It is with the Greeks that geometry acquired a drastically differentcharacter. With Thales as a pioneer and then with the school of Pythagoras,geometry was constructed as an axiomatic, deductive discipline. Since, infact, this point of view is still prevailing as a way to sanction the resultsof the mathematician’s creative imagination, let us describe and explain itsmain features.

In order to establish human discourse, the meaning of the terms em-ployed has to be specified and universally accepted. To determine themeaning of a word, this word has to be explained in terms of other wordsand this process will never terminate unless we reach a word whose mean-ing is generally accepted. A word whose meaning is generally accepted iscalled a primitive term or an undefined term. The terms point, straight line,and plane are examples of primitive terms used in geometry. Other termsof the system in question are defined in terms of previously defined termsand the primitive terms. The following example gives a definition phrasedin three different ways.

Example 1. The following definitions are exactly the same:

• The midpoint of a line segment is the point that divides the linesegment into two equal parts.

• If a point divides a line segment into two equal parts, then thatpoint is called the midpoint of that line segment.

• A point is the midpoint of a line segment if and only if it dividesthe line segment into two equal parts.

All definitions are really if and only if statements,* even if we are not carefulto state them in that way. The context should tell you when a statement isreally a definition.

*We will discuss if and only if statements more in Section 2 of Chapter 2.


When starting from scratch—from the level of undefined terms—itis not so easy to give definitions of even simple objects. However, this isnecessary since we cannot assume that terms are understood the same wayby different people. Take the basic geometric notion of a triangle. SeeFigure 2.

Figure 2: Triangle vs. triangular region

You may think everyone knows what a triangle is. Try this experiment:ask a few non-mathematicians “What is a triangle?” You will probably getsome different answers, perhaps including drawings. The most popularanswer may describe a triangular region, which a geometer would call aregion bounded by a triangle. Clearly, we need definitions, even for simplemathematics objects.

Exercise 3. Define the concept of a triangle. Do you need more primitive

terms than the ones mentioned above?

There follows another basic term to define.

Exercise 4. Define the concept of an angle. How might addition of two

angles be defined?

A key concept in mathematics is that of a proof. Suppose you statean opinion on an issue using a single sentence, for example “freedom ispreferable to happiness” and your interlocutor asks you to justify it. Justafter you have pronounced your next sentence he asks you to justify thissentence also. It is clear that this process will never end, unless you andyour interlocutor agree upon the truth of some sentence. Such a sentenceis called an axiom.

Exercise 5. Are any axioms involved in the definition of a triangle?

We list a few axioms from “The Elements,” a book written around 300B.C. by Euclid and preserving the geometric achievements of the ancientsin their final form.

Axiom 1. If two straight lines intersected by a transversal make the sumof the measures of the interior angles on the same side lessthan 180◦, then the two straight lines meet on that same side.

Section 4. Parallel Mathematical Universes 7

Axiom 2. Things which are equal to the same thing are equal to oneanother.

Axiom 3. If equals are added to the same thing, the sums are equal.Axiom 4. If equals are subtracted from equals, the differences are equal.Axiom 5. All right angles are equal to one another.

A proof consists of a series of logical steps by which we deduce thetruth of a statement from the axioms that are explicitly stated. A statementwhich can be proved is called a theorem. When we present a proof ofa theorem, we present only a skeleton; this skeleton indicates the ideasnecessary to give a complete proof. Such skeletal proofs rarely go backto the level of axioms, rather they usually depend on theorems which arealready proved.

Examples of theorems are:

(1) The diagonals of any rectangle are congruent.(2) The square of an even number is an even number.

Frequently, a theorem is called a proposition or a lemma or a corol-lary. Logically, all of these are the same. A statement which is calleda proposition is usually a less important result than a statement which iscalled a theorem. A lemma is usually only important as a logical step in theproof of some theorem.* A corollary is a theorem which follows, usuallyeasily, from the preceding theorem.

But what do we mean by logical steps? Mathematicians have contrivedvarious methods of deducing a desired result, that is, they have deviseddifferent kinds of proof. We illustrate the methods with specific examplesin the next chapter. However, there is more to the methodology of doingproofs than the types of proofs. This involves the very thought processesthemselves. The only way to develop and refine these processes is to usethem frequently.

4. Parallel Mathematical Universes.You may think that there is only one mathematics, that is, one “correct”

set of axioms from which to start. Surprisingly, this is quite wrong, in twodifferent ways! The removal, addition or other change of just one axiom candramatically change the mathematics. Also, two very different collectionsof axioms can lead to the same mathematics.

You may already know the story of Euclid’s “Parallel Postulate” fromyour high school geometry class. (By the way, we regard the words axiomand postulate as synonyms, though Euclid and others regard them as tech-nically different.) In plane geometry, the Parallel Postulate is equivalent to

*It is interesting to note that posterity may judge—sometimes has judged—a lemma tobe a theorem. For example, Zorn’s Lemma and the Urysohn Lemma are famous theorems inset theory and topology, respectively, that were considered lemmas by their original authors.


the statement that there is a unique line that is parallel to a given line passingthrough a given point not on the original line. One may instead assumethat no parallel line may exist or that more than one parallel line may exist.Such changes in the axioms lead to new and different geometries; these arecalled non-Euclidean geometries. In fact, non-Euclidean geometries areas legitimate as Euclidean geometry. Despite the fact that Euclid and theancient Greeks tried to model the physical universe with their geometry,in physics, for example in relativity theory, it is a non-Euclidean geometrywhich is the correct model.

Recall your high school geometry class. You may remember thatthere were quite a few axioms. Perhaps the notions of betweenness andcongruence were left undefined, while the ideas of measurement were de-veloped from the axioms. You may remember developing the ideas ofsegment length and angle measure from the axioms. It is quite possible todevelop Euclidean geometry by starting from different collections of unde-fined terms and axioms. The following lengthy example gives a differentset of axioms with the opposite approach.

Example 6. In 1932, G. D. Birkhoff published a paper* giving a collectionof four axioms for Euclidean plane geometry. Birkhoff assumes you haveconstructed the set of real numbers. The terms set, point, line, on, linemeasure (or distance or length), and angle measure are undefined. Birkhoffgave his axioms the following names:

I. Postulate of Line MeasureII. Point-Line Postulate

III. Postulate of Angle MeasureIV. Postulate of Similarity

Let us consider the statements of these four axioms (or postulates)individually; we will state each one and examine its content. You shouldremember the big picture as you read this; ask yourself if this really givesyou the same plane geometry that you already know. (We follow the usualcustom of using the term “line” to mean “straight line.”)

Postulate of Line Measure. Given any line, there is a one-to-one corre-spondence from the set of points of this line onto the set of real numbers,where every point A on this line corresponds to the real number xA, so thatthe length of the line segment AB (or distance between A and B) is givenby

d(A, B) = |xB − xA|for all points A, B.

*A Set of Postulates for Plane Geometry (based on scale and protractor) appeared involume 33 of the journal Annals of Mathematics.

A

Ax

B

Bx

C

CxBxCx –| |AxBx –| |

A

as

r

B

O

s

ar

as-ar (mod 2π)


The idea of this postulate is that every line comes with coordinates.Consider Figure 7.

Figure 7: Birkhoff’s Postulate of Line Measure

You are familiar with thinking of the set of real numbers as a line.The Postulate of Line Measure indicates that every line can be thought ofas a number line. The length of a segment is then the distance between itsendpoints. The absolute value is necessary to avoid negative length sincewe do not know, a priori, which endpoint will have the larger coordinate;of course, |xA − xB | = |xB − xA|.Point-Line Postulate. One and only one line contains two given distinctpoints.

The Point-line Postulate should be familiar to you; it can be restatedas “Two points determine a line.”

The first two postulates allow us to define the distance between anytwo distinct points, A and B, of the plane. By the Point-Line Postulate,there exists a line containing A and B. The Postulate of Line Measure thengives the distance between the points. This is well-defined since, by thePoint-Line Postulate, there is only one line containing the two points.

Before proceeding with a postulate on measuring angles, considerFigure 8.

Figure 8: Birkhoff’s Postulate of Angle Measure

A

B

O�BOA

�AOB

A

B

O

�BOA

�AOB


Postulate of Angle Measure. Given any point O, there is a one-to-onecorrespondence from the set of rays with endpoint O onto the set of realnumbers (mod 2π ),* where every ray r with endpoint O corresponds to areal number ar (mod 2π ), so that, if A �= O and B �= O are points on raysr and s, respectively, then the measure of angle �AO B is given by

m�AO B = as − ar (mod 2π ).

Furthermore, if the point B on ray s varies continuously in a line � notcontaining the vertex O, the number as varies continuously also.

Figure 8 may help you to understand the Postulate of Angle Measurebetter. The angle measure defined here is a little different than you maythink. In general, m�AO B �= m�BO A; in fact, the sum of m�AO B andm�BO A is 0 (mod 2π ), which equals 2π (mod 2π ). See Figure 9.

Figure 9: m�AO B + m�BO A = 0 = 2π (mod 2π )

Before proceeding to the last of Birkhoff’s postulates, which is abouttriangles, we should make sure we understand how triangles are defined.

Exercise 10. Define triangle using only Birkhoff’s first three postulates.

Postulate of Similarity. Given two triangles, ABC and A′ B ′C ′, anda constant k > 0, if

d(A′, B ′) = k · d(A, B), d(A′, C ′) = k · d(A, C)

and m�B ′ A′C ′ = ±m�B AC,

*You may not be familiar with the phrase “mod 2π .” However, you are probably familiarwith the concept from trigonometry. In trigonometry, one may consider angles with negativemeasure or with measure larger than 2π . The sine or cosine of these angles only dependon the equivalent angle between 0 and 2π or, more precisely, in the interval [0, 1). Forexample, sin(−π) = sin π = sin(3π) and cos(−π) = cos π = cos(3π). Therefore, we write−π = π = 3π (mod 2π ); actually we usually read the equal sign as either “is congruent to”or “is equivalent to” rather than “is equal to.” You may recall that the symbol ≡ is usually usedinstead of = in this context. We consider this idea more carefully in Section 3 of Chapter 6.

A

BCC′

A′

B ′


then

d(B ′, C ′) = k · d(B, C), m�C ′ B ′ A′ = ±m�C B A

and m�A′C ′ B ′ = ±m�AC B.

Figure 11: Birkhoff’s Postulate of Similarity

Figure 11 may help you to understand the Postulate of Similaritybetter. This is the familiar Side-Angle-Side statement (usually called SAS)for similar triangles. In Figure 11, k = 2; that is, A′ B ′ is twice as longas AB, A′C ′ is twice as long as AC and �B ′ A′C ′ is congruent to �B AC .The triangles are congruent if k = 1.

Of course, the statement at the beginning of this example, that theseare axioms for Euclidean plane geometry, must be proved. We will not dothis here.

Exercise 12. Using Birkhoff’s axiomatic system in Example 6, define the

concepts of betweenness and congruence (for line segments and angles).

We will see other axioms in Chapter 7 that play similar roles. Theseare the Axiom of Choice and the Continuum Hypothesis. Mathematiciansare free to accept or reject these as axioms of their mathematics; a theoremof one mathematics may be false in another, which is based on a differentcollection of axioms.

12

Chapter 2

Statements in Mathematics

1. Mathematical Statements.Whatever mathematics may be as a mental activity, it is communicated

as a language. Therefore, it has its specific syntax, its own technical terms,and its own conventions. Mathematics is also an exact science, which meansthat we are obliged to express our mathematical thoughts with high preci-sion. A deviation from the norm may easily lead to a complete distortionof the intended meaning.

In mathematics, we assert the truth of certain statements. Other state-ments are to be proved or disproved. This is a fundamental dichotomy:

No mathematical statement is both true and false.

Viewed axiomatically, you can take true and false to be undefined terms;the above dichotomy can be taken as an axiom. We should clarify what ismeant by a mathematical statement.

Definition 1. A declarative sentence is a logical statement iff, according toits logical content, it is unambiguously either true or false. A mathematicalstatement is a logical statement used in mathematical discourse.

By statement, without the adjective logical or mathematical, we willalways mean a logical or, usually, a mathematical statement. Let us considerthis carefully with some examples.

The sentenceThe number 2 is positive.

is a true mathematical statement and the sentence

The number 2000 is divisible by 3.

is a false mathematical statement. Note that the previous two sentences areboth declarative.

Section 1. Mathematical Statements 13

By definition, sentences that are not declarative cannot be logical state-ments; such sentences are neither true nor false. An interrogative sentencesuch as

Who is there?

is not a logical statement. An imperative sentence such as

Go away.

is not a logical statement. An exclamatory sentence such as

Wow!

is not a logical statement. Of course, a sentence like

Hello.

is not a logical statement either.A declarative sentence may be a logical statement even though its truth

value is not known. This is very common in the most advanced areas ofmathematics; it is the fuel of mathematical research. Consider the followingsentence:

The 1010 digit in the decimal expansion of the number π is 3.

This is a mathematical statement. Even if no one knows whether this istrue or false, it must be exactly one or the other. This is what we mean byunambiguous in Definition 1.

Not every declarative sentence is a statement. Opinions are not logicalstatements. For example, the declarative sentence

Vanilla is the best ice cream flavor.

is not a logical statement since it does not have a truth value; it is true thata person may have an opinion on this subject, but this does not make thestatement true or false. On the other hand,

Vanilla is Jon’s favorite ice cream flavor.

may be a statement for a specific person named Jon. The sentence

Socrates was a good philosopher.

14 Chapter 2. Statements in Mathematics

is a less clear example. We should ask what constitutes a good philosopher;the sentence about the famous ancient Greek philosopher, Socrates, is alogical statement if we give precise criteria for the adjective good.

The examples given above are usually called simple statements sincethey cannot be broken into pieces that are themselves statements. Someauthors call simple statements atoms or atomic statements.

You can probably think of examples that are not so simple, such ascompound statements involving words like not, and, or, if, then, implies, andequivalent. We will consider these terms, called connectives, in Section2. The same authors who call simple statements either atoms or atomicstatements call compound statements molecules or molecular statements.

You might also imagine statements involving variables. Consider thesentences

The square of every real number is nonnegative.

and

There exists a real number whose square is not positive.

Though these are true statements, they are rather different from our simplestatements. Both contain variables. The former states a fact about everyreal number and the latter states a fact about at least one real number (specif-ically, 0). Predicates—sentences involving variables—that are quantified,as in the examples above, will be studied in Section 5.

You may say that you know exactly what the words not, and, or,etc. mean and how their use will change the meaning of a statement. How-ever, the colloquial usage of these terms does not entirely agree with themathematical usage. For this reason, we will take a rather pedantic ap-proach and define these words in terms of truth values—the truth valuesof compound statements formed by connectives are functions of the truthvalues of the simple statements from which they are formed.

Linguistics also comes into play. Translating between mathematical,or logical, language and a natural language, such as English, is not alwaysstraight forward. We will start by using statements in symbolic logic andtheir English translations, side by side. After this chapter, we will abandonthe logical notation completely; however, we recommend that you adoptsome of it as a type of short hand for your own writing of mathematics. Wewill remind you of this in Remark 37 at the end of this chapter.

2. Mathematical Connectives.We are now ready to study the different connectives used in creating

compound statements. Remember that these are being defined in a for-mal way, based on truth values, rather than a linguistic way, based on thecolloquial usage of English.

Section 2. Mathematical Connectives 15

Negation (not)

You may recall that an integer is called an even number when it equalstwice some integer and is called an odd number when it equals one plus twicesome integer. The statement “3 is an odd number.” is a true mathematicalstatement; the statement “3 is not an odd number.” is a false mathematicalstatement. How are these two statements related? You might say that thesecond is the negation of the first. This is correct. However, we will definenegation not in colloquial terms, but in terms of truth values.

Definition 2. For a mathematical statement p, the negation of p is a math-ematical statement, denoted ¬p, whose truth value is the opposite of thetruth value of p.

In general, we denote the negation of a statement, p, by not p; somedenote it symbolically as either ∼ p or −p, rather than ¬p. We canillustrate Definition 2 using the following truth table:

p ¬pT FF T

In a truth table, one reads the truth values—T for true and F for false—of the column headings across a given row. In the truth table above for notp, the first row indicates that whenever p is true, then not p is false whilethe second row indicates that whenever p is false, then not p is true. Truthtables give us a nice shorthand for expressing the various truth values ofrelated statements, starting with simple statements like p.

Colloquially, we generally negate a statement by using the word not.Of course, it isn’t* as simple as changing p to not p. For example, supposep is the statement “3 is an odd number.” For not p, we write “3 is not anodd number.” rather than “Not 3 is an odd number.”

Conjunction (and)

Two mathematical statements can be combined to form new, morecomplicated statements. One way to do that is to connect two statementswith the word and. Let us consider this connective, defining it by a truthtable.

Definition 3. For mathematical statements p and q, the conjunction of pand q is the mathematical statement, denoted p ∧ q, whose truth valuevaries according to the following truth table:

*Sorry for the contraction; we didn’t want the word not to appear here.


p q p ∧ qT T TT F FF T FF F F

We will generally write p and q for the conjunction p ∧ q. Someauthors use the notation p · q or p & q instead of p ∧ q.

An example of conjunction is the statement “2 is positive and 3 isnegative.” Of course, this conjunction is false since “3 is negative.” is afalse statement.

The conjunction of two statements is a true statement exactly whenboth statements are true and is a false statement when at least one statementis false. This agrees with the truth table in Definition 3.

Disjunction (or)You might think we were excessively pedantic in defining negation

and conjunction of mathematical statements. The next two definitions maychange your mind. Consider, for example, the statement

2 is positive or 3 is positive.

Is this statement true or false? You might want to say this is false becauseit is not true that only one or the other is true. For a less mathematicalexample, consider the statement

I will take a bath or I will take a shower.

Would you expect that the speaker may take both a bath and a shower?

Definition 4. For mathematical statements p and q, the disjunction of pand q is a mathematical statement, denoted p ∨ q, whose truth value variesaccording to the following truth table:

p q p ∨ qT T TT F TF T TF F F

We will generally write p or q for the disjunction p∨q. You should beextremely careful here, since the word or is not used with its most commonmeaning. When someone says

I will give my ticket to Alex or to Jamie.


it is implied that only one person will receive the ticket: either Alex orJamie, but not both Alex and Jamie. In mathematics, we use the word orin the sense of at least one.

The disjunction of two statements is true if at least one of the statementsis true. In mathematics, the word or is equivalent to the hideous wordand/or. As we use it in mathematics, or is referred to as the mathematicalor and the inclusive or (since the possibility of both is included). Similarly,the or of everyday language is referred to as the exclusive or (since thepossibility of both is excluded).*

Example 5. It is usually easy to recognize negation, conjunction, anddisjunction by use of keywords: the word not in negations, the word and inconjunctions, and the word or in disjunctions. However, notation can hidethese constructions. Consider the following three true statements.

• 1 ≤ 1• 2 < 3 < 4• 5 ≤ 6 ≤ 6

Reading the symbol ≤ as “is less than or equal to” shows that the firststatement is a disjunction; it can be interpreted as 1 < 1 or 1 = 1. Thesecond statement is the conjunction 2 < 3 and 3 < 4. The third statementis a conjunction of two disjunctions. Implication (implies/if, then)

We will consider two additional types of compound statements, im-plications and equivalences. These are also called conditionals and bicon-ditionals, respectively, by some authors. The next construct, implication,also disagrees with the common colloquial interpretation.

Definition 6. For mathematical statements p and q, the implication (orconditional statement) denoted p ⇒ q is a mathematical statement whosetruth value varies according to the following truth table:

p q p ⇒ qT T TT F FF T TF F T

We will generally write either if p, then q or p implies q for theimplication p ⇒ q; some authors use the notaion p → q or p ⊃ q insteadof p ⇒ q. For the implication p implies q, the statement p is called the

*Some people go to the extreme of using the new word xor for the exclusive or. We donot like it and you may (or may not) like it. Is xor any better (or worse) than and/or?


hypothesis or antecedent and the statement q is called the conclusion orconsequent of the implication.

Usually, in ordinary language, it is understood, for a statement of theform if p, then q , that there exists a causal connection between p and q.We cite as an example the following statement:

If you leave me, then I will be badly hurt.

In mathematical language, a conditional statement is false only when thehypothesis is true and the conclusion is false. Since a false hypothesisalways yields a true implication, no sense of causality can be interpreted.

Example 7. Cinsider the following four examples:

• “If 2 is positive, then 3 is positive.” is a true statement since both thehypothesis and the conclusion are true statements.

• “If 2 is negative, then 3 is positive.” is a true statement since thehypothesis is false.

• “If 2 is negative, then 3 is negative.” is a true statement since thehypothesis is false.

• “If 2 is positive, then 3 is negative.” is a false statement since thehypothesis is true, but the conclusion is false.

Make sure you read the middle two examples carefully and understand whythey are true! Do not let the trivial nature of the hypotheses and conclusionsfool you: once the truth values of the hypothesis and the conclusion of anygiven implication are known, the truth value of that implication is triviallyknown as well.

An implication if p, then q is called vacuously true if its hypothesis,p, is false. For a vacuously true implication, it does not matter whether itsconclusion is true or false. Look again at Example 7 and notice that wedid not indicate whether the conclusion was true or false in the middle twoexamples since it did not matter.

We have already said that if p, then q and p implies q are used in-terchangeably to mean p ⇒ q. In fact, all of the following are equivalentEnglish sentences:

(1) If p, then q.(2) p implies q.(3) p only if q.(4) q if p.(5) p is sufficient for q.(6) q is necessary for p.

Forms (3) and (4) may seem peculiar to you; think of them, for a trueimplication p implies q , as “p is true only if q is true” and “q is true if


p is true.” (You should compare these with the truth table for p ⇒ q.)The forms (5) and (6) are more esoteric. We have a prejudice against thislanguage unless used together, as explained after the following example.

Example 8. Consider the sentence

For all real numbers a and b, if a = 0, then ab = 0.

This is a statement of a different type since it contains variables. Noticethat the truth value of “a = 0” depends on the value of the variable a. Wewill discuss such statements further in Section 5; we use it here because itbetter illustrates the different forms of implications mentioned above. Thefollowing forms should strike you as equivalent.

• For all real numbers a and b, a = 0 only if ab = 0.• For all real numbers a and b, ab = 0 if a = 0.• For all real numbers a and b, a = 0 is sufficient for ab = 0.• For all real numbers a and b, ab = 0 is necessary for a = 0.

Equivalence (equivalent/if and only if/necessary and sufficient)Our final construct, equivalence, probably agrees with your colloquial

use of that word.

Definition 9. For mathematical statements p and q, the equivalence of pand q, denoted p ⇔ q, is a mathematical statement whose truth valuevaries according to the following truth table:

p q p ⇔ qT T TT F FF T FF F T

Instead of writing p ⇔ q for an equivalence, we will generally writeone of the following:

• p is equivalent to q,• p if and only if q,• p is necessary and sufficient for q.

For brevity, many people write p iff q in place of p if and only if q. Someauthors use p ↔ q or p ≡ q instead of p ⇔ q.

Remark 10. You may recall that, in Section 3 of Chapter 1, we explainedthat definitions are often stated in the form of if and only if statements.Of course, definitions are not logical statements whose truth value canbe determined.* Rather, for a definition in the form of an if and only if

*You may choose to think of all mathematical terms as undefined, in which case defini-tions become axioms. This make the biconditional appearance of definitions more reasonable.


statement, the first half of the statement names the term being defined bythe other half. While we try to use the word iff in our definitions, mostauthors simply write if. For example, we could define even integers asfollows: an integer is even iff it is two times some integer. 3. Symbolic Logic.

We could make an entire course out of symbolic logic. Rather thando that, we will discuss some of the more important compound statements,those we will use from time to time. Before doing this, let us discuss howstatements of symbolic logic are parsed. Note that we will use the symbolicnotation throughout this section and return to English statements in the nextsection.

Remark 11. You may remember rules for order of operations in arithmetic.For example

1 + 2 × 3 = 7 �= 9

since multiplications are performed before additions. There are also rulesfor order of operations in logic. Reading from left to right, do things in thefollowing order:

(1) parentheses (or brackets, etc.)(2) negation(3) conjunctions and disjunctions (in any order)(4) implication(5) equivalence

Groupings by parentheses, brackets, etc. are treated hierarchically. That is,nested groupings are evaluated from the inside towards the outside (just asfor algebraic expressions). We remember this as working from the insideout.

Therefore, ¬p∨q is the same as (¬p)∨q and different from ¬(p∨q).Remember that between matched pairs of parentheses, you must apply thesame rules. For a more complicated example, the following two compoundstatements based on simple statements a, b, c, . . . , i are equivalent:

a ∨ ¬b ∧ c ⇒ ¬d ∧ e ⇔ ¬ f ∧ g ∨ h ⇒ i,⟨{[a ∨ (¬b)

] ∧ c

}⇒ [

(¬d) ∧ e]⟩ ⇔

⟨{[(¬ f ) ∧ g

] ∨ h

}⇒ i

⟩.

Notice that, in order to evaluate this, we must start with the innermost pairsof parentheses, then the square brackets, then the curly braces, and then,finally, the angle brackets.

Section 3. Symbolic Logic 21

Example 12. Consider the disjunction p ∨ ¬p. If p is true, then thedisjunction is true.* On the other hand, if p is false, then ¬p is true andthe disjunction is true. This can be expressed by the following truth table:

p ¬p p ∨ ¬pT F TF T T

Notice how the column for p ∨ ¬p has all T’s in its column; that is,p or not p is always true.† Definition 13. A compound statement is a tautology iff it is true for allpossible truth values of its component statements.

Tautologies are easily recognized in truth tables: their columns con-tains only T’s. Next we examine the other extreme.

Example 14. Consider the conjunction p ∧ ¬p. If p is false, then theconjunction is false.‡ On the other hand, if p is true, then ¬p is false andthe conjunction is false. This can be expressed by the following truth tablewith only F’s in the column for p ∧ ¬p.

p ¬p p ∧ ¬pT F FF T F

Definition 15. A compound statement is a contradiction iff it is false forall possible truth values of its component statements.

Contradictions are easily recognized in truth tables: their columnscontains only F’s. Contradictions are actually quite important in math-ematics. In fact, we will see in the next chapter that contradictions canactually be useful in proving theorems.

Let us look at a couple additional examples.

Example 16. Consider the implication p ⇒ p. The following truth tableshows that this is a tautology.

p p ⇒ pT TF T

*The fact that ¬p is false in the case when p is true does not matter.†The moral of this example is that you should never ask your mathematics instructor

a question like “Will this topic be covered on the exam or not?” because you are likely toreceive the perfectly correct, yet uninformative, response “Yes.”

‡The fact that ¬p is true in the case when p is false does not matter.


Example 17. Consider the equivalence ¬¬p ⇔ p. This is a tautology:

p ¬p ¬¬p ¬¬p ⇔ pT F T TF T F T

The following shows that some data can be extraneous.

Exercise 18. Use truth tables to show that the following are tautologies:

(a) p ∧ q ⇒ p (b) p ⇒ p ∨ q

The following gives us an alternative definition of implication.

Exercise 19. Use a truth table to show that p ⇒ q ⇔ ¬p ∨ q.

The following exercise may remind you of the concept of equality:x = x , x = y if and only if y = x , and x = y = z implies x = z.

Exercise 20. Use truth tables to show that the following are tautologies:(a) p ⇔ p(b) (p ⇔ q) ⇔ (q ⇔ p)

(c) (p ⇔ q) ∧ (q ⇔ r) ⇒ (p ⇔ r)

In the language of Chapter 6, the previous three exercises say thatthe relation defined on any set of statements by equivalence is, indeed, anequivalence relation.

Recall that p ⇒ q can be read as p only if q and q ⇒ p can be readas p if q. It seems reasonable that p if and only if q should be equivalentto p if q and p only if q . This suggests the following important (though,perhaps, obvious) result; we will use this very often.

Exercise 21. Use a truth table to show that (p ⇔ q) ⇔ (p ⇒ q)∧(q ⇒ p)

is a tautology.

One can use truth tables in this way to prove many such statementsin logic. Remember that this is just a shorthand. To write out the proof ofExercise 21 in words would go something like this:

Proof. We consider four cases:

(1) p is true and q is true,(2) p is true and q is false,(3) p is false and q is true, and(4) p is false and q is false.

Section 3. Symbolic Logic 23

First, suppose p is true and q is true. Hence, if p, then q is true and if q,then p is true. Therefore, the conjunction if p, then q, and if q, then p istrue. Moreover, since p and q are true, p is equivalent to q is true. Thiscompletes the first case.Second, . . . . �

Exercise 22. Complete the proof, started above, of Exercise 21 using

words.

Let us return to conditional statements. Given statements p and q, wecan make implications p implies q as well as q implies p. We can alsotake the negations of p and q and form other implications. Three of theseconstructions are important enough to have names.

Definition 23. The converse of the implication p ⇒ q is the implicationq ⇒ p.

That is, the converse of the statement if p, then q is the statement if q,then p. If a statement is true, then its converse may be either true or false;you can examine the possibilities in the following exercise.

Exercise 24. Give examples of each of the following, if possible:(a) a true implication whose converse is true,(b) a true implication whose converse is false,(c) a false implication whose converse is true,(d) a false implication whose converse is false.

Let p and q be simple statements. Remember that the implicationif p, then q can only be false when p is true and q is false. Similarly,if q, then p can only be false when q is true and p is false. So, as youshould have discovered in the previous exercise, it is impossible for bothsuch an implication—between simple statements—and its converse to befalse. If you thought otherwise, you probably introduced a variable in yourstatements. Soon, we will introduce variables and quantifiers; this willmake things more difficult, but far more interesting.

Definition 25. The contrapositive of the implication p ⇒ q is the impli-cation ¬q ⇒ ¬p.

So, the contrapositive of the statement if p, then q is the statement ifnot q, then not p.

Example 26. Consider the statement

If the moon is made of cheese, then 1 = 2.


The contrapositive of this statement is

If 1 �= 2, then the moon is not made of cheese.

You should recognize that both of these statements are true! The truth value of a contrapositive statement is always the same as the

truth value of the original statement.

Exercise 27. Prove that any implication is equivalent to its contrapositive.

That is, p ⇒ q ⇔ ¬q ⇒ ¬p.

Definition 28. The inverse of the implication p ⇒ q is the implication¬p ⇒ ¬q.

The inverse of the statement if p, then q is the statement if not p, thennot q; it is the contrapositive of the converse of if p, then q. Therefore, theconverse and the inverse of a given statement are equivalent:

q ⇒ p ⇔ ¬p ⇒ ¬q.

Exercise 29. (a) Use a truth table to show that, given an implication, itsconverse and its inverse are equivalent; that is, q ⇒ p ⇔ ¬p ⇒ ¬q.(b) Are an implication and its inverse equivalent? Prove your answer.

So, from every given implication we can derive three implicationscalled the converse, the contrapositive, and the inverse of the original im-plication:

Original: p �⇒ q, if p, then q

Converse: q �⇒ p, if q, then p

Contrapositive: ¬q �⇒ ¬p, if not q, then not p

Inverse: ¬p �⇒ ¬q, if not p, then not q

4. Compound Statements in English.Of course, statements can get quite complicated and writing statements

in English is further complicated by the fact that we do not use parenthesesto indicate order of thoughts; commas can play an important role. Youmay have noticed that we used italics and quotation marks to try to clarifysentences in this chapter; this is a luxury upon which we should not depend.

For example, consider the sentence:

1 is negative and 2 is negative or 3 is positive.

Section 5. Predicates and Quantifiers 25

If we let p denote the statement 1 is negative, let q denote the statement 2is negative and let r denote the statement 3 is positive, then our compoundstatement looks like

p ∧ q ∨ r.

For symbolic logic, order of operations dictates that this is equivalent to

(p ∧ q) ∨ r.

Since p and q are false and r is true, this statement is true. Notice thatp ∧ q ∨ r is different from p ∧ (q ∨ r); in fact, p ∧ (q ∨ r) is false.

Now how do we make this distinction in written English? What doyou think of the following two statements?

1 is negative and 2 is negative, or 3 is positive.

1 is negative, and 2 is negative or 3 is positive.

There certainly is a lot of weight riding on those two little commas. Noticethat the former should be interpreted as (p ∧ q) ∨ r while the latter shouldbe interpreted as p ∧ (q ∨ r).

Now consider spoken English. It is very difficult to hear those littlecommas!

An alternative is to rephrase the sentences completely. This is not sosimple. For example,

The conjunction of the statement 1 is negative and the

disjunction of the statements 2 is negative and 3 is positive.

While this may be clearer, it is not a very satisfying solution. Look at thesymbolic statement in Exercise 21. In words, we could say something likethe following:

p if and only if q is equivalent to the conjunction

of the implications if p, then q and if q, then p.

5. Predicates and Quantifiers.Take another look at Example 8. Also, did you have any difficulties

with Exercise 24? These were tricky since they got you thinking (perhaps,in the case of Exercise 24) about introducing variables into statements. Inthis section, we introduce the use of variables in both logical and mathe-matical statements. The statements of most mathematical theorems involve


variables. By the way, variables are sometimes called free variables, to em-phasize that they are free to change.

Consider the statement

The square of −2 is positive.

We know that this statement is true. Now consider the sentence

The square of a negative number is positive.

Is this a logical statement? You might think that it is, since you see thatit is true for every real number. However, this sentence is not a logicalstatement since there is an unresolved variable involved.

Now, let us consider the following sentence:

The number x is positive.

Is that a logical statement? No, again, since there is an unresolved variableinvolved. Again, no answer can be given as to the truth value of the sentence,unless x is given a specific value. For example, if x = 2, the sentencebecomes a true statement, and, if x = −4, the sentence becomes a falsestatement.

Any declarative sentence that contains one or more unresolved vari-ables is called a predicate iff it is a statement whenever specific values aresubstituted for the variables. The word unresolved means that no values,or possible set of values, are given for the variables. We will return to thebusiness of resolving, or substituting for, the variables shortly.

Two other examples of predicates are x ≥ 0 and x + y = 0; the formerhas one variable and the latter has two variables. Using functional notation,we could denote these predicates by P(x) and Q(x, y), respectively.

Just as in the case of logical statements, we can negate a predicate.For instance, if P(x) is the predicate x ≥ 0, then not P(x) is the predicatex �≥ 0.*

Predicates also can be combined to form new predicates as is illustratedin the following exercise.

Exercise 30. Equations and inequalities involving variables are predicates.(a) Determine all real numbers x for which the conjunction x + 3 > 6 andx2 < 3 is a true statement.(b) Similarly, determine all real numbers x and y for which the disjunctionxy = 0 or 2x − 3y = 0 is a true statement.

*Thinking of the real numbers, we could have said x < 0, but, technically, that requiresproof!


Substituting a single value for a variable is not generally of interestsince it can be done directly, thereby turning a predicate into a statement.For example, instead of writing a predicate with a substitution such as

x2 > 3 for x = 2.

we could just write the statement

22 > 3.

However, there are two very important ways of creating statements out ofpredicates; this is called quantifying a predicate.

Consider the sentence:

For all real numbers, x2 = 4.

This sentence should be a false statement since 52 �= 4. This is one kind ofquantifier.

Definition 31. Let S be a set and let P(x) be a predicate. The universalquantifier is a phrase of the form for all x ∈ S, denoted ∀x ∈ S. Theuniversally quantified predicate sentence

∀x ∈ S, P(x)

is a mathematical statement which is true iff P(x) becomes a true statementwhenever an element of S is substituted for x in the predicate P(x).

We will generally write for all x ∈ S, P(x) for ∀x ∈ S, P(x). Thephrases for every x ∈ S and for any x ∈ S are also used for universalquantifiers. We recommend that you be careful with the use of the wordany since it may be misinterpreted as some rather than every. The phrasefor all real numbers x is, of course, equivalent to for every x ∈ R.

Example 32. Consider the following statements. Which are true and whichare false?

(a) For all real numbers x , (x + 3)2 + (x − 2)2 > 0.(b) For all real numbers x ,

√x2 = x .

For (a), the squares are nonnegative. (x + 3)2 = 0 only if x = −3and (x − 2)2 = 0 only if x = 2. So they cannot both be equal to 0 for thesame x . Therefore, the statement in (a) is true.

For (b), the square root notation denotes the nonnegative square root.

Since, for x = −1,√

(−1)2 = √1 = 1 �= −1, statement (b) is false.


Remark 33. Sometimes a universal quantifier is hidden in a statement. It issometimes preferable, for purposes of logical manipulation, to restate sucha statement, as is done in the next example. Sometimes the set defining theuniversal quantifier is hidden. Consider the statement

For all x , there exists a positive square root.

This is true for the set of positive real numbers, but is false for the set ofreal numbers. When the set is omitted, we will assume that it is the set ofreal numbers. Example 34. In geometry, the Isosceles Triangle Theorem may be stated:

The angles at the base of an isosceles triangle are congruent.

This could be restated as follows:

For every triangle T , if T is isosceles,

then the angles at the base are congruent.

You might even wonder exactly what set of triangles we are talking about!

Next, consider the sentence:

There exists x , such that x2 = 4.

This sentence is a true statement since 22 = 4; it is also true since (−2)2 =4. The expression “there exists” is a new type of quantifier.

Definition 35. Let S be a set and let P(x) be a predicate. The existentialquantifier is a phrase of the form there exists x ∈ S, denoted ∃x ∈ S. Theexistentially quantified predicate sentence

∃x ∈ S, P(x)

is a mathematical statement which is true iff P(x) becomes a true statementfor at least one element of S substituted for x in the predicate P(x).

We will generally write there exists x ∈ S such that P(x) for ∃x ∈S, P(x). The phrase for some x ∈ S is also used for existential quantifiers.Again, we remind you to be careful with the use of the word any.


Example 36. Which of the following statements about sets of real numbersare true and which are false?

(a) There exists x such that (x + 5)2 = 0.(b) There exists x such that |x − 1| = x .(c) There exist x, y such that x2 + y2 = 1.

Statement (a) is true since x = −5 satisfies the equation. Statement(b) is true since x = 1

2 satisfies the equation. Statement (c) is true since,

for example, x = y = 12

√2, satisfies the equation. Notice that the set of

values for x is important: if the word real were changed to natural, thenall three statements above would be false.

Let us see how we negate a sentence which is introduced by a universalquantifier. Consider the statement: “For all x and y, x + y = 0.” In fact,this statement is false. What is the negation of this sentence? If not allpairs of numbers add up to zero, there exist numbers x and y, whose sumis not zero, so the negation of our original sentence is “There exist x and ysuch that x + y �= 0.” which is a true statement. In general, the negation ofthe statement

For all x , P(x).

is the statement

There exists an x such that not P(x).

The negation of predicates with more than one variable is defined in ananalogous way.

To see how we negate a sentence which is introduced by the existentialquantifier, we look at the following example. Consider the statement:“There exists a number x , such that x2 < 0.” which is a false statement. Ifthere does not exist a number whose square is negative, then the square ofall numbers will be nonnegative, so the negation of the original sentence is“For all numbers x , x2 ≥ 0.” In general, the negation of the statement

There exists x such that P(x).

is the statementFor all x , not P(x).

Remark 37: Logical notation. Although we have mentioned the symbols¬, ∧, ∨, ⇒, ⇔, ∀, and ∃, we will not use them in this text after this chapter.Many people use them in writing mathematics by hand. The symbols forimplications, equivalences, and quantifiers are especially useful. Somepeople also use � or s.t. for “such that,” ∵ for “since” or “because” and ∴for “therefore” or “thus” or “hence.”


We prefer to use double barred arrows (⇒, ⇐, ⇔) for logical opera-tions since single barred arrows, especially →, are used frequently in thenotation for functions.

We recommend that you use whatever shorthand you feel comfortablewith. Moreover, we want to remind you to keep in mind your audience,those who will read what you write!


Supplemental ExercisesDefinition Review. There were 13 definitions in this chapter. You can takethe truth values true and false to be undefined. Let p and q be logical ormathematical statements (or predicates). Define each of the following andgive an example of each:

(1) p is a logical statement(1) p is a mathematical statement(2) ¬p is negation of a statement p(3) p ∧ q (or p and q) is conjunction of statement p and q(4) p ∨ q (or p or q) is disjunction of statement p and q(6) the implication p ⇒ q (or p implies q or if p, then q)(9) the equivalence p ⇔ q of statements p and q

(13) a statement is a tautology(15) a statement is a contradiction(23) the converse of an implication p ⇒ q(25) the contrapositive of an implication p ⇒ q(28) the inverse of an implication p ⇒ q(31) the universal quantifier(31) an universally quantified predicate sentence(35) the existential quantifier(35) an existentially quantified predicate sentence

In Exercises S1—S35 below, use truth tables to show that each statementis a tautology involving statements p, q, and r .

Exercise S1. p ∧ (p ⇒ q) ⇒ q

Exercise S2. ¬p ∧ (q ⇒ p) ⇒ ¬q

Exercise S3. ¬p ∧ (p ∨ q) ⇒ q

Exercise S4. p ⇒ (q ⇒ p ∧ q)

Exercise S5. (p ⇒ q) ∧ (q ⇒ r) ⇒ (p ⇒ r)

Exercise S6. (p ∧ q ⇒ r) ⇔ (p ⇒ [q ⇒ r ])

Exercise S7. (p ∧ q ⇒ r) ⇔ (p ⇒ [q ⇒ r ])

Exercise S8. (p ⇒ [q ∧ ¬q]) ⇒ ¬p

Exercise S9. (p ⇒ q) ⇒ (p ∨ r ⇒ q ∨ r)

Exercise S10. (p ⇒ q) ⇒ ([q ⇒ r ] ⇒ [p ⇒ r ])

Exercise S11. (p ⇔ q) ∧ (q ⇔ r) ⇒ (p ⇔ r)

Exercise S12. (p ⇔ q) ⇔ (q ⇔ p)


Exercise S13. (p ⇒ q) ∧ (r ⇒ q) ⇔ (p ∨ r ⇒ q)

Exercise S14. (p ⇒ q) ∧ (p ⇒ r) ⇔ (p ⇒ q ∧ r)

Exercise S15. (p ∨ q) ⇔ (q ∨ p)

Exercise S16. (p ∧ q) ⇔ (q ∧ p)

Exercise S17. (p ∨ q) ∨ r ⇔ p ∨ (q ∨ p)

Exercise S18. (p ∧ q) ∧ r ⇔ p ∧ (q ∧ p)

Exercise S19. p ∨ (q ∧ r) ⇔ (p ∨ q) ∧ (p ∨ r)

Exercise S20. p ∧ (q ∨ r) ⇔ (p ∧ q) ∨ (p ∧ r)

Exercise S21. p ∨ p ⇔ p

Exercise S22. p ∧ p ⇔ p

Exercise S23. ¬(p ∨ q) ⇔ ¬p ∧ ¬q

Exercise S24. ¬(p ∧ q) ⇔ ¬p ∨ ¬q

Exercise S25. p ⇒ q ⇔ ¬p ∨ q

Exercise S26. p ⇒ q ⇔ ¬(p ∧ ¬q)

Exercise S27. p ∨ q ⇔ ¬p ⇒ q

Exercise S28. p ∨ q ⇔ ¬(¬p ∧ ¬q)

Exercise S29. p ∧ q ⇔ ¬(p ⇒ ¬q)

Exercise S30. p ∧ q ⇔ ¬(¬p ∨ ¬q)

Exercise S31. (p ⇔ q) ⇔ (p ⇒ q) ∧ (q ⇒ p)

In Exercises S32—S35, let F and T represent statements that arealways false or true, respectively.

Exercise S32. p ∨ F ⇔ p

Exercise S33. p ∧ F ⇔ F

Exercise S34. p ∨ T ⇔ T

Exercise S35. p ∧ T ⇔ p

Exercise S36. Which of the following statements are true and which arefalse? Explain.

(a) For all x , (x + 3)2 + (x − 2)2 > 0.(b) For all x ,

√x2 = x .

(c) For all x and y, |x − y| = |y − x |.Exercise S37. Show that q ∧ (p ⇒ q) ⇒ p is not a tautology. This is

sometimes referred to as the fallacy of asserting the conclusion.

33

Chapter 3

Proofs in Mathematics

1. What is Mathematics?Using the ideas of the previous section, we are now in a position where

we can understand a given statement. The next step would be to prove ordisprove that statement. Is this mathematics? Well, not quite.

Mathematics is, foremost, a creative activity. Theorems are not handedout by God (or professors) to the faithful (students?) with the expectationthat the supplicants (students??) will supply the appropriate proof beforeawaiting the next theorem from on high.* Theorems are a product ofhuman endeavor, dependent on the human experience and knowledge. Todo mathematics is to create mathematics, theorems along with proofs. Yes,perhaps the scholastic experience is slightly different. Most professors andtheir texts play the roles of gods and kings, expecting their slaves (students!)to do the grungy work.

Before you proceed you should give the preceding some thought.This text deals with finding proofs, not theorems. From elementary schoolthrough the master’s degree, students of mathematics rarely encounter theother side, the creation of the theorems themselves. Finding theorems isthe primary goal of a doctoral dissertation and of mathematical research.

Most times you are given a proof immediately after the theorem. Yourjob is then to comprehend both the theorem and its proof. In order to learnhow to do proofs, you should think about how you would prove the theoremfirst. To start thinking like a mathematician you must read slowly, pausingto think often. If you pause now and then to ask the question “What canI try to prove next?” you will be taking a large step towards becoming amathematician and you will be better able to construct proofs.

Two things that you should do frequently, which are helpful both infinding theorems and proofs (not surprising since they go hand in hand),are to think of examples and draw pictures. In fact, these are not unrelatedactivities. Often, working out an example will help you to better understand

*Some philosophers certainly disagree with this point of view.

34 Chapter 3. Proofs in Mathematics

what makes a theorem true, that is, how to prove it. Pictures are usually anextension of that process. Just as important is to change the hypotheses andconsider examples where the conclusion fails. These “counterexamples”again can help you to isolate the missing ideas which make up the proof.

You may be thinking that you have seen lots of pictures in mathematicstexts that you did not find the least bit illuminating. That is quite under-standable. The important part of the picture is usually in its construction,not in its final presentation. It is like looking up the answer to an exercisein the back of a text. Usually, the answer alone does not illuminate the pathfrom the problem to the answer.

If you get nothing else out of this chapter, we hope you will understandthe following two things. First, you must spend far more time thinking thanreading. (It is not uncommon for a mathematician to take hours, even days,to read and comprehend a single page!) Second, you should always thinkof examples (and counterexamples), drawing pictures as a guide. (Yes, youcan draw mental images, but it is surprising how different things can lookon paper.)

In the remainder of this chapter, we will examine various techniquesof proof.

2. Direct Proof.The first type of proof we will consider is the direct proof. When

someone says that a proof is “by brute force” or by “following your nose,”they are referring to a direct proof. While “brute force” generally impliessome sort of explicit computation, all of these descriptions describe the basicnature of a direct proof. Many mathematicians will say that they “construct”a proof. We warn you that this naıve use of the word “construct” may bemisunderstood.* The basic idea of direct proof is simple: start from thehypotheses, make a sequence of implications, and arrive at the conclusion.

An Example of Direct ProofConsider the following proposition from geometry. You may refer

back to Section 3 of Chapter 1 for our discussion of the axioms of Euclid;the axiom numbers used below refer to the list of axioms in that section.

Proposition 1. If two lines intersect, then they make the vertical anglesequal to one another.

Before writing down a proof, let us start with a picture. ConsiderFigure 2, which shows two intersecting lines.

*“Constructive proof” may refer to the constructivist philosophy of mathematics. Noteveryone subscribes to this philosophy. In this philosophy, the existence of a mathematicalobject is accepted only when it can be constructed (for instance, only when an explicit exampleis known).

A

C

D

B

E

Section 2. Direct Proof 35

Figure 2: Vertical angles

If nothing else, the figure gives us a shorthand notation for our proofby giving names to the angles. By the way, if you do not remember thedefinition of vertical angles, you should be able to figure it out from theproposition and the figure!

Proof. Let B AD and C AE be two lines intersecting at the point A. Thenthe angles �B AD and �C AE are both equal to two right angles.† Ifwe subtract from both angles the angle �C AD, by Euclid’s Axiom 4, weconclude that �B AC = �D AE , which is what we wanted to prove. Usingthe same line of reasoning, we can prove that �C AD = �E AB. �

Exercise 3. Complete the proof of Proposition 1: prove that �C AD =�E AB.

If you prefer to think about the measures of angles (perhaps followingBirkhoff’s axioms in Example 6 of Chapter 1), then you might rewrite theabove proof as follows:

Proof. Let B AD and C AE be two lines intersecting at the point A. Then

m�B AD = π and m�C AE = π.

Therefore, m�B AD = m�C AE and

m�B AD − m�C AD = m�C AE − m�C AD.

Since

m�B AC = m�B AD − m�C AD

and

m�D AE = m�C AE − m�C AD,

†By Axiom 5 and the familiar construction of perpendicular lines, the angle at a pointon a line is equal to two right angles. In more modern language, we can say these anglesmeasure π or 180◦.


we have proved that m�B AC = m�D AE . Using the same line of reason-ing, he reader can prove that m�C AD = m�E AB. �Proof by Brute Force: Another Example of Direct Proof

Let A = { 1, 3, 5, 7, . . . } be the set of odd natural numbers. We know,by definition, that every odd natural number can be expressed in the form2n + 1 for some nonnegative integer n.

Proposition 4. The square of any odd integer is odd.

Proof. Consider the odd natural number x = 2n +1 for some nonnegativeinteger n. Then

x2 = (2n + 1)2 = 4n2 + 4n + 1 = 2(2n2 + 2n) + 1 = 2k + 1,

where k is the natural number 2n2 + 2n. Hence, x2 is odd. �

Exercise 5. Prove that the square of an even integer is even.

Exercise 6. (a) Prove that the sum of two even integers is even.

(b) Prove that the sum of an even integer and an odd integer is odd.

Exercise 7. Prove that the cube of an odd integer is odd.

Exercise 8. Prove that the sum, the difference, the product, and the quotient

(when it is defined) of two rational numbers is a rational number.

Proof by cases.In proving a proposition, we sometimes have to distinguish cases. The

following example is suggestive.

Example 9. To prove that (−1)n + (−1)n+1 = 0 for all integers n, wedistinguish two cases: n is either even or odd.

Case 1: suppose n is even. Then n + 1 is odd. Hence, (−1)n = 1 and(−1)n+1 = −1. Therefore, the sum equals 0.

Case 2: suppose n is odd. Then n + 1 is even. Hence, (−1)n = −1and (−1)n+1 = 1. Therefore, the sum equals 0.

We give another proof of (−1)n + (−1)n+1 = 0 for all integers n:

(−1)n + (−1)n+1 = (−1)n + (−1)n · (−1) = (−1)n · (1 + (−1)) = 0.

The moral of this example is that there are generally many ways to prove atheorem. The usual problem is to find some proof. Some mathematicians

Section 3. Contraposition and Proof by Contradiction 37

like to find “the best” proof of every theorem; of course, this is subjective.The twentieth century mathematician P. Erdos liked to say that God has “TheBook,” which contains all of the theorems, along with their best proofs.

Recall the definition of the absolute value, |x |, of a real number, x . Ofcourse, we can define |x | to equal

√x2. However, the first definition you

saw was probably something like

|x | ={

x , if x ≥ 0

−x , if x < 0

where −x plays the role of “dropping the negative sign” when x is negative.Since this is a definition by cases, proofs of statements involving absolutevalues often consider cases.

Example 10. Let us prove that |x | ≥ x for all x ∈ R. We consider twocases according to the definition of absolute value. If x ≥ 0, then |x | = xby definition; hence |x | ≥ x . On the other hand, if x < 0, then |x | = −xis positive; hence |x | ≥ x .

Exercise 11. Prove that for all x ∈ R, |2x + 1| > x .

We would like to point out that “proof by cases” is not a proof techniqueat all. Instead, “proof by cases” is a method of reducing a problem to easier,smaller problems. In fact, the methods of proof used to prove each of theindividual cases may be different.

Exercise 12. Prove that, for all real numbers a, if a �= 0, then a2 > 0.

3. Contraposition and Proof by Contradiction.We noted that a statement is equivalent to its contrapositive. Therefore,

to prove a statement, we can prove its contrapositive. We call this proof bycontraposition.

An Example of Proof by ContrapositionConsider the converse of Exercise 5.

Proposition 13. If the square of an integer is even, then the integer is even.

Proof. By contraposition, this statement is equivalent to saying “If an in-teger is odd, then its square is odd,” which is Proposition 4. �

The following exercises are examples of statements whose proofs aremost easily done by contraposition.

Exercise 14. Prove that if 2n1+n2 is irrational, then n is irrational.


Exercise 15. Prove that, for all real numbers a, if a2 > 0, then a �= 0.

Look at Exercise 7. The converse can be proved by contraposition.

Exercise 16. Prove that the cube of an integer n is odd if and only if n is

odd.

A similar idea to contraposition is proof by contradiction. In proof bycontraposition, we used the fact that an implication and its contrapositiveare logically equivalent. Mechanically, this amounts to the following: toprove that p implies q, we assume that the negation of q is true (i.e., q isfalse) and proceed o show that the negation of p is true (i.e., p is false).Seen as a direct proof, the hypothesis p has been contradicted since bothp and its negation are apparantly true. We use this idea to generalize thisproof technique.

In proving an implication, p implies q, by contradiction, if we firstsuppose the hypotheses of a statement, p, and we assume the negation ofits conclusion, q, and use these to derive a contradiction (to a hypothesisor to our assumption or to any other true statement), then we have provedthe original implication. Note that the case of deriving the negation ofthe hypotheses is exactly proof by contraposition. To understand that thismakes sense, consider that a contradiction should never arise logicallyunless there is a false assumption and the only assumption we made wasthe negation of the conclusion of our implication.

Two Examples of Proof by ContradictionLet us consider two examples of proof by contradiction. First, Propo-

sition 17 shows another way to approach Exercise 6(b).

Proposition 17. The sum of an odd integer and an even integer is odd.

Proof. Let x be odd and y be an even integer. Assume that x + y isnot odd, that is, x + y is even. Then x = 2n + 1 for some integer n,y = 2m, for some integer m and x + y = 2p for some integer p. Hence,x + y = 2n + 1 + 2m = 2p. This is equivalent to 2(p − m − n) = 1. Butthis is a contradiction since the number on the left-hand side is even, but 1is odd. Therefore, x + y is odd. �

This proof would not make Erdos’ The Book (see Example 9). Thedirect proof is shorter and neater.

The second example of a proof by contradiction is given in Theorem18. This is a famous theorem.

Theorem 18.√

2 is irrational.

Before giving a proof of this proposition, you might wonder how weknow that there exists a real number called

√2. This is actually proved by

Section 3. Contraposition and Proof by Contradiction 39

constructing the set of real numbers in Part III. The following proof showsthat this real number is indeed irrational.

Proof. Assume that√

2 is rational. Then there exist natural numbers mand n such that √

2 = m

n.

Without loss of generality, we assume that the fraction mn is reduced to

lowest terms; hence, m and n are not both even. Squaring both sides weget

m2

n2= 2.

This is equivalent to the equality

2n2 = m2.

Since the left-hand side is an even number, m2 is even. By Proposition 13,m is even. This means that

m = 2k

for some natural number k. Hence

2n2 = (2k)2 = 4k2,

which implies thatn2 = 2k2.

Therefore, n2 is even and so n is even. This is a contradiction since weassumed that m and n are not both even. Hence,

√2 is irrational. �

Remark 19. The following remarks concern the style of writing proofs bycontradiction.(1) Many mathematicians start their proofs by indicating how they plan toproceed. A good way to start a proof by contradiction is a statement suchas “This proof is by contradiction.”(2) Although the words “assume” and “suppose” are synonyms, you mayfind it helpful to use “assume” when trying to construct a proof by contra-diction and “suppose” for all hypotheses. In this way the word “assume” isa reminder that you are trying to find a contradiction. We are divided on thisadvice; one of us tries to follow it regularly while the other is indifferent.(3) It is common to leave out the punch line. What is the punch line? It isthe declaration of a contradiction and the negation of the original assump-tion. Sometimes such proofs end with the simple declarative statement


“Contradiction.” This may leave the reader with a little bit to think about!For example, a much terser proof of Proposition 17 may look as follows.

Proof. Suppose x = 2n+1 is odd and y = 2m is even. Assume x+y = 2pis even. Hence, x + y = 2(n + m) + 1 = 2p. Contradiction. �We strongly suggest that you write as much as is necessary so that yourfellow students could understand your proofs.

We will need the following important theorem to generalize Proposi-tion 13 and Theorem 18; its proof is postponed until the Exercise 42. Recallthat a integer p is prime iff p is neither 0 nor ±1, and its only divisors are±1 and ±p.

Theorem 20 (Fundamental Theorem of Arithmetic). Each natural num-ber other than 1 is the product of a uniquely determined finite collection ofpositive primes.

For example,

12 = (2)(2)(3) = (2)(3)(2) = (3)(2)(2);the factorization is unique in that each includes two 2’s and one 3. We thinkof 2 as a “product” of only one prime, namely 2. That is, every integer mgreater than 1 can be written uniquely in the form

m = pn11 pn2

2 · · · pnkk ,

wherep1 < p2 < · · · < pk

are k distinct positive prime numbers and n1, n2, . . . , nk are natural num-bers.

Consider the following generalization of Proposition 13; they are thesame when n = 2 and p = 2.

Theorem 21. Let n be a natural number, m be an integer greater than 1,and p be a positive prime number. If the mn is divible by p, then m is alsodivisible by p.

Proof. By the Fundamental Theorem of Arithmetic, there exists a factor-ization

m = pn11 pn2

2 · · · pnkk

as a product of positive prime numbers where p1 < p2 < · · · < pk . Bythe definition of prime, if m is divisible by p, then there exists j (between1 and k inclusive) such that pj is divisible by p and, hence, p = pj . Now

mn = (pn11 pn2

2 · · · pnkk )n = pn1n

1 pn2n2 · · · pnk n

k .

Section 4. Proof by Induction 41

By the Fundamental Theorem of Arithmetic, the uniqueness of the factor-ization of mn guarantees that mn is divisible by p = pk . �

Use these ideas for the following exercise.

Exercise 22. Prove that√

3 and√

5 are irrational.

The following exercises are two more examples of statements whoseproofs are best done by contradiction (or contraposition).

Exercise 23. Let a and b be real numbers. Prove that a2 + b2 = 0 if and

only if a = b = 0.

Exercise 24. Let x be a real number. Prove that if x2 < x , then x < 1.

The next exercise is a type of problem in combinatorics. Be carefulwith counting problems as they are notoriously difficult.

Exercise 25. Prove that if at least two people are present at a party, at leasttwo of them know the same number of participants. (Assume that eithereveryone or no one counts themselves as someone they know and that ifone person knows another, then the latter also knows the former.)

4. Proof by Induction.Before we discuss the next example we need some preliminaries. Sup-

pose we want to prove that n2 ≥ n for all natural numbers n. In fact, wehave to prove infinitely many statements:

12 ≥ 1

22 ≥ 2

32 ≥ 3

...

We can denote these statements by P(1), P(2), P(3), etc. The followingtells us one way to prove such a sequence of statements.

Theorem 26 (Principle of Induction). Statements P(n) are true for alln ∈ N if

(1) the first statement, P(1), is true and(2) whenever a statement in the sequence, P(n), is true, then the suc-

ceeding statement, P(n + 1), is also true.


Notice that we called the preceding a theorem. In fact, it is very closeto being an axiom, usually called the Axiom of Infinity. We state this axiombelow in a somewhat different (also more readable and less precise) formthan you would encounter in an axiomatic approach to set theory.

Axiom 27 (Axiom of Infinity). There exists a set S of natural numberssuch that

(1) 1 ∈ S and(2) if n ∈ S, then n + 1 ∈ S.

If we have a sequence of statements P(1), P(2), P(3), P(4), . . . , thenwe can define a set

S = { n | P(n) is true } .

This axiom characterizes the set of natural numbers: S = N. For ourpurposes, we may as well assume that the Principle of Induction is itself anaxiom. Let us see how this gives us an important proof technique.

The second step in the Principle of Induction can be restated as

P(n) implies P(n + 1) for all n ∈ N.

This step is called the inductive step and P(n) is called the inductive hy-pothesis (or induction hypothesis). Sometimes, it is more advantageous touse the implication P(n − 1) implies P(n); in this case, P(n − 1) is calledthe inductive hypothesis. The idea is actually quite natural: if P(1) is trueand if P(n) → P(n + 1) for all n ∈ N, then

P(1) is true → P(2) is true → P(3) is true → P(4) is true → . . . .

Remark 28. We like to think of a proof by induction as building a machine:you need an “ON” switch to get it going and then the machine needs tocrank out the desired product. The switch is part (1) above and the machinemechanism, or crank, is part (2).

We are now ready to apply the Principle of Induction to prove theassertion that n2 ≥ n, for all n ∈ N. Actually we will consider threeexamples of proof by induction in Propositions 29, 30 and 31.

Proposition 29. n2 ≥ n for all n ∈ N.

Proof. The proof uses the Principle of Induction. Let P(n) denote thestatement “n2 ≥ n.” We first prove P(1). For n = 1, we have 12 ≥ 1,which is obvious.

Next we suppose that P(n) is true for some fixed n, that is, n2 ≥ n;this is the inductive hypothesis. Next we prove that P(n + 1) is true, thatis, (n + 1)2 ≥ n + 1. We compute (n + 1)2 = n2 + 2n + 1 and, by the


inductive hypothesis, n2 + 2n + 1 ≥ n + 2n + 1 ≥ n + 1 since 2n ≥ 0.*Therefore, (n + 1)2 ≥ n + 1. �

You should take a close look at the proof of Proposition 29. Do yousee that the work in the second paragraph is necessary? Many students whoare seeing induction proofs for the first time want to simply assert the truthof P(n + 1) by substituting n + 1 into the inductive hypothesis. Of course,this is wrong.

Proposition 30. 1 + 2 + 3 + · · · + n = n(n+1)

2 for all n ∈ N.

You may already be familiar with summation notation. In general,

b∑k=a

f (k)

indicates a sum of terms of the form f (k), for some function f , where wesubstitute sequential values from a to b in place of k. The statement inProposition 30 can be rewritten as

n∑k=1

k = n(n + 1)

2.

Proof of Proposition 30. The proof is by induction on n. First we provethe statement for n = 1: we have 1 = 1·(1+1)

2 which is obviously true.

Next, we suppose that is 1 + 2 + 3 + · · · + n = n(n+1)

2 . Add n + 1 toboth sides of our inductive hypothesis and simplify the right-hand side toget

1 + 2 + 3 + · · · + n + (n + 1) = n(n + 1)

2+ (n + 1)

= n(n + 1) + 2(n + 1)

2

= (n + 1)(n + 2)

2. �

The above proof would probably not make Erdos’ book. Never beafraid to try different techniques in proving a result. The following proof

*In fact, 2n ≥ 2 since n ≥ 1, but we do not need this in the proof. Of course, wecould say that n2 > n for n > 1. Since 12 = 1, this would slightly improve the statement ofProposition 29.


of Proposition 30 seems more intuitive. It often seems that inducive proofsobscure the motivation of mathematical statements.

Another Proof of Proposition 30. Let x = 1 + 2 + 3 + · · · + n. Thenx = n + (n − 1) + (n − 2) + · · · + 1 also. Therefore,

2x = x + x

= (1 + 2 + 3 + · · · + n) + (n + (n − 1) + (n − 2) + · · · + 1)

= (1 + n) + (2 + n − 1) + (3 + n − 2) + · · · + (1 + n)

= n(n + 1).

Hence,

x = n(n + 1)

2. �

Now there is actually nothing special about starting an induction prooffrom n = 1; an induction can start with any integer and continue with allintegers greater than the original integer. (Even wilder possibilities exist!)To understand this you should realize that it is just a matter of relabelingwhich integer is “the first.” Consider the following proposition.

Proposition 31.n∑

k=0rk = 1 − rn+1

1 − rfor n = 0, 1, 2, . . . and r �= 1. [For

r = 0, use the convention r0 = 1.]

Proof. The proof is by induction. Let r be a fixed real number different

from 1. For n = 0, 1, 2, . . . , let P(n) be the statementn∑

k=0rk = 1 − rn+1

1 − r.

We want to prove P(0); that is, we start with n = 0.0∑

k=0rk = r0 =

1 = 1 − r

1 − r= 1 − r0+1

1 − r.

Next, we want to show, for every n = 0, 1, 2, . . . , that P(n) implies

P(n + 1). Let n be a fixed nonnegative integer. Suppose thatn∑

k=0rk =

1 − rn+1

1 − r; this is the inductive hypothesis. We must show that the inductive

hypothesis impliesn+1∑k=0

rk = 1 − r (n+1)+1

1 − r= 1 − rn+2

1 − r.

Nown+1∑k=0

rk =(

n∑k=0

rk

)+ rn+1 = 1 − rn+1

1 − r+ rn+1 by the inductive

hypothesis. Moreover,1 − rn+1

1 − r+ rn+1 = 1 − rn+1 + (1 − r)rn+1

1 − r=


1 − rn+1 + rn+1 − rn+2

1 − r= 1 − rn+2

1 − r. This is what we needed to show and

completes the proof. �

Remark 32. The following remarks concern the style of writing proofs byinduction.(1) Since a proof by induction has two steps, two paragraphs make it morereadable.(2) Usually, the first step is much easier than the inductive step.(3) Many authors like to use a new variable when considering the inductivestep in the proof. For example, for the inductive step in the proof of n2 ≥ nfor all n ∈ N they will use k2 ≥ k to show that (k + 1)2 ≥ k + 1. We arenot convinced that this makes proofs by induction any easier to understand.(4) During a lecture, a mathematician may give a shorthand proof whichlooks like the following.

Short proof of Proposition 29. n = 1: 12 ≥ 1.n + 1: (n + 1)2 = n2 + 2n + 1 ≥ n + 2n + 1 ≥ n + 1. �

Short proof of Proposition 30. n=1: 1 = 1·(1+1)

2 .

n + 1: (1+2+3+· · ·+n)+ (n +1) = n(n+1)

2 + (n +1) = n2+3n+22 =

(n+1)(n+2)

2 . �

Short proof of Proposition 31. n=0:0∑

k=0rk = r0 = 1 = 1−r

1−r .

n+1:n+1∑k=0

rk =(

n∑k=0

rk

)+rn+1 = 1−rn+1

1−r +rn+1 = 1−rn+1+(1−r)rn+1

1−r =1−rn+2

1−r . �

Can you follow the three proofs above? Do you find the original,wordy proofs more readable? Remember that written words do help thereader to understand your proofs. We do not support shorthand for its ownsake, but it is acceptable as long as there is no loss of understanding suchas in a lecture where the writer is there to answer questions!

The following statement and proof are wrong. [Recall that, for anynatural number n, n factorial is the number n! = n(n − 1)(n − 2) · · · (1).]

Absurdum. 2n < n! for all n ∈ N.

Proof? The proof is by mathematical induction. Suppose that 2k < k! istrue; this is our inductive hypothesis. We need to show this implies that2k+1 < (k + 1)!. We see that

(k +1)! = (k +1)(k!) > (k +1)(2k) ≥ (1+1)(2k) = (2)(2k) = 2k+1. �


Exercise 33. (a) Explain why the statement above is absurd and why itsproof is wrong.(b) Prove that 2n < n! for all n = 4, 5, 6, 7, 8, . . . .

Once more, the following statement and proof are wrong.

Absurdum. For n = 1, 2, 3, . . . , every bag containing n solid-coloredballs contains balls of only one color.

Proof? n=1: A bag with one ball clearly has balls of only one color.n+1: Take a bag with n + 1 balls. Remove one ball. By hypothesis,

the bag now has balls of only one color. Replace the first ball and removea different ball. Again, the bag now has balls of only one color. Since theother balls which never left the bag did not change color, all the balls mustbe the same color. �

Exercise 34. Explain why the statement above is absurd and why its proof

is wrong.

Now it’s time for you to try some proofs by induction.

Exercise 35. Prove that 1 + 3 + 5 + · · · + (2n − 1) = n2 for all n ∈ N.

Exercise 36. Prove that, for all natural numbers n,n∑

k=1

1

k(k + 1)= n

n + 1.

Exercise 37. Prove that, for all natural numbers n,n∑

k=1

k2 = n(n+1)(2n+1)

6 .

In the following exercise, there are two variables, p and n. Rememberthat induction is on one variable which varies through integer values.

Exercise 38. Prove that for every real number p > −1 and any natural

number n, (1 + p)n ≥ 1 + np.

We now state the Principle of Complete Induction, which is an equiv-alent version of the Principle of Induction.

Theorem 39 (Principle of Complete Induction). Statements P(n) aretrue for all n ∈ N if

(1) P(1) is true and(2) for every n ∈ N if P(k) is true for k = 1, 2, 3, . . . , n, then P(n + 1)

is also true.


The next result is quite amazing at first glance; we prove it by completeinduction. The Fibonacci sequence is defined inductively as follows: set

an ={

1, if n = 1, 2

an−1 + an−2, if n = 3, 4, 5, . . .

That is, the Fibonacci sequence is

{ 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, . . . }.

You probably do not see a formula to determine the thousandth term in thissequence as you would for an arithmetic or geometric sequence. However,there is the following formula.

Proposition 40. The nth term of the Fibonacci sequence is

an =(

1 + √5)n

−(

1 − √5)n

2n√

5.

What makes this statement amazing is the appearance of irrationalnumbers—square roots of 5—in this sequence of sums of pairs of naturalnumbers! The number

1 + √5

2

is a very special number called the golden ratio. It has interesting properties

and comes up in surprising situations, like this one. The number 1−√5

2 isreferred to as the conjugate of the golden ratio.

Before giving the proof, let us consider what must be done. It is asimple computation to check the formula for a1 = 1. Consider the inductivestep using complete induction: we want to use the inductive hypothesis

ak =(

1 + √5)k

−(

1 − √5)k

2k√

5,

for all k = 1, 2, 3, . . . , n − 1, to prove that

an =(

1 + √5)n

−(

1 − √5)n

2n√

5.

We know thatan = an−1 + an−2.


By the inductive hypothesis for k = n − 1 and k = n − 2, we have

an =(

1 + √5)n−1

−(

1 − √5)n−1

2n−1√

5+

(1 + √

5)n−2

−(

1 − √5)n−2

2n−2√

5.

So we wish to verify that

(1 + √

5)n

−(

1 − √5)n

2n√

5

?=(

1 + √5)n−1

−(

1 − √5)n−1

2n−1√

5

+(

1 + √5)n−2

−(

1 − √5)n−2

2n−2√

5.

Using algebra, you might derive the following equivalent equations:

(1 +

√5)n

−(

1 −√

5)n ?= 2

[(1 +

√5)n−1

−(

1 −√

5)n−1

]

+ 4

[(1 +

√5)n−2

−(

1 −√

5)n−2

](

1 +√

5)n

− 2(

1 +√

5)n−1

− 4(

1 +√

5)n−2

?=(

1 −√

5)n

− 2(

1 −√

5)n−1

− 4(

1 −√

5)n−2

(1 +

√5)n−2

[(1 +

√5)2

− 2(

1 +√

5)

− 4

]

?=(

1 −√

5)n−2

[(1 −

√5)2

− 2(

1 −√

5)

− 4

](

1 +√

5)n−2 [

1 + 2√

5 + 5 − 2 − 2√

5 − 4]

?=(

1 −√

5)n−2 [

1 − 2√

5 + 5 − 2 + 2√

5 − 4]

(1 +√

5)n−2[0]?= (1 −

√5)n−2[0]

0?= 0

You know that since the last of these equations is true, all of the previous,equivalent equations are true also. Of course, you never see a computationwith question marks like this in a textbook and you should be carefulin doing so even as a side computation. It is important that every step


is reversible! You can rewrite this computation in a better form, goingforwards rather than backwards. We will give a less messy proof below.

Does this complete the proof? No, since our proposed inductive stepnever settles the case of n = 2. This is because we used the fact that

an = an−1 + an−2

which is only valid for n ≥ 3. The proof is completed by checking thiscase.

Let us now rewrite the proof, following the ideas just discussed. Beforedoing so, let us consider the golden ratio and its conjugate again. Note thatthese are exactly the two zeros of the quadratic polynomial x2 − x − 1. Let

ϕ = 1 + √5

2and ϕ = 1 − √

5

2.

So ϕ2 − ϕ − 1 = 0 and ϕ2 − ϕ − 1 = 0. That is,

ϕ2 = ϕ + 1 and ϕ2 = ϕ + 1.

Proof of Proposition 40. We wish to show that

an =(

1 + √5)n

−(

1 − √5)n

2n√

5= 1√

5

(ϕn − ϕn

).

For n = 1 and n = 2, we compute

1√5

(ϕ1 − ϕ1

) =(

1 + √5)

−(

1 − √5)

2√

5= 1 = a1

and

1√5

(ϕ2 − ϕ2

) = 1√5

((ϕ + 1) − (ϕ + 1)

) = 1√5

(ϕ − ϕ

) = 1 = a2.

For an integer n ≥ 3, suppose that the induction hypothesis that

ak =(

1 + √5)k

−(

1 − √5)k

2k√

5= 1√

5

(ϕk − ϕk

)


holds for all k = 1, 2, 3, . . . , n − 1. By the definition of the Fibonaccisequence and the induction hypotheses for k = n − 1 and k = n − 2,

an = an−1 + an−2 = 1√5

(ϕn−1 − ϕn−1

) + 1√5

(ϕn−2 − ϕn−2

).

To show that an = 1√5

(ϕn − ϕn

), we need to show that

1√5

(ϕn − ϕn

) = 1√5

(ϕn−1 − ϕn−1

) + 1√5

(ϕn−2 − ϕn−2

).

We compute that

(ϕn−1 − ϕn−1) + (ϕn−2 − ϕn−2) = ϕn−1(ϕ + 1) − ϕn−1(ϕ + 1)

= ϕn−1(ϕ2) − ϕn−1(ϕ2)

= ϕn − ϕn.

Dividing both sides by√

5 yields the desired result. �

Let’s look at two erroneous attempts at using complete induction.

Absurdum. For all n ∈ Z+, n = 0.

Proof? The proof is by complete induction. Let P(n) be the predicaten = 0. Since the least number in Z+ is 0, the first step is for n = 0. Thisalready proves the first step.

For the inductive step, we hypothesize that, for k = 0, 1, 2, . . . , n,P(k) is true. That is, for k = 0, 1, 2, . . . , n, k = 0. We want to use thisto show that n + 1 = 0. Since, by hypothesis, 1 = 0 and n = 0, thenn + 1 = 0 + 0 = 0. This proves the inductive step and completes the proofof the statement. �

Of course, there is a problem in the proof, but where is it? You maynotice that the hypothesis 1 = 0 is nonsense. However, that alone does notexplain how the logic of the proof is faulty. You may also notice that weproved only one starting step (n = 0), but use two hypothses (1 = 0 andn = 0) in proving the inductive step. This brings us closer to the error.

The very first application of the inductive step is when n + 1 = 1or, equivalently, when n = 0. The induction hypothesis is k = 0 fork = 0, 1, 2, . . . , n, which is just 0 = 0 when n = 0. That is, 1 = 0 is notpart of the induction hypothesis when n = 0.

The next proof attempt, in short-hand form, suffers a similar fate.

Section 5. Examples and Counterexamples 51

Absurdum. For all n ∈ Z+, 10n = 1.

Proof? n=0: 100 = 1.

n+1: 10n+1 = 102n−(n−1) = 102n

10n−1= (10n)2

10n−1= 12

1= 1 since

10n = 10n−1 = 1. �

Exercise 41. Explain why the statement above is absurd and why its proof

is wrong.

You can use the Principle of Complete Induction to prove the follow-ing, which is the existence part of Theorem 20, the Fundamental Theoremof Arithmetic.

Exercise 42. Prove that every integer n greater than 1 equals a product of

one or more positive prime numbers.

The uniqueness part of the Fundamental Theorem of Arithmetic isnot so easy to write down. Think about why it is true and compare yourthoughts with the following.

Proof of the Fundamental Theorem of Arithmetic. In Exercise 42, weproved the existence of such a factorizaion. Now, suppose there are twodistinct postive prime factorizations of an integer n with n ≥ 2:

n = p1 p2 p3 · · · pa = q1q2q3 · · · qb;that is, n is factored into a primes on one hand and b primes on the other.Since p1 is a divisor of n = q1q2 · · · qb, p1 must be a divisor of a prime qj ;in fact, p1 = qj . Next rearrange the factorization into qk’s as q ′

1q ′2 · · · q ′

bwith q ′

1 = qj . Cancelling p1 = q ′1 yields

p2 p3 · · · pa = q ′2q ′

3 · · · q ′b.

Continue to cancel primes in this way until one side of the equation equals1. Since 1 is neither prime nor the product of primes greater than 1, bothsides must equal 1. Therefore, a = b. This shows that the factorizationis unique up to the order of the factors. The Fundamental Theorem ofArithmetic is proved. �

5. Examples and Counterexamples.Thus far we have considered methods of proof. We will now consider

a method for disproving statements. Let us consider the following claim:

There do not exist integers a, b, and c, such that a2 + b2 = c2.


Obviously this claim is false, since for a = 3, b = 4, and c = 5 the aboverelation holds. In this case, we say that we have provided a counterexample.

Exercise 43. Find a counterexample to the assertion that the absolute value

of every real number is positive.

Exercise 44. Find a counterexample to the assertion that every continuous

function is differentiable.

As we mentioned in Section 1, we consider examples and counterex-amples to be extremely important. You should always consider them; oftenthey will help you to find a proof of a statement. You should keep inmind that you can never prove or disprove a statement simply by giving anexample unless

• you examine all possible examples (this is a proof by cases), or• the statement is an existence theorem, or• the statement is false in at least one case (this is a counterexample).

A counterexample to a statement is also a proof of the negation of thestatement. Remember that the negation of a statement of the form “for allx , P(x)” is a statement of the form “there exists an x such that not P(x).”So a counterexample for such a statement gives an example that proves acorresponding existence theorem.

For example, consider the following existence theorem.

Proposition 45. There exists a continuous function f which is not differ-entiable.

Proof. Your counterexample in Exercise 44 proves this proposition. �


Supplemental ExercisesExercise S1. Prove that 2n + 2 ≤ 2n+1 for all n ∈ N. Use this result to

prove that 2n + 1 ≤ 2n for n = 3, 4, 5, . . . .

54

How to THINK about mathematics: A Summary

• Before trying to prove a statement:

• Construct examples (under the same and modified hypotheses)

• Construct counterexamples (under the same and modified hy-potheses)

• Construct pictures

• Consider different proof techniques:

• Direct Proof (including Brute Force)

• Proof by Contraposition (or proof by proof of the contrapositive)

• Proof by Contradiction

• Proof by Induction (including Complete Induction)

• Consider proof by cases and breaking a proof into smaller pieces

How to COMMUNICATE mathematics: A Summary

• Discuss mathematics with people—all parties should be honest andactive

• Organize your ideas (rewrite your proofs several times, if necessary)

• Remember to use enough words (a reader of your written proofscannot ask you questions)

• Use notation only as a shorthand or to clarify your proofs

55

PART II

SET THEORY

56

Chapter 4Basic Set Operations

1. Introduction.Perhaps our ability to distinguish and to classify objects constitutes

the essential part of our intellectual armor as human beings. These objectsmay be tangible—inanimate objects, distant galaxies, living beings, ourbeloved persons—or they may even be ideas, feelings or figments of ourimagination. Very often we view a certain collection of these objects asa unity. When Aristotle claimed that man is a rational animal, he meantto say that all human beings are rational. In fact, of all the animals, heisolated those, and only those, who shared a common characteristic. Ofcourse, categorization is not always an easy process. Very often, it stumblesover insurmountable ambiguities, especially due to the fact that universalagreement on the meaning of terms is not always achieved.

If we restrict reality to the mathematical realm, it turns out that wecan come up with an elegant and powerful theory. The key concept is thatof a set. To cite Cantor’s so-called naıve definition, a set is any collectionof definite, distinguishable objects of our intuition or of our intellect to beconceived as a whole.

You are probably already familiar with the notation used to describesets. We may list the elements of sets explicitly as in

T = { earth, wind, fire, water } and E = { 2, 4, 6, . . . }.The set T has exactly four elements. E , the set of all even natural numbers,has infinitely many; the three dots indicate that the list continues indefinitelyin the same pattern.

We may specify E in different ways as follows:

E = { n ∈ N | 2 divides n }= { 2n | n ∈ N } .

This kind of notation is called set-builder notation. It is common to readthis notation by substituting “such that” for the vertical bar. Some authors

Section 2. Subsets 57

prefer to use a colon in place of the vertical bar. Notice that the set Nappears on different sides of the vertical bar. We must specify a set forevery variable.

2. Subsets.To study set theory, as with any new mathematics, we start with some

undefined terms and some definitions.

Undefinition* 1. The terms set, element, and is an element of (representedby the symbol ∈) are taken as undefined.

For example, the notation x ∈ S means that x is an element of a set, S.We will be sloppy with this notation when we write something like x, y ∈ Sto mean x ∈ S and y ∈ S.

To make set theory interesting, we need an axiom which tells us thatsome set actually exists. The existence of the following set is assumedaxiomatically.

Definition 2. A set is called empty iff the set contains no elements. Theempty set is denoted in two ways, ∅ and { }.

You should be very careful about notation! The sets { {} } and { ∅ } arenot empty. Both have exactly one element, namely, the empty set.

You may have noticed that we wrote “the” empty set instead of “an”empty set in Definition 2. This will be justified in Proposition 12; there isonly one empty set.

Definition 3. A set S is called a subset of a set X iff every element of S isalso an element of X . We write either S ⊂ X or X ⊃ S.

That is, S is a subset of X iff s ∈ S implies s ∈ X .

Important Remark 4. You should be careful with the notation used in thistext. Many authors, especially of more elementary texts, use the notationS ⊆ X to describe subsets. Those authors use S ⊂ X to mean that S isa proper subset of X , that is, S is a subset of X , but is not equal to X†.We will use the notation S � X for proper subsets. It is more common todiscuss subsets than proper subsets; we prefer to use the simpler notationfor the more common usage. In more advanced mathematical texts, it isnormal to use ⊂ for all subsets.

*Since a sentence introducing defined terms is called a definition, we will call a sentenceintroducing undefined terms an undefinition. Or perhaps we should call it a primativum sinceprimitive terms are introduced?

†You may take the concept of set equality for granted. However, we will define this inDefinition 9 later in this section.

B

AU

x

58 Chapter 4. Basic Set Operations

There is a moral to this story: whenever you pick up a new mathe-matics text, you should be sure that you understand the notation and theterminology since usage may differ among different authors.

In constructing proofs, it is often helpful, as we noted in Section 1 ofChapter 3, to draw pictures. For proofs about sets, Venn diagrams, such asin Figure 5, may help you to construct proofs.

Figure 5: Example of a Venn diagram

There is no geometry in Figure 5; the circular nature of the actualshapes is irrelevant. The sets A and B are subsets of some set U . Also, Ais a subset of B. Sometimes it is useful to indicate elements explicitly. Theelement x is an element of B. However, x is not an element of A, whichwe denote by x /∈ A. Remember this tool when you construct proofs inthis chapter.

Absurdum. Let X be a set. The empty set is not a subset of X.

Proof? For the empty set to be a subset of X , every element of the emptyset must belong to X . Since the empty set has no elements, this conditionis not satisfied. Therefore, the empty set is not a subset of X . �

Exercise 6. Find the fallacy in the “Proof?” of the Absurdum and prove

that ∅ ⊂ X .

Remember that the notation A ⊂ B does not imply that A �= B(not that we have even defined the equality—or inequality—of sets yet).Keeping this in mind, prove the following:

Exercise 7. Let X be a set. Prove that X ⊂ X .

The following is an important property called the transitive propertyof subsets.

Exercise 8. Let A, B, and C be sets. Prove that if A ⊂ B and B ⊂ C , then

A ⊂ C .

If you ask yourself what it means for two sets A and B to be equal,you might say that A and B have exactly the same elements. We make thisprecise in the following definition.

Section 3. Intersections and Unions 59

Definition 9. Two sets A and B are equal iff the following equivalenceholds:

x ∈ A if and only if x ∈ B.

Of course, when A equals B we write A = B.Proving that two sets are equal can be quite easy, as in the following

example.

Example 10. Let A = { 2, 4, 6 } and B = { 2n | n ∈ Z3 }. Since Z3 ={ 1, 2, 3 }, we see that the elements of B are 2(1) = 2, 2(2) = 4, and2(3) = 6. But these are exactly the elements of A! Hence, A = B.

Of course, proving that two sets are equal is generally much moredifficult than in Example 10! The following is a useful way of thinkingabout equality of sets.

Exercise 11. Let A and B be sets. Prove that A = B if and only if A ⊂ B

and A ⊃ B.

In other words, to prove that two sets A and B are equal, we mustshow that a ∈ A implies a ∈ B, and b ∈ B implies b ∈ A.

You may have wondered why we wrote “the empty set” (and not “anempty set”) in and after Definition 2. The following proposition shows thatthere is a unique empty set.

Proposition 12. If A and B are empty sets, then A = B.

Proof. By Exercise 11, to show that A = B, it suffices to show that A ⊂ Band B ⊂ A. By Exercise 6, since A is empty, A ⊂ B. Again, by Exercise6, since B is empty, B ⊂ A. �

In the following exercise, we examine the important properties ofequality, called the reflexive, symmetric, and transitive properties. Thesethree properties virtually characterize equality; this will be explored indetail in Chapter 6.

Exercise 13. Let A, B, and C be sets. Prove that(a) A = A,(b) if A = B, then B = A, and(c) if A = B and B = C , then A = C .

3. Intersections and Unions.Next we consider two basic set operations, intersection and union.

We will do this first for two sets and then for arbitrarily many sets. Theintersection of two sets A and B should be the set consisting of all elementscommon to both of the sets A and B.

A B

C

D E

U U

x y


Definition 14. Let A and B be sets. The intersection of A and B is the set

A ∩ B = { x | x ∈ A and x ∈ B } .

That is, x ∈ A ∩ B if and only if x ∈ A and x ∈ B.Consider how you see intersections in Venn diagrams, such as in Figure

15.

Figure 15: Intersections in Venn diagrams

The left figure shows two sets, A and B with the intersection A ∩ Bshaded. The right figure shows three sets intersecting each other in themost generic way possible. Note that x, y ∈ C ∩ E , x ∈ (C ∩ E) ∩ D buty /∈ (C ∩ E) ∩ D.

Let us consider two easy intersection properties.

Exercise 16. Let A be a set. Prove that A ∩ ∅ = ∅.

Exercise 17. Let A be a set. Prove that A ∩ A = A.

Definition 18. Two sets A and B are disjoint iff A ∩ B = ∅.

By Exercise 16, every set is disjoint from the empty set. By Exercises12 and 17, only the empty set is disjoint from itself.

Next, we consider unions. The union of A and B should be the setconsisting of all elements of the sets A and B.

Definition 19. Let A and B be sets. The union of A and B is the set

A ∪ B = { x | x ∈ A or x ∈ B } .

That is, x ∈ A ∪ B if and only if x ∈ A or x ∈ B. Remember thator is not exclusive; x ∈ A or x ∈ B is true when x ∈ A and x ∈ B. Anelement may appear only once in a set. So {1, 2} ∪ {2, 3} = {1, 2, 3}; infact, {1, 2, 2, 3} is not a set.

The Venn diagrams in Figure 20 show the unions of two and three sets,respectively.

A B

C

D E

U U


Figure 20: Unions in Venn diagrams

Here are some easy exercises.

Exercise 21. Let A be a set. Prove that A ∪ ∅ = A.

Exercise 22. Let A be a set. Prove that A ∪ A = A.

The following exercise relates intersections and unions.

Exercise 23. Let A and B be sets. Prove* A ∩ B ⊂ A ⊂ A ∪ B. Prove

that this implies that A ∩ B ⊂ A ∪ B.

Remember that A ∩ B ⊂ A ∪ B does not necessarily mean that A ∩ Bis a proper subset of A∪ B; this should remind you of the non-exclusivenessof the logical “or” operation.

Exercises 16 and 17 are generalized by Exercise 24; Exercises 21 and22 are generalized by Exercise 25.

Exercise 24. Let A and B be sets. Prove that A ∩ B ⊃ A if and only if

A ⊂ B. Conclude that A ∩ B = A if and only if A ⊂ B.

Exercise 25. Let A and B be sets. Prove that A ∪ B ⊂ B if and only if

A ⊂ B. Conclude that A ∪ B = B if and only if A ⊂ B.

The following shows that the equality of subsets is truly exceptional.

Exercise 26. Let A and B be sets. Prove that A ∩ B = A ∪ B if and only

if A = B.

In the following exercises, you will prove the commutative, associa-tive, and distributive properties of intersections and unions.

Exercise 27. Prove the commutative properties for intersections and unions.That is, for sets A and B show that

A ∩ B = B ∩ A and A ∪ B = B ∪ A.

A B


Exercise 28. Prove the associative properties for intersections and unions.That is, for sets A, B, and C show that

(A ∩ B) ∩ C = A ∩ (B ∩ C) and (A ∪ B) ∪ C = A ∪ (B ∪ C).

Exercise 29. Prove the distributive properties. That is, for sets A, B, andC show that

A∩(B∪C) = (A∩ B)∪(A∩C) and A∪(B∩C) = (A∪ B)∩(A∪C).

Notice that, by Exercises 27 and 29, we also have the following dis-tributive properties:

(A∪ B)∩C = (A∩C)∪(B∩C) and (A∩ B)∪C = (A∪C)∩(B∪C).

Some people use terms like “left” or “right” distributive properties; we donot use these terms since we can never remember which is which!

Let us now return to the definitions of intersection and union, this timefor a collection of sets. The intersection of a collection of sets should be theset consisting of the common elements of all of the sets in the collection.

Definition 30. Let A be a collection of sets. The intersection of the sets inA is the set

{ a | a ∈ A for all A ∈ A } .

Some authors simply say the intersection of A. The notation is usually

⋂A∈A

A or⋂

A

Recall Definition 18 of a disjoint pair of sets. Before generalizing thatconcept to collections of sets, consider Figure 31. Figure 31 shows twodifferent collections, A and B, of subsets of R2 which exhibit some kindof “disjointness” property; in both A and B, the circular and triangularregions are disjoint as in Definition 18.

Figure 31: Disjointness properties of collections of sets

0 11/21/31/4x


Definition 32. Let A be a collection of sets. The sets in A are pairwisedisjoint iff A1 and A2 are disjoint for all A1, A2 ∈ A such that A1 �= A2.

Definition 33. Let A be a collection of sets. The sets in A are disjoint iff⋂A = ∅.

The above definition of a disjoint collection is not universally usedsince pairwise disjointness is usually desired. Hence, some authors usedisjoint to mean pairwise disjoint. Consider Figure 31 again; which collec-tion(s) is (are) pairwise disjoint and which collection(s) is (are) disjoint?

Often a collection of sets is indexed by some parameter. For example,the following is such a collection of intervals:

{(0, 1),

(0, 1

2

),(0, 1

3

), . . .

} = { (0, 1

n

) ∣∣ n ∈ N}.

This family is indexed by N. For operations such as intersections of indexedsets the following notations are equivalent:

∞⋂n=1

(0, 1

n

) = ⋂n∈N

(0, 1

n

) = (0, 1) ∩ (0, 1

2

) ∩ (0, 1

3

) ∩ . . . .

Example 34. Consider Figure 35, showing the nested intervals (0, 1),(0, 1

2

),(0, 1

3

), . . . .

Figure 35: Nested intervals with empty intersection

∞⋂n=1

(0, 1

n

) = ∅ since 1/n can be made arbitrarily small by choosing

a sufficiently large n. To be more precise, start by assuming that there

exists x ∈∞⋂

n=1

(0, 1

n

). Hence, x > 0 and, for all n ∈ N, x < 1/n. But

this is clearly false since there are certainly natural numbers greater than1/x . This “obvious” statement is called the Archimedean Principle; wewill study it more carefully in Part III.

Exercise 36. Find∞⋂

n=1

[0, 1

n

)and

∞⋂n=2

[1n , 1

).


Exercise 37. Let A be a collection of sets containing at least two sets.(a) Prove that if A is a pairwise disjoint collection, then A is a disjointcollection.(b) Give a counterexample for the converse of the statement in (a).

The definition of the union of a collection of sets is similar to that ofthe intersection. The union of the sets in a collection is the set consistingof all of the elements of all of the sets in the collection.

Definition 38. Let A be a collection of sets. The union of the sets in A isthe set

{ a | there exists A ∈ A such that a ∈ A } .

In analogy with intersections, some authors simply say the union ofA. The notation is usually

⋃A∈A

A or⋃

A.

We can rephrase this definition to say that something is an element of theunion of A if and only if there is some set in A that contains it.

Exercise 39. Find∞⋃

n=1

(0, 1

n

)and

∞⋃n=2

[1n , 1

).

The distributive properties can be generalized for collections of sets.

Exercise 40. Let A be a collection of sets and let B be a set. Prove:

B ∩( ⋃

A∈A

A

)= ⋃

A∈A

(B ∩ A) and B ∪( ⋂

A∈A

A

)= ⋂

A∈A

(B ∪ A).

What happened to the associative and commutative properties? Youcan think of these properties as coming for free in the definitions of inter-section and union for collections.

Remark 41. Let us consider the definitions of intersection and union forsmall collections A. For example, suppose A = { A, B }. In this case, weare back to our first definition of intersection and union:

∩A = A ∩ B and ∪ A = A ∪ B.

Suppose we have a collection A = { A } of only one set. Then

∩A = A and ∪ A = A.

Section 4. Differences and Complements 65

The situations for collections with two sets and with one set should be clearto you. Now consider what happens when A = ∅. Since there does notexist a set in the empty collection, no element can be contained in a set inthe empty collection A. Therefore,

∪A = ∅.

What about ∩A? Which elements are contained in every set of the emptycollection A? Every element satisfies this condition vacuously! That is, thestatement if A ∈ A, then x ∈ A is true for all elements x . So what is thisset of all possible elements? If we assume that every set we are consideringis a subset of some universal set U , then

∩A = U.

The necessity of universal sets comes up often, e.g., Definition 49 of thecomplement of a set; it is also useful for avoiding paradoxes such as wedscuss in Section 6. Also, you may have wondered whence the elementsin Definitions 30 and 38 came. 4. Differences and Complements.

You have probably seen the operation of taking complements of sets.Before defining this concept we will first look at the operation of set dif-ference.

Definition 42. Let A and B be two sets. The set difference A − B is theset of all elements of A that are not elements of B.

That is,A − B = { x | x ∈ A and x /∈ B } .

Set difference is characterized by the following property.

Exercise 43. For all sets A and B, prove that A − B is the largest subset ofA that is disjoint from B. (Being the largest such set means that if C ⊂ Aand C ∩ B = ∅, then C ⊂ A − B.)

Set difference is not a commutative operation.

Exercise 44. (a) Give a counterexample to A − B = B − A.

(b) Give an example where A − B = B − A.

The following exercise relates intersection, union, and difference.

Exercise 45. Prove that A−B and A∩B are disjoint and (A−B)∪(A∩B) =A for all sets A and B.


Set difference is not an associative operation either.

Exercise 46. Give a counterexample to A − (B − C) = (A − B) − C .

The following exercise concerns properties which are referred to asDe Morgan’s Laws.

Exercise 47. Let A, B, and C be sets. Prove that

A−(B∩C) = (A−B)∪(A−C) and A−(B∪C) = (A−B)∩(A−C).

This can be generalized to collections of sets; the following are alsocalled De Morgan’s Laws.

Exercise 48. Let A be a set and let B be a collection of sets. Prove that

A −( ⋂

B∈B

B

)= ⋃

B∈B

(A − B) and A −( ⋃

B∈B

B

)= ⋂

B∈B

(A − B).

So what is the complement of a set? This is not so obvious. We relyon the existence of a universal set, of which all sets are subsets.

Definition 49. Suppose that U is the universal set. The complement of Ais U − A.

The complement is denoted differently by many authors; A′, Ac, �Aand, of course, U − A are all commonly used to denote the complement.We will usually use U − A and occasionally use A′ when U is clear fromthe context.

Exercise 50. Suppose that A and B are subsets of a universal set U . Provethat

A − B = A ∩ B ′.


(B ′)′ = B and A′ − B ′ = B − A.


(A − B) ∪ (B − A) = (A ∩ B)′ − (A ∪ B)′.

The set (A−B)∪(B− A) is sometimes called the symmetric differenceof A and B.

Section 5. Power Sets 67

5. Power Sets.A construction with which you may not be familiar is the power set

of a set. It is an elementary construction and it is useful in more advancedtopics in mathematics. We will use it again when we discuss different typesof infinities in Chapter 7.

Definition 53. Let X be a set. The power set of X , denoted P(X), is theset of all subsets of X .

For example,P({ 1 }) = { ∅, { 1 } }

andP({ 1, 2 }) = { ∅, {1}, {2}, { 1, 2 } }

.

Of course, since ∅ and X are always subsets of a set X , the power set of Xalways contains the sets ∅ and X .

Exercise 54. What is P(∅)? Prove your result.

We have noted that

P(Z1) = P({ 1 }) = { ∅, { 1 } }

has two elements and that

P(Z2) = P({ 1, 2 }) = { ∅, {1}, {2}, { 1, 2 } }

has four elements. This can be generalized.

Exercise 55. How many elements does P(Zn) have? Prove your result.

Exercise 55 can be proved by induction and by a direct counting ar-gument. Try to give both proofs.

Exercise 56. Prove: if S ⊂ X , then P(S) ⊂ P(X).

It follows from Exercise 56 that if S = X , then P(S) = P(X).

Exercise 57. Let A and B be sets. Prove, or disprove by a counterexample,each of the following:

P(A) ∩ P(B) ⊂ P(A ∩ B) and P(A) ∩ P(B) ⊃ P(A ∩ B).



P(A) ∪ P(B) ⊂ P(A ∪ B) and P(A) ∪ P(B) ⊃ P(A ∪ B).


P(A) − P(B) ⊂ P(A − B) and P(A) − P(B) ⊃ P(A − B).

6. Russell’s Paradox.We have mentioned the term “collections” of sets in this chapter. We

should be careful here. If we were to develop the theory of sets axiomati-cally, we would need to define this term. Alternatively, we might want toreplace the term “collection” with “set.” The use of sets of sets (or collec-tions of sets) has to be dealt with very carefully, however. To see that thisis so, let us consider the following paradox, which is usually attributed toBertrand Russell.

Let us consider the set of all sets. Some sets may contain themselvesas elements while others may not. Think about this for a moment. Mostsets do not contain themselves. For example, the set of natural numbersdoes not contain itself; N is not an element of N since every element of Nis a number and not a set of numbers.* For another example, consider theset of all sets. This set must contain itself. There is no paradox here, yet.

Now consider the set R of all sets which do not contain themselvesas elements. Now we ask the question, is R contained in R? If R is anelement of R, then, by definition of the set R, R cannot contain R. Thisis a contradiction. In case you think we were unwise to start with theassumption that R is an element of R, consider the other case. If R is notan element of R, then, again by the definition of R, R must be an elementof R. Again this is a contradiction.

Much effort was spent by mathematicians in trying to remedy thissituation. The usual remedy is to use axioms which do not allow theexistence of the set of all sets. One axiom allows for the specification ofsubsets as we described in the introduction. For example, set difference isallowed by this axiom. Other axioms allow for unions, intersections, andpower sets. That is, such a formal system of axioms allows us to do whatwe have done in this chapter.

If this all strikes you as a bit abstract, consider the following amusingversion of Russell’s paradox. If the barber in a certain village shaves all

*You may ask what a number is; indeed Chapter 10 will blur the distinction betweennumbers and sets.

Section 6. Russell’s Paradox 69

those, and only those, who do not shave themselves, then who shaves thebarber?


Supplemental Exercises

Definition Review. There were 13 definitions in this chapter. Recall thatthe terms set, element, and is an element of are undefined. Let A and B besets and let C be a collection of sets; assume all of these sets are subsetsof a universal set U . Define each of the following and give an example ofeach:

(2) A is empty(3) A is a subset of B(9) A equals B

(14) the intersection of A and B(18) A and B are disjoint(19) the union of A and B(30) the intersection of C

(32) C is a pairwise disjoint collection(33) C is a disjoint collection(38) the union of C

(42) the difference of B from A(49) the complement of A(53) the power set of A

Exercise S1. List the elements of each of the following four sets.

A ={

n ∈ N

∣∣∣∣ 12

n∈ N

}

B ={

(x, y)

∣∣∣∣ x ∈ N, y ∈ N, x + y = 5

}

C ={

(x, y) ∈ R2

∣∣∣∣ |x + y| ≤ 0

}

D ={

x ∈ Z

∣∣∣∣ x + 2

x − 3< 0

}

Exercise S2. After Definition 2, we remarked that { {} } and { ∅ } are notempty. Verify this by finding an element of each. Are these two sets equalor not? Explain!

Exercise S3. Let A and B be sets. Can A ∩ B � A? Can A ∩ B � B?

Explain!

Exercise S4. Prove that{

x3 | x ∈ R} = R.


Exercise S5. Determine which of the following sets are equal and whichare proper subsets of which.

A ={

x ∈ R∣∣∣ √x2 = x

}

B ={

x ∈ R

∣∣∣∣ 1 + √x

1 − √x

∈ R}

C ={

x ∈ R

∣∣∣∣ x ≥ 0

}

D ={

x ∈ R

∣∣∣∣ x3

(x − 1)2

}

E ={

x2

∣∣∣∣ x ∈ R}

Exercise S6. Prove or disprove that if A �= B and B �= C , then A �= C .

Exercise S7. Prove that if A ⊂ B, B ⊂ C , C ⊂ D and D ⊂ A, then

A = B = C = D.

Exercise S8. Let

A ={

x ∈ R

∣∣∣∣ x = a

b+ b

a, for some a, b ∈ R∗

}

and B ={

y ∈ R

∣∣∣∣ y = sin t, for some t ∈ R}

.

Prove that A ∩ B = ∅.

Exercise S9. Find a condition such that A ∩ C = B ∩ C implies A = B.

Also, find an example where A ∩ C = B ∩ C , but A �= B.

Exercise S10. Let A = { 1, 2, 3, 4, 5, 6 } and B = { 2, 4, 6, 8 }. Find allthe subsets of A ∩ B. Also, find all the subsets of A ∪ B, containing exactlythree elements.

Exercise S11. Let A = { 4, 5 }. Suppose that

A ∪ B = { 1, 2, 3, 4, 5, 8, 9, 10 } and B ∪ C = { 4, 5, 6, 7, 8, 9, 10 }.

Can we determine uniquely the sets B and C? If not, what is the minimaladditional information needed, concerning unions or intersections of thesets, for these sets to be determined uniquely?


Exercise S12. Let

A = {n ∈ N

∣∣ n2 − 8n + 7 > 0},

B = { x ∈ N | x is a prime number }and C = {

n2 | n ∈ N}.

Determine the sets A ∩ B, (A ∩ B) ∪ (A ∩ C) and (A ∩ C) − (A ∪ B ∪ C).

Exercise S13. Find a condition such that A ∪ C = B ∪ C implies A = B.

Prove it! Also, find an example where A ∪ C = B ∪ C , but A �= B.

Exercise S14. Prove that A ∪ B = ∅ if and only if A = B = ∅.

Exercise S15. Prove that A ∪ (A ∩ B) = A

Exercise S16. Prove that if C ⊂ A, then (A ∩ B) ∪ C = A ∩ (B ∪ C). Is

the converse true?

Exercise S17. Prove or disprove that if C ⊂ B, then (A ∪ B) ∩ C =A ∪ (B ∩ C).

Exercise S18. Let A1, · · · , An be n sets. Prove that, for all 1 ≤ i, j ≤ n,

Ai ∩ Aj ⊃ A1 ∩ · · · ∪ An and Ai ∪ Aj ⊂ A1 ∪ · · · ∪ An .

Exercise S19. Consider the following three collections of sets. Which aredisjoint? Which are pairwise disjoint?

A = { {0, 1}, {1, 2}, {2, 3} }B = { (0, 1), (1, 2), (2, 3) }C = { [0, 2], [1, 3], [2, 4] }

Exercise S20. Let C be a collection of sets. Prove that if C contains exactly

two sets, then C is disjoint if and only if C is pairwise disjoint.

Exercise S21. Let C be a collection of sets. Prove that if S ∈ C, then

∩C ⊂ S ⊂ ∪C.

Exercise S22. Let C be a collection of sets. Prove that ∩C = ∪C if and

only if S1 = S2 for all S1, S2 ∈ C.

Exercise S23. Prove that A − (A − B) = A ∩ B.

Exercise S24. Prove that A ∩ (B − C) = (A ∩ B) − (A ∩ C).

Exercise S25. Prove that (A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B).


Exercise S26. If A − B = B − A, what can you conclude about the two

sets? Prove your answer. Also, answer the same question if A−B ⊂ B−A.

Exercise S27. Prove that if A ∪ B = A ∩ B, then A − B = ∅.

Exercise S28. Prove that (A − B) ∩ (C − A) = ∅.

Exercise S29. Prove that the following are equivalent conditions on sets Aand B:

(1) A − B = B − A;(2) A = B;(3) A − B = B − A = ∅.

Exercise S30. Prove that (A′)′ = A.

Exercise S31. If A = {x ∈ R

∣∣ x2 > x}, express A′ as an interval.

Exercise S32. Prove that B ⊂ (A ∩ B ′)′.

Exercise S33. Let A and B be sets such that P(A) = P(B). Prove that

A = B.

Exercise S34. Show that (A ∩ B ∩ C)′ = A′ ∪ B ′ ∪ C ′. Is it also true that

(A ∪ B ∪ C)′ = A′ ∩ B ′ ∩ C ′?

Exercise S35. Prove that if (A ∪ B)′ = A′ ∪ B ′, then A = B. Does

(A ∩ B)′ = A′ ∩ B ′ imply that A = B?

Exercise S36. Prove that (A′ ∪ B ′) ∪ (A′ ∪ B)′ = A.

Exercise S37. Can there exist a set whose power set has 4k + 1 elements,

where k ∈ N?

Exercise S38. Prove that (A − B) ∩ (B − A) = ∅ for all sets A and B.

Exercise S39. Prove that P(A − B) ⊂ (P(A) − P(B)) ∪ { ∅ }.Project. We have taken a relatively “naıve” approach to set theory. It ispossible to be much more “axiomatic.” Find a text on axiomatic set theoryand outline the development of the topics of this chapter in an axiomaticform, giving undefined terms, axioms, definitions, etc. Pay particular atten-tion to the axioms that allow new sets to be created from known sets, suchas by unions and intersections. Is a universal set needed in your axiomaticset theory? How is Russell’s Paradox avoided in your axiomatic set theory?

74

Chapter 5Functions

1. Functions as Rules.Much of mathematics concerns the study of functions; the notion of

“function” is perhaps the single most important concept in all of mathe-matics. Let us start with an informal definition of the term function.

Probably you have seen (but perhaps forgotten?) a definition of theword function which goes something like this: a function is a pair of setsand a rule which assigns a unique element of the second set to each elementof the first set. You may also recall that the first set is called the domainof the function and, again, you may recall (somewhat incorrectly!) thatthe second set is called the range of the function. Actually, the second setshould be called the codomain of the function. Probably your facility withfunctions is better than your recollection of the actual definition of the termfunction.

In the next section, we will define functions as certain types of sets.From a rigorous (logical, axiomatic) point of view, this is preferable. How-ever, you may (should?!) continue to think of a function as a rule thatassigns to each value of its domain a unique value in its range.

Before you continue reading, think for a moment about a function andits graph. If, for a graph, you are thinking of a pictoral representation, thenwhat does that picture represent? Is there a difference between that object,called the graph of a function, and the function itself?

2. Cartesian Products, Relations, and Functions.Before giving our definition of the word function, we will remind you

of the cartesian product of two sets. Often, cartesian products are simplycalled products.

Definition 1. Let A and B be sets. The cartesian product of A and B,denoted by A × B, is the set of all ordered pairs* (a, b) such that a ∈ Aand b ∈ B.

*While it is possible to define the term ordered pair by something like (a, b) ={ a, {a, b} }, you may accept this as undefined. Since the notation (a, b) also representsan open interval, some authors use the notation a × b for an ordered pair.

0-1 1-2 2

1

2

3

P(x,y)

-3

-2

-1

3-3x

y

Section 2. Cartesian Products, Relations, and Functions 75

Of course, you should have in mind the usual example and its model.R × R, which is usually denoted as R2, is the setting for most functionsyou have studied; the model of R2 is the plane which you usually draw byindicating the two coordinate axes. A point P in this plane corresponds toan ordered pair (x, y); the vertical line through P intersects the horizontalaxis at x while the horizontal line through P intersects the vertical axis aty. See Figure 2.

Figure 2: A point of the plane models an ordered pair in R2

It is possible to construct the products of other sets. Using the kind ofgeometric intuition that we used for R2, we may model various products.For example, R3 = R × R × R is modeled by (three dimensional) space.†

Similarly, N2 = N × N is modeled by a lattice of isolated points andR × {1, 2} and {1, 2} × R are modeled by pairs of lines. We draw picturesof these models as if drawing the sets themselves.

Exercise 3. Draw pictures of (the models of) R3, N2, R × {1, 2} and

{1, 2} × R.

We did not require non-empty sets in Definition 1; of course the productof empty sets is again empty. Empty products are not very important, butmay come up on occasion.

Exercise 4. Let A and B be sets. Prove that A × B = ∅ if and only if

A = ∅ or B = ∅.

†You may wonder if R×R×R means (R×R)×R or R×(R×R); it turns out that it doesnot matter however. The reason for this has to do with the concept of isomorphism, which wewill introduce in Chapter 8. The general idea is that every element ((x, y), z) ∈ (R × R) × Rcorresponds to exactly one element (x, (y, z)) ∈ R × (R × R); we can simply think of bothof these points as (x, y, z) ∈ R3. In fact, we will give a different definition of products ofcollections of sets in Definition 52 of Chapter 7.

0-1 1-2 2

1

2

3

-3

-2

-1

3-3x

y

0-1 1

1

-1

x

y

76 Chapter 5. Functions

Of course, the operation of taking products is not commutative. Forexample, if A = { 1 } and B = { 2 }, then A × B = { (1, 2) } while B × A ={ (2, 1) }. The following exercise relates to this.

Exercise 5. Let A and B be sets. Prove that A × B and B × A are disjoint

if and only if A and B are disjoint.

Before giving a definition of the term function, we define a simplerterm: relation.

Definition 6. Let A and B be sets. A relation from A to B is a subsetR ⊂ A × B. This relation is denoted as R(A, B) or, when A and B areunderstood, simply as R. If A = B, then R ⊂ A × A = A2 and we call Ra relation on A.

An ordered pair (a, b) in a relation R is usually denoted by a R bwhich is read “a is R-related to b” or, more simply, as “a is related to b.”That is, if we write a R b, we mean (a, b) ∈ R; similarly, a /R b will meanthat (a, b) /∈ R.

Figure 7: The circle x2 + y2 = 1 in R2 and < on R

Let us consider two examples of relations that you have seen before.They are represented by Figure 7.

Example 8. The equation x2 + y2 = 1 determines a subset of the productR2, that is, a relation on R.

Similarly, any equation in two variables determines a relation.

Example 9. We define a relation <(R, R), in the notation of the definition.This is the usual relation “less than” on the set of real numbers. We usethe notation x < y to mean that (x, y) is an element of this relation. Wewill refer to this as the relation < on R and never use the hideous notation<(R, R) again!


In Chapter 6, we will study relations on a set more closely.For the remainder of this chapter, we will consider functions. You

may already know that every function is a relation. Also, the converse isfalse: you may recall that the circle of Example 8 is not the equation of afunction. No doubt you would suspect that the relation of Example 9, asdepicted in Figure 7, is not a function either.

We are now ready to give a formal definition of the term function. Thisdefinition will be rather different from the informal definition in Section1. In fact, the object we will define is often called the graph of a functionrather than a function.

Definition 10. Let X and Y be sets. A function from X to Y is a relationf from X to Y with the property that for every x ∈ X there exists a uniquey ∈ Y such that the ordered pair (x, y) is in f . X is called the domain ofthe function and Y is called the codomain of the function.

That is, a function f from X to Y is a subset of X × Y such that forevery x ∈ X there exists a unique y ∈ Y such that (x, y) ∈ f . When wesketch the graph of a function, the domain is usually on the horizontal axiswhile the codomain is on the vertical axis.

Consider again the circle of Example 8; this is a relation on R. It failsto define a function from R to R in two ways. First, the domain is not Rsince x must be between −1 and 1, inclusive. Second, domain values donot, in general, correspond to unique codomain values; for example,

(35 , 4

5

)and

(35 , − 4

5

)are both on the circle. So, this circle is not a function with

domain [−1, 1] either.

Remark 11. Here are some important words about notation. A function, asin Definition 10, is denoted by f : X → Y . When X and Y are understood,we may simply denote the function by f . If (x, y) ∈ f , then we denote theelement y by f (x); that is, f (x) = y and (x, y) = (x, f (x)). Be careful,while most mathematicians are happy to call a function f , many frown atcalling a function f (x) since f (x) is an element of the codomain. Thenotation y = f (x) is better and is very commonly used, as in the functiony = x2 or f (x) = x2.

You may have noticed that we used a lower case f to represent a set;of course, there is no rule which requires the name of a set to be an uppercase Latin letter. Some authors substitute the words source and target fordomain and codomain, respectively. When specifying a function f withoutexplicitly giving the domain and codomain, we will use the notations D f

and C f for the domain and codomain, respectively.This formal definition certainly looks very different from the informal

definition, but it is not so very different at all. The domain and codomainare actually exactly the same in both the informal and the formal definitions.

f(x) = x2

1R R

-1

2.5

1

4

.25

0-1 1-2 2

1

2

3

x

f(x)


The difference is in the interpretation of f , which is either a rule or a set.The primary difference may be in the way one thinks of a function. Infact, we ignore this ambiguity and think of a function in whatever way isconvenient for us at the time; consider Figure 12.

Figure 12: A function thought of as a rule and as a subset of R2

So what happened to the range, which you may remember from earlierdays?

Definition 13. Let f : X → Y be a function and S ⊂ X . The image of Sunder f is the set

f (S) = {y ∈ Y | (x, y) ∈ f for some x ∈ S

}.

Definition 14. The range of a function f : X → Y is the set f (X), theimage of the domain.

Exercise 15. Let f : X → Y be a function and S ⊂ X . Prove that

f (S) = { f (x) | x ∈ S }.The following examples should help to clarify these matters.

Example 16. Consider again the function f (x) = x2. Figure 12 showsa picture of this function. What are the domain and codomain for thisfunction? Of course, the domain is R. By convention, when the domain isnot given explicitly, then the domain is the largest subset of real numbers(or whatever set fits the situation) for which the rule defining the functionmakes sense. The range of this function is [0, ∞). Since the codomainmust contain the range, any set containing the set [0, ∞) would suffice. Inthe case of real-valued functions, that is, functions whose range is a subsetof R, we use the convention that the codomain is R. That is, we can writef : R → R. Now what is f ? Since the domain of f is R and the codomainis R, the function f is a subset of the product R2 = R × R. We can writeit as

f = {(x, x2) | x ∈ R

}.


Example 17. Consider the familiar function f (x) = √x . The domain and

range of this function is [0, ∞) and the codomain of this function is R; sowe can write f : [0, ∞) → R. The range of this function is [0, ∞). f isthe following subset of [0, ∞) × R:

f = {(x,

√x) | x ∈ [0, ∞)

}.

Exercise 18. Draw a picture of f from the preceding example.

In general, if we know the rule defining a function f and its domainX , then we know that

f = {(x, f (x)) | x ∈ X

}.

Notice that we did not specify the codomain of the function f . Hence, wedo not know the product, in the definition of function, that is supposed tocontain the set f . However, this is generally good enough. In practice, wemay be given a formula defining a function where the domain and rangeare implied and the codomain may be inferred. (Perhaps this is why youremembered the range rather than the codomain at the beginning of thischapter?)

Exercise 19. Determine the domain, range, and function defined by the

rule y = xx2−1 . Draw a picture of f .

When are two functions equal? Since functions depend on the domainX , the codomain Y , and the set of ordered pairs in X × Y , it makes senseto say two functions are equal if their domains, codomains, and sets ofordered pairs are equal. The equality of the sets of ordered pairs impliesthe equality of the domains. [You should check this!] We therefore givethe following definition.

Definition 20. The functions f : X → Y and g : U → V are equal ifY = V and f = g as sets of ordered pairs.

In practice, to check that two functions f and g are equal, we checkthat f (x) = g(x) for all x . You should be asking yourself “For all x inwhat set?” In fact, one must first check that the domains of f and g areequal and then that f (x) = g(x) for all x ∈ D f = Dg; that is, we mustcheck that X = U in the notation of Definition 20.


Example 21. Let f : R → R be defined by f (x) = x and let g : R+ → Rbe defined by g(x) = x . Of course, these are very different functions; theyare not equal. However, note that f (x) = g(x) for all x ∈ R+ = Dg . Onthe other hand, f (−1) = −1 while g(−1) is not defined since −1 /∈ Dg .Another way of saying this is that (−1, −1) ∈ f but (−1, −1) /∈ g.

In Example 21, the functions f and g are very similar. You can thinkof g as being f with a restricted domain. This idea is important enough tomerit a name and some notation.

Definition 22. Let f : X → Y be a function and let S ⊂ X . The functiong : S → Y defined by g(x) = f (x) for all x ∈ S is called the restrictionof f to S and is denoted f |S .

We continue with exercises and an example about the image of setsunder a function.

Exercise 23. Let f be a function and let A ⊂ B ⊂ D f . Prove that

f (A) ⊂ f (B).

Exercise 24. Let f be a function and let A and B be subsets of D f . Prove

that f (A ∪ B) = f (A) ∪ f (B).

The situation for intersections is more complicated.

Example 25. Let f : R → R be defined by f (x) = |x | and consider theintervals A = (−2, −1) and B = (1, 2). Since A ∩ B = ∅, f (A ∩ B) = ∅.Also f (A) ∩ f (B) = (1, 2). Therefore, f (A ∩ B) �= f (A) ∩ f (B).

Another example where f (A ∩ B) �= f (A) ∩ f (B) is f : R → Rdefined by f (x) = x2 with A = { −1 } and B = { 1 }. Then A ∩ B = ∅,but f (A) = f (B) = { 1 }.

In both cases, f (A ∩ B) � f (A) ∩ f (B). However, containment in one direction always holds for intersections.

Exercise 26. Let f be a function and let A and B be subsets of D f . Prove

that f (A ∩ B) ⊂ f (A) ∩ f (B).

3. Injective, Surjective, and Bijective Functions.In this section we will look at some properties of functions which you

may find familiar.

Definition 27. A function f : X → Y is one-to-one if and only if x1 andx2 are distinct elements in X implies f (x1) and f (x2) are distinct elementsin Y .

Section 3. Injective, Surjective, and Bijective Functions 81

That is, f : X → Y is one-to-one if, for all x1, x2 ∈ X , x1 �= x2

implies f (x1) �= f (x2). For a one-to-one function we often write that itis “1-1.” A one-to-one function is also called an injective function or aninjection.

The following exercise gives us an equivalent definition of one-to-one.

Exercise 28. Prove that a function f : X → Y is one-to-one if and only

if, for all x1 and x2 in X , f (x1) = f (x2) implies x1 = x2.

One-to-one functions are better behaved than generic functions. Com-pare Exercises 24, 26, and the following exercise.

Exercise 29. Let f be a function. Prove that f is one-to-one if and only if

f (A ∩ B) = f (A) ∩ f (B) for all subsets A and B of D f .

Definition 30. A function f : X → Y is onto if and only if f (X) = Y ,that is, the range of f is the codomain of f .

An onto function is also called a surjective function or a surjection.The following exercise gives us an equivalent definition of onto.

Exercise 31. Prove that a function f : X → Y is onto if and only if, for

each y ∈ Y , there exists x ∈ X such that f (x) = y.

Definition 32. A function f : X → Y is bijective if and only if it isone-to-one and onto.

Bijective functions are also called one-to-one correspondences. Theauthors generally prefer bijective function or bijection to one-to-one cor-respondence; however, we prefer the latter in connection with cardinality,the topic of Chapter 7.

The following will remind you of the tenuous role of the codomain ascompared with the range.

Exercise 33. Let f : X → Y be a function.(a) Prove that there exists a surjective function from X onto f (X).(b) Prove that if f is one-to-one, then there exists a bijective function fromX onto f (X).

The following exercise gives us an equivalent definition of bijective.

Exercise 34. Prove that a function f : X → Y is bijective if and only if,

for each y ∈ Y , there exists a unique x ∈ X such that f (x) = y.


It is possible for a function to be bijective or to have exactly oneof the properties of one-to-one or onto or to have neither property. Forexample, f (x) = tan x , with the conventional codomain of R is onto, butnot one-to-one since tan 0 = tan π .

Exercise 35. Find three examples of functions with codomain R such thatthe first is bijective, the second is one-to-one but not onto, and the third isneither one-to-one nor onto.

4. Compositions of Functions.An important operation we perform on functions is that of composi-

tion. The most common definition that mathematicians give for the com-position of functions requires that the codomain of the first function equalsthe domain of the second function as in

f : X → Y and g : Y → Z .

Since the range of a function is more essential than the codomain of thatfunction, the situation for composition can be relaxed to allow the range ofthe first function to be a subset of the domain of the second function. Youwill recall defining the function g ◦ f : X → Z by

(g ◦ f )(x) = g( f (x)).

Below, we define the composition of two functions as a subset of a cartesianproduct. A more general definition, which does not require the range of thefirst function to be contained in the domain of the second function, is givenin Section 6; it evolves from the definition of composition of relations.

Definition 36. Let f : X → Y and g : W → Z be functions withf (X) ⊂ W . The composition of f followed by g is the function defined by

g ◦ f = {(x, z) ∈ X × Z | there exists y ∈ f (X)

such that f (x) = y and g(y) = z}.

The domain of g ◦ f is X , the domain of f , and the codomain of g ◦ f isZ , the codomain of g.

Definition 36 is actually quite reasonable if you consider things. Theidea is that f sends an x ∈ X to some y ∈ Y . In order to be able to apply g,we must have y ∈ W also; since f (X) ⊂ W this condition is automaticallysatisfied. Finally, g sends y to some z ∈ Z . Notice that if Y ⊂ W (forexample, when Y = W ), then f (X) ⊂ W .

Section 4. Compositions of Functions 83

Exercise 37. For f : X → Y and g : W → Z with f (X) ⊂ W , verifythat

g ◦ f = { (x, g( f (x))) | x ∈ X } .

You may already have noticed that given two functions f and g, thereare often two composite functions: f ◦ g and g ◦ f . In general, f ◦ g �=g ◦ f , as the following examples demonstrate. This is true even when bothcomposite functions exist.

Example 38. Define f : R → R and g : R → R by f (x) = 2x andg(x) = 2x + 1 for all x ∈ R. Then f ◦ g and g ◦ f are both functions fromR to R. However,

( f ◦ g)(x) = f (g(x)) = f (2x + 1) = 2(2x + 1) = 4x + 2

while

(g ◦ f )(x) = g( f (x)) = g(2x) = 2(2x) + 1 = 4x + 1.

So f ◦ g �= g ◦ f . Therefore, composition is not a commutative operation.

The following example is much more peculiar, yet more typical.

Example 39. Let f (x) = 2 + √x and g(x) = √

1 − x . Then, f (g(x)) =2 +

√√1 − x = 2 + 4

√1 − x and hence

f ◦ g ={

(x, 2 + 4√

1 − x) | x ∈ (−∞, 1]}

.

Note that the range of g equals the domain of f : the interval [0, ∞). Thisis consistent with Definition 36.

On the other hand, g( f (x)) =√

1 − (2 + √x) =

√−1 − √

x is notdefined for any real number x . Hence, g ◦ f does not exist. Another wayof saying this is that g ◦ f does not exist since the domain of g, whichequals (−∞, 1], does not contain the range of f , which equals [2, +∞).In fact, the domain of g and the range of f are disjoint sets. So, certainlyf ◦ g �= g ◦ f .

Examples 38 and 39 show that composition depends on the orderof the functions, that is, composition is not commutative. The secondexample also shows that one of the compositions may be more interestingthan the other. In fact, usually we will have the following situation: givenf : X → Y and g : W → Z , if Y ⊂ W (or Y = W ), then we have thecomposite g ◦ f : X → Z .


Given the noncommutative nature of compositions, you may be sur-prised to learn that composition is associative.

Exercise 40. Show that composition of functions is associative.

Compositions of one-to-one and onto functions are well-behaved.

Exercise 41. Prove that the composition of two one-to-one functions is

again one-to-one.

Exercise 42. Prove that if f : X → Y and g : Y → Z are onto functions,

then the composite function g ◦ f is onto.

Recall that in Definition 36, g ◦ f was defined whenever the range off was a subset of the domain of g. In Exercise 42, we strengthened thisrequirement to equality: Y = R f = Dg . The following example showsthat this stronger requirement is necessary.

Example 43. Define f : R+ → R+ and g : R → R be defined byf (x) = x and g(x) = x for all x in their respective domains. Bothfunctions are onto. By Definition 36, g ◦ f has domain R+ and codomainR; g ◦ f is not onto.

From Exercises 41 and 42 we see that the composite g ◦ f of twobijections f : X → Y and g : Y → Z is a bijection.

The converses of Exercises 41 and 42 are interesting and are consideredin Exercises 44 and 45. Let us consider one-to-one composite functionsfirst.

Exercise 44. Suppose that g ◦ f is the composition of f : X → Y andg : Y → Z . Suppose that g ◦ f is one-to-one.(a) Prove that f is one-to-one.(b) Find a counterexample to show that g need not be one-to-one.(c) Prove that if f is onto, then f is bijective and g is one-to-one.

Now let us consider onto composite functions.

Exercise 45. Suppose that g ◦ f is the composition of f : X → Y andg : Y → Z . Suppose that g ◦ f is onto.(a) Prove that g is onto.(b) Find a counterexample to show that f need not be onto.(c) Prove that if g is one-to-one, then g is bijective and f is onto.

Section 5. Inverse Functions and Inverse Images of Functions 85

5. Inverse Functions and Inverse Images of Functions.The study of inverses is common in mathematics. It requires identities

and composites.

Definition 46. A function f is the identity function on X iff D f = C f = Xand

f = { (x, x) | x ∈ X } ⊂ X × X.

That is, f : X → X , defined by f (x) = x for all x ∈ X , is the identityfunction on X . We use IX to denote the identity function on X .

Exercise 47. Prove that IX is the unique function from X to X whichsatisfies f ◦ IX = f and IX ◦ g = g for all functions f : X → Y andg : W → X . (That is, show that if i : X → X satisfies f ◦ i = f andi ◦ g = g for all functions f : X → Y and g : W → X , then i = IX .)

Definition 48. Let f and g be functions. f and g are inverse functions ifff ◦ g = IDg and g ◦ f = ID f . We also say that g is an inverse of f . Afunction f is invertible iff an inverse function g exists.

Notice that this definition implies a relationship between the domainsand codomains of the inverse functions, as given in the following exercise.

Exercise 49. Prove that if f is an invertible function and g is an inverse of

f , then D f = Cg and Dg = C f .

It is important to include both f ◦ g = IDg and g ◦ f = ID f in thedefinition of inverse functions, as Example 50 will show.

Example 50. Let f : R+ → R be defined by f (x) = x and let g : R →R+ be defined by g(x) = |x |. Then (g ◦ f )(x) = g(x) = x for all x ∈ R+;that is, g ◦ f = IR+ . However, ( f ◦ g)(x) = |x | for all x ∈ R, which is notIR. You may notice that f is one-to-one, but not onto, and that g is onto,but not one-to-one.

Exercise 51. Prove that a function is invertible if and only if it is bijective.

If it sounds strange to you to read “an inverse” rather than “the inverse”in the definition of inverse functions, the following exercise will help. Ofcourse, we will always write “the inverse” after the following exercise.

Exercise 52. Prove that every invertible function has a unique inverse

function.


Since inverse functions are unique, when they exist, we have a dedi-cated notation for them: the inverse of f is denoted by f −1.

When f is invertible, we have an inverse function f −1 and it makessense to consider images of the inverse function such as f −1(S). We callthis the inverse image of S. This concept can be generalized to noninvertiblefunctions.

Definition 53. Let f : X → Y be a function and T ⊂ Y . The inverseimage of T under f is

f −1(T ) = { x ∈ X | (x, y) ∈ f for some y ∈ T } .

Remark 54. Be very careful here! The use of the superscript on f −1(T )

should not make you think the function is invertible. Also, when T is asingleton set { y }, we write f −1(y) instead of f −1({ y }).

The use of inverse images gives us a nice way of describing one-to-oneand onto functions.

Exercise 55. Let f : X → Y be a function. Prove the following:(a) f is one-to-one if and only if f −1(y) contains at most one element forevery y ∈ Y .(b) f is onto if and only if f −1(y) contains at least one element for everyy ∈ Y .(c) f is bijective if and only if f −1(y) contains exactly one element forevery y ∈ Y .

Inverse images behave nicely with respect to set operations.

Exercise 56. Let f : X → Y be a function with A ⊂ Y and B ⊂ Y . Provethe following:(a) f −1(A ∩ B) = f −1(A) ∩ f −1(B),(b) f −1(A ∪ B) = f −1(A) ∪ f −1(B),(c) f −1(A − B) = f −1(A) − f −1(B).

Forward and inverse images can be combined. In general, the new setis different from the original.

Exercise 57. Let f : X → Y be a function with S ⊂ X .(a) Prove that f −1( f (S)) ⊃ S.(b) Give a counterexample to f −1( f (S)) = S.(c) Prove that if f is one-to-one, then f −1( f (S)) = S.(d) Is the converse of the statement in (c) true? Prove it or find a counter-example.

Section 6. Another Approach to Compositions 87

Exercise 58. Let f : X → Y be a function with T ⊂ Y .(a) Prove that f ( f −1(T )) ⊂ T .(b) Give a counterexample to f ( f −1(T )) = T .(c) Prove that if f is onto, then f ( f −1(T )) = T .(d) Is the converse of the statement in (c) true? Prove it or find a counter-example.

Compare parts (d) of Exercises 57 and 58 with Supplemental ExerciseS17.

6. Another Approach to Compositions.In Section 4, we looked at compositions of functions where the range

of the first function was contained in the domain of the second function.In this section, we will remove this requirement. The moral of this sectionis that the easier (i.e., more general) the definitions are, the harder (or lessgeneral) the theorems are. We consider first the composition of relations.

Definition 59. Let f be a relation from X to Y and let g be a relation fromW to Z . The composition of f followed by g is the relation from X to Zdefined by

g ◦ f = {(x, z) ∈ X × Z | there exists y ∈ Y ∩ W


It is more common to compose functions than relations. However, wewill use this idea to generalize composition of functions.

Definition 60. Let f : X → Y and g : W → Z be functions. Thecomposition of f followed by g is the function defined by

g ◦ f = {(x, z) ∈ X × Z | there exists y ∈ Y ∩ W


The codomain of g ◦ f is Z , the codomain of g. The domain of g ◦ f isthe largest subset of X for which it makes sense.

Definition 60 differs from Definition 36 in two ways: first, the require-ment that f (X) ⊂ W was removed and, second, the domain is a subset ofX , but not necessarily all of X . In fact, the domain may turn out to be theempty subset of X ! You could complain that the phrase “the largest subsetof X for which it makes sense” is not very precise. An example shouldhelp.


Example 61. Let us consider Example 39 again. Let f (x) = 2 + √x and

g(x) = √1 − x . Since (g ◦ f )(x) =

√1 − (2 + √

x) =√

−1 − √x is

not defined for any real number x , the domain of g ◦ f is empty. Hence,g ◦ f = ∅! This is OK; the definition of a function does not require anonempty domain and ∅ × Z = ∅ for every set Z .

The following is similar.

Exercise 62. Suppose that w, x , y, and z are distinct; in particular, y �= z.Let X = { x }, Y = { y }, W = { w }, and Z = { z }. Define f : X → Y andg : W → Z by f (x) = y and g(w) = z. Describe g ◦ f and f ◦ g. Do fand g commute under composition? (That is, does g ◦ f equal f ◦ g?)

In general, what is the domain of the composite function g ◦ f ? Thecomposition of f followed by g may easily be rewritten as

g ◦ f = {(x, g( f (x)))

∣∣ x ∈ D f and f (x) ∈ Dg}.

[You should check this!] This tells us that the domain is the set

Dg◦ f = {x ∈ D f

∣∣ f (x) ∈ Dg}.

We have already seen (Exercise 38, Exercise 39, Example 61, andExercise 62) that composition is not commutative. Composition is stillassociative in this setting, but the proof requires you to consider the domainscarefully. Here we have an example where our more general definitionmakes a theorem more difficult to prove.

Exercise 63. Show that composition of functions is associative.

Next, consider the relationship of compositions with the properties ofbeing one-to-one and onto. The following exercise gives us examples ofhow generalizing a definition can turn a true theorem into a false statement.

Exercise 64. For each of the following, give two functions f and g satis-fying the conditions:(a) f and g are onto, but the composite function g ◦ f is not onto.(b) neither f nor g is one-to-one, but the composite function g ◦ f is one-to-one.(c) f is a bijection, g is not one-to-one, but the composite function g ◦ fis one-to-one.(d) g is a bijection, f is not onto, but the composite function g ◦ f is onto.

Recall that the definition of inverses is dependent on composition offunctions. You may wish to return to the exercises in Sections 4 and 5, thistime using Definition 60 for the definition of composition of functions.


Supplemental ExercisesDefinition Review. There were 16 definitions in this chapter. Let A, B,X , Y , S, and T be sets with S ⊂ X and T ⊂ Y . Let f : X → Y andg : W → Z be functions (or relations). Assume all of these sets are subsetsof a universal set U . Define each of the following and give an example ofeach:

(1) the cartesian product of A and B(6) a relation from A to B(6) a relation on A

(10) a function f from X to Y(10) the domain of a function f(10) the codomain of a function f(13) the image of S under f(14) the range of a function f(20) functions f and g are equal(22) the restriction of f to S(27) f is one-to-one (or injective)(27) f is an injection(30) f is onto (or surjective)(30) f is a surjection(32) f is bijective (or a one-to-one correspondence)(32) f is a bijection(36) composition of a function f followed by a function g ( f (X) ⊂ W )(36) the domain of g ◦ f ( f (X) ⊂ W )(36) the codomain of g ◦ f ( f (X) ⊂ W )(46) f is the identity function on X(48) f and g are inverse functions(48) f is an invertible function(53) the inverse image of T under f(59) composition of a relation f followed by a relation g(60) composition of a function f followed by a function g(60) the domain of g ◦ f(60) the codomain of g ◦ f

Exercise S1. Find the range of the following functions

(1) f (x) = x2

x2+x−2

(2) g(x) = |x3 + x + 1|(3) h(x) = 2 sin x2

x2+1 .


Exercise S2. How many 1 − 1 functions can be defined from {1, 2, 3} to

{1, 2, 3, 4}?Exercise S3. Determine which of the following functions is 1 − 1

(1) f (x) = 3x+12x−3

(2) g(x) =√

1 + 1x

(3) h(x) = |2x + 1|(4) t (x) = √−x6

Exercise S4. Determine which of the following functions is a bijectionfrom R to R

(1) f (x) = ax + b(2) g(x) = |x |(3) h(x) = x3 + 1

Exercise S5. Let f : R → R be a function, with f (x) = 2x + 1. Prove

that f (N) ⊂ f (R). Find a function g : R → R, such that g(N) = g(R).

Exercise S6. If ( f ◦ g)(x) = x2 + 3 and g(x) = 2x − 1, find the function

f .

Exercise S7. If ( f ◦ g)(x) = 3x + 2 and f (x) = 4x − 1, find the function

g.

Exercise S8. Determine the values of a and b, for which the function

f (x) = ax + b is invertible and find its inverse.

Exercise S9. Determine the values of a, b, and c, for which the function

f (x) = ax2 + bx + c is invertible and find its inverse.

Exercise S10. Find a restriction f |S of the function f (x) = 2x2 − 5x + 1which is invertible. Prove or disprove that for each function f , there existsa restriction of f which is invertible.

Exercise S11. Let f (x) = ax + b and g(x) = 2x − 3. Determine a and

b, such that f ◦ g = g ◦ f .

Exercise S12. Let f (x) = 5x − 4. Find f −1([2, 3]) and f −1(R).

Exercise S13. Let f : R − {−2} → R be a function, with f (x) = 2x−7x+2 .

Prove that f is not onto. Restrict the codomain so that f is onto, prove thatwith the restricted codomain is invetrible and find its inverse.

Exercise S14. Let f : A → A and g : A → A be invertible functions.

Show that ( f ◦ g)−1 = g−1 ◦ f −1.


Exercise S15. Let f (x) ={

x, x ∈ Q

x + 1, x /∈ Q. Prove that f is invertible

and find its inverse.

Exercise S16. Let f (x) ={

x2, x ∈ Q

x + 1, x /∈ Q. Is f invertible?

Exercise S17. Let f : X → Y be a function.(a) Prove that if f −1( f (S)) = S for every subset S of X , then f is one-to-one.(b) Prove that if f ( f −1(T )) = T for every subset T of Y , then f is onto.(Compare with Exercises 57 and 58.)

92

Chapter 6

Relations on a Set

1. Properties of Relations.As we saw in Chapter 5, a function is a specific kind of relation. Recall

that a relation from a set A to a set B is a subset R ⊂ A × B and if A = B,we call R a relation on A. You may have noticed that the examples ofrelations we looked at in Chapter 5 are relations on a single set. In fact,we are more interested in relations on a set and consider various propertiesthat a relation may or may not have.

Definition 1. Let R be a relation on a set S.(a) R is reflexive iff s R s for all s ∈ S.(b) R is nonreflexive iff s /R s for all s ∈ S.(c) R is symmetric iff s R t implies t R s for all s, t ∈ S.(d) R is asymmetric iff s R t implies t /R s for all s, t ∈ S.(e) R is antisymmetric iff s R t and t R s implies s = t for all s, t ∈ S.(f) R is transitive iff s R t and t R u implies s R u for all s, t, u ∈ S.(g) R is connected iff s R t or t R s or s = t for all s, t ∈ S.(h) R is trichotomous iff exactly one of s R t , t R s, s = t is true for alls, t ∈ S.

In the definition of connected, we say that s and t are comparable iffs R t or t R s or s = t . There are many relationships between the propertiesin Definition 1; the most obvious is that if a relation is trichotomous, thenit must be connected.

Example 2. Let us consider the relation in Example 8, given by x2 + y2 =1. It is not reflexive since, for example, 1 is not related to 1, that is,12 +12 = 2 �= 1. It is not nonreflexive either since ( 1√

2)2 +( 1√

2)2 = 1. It is

symmetric by the commutative property of addition. It is not antisymmetricsince 02 +12 = 12 +02 = 1, but 0 �= 1. It is not transitive since 1 is relatedto 0 which is related to 1 but 1 is not related to 1. It is neither connectednor trichotomous since 1 and 1

2 satisfy none of the conditions.

Section 2. Order Relations 93

Exercise 3. Consider whether the relation < on R satisfies or fails to satisfy

each of the eight properties of relations given above. Prove your answers.

Exercise 4. Consider whether the relation ⊂ on P(U ) for some nonemptyU satisfies or fails to satisfy each of the seven properties of relations givenabove. Prove your answers.

Exercise 5. Prove that a relation on a nonempty set cannot be both reflexive

and nonreflexive but can be both symmetric and antisymmetric.

Exercise 6. Give an example of a relation on a nonempty set which is

connected but not trichotomous.

Exercise 7. Give examples of relations which have any combination of the

properties given above.

2. Order Relations.One natural kind of relation is of the type which puts an order on

the set. The relations < on R and ⊂ on P(U ) (which were considered inExercises 3 and 4) are examples of such order relations. We start with twodefinitions that model these two examples, but are more general.

Definition 8. A relation ≺ on a set S which is nonreflexive and transitiveis called a strict partial ordering. A strictly partially ordered set is a set Stogether with a strict partial ordering ≺ on S.

Definition 9. A relation " on a set S which is reflexive, antisymmetric andtransitive is called a partial ordering. A partially ordered set is a set Stogether with a partial ordering " on S.

Using the highlighted characters, a partially ordered set is sometimescalled a poset.

Exercise 10. Consider = as a relation on R. Does it define a strict partial

ordering? Does it define a partial ordering? Prove your answers.

The example in Exercise 10 is rather trivial since most elements arenot comparable. It is sometimes called a totally disordered set.

Exercise 11. Consider R with < and P(U ) with ⊂ again. Which is astrictly partially ordered set? Which is a partially ordered set? Prove youranswers.

{1,2}

{1} {2}

∅

{1,2}

{1} {2}

∅

{1,2,3}

{1,3} {2,3}

{3}

94 Chapter 6. Relations

Sometimes, especially for finite partially ordered sets in which not allelements are comparable, it is useful to make an illustration, called a Hassediagram, of the set with its relation as Figure 12 illustrates.

Figure 12: The Hasse diagram for P({1, 2}) with ⊂

Figure 12 shows the elements of the P({1, 2}) with the relation ⊂. Theelements of P({1, 2}) are shown as the vertices of a graph. Arrows showthe order relation; if there is a directed path from one vertex to anotherusing one or more arrows, then the vertex at the tail of the (first) arrow isrelated to the vertex at the head of the (last) arrow. For example, the arrowfrom ∅ to {1} indicates that ∅ ⊂ {1} and the arrow from ∅ to {2} togetherwith the arrow from {2} to {1, 2} indicate that ∅ ⊂ {1, 2}.

It is possible to give a precise definition of a Hasse diagram in termsof directed graphs, which are discussed in Chapter ???.

Using the convention that arrows always go up simplifies the presenta-tion. Figure 13 shows the more complicated Hasse diagram for P({1, 2, 3});you should see it as a cube.

Figure 13: The Hasse diagram for P({1, 2, 3}) with ⊂

The next exercise shows that strict partial orderings and partial order-ings are intimately related.


Exercise 14. (a) Suppose that " is a partial ordering on S and define therelation ≺ on S by s ≺ t if and only if s " t and s �= t . Show that ≺ is astrict partial ordering on S.(b) Suppose that ≺ is a strict partial ordering on S and define the relation" on S by s " t if and only if s ≺ t or s = t . Show that " is a partialordering on S.

Remark 15. According to Exercise 14, it does not matter whether the orderon a partially ordered set S is derived from a partial ordering " on S or astrict partial ordering ≺ on S. Put another way, whenever either type oforder relation is present, both types are actually present. Hence, one canuse whichever is more convenient. A mathematician may be sloppy andwrite “a partial ordering ≺” when “a strict partial ordering ≺” is meant.

Definition 16. A strict partial ordering ≺ on S which is connected is calleda strict total ordering. A strictly totally ordered set is a set S together witha strict total ordering ≺ on S.

Definition 17. A partial ordering " on S which is connected is called atotal ordering. A totally ordered set is a set S together with a total ordering" on S.

Exercise 18. (a) Prove that every strict total ordering ≺ on a set S is tri-chotomous.(b) Suppose that R is a relation on S which is transitive and trichotomous.Prove that R is a strict total ordering.

That is, a relation is a strict total ordering if and only if it is transitiveand trichotomous.

In our notation, we will always assume that we can go back and forthbetween ≺ or ", as in Exercise 14. Perhaps you were wondering if wecan use # or $ when we have ≺ or ", respectively? The answer is yes, ofcourse. That is, x # y is defined as y ≺ x and x $ y is defined as y " x .

The following definitions are related, but not the same.

Definition 19. Let S be a partially ordered set with partial ordering ". Anelement m ∈ S is maximal iff there exists no s ∈ S such that m ≺ s.

Definition 20. Let S be a partially ordered set with partial ordering ". Anelement M ∈ S is greatest iff s " M for all s ∈ S.

Definitions 19 and 20 are different. The key thing to remember is that,in a partially ordered set, all elements need not be related. The followingexample should help you to see the difference between maximal and greatestelements.


Example 21. Let A = { 1, 2 }. Consider S = P(A) and T = S − { A }with the partial ordering ⊂. S has A as its only maximal element and itsonly greatest element. T has two maximal elements, { {1} } and { {2} }, butno greatest element.

Example 22. Consider N with the partial ordering ≤. N has neither a max-imal element nor a greatest element. [Though this is intuitively obvious,its proof is surprisingly difficult. The proof depends on the Principle ofArchimedes, which we postpone until Chapter 10.]

Of course, maximal and greatest elements are intimately related as wewill see from the following four exercises.

Exercise 23. Let S be a partially ordered set with partial ordering ". Prove

that if S has a greatest element, then this greatest element is unique.

Exercise 24. Let S be a partially ordered set with partial ordering ". Prove

that the greatest element of S, if it exists, is a maximal element.

Exercise 25. Let S be a partially ordered set with partial ordering ". Provethat if S has more than one maximal element, then S does not have a greatestelement.

Exercise 26. Let S be a totally ordered set with total ordering ". Prove

that a maximal element of S, if it exists, is the greatest element.

In a totally ordered set, by Exercises 24 and 26, maximal and greatestelements are identical and, by Exercise 23, if a maximal element exists,then it is unique.

We can define minimal elements and the least element of a partiallyordered set in analogy with the definitions of maximal and greatest elements.Analogous results can be proved about these.

Next we consider subsets in a partially ordered set. Of course, thepartial ordering on the ambient set induces a partial ordering on the subset.

Definition 27. Let A be a subset of S, a partially ordered set with partialordering ". An element m ∈ S is an upper bound on A iff a " m for alla ∈ A. A is bounded above iff A has an upper bound.

Definition 28. Let A be a subset of S, a partially ordered set with partialordering ". An element M ∈ S is a least upper bound on A iff M is anupper bound on A and M ≤ s for all upper bounds s on A.

A least upper bound is also called a supremum.


Example 29. Let A = {x

∣∣ 0 < x2 < 2}. Clearly, 2 is an upper bound

on A. What is the least upper bound of A? It turns out that we have notbeen precise enough to answer this question! If we consider A as a subsetof R, then

√2 is the least upper bound. What if we consider A as a subset

of Q? In that case, there is no least upper bound! You may recall that√2 is irrational (see Proposition 18 of Chapter 3). You can think of the

decreasing sequence of rational upper bounds

2, 1.5, 1.42, 1.415, 1.4143, 1.41422, 1.414214, . . . .

This sequence converges to√

2 in R. Since this limit is unique, it can beshown that there is no rational least upper bound. We will not go into thedetails.

Example 30. Consider N with the partial ordering ≤. N has no upperbounds. Also the set E of all even natural numbers has no upper bounds.[Again, while this is intuitively obvious, its proof is surprisingly difficultand is postponed until Exercise 3 of Chapter 8.]

Exercise 31. Let S be a partially ordered set with partial ordering " andA ⊂ S. Prove that if A has a least upper bound, then A has a unique leastupper bound.

Exercise 32. For both of the following, give an example of a nonemptysubset A of a partially ordered set S such that:(a) the least upper bound of A exists and equals the greatest element of A,(b) the least upper bound of A exists but is not the greatest element of A.Prove your answers.

Exercise 33. Let A be a subset of a partially ordered set S with partialordering ". Prove that if A has a greatest element, then the least upperbound of A equals the greatest element of A.

We can define lower bounds and the greatest lower bound of a subsetof a partially ordered set in analogy with the definitions of upper bound andleast upper bound. Analogous results can be proved about these.

Definition 34. Let S be a partially ordered set with partial ordering ".S has the least upper bound property iff every nonempty subset which isbounded above has a least upper bound.

The least upper bound property is important. It distinguishes therational numbers from the real numbers and will be discussed more inChapters 8 and 13.


Example 35. Let us again consider A = {x ∈ Q

∣∣ 0 < x2 < 2}. We

indicated that√

2 is the least upper bound of A as a subset of R and that Ahas no least upper bound as a subset of Q. Hence Q does not have the leastupper bound property.

We mention here one more definition related to order.

Definition 36. A totally ordered set S is well-ordered iff every nonemptysubset of S has a least element. The ordering is called a well-ordering.

The fact that N is well-ordered is proved in Exercise 7 of Chapter10. The question of well-ordering a set in general is quite difficult. Thestatement that every set has a well-ordering is actually equivalent to theAxiom of Choice.

3. Equivalence Relations.

Definition 37. A relation ≈ on a set S which is reflexive, symmetric andtransitive is called an equivalence relation.

The prototypical example is equality on a set. There are other, moreinteresting, examples.

Example 38. Congruence and similarity are equivalence relations on theset of triangles.

Example 39. Let k ∈ N. Define the relation ≡ on N by n ≡ m if and onlyif n−m

k ∈ Z. We usually say that n is congruent to m modulo k and wewrite “n ≡ m (mod k).” This equivalence relation is well known to smallchildren, who learned it in the form that 3 hours after 11 o’clock comes 2o’clock; in this case k = 12 and the statement is that 11 + 3 ≡ 2 (mod 12).

In Chapter 7, given two sets A and B, we will say that A has the samecardinality as B if and only if there exists a bijection f : A → B.

Exercise 40. Prove that “has the same cardinality as” is an equivalence

relation on the power set of a given set.

Partitioning the set of clothes in your house into categories such ascoats, shirts, pants, underwear, socks and shoes is an example of an equiv-alence relation at work. Rather than giving the relation in terms of pairs ofobjects, it is given in terms of the subsets which supply the entries of thesepairs. This leads us to the following definition.

Definition 41. Let ≈ be an equivalence relation on S and let x ∈ S. Theset

[x]≈ = { s ∈ S | x ≈ s }

Section 3. Equivalence Relations 99

is called the equivalence class of x under ≈.

The notation for the equivalence class of x is sometimes written simplyas [x] when the equivalence relation is understood.

Exercise 42. Let ≈ be an equivalence relation on a set S and let s, t ∈ S.

Prove that [s] = [t] if and only if s ≈ t .

Exercise 43. Let ≈ be an equivalence relation on a set S and let s, t ∈ S.

Prove that if [s] �= [t], then [s] ∩ [t] = ∅, that is, [s] and [t] are disjoint.

We will rephrase the previous exercises using the following language.

Definition 44. Let S be a nonempty set. A partition of S is a set ofnonempty, pairwise disjoint subsets of S whose union is S.

Definition 45. Let ≈ be an equivalence relation on a nonempty set S.Denote by S/≈ denote the set of all equivalence classes of S under ≈.

Assuming we read ≈ as “equivalence,” we read S/ ≈ as “S modequivalence.” After the following exercise, we call S/≈ the partition of Sinduced by ≈.

Exercise 46. Prove that if ≈ is an equivalence relation on a nonempty set

S, then S/≈ is a partition of S.

We have gone from equivalence relations to partitions. Next we go inthe other direction.

Definition 47. Let C be a partition of a nonempty set S. Define a relationR on S by s R t if and only if there exists C ∈ C such that s, t ∈ C . R iscalled the equivalence relation induced by the partition C and is denotedby S/C.

We read S/C as “S mod C” and s S/C t as “s is S mod C related to t .”Of course, it must be verified that S/C actually is an equivalence relation.

Exercise 48. Let C be a partition of a nonempty set S. Prove that S/C is

an equivalence relation.

Of course, since we can pass from an equivalence relation to a partitionand then back to an equivalence relation, we can ask how the originalrelation and the induced relation are related. Actually they are the same.One can do this starting with a partition as well.


Theorem 49. If ≈ is an equivalence relation on a nonempty set S, thenS/(S/≈) = ≈.

Proof. By Exercise 46, S/≈ is a partition of S and, by Exercise 48, S/(S/≈) is an equivalence relation on S. Since a relation on S is a subset of S × S,it suffices to show that S/(S/≈) ⊂ ≈ and ≈ ⊂ S/(S/≈).

Suppose (s, t) ∈ S/(S/≈); that is, s S/(S/≈) t . By Definition 47,there exists C ∈ S/≈ such that s, t ∈ C . By Definition 45, there existsu ∈ S such that C = [u]. Since s, t ∈ [u], u ≈ s and u ≈ t . Hence, s ≈ t .Therefore, S/(S/≈) ⊂ ≈.

The remainder of the proof is left for Exercise 50. �

Exercise 50. Complete the proof of Theorem 49: prove that ≈ ⊂ S/(S/≈).

Theorem 51. If C is a partition of a nonempty set S, then S/(S/C) = C.

Proof. By Exercise 48, S/C is an equivalence relation on S, and, by Ex-ercise 46, S/(S/C) is a partition of S. Since a partition of S is a subset ofP(S), it suffices to show that S/(S/C) ⊂ C and C ⊂ S/(S/C).

Suppose C ∈ C. Choose an element s ∈ C . Since S/(S/C) is apartition of S, there exists C ′ ∈ S/(S/C) such that s ∈ C . C ′ = [s] withrespect to the equivalence relation S/C. We must show that C = [s].

First, fix t ∈ C . By Definition 47, s S/C t . By Exercise 42, t ∈ [t] =[s]. So C ⊂ [s].

Next, fix t ∈ [s]. So s S/C t . By Definition 47, there exists C ′′ ∈S/(S/C) such that s, t ∈ C ′′. Since S/(S/C) is a partition and s ∈ C ∩ C ′′,C ′′ = C . So t ∈ C and [s] ⊂ C .

Hence, C ∈ S/(S/C). Therefore, C ⊂ S/(S/C).The remainder of the proof is left for Exercise 52. �

Exercise 52. Complete the proof of Theorem 51: prove that S/(S/C) ⊂ C.


Supplemental ExercisesDefinition Review. There were 12 definitions in this chapter. Let S be aset with subset A and let R, ≺, ", and ≈ be relations on S. Define each ofthe following and give an example of each:

(1) R is reflexive(1) R is nonreflexive(1) R is symmetric(1) R is asymmetric(1) R is antisymmetric(1) R is transitive(1) R is connected(1) R is trichotomous(8) ≺ is a strict partial ordering(8) S is a strictly partially ordered set(9) " is a partial ordering(9) S is a partially ordered set

(16) ≺ is a strict total ordering(16) S is a strictly totally ordered set(17) " is a total ordering(17) S is a totally ordered set(19) m ∈ S is a maximal element(20) m ∈ S is a greatest element

For 27–34, let S be a partially ordered set with partial ordering ".

(27) m ∈ S is an upper bound on A(27) A is bounded above(28) m ∈ S is a least upper bound on A(34) S has the least upper bound property

For 36, let S be a totally ordered set with total ordering ".

(36) S is well-ordered(36) " is a well-ordering

(37) ≈ is an equivalence relation(41) the equivalence class of x under ≈, an equivalence relation(44) a partition of a nonempty set, S

For 45, let ≈ be an equivalence relation on S.

(45) S/≈, the set of all equivalence classes of S under ≈For 47, let C be a partition of S.

(47) S/C, the equivalence relation induced by C


Exercise S1. Let X be the set of all 2 × 2 matrices with real entries.

Recall that the determinant of a matrix

(a bc d

)equals ad − bc; denote

the determinant of a matrix M by |M |. Define a relation < on X by A < Biff |A| < |B|. Prove that < is a strict partial ordering on X . Is < a totalordering? Explain your answer.

a

b

c

δε

ζ

103

Chapter 7Cardinality of Sets

1. Introduction.If we consider the two sets

A = { a, b, c } and B = { δ, ε, ζ },

it is clear that they have the same “size.” Perhaps we should say what“size” means exactly. If you think about this for a moment, you may saysomething like “Of course, both sets have exactly three elements.” Youwould be showing your heritage and knowledge. The idea of counting is avery advanced concept compared to what is needed here.

Figure 1: A one-to-one correspondence

A more primitive idea would be to compare the two sets by matchingtheir elements as in the following:

a ↔ δ b ↔ ε c ↔ ζ.

Perhaps you recognize that this is just the concept of producing a one-to-one correspondence, or bijective function, between the two sets. We thinkthis is a fancy description for a fairly simple idea.

The comparison of infinite sets such as N, Z, Q, R and C is trickier. Infact, many people have been intimidated by the notion of infinity. The ideasof one-to-one correspondence and counting, which we considered for finitesets, lead to analogous ideas about infinite sets. In the infinite case, thenotion of one-to-one correspondence is virtually no more difficult, while

1 2 3 4 5 6 7 8 9 10 ... N

2 4 6 8 10 12 14 16 18 20 ... E

104 Chapter 7. Cardinality of Sets

the notion of “infinite numbers” to represent the “size” of infinite sets is farmore difficult.

To see why people have been intimidated by infinity for a long time,consider the paradox of Zeno, an ancient Greek philosopher. Accordingto Zeno, you can never reach the end of a racecourse since you must firstcover the first half before you can cover the whole course, and half of theremainder before covering the second half of the course, and so on, adinfinitum. That is, you cannot start from 0, go through 1

2 , 34 , 7

8 , 1516 , etc., and

arrive at 1 in finite time. This paradox, though often forgotten or ignored,has been around for nearly 2500 years. It points to the very foundations ofmathematics and issues unresolved until the end of the nineteenth century.

2. Finite Sets.We would like to formalize the definition of the “size” of a set, but

rather we will do something a little different.

Definition 2. Sets A and B have the same cardinality iff there exists aone-to-one correspondence f : A → B.

As indicated above, the sets A = { a, b, c } and B = { δ, ε, ζ } havethe same cardinality since f : A → B defined by

f (a) = δ f (b) = ε f (c) = ζ

is a one-to-one correspondence. But you should notice that we did notsay what “cardinality” is, only what “same cardinality” is. That is, we aresaying that A and B have the same number of elements without ever sayingwhat that number is!

Example 3. Infinite sets are more interesting. Consider

N = { 1, 2, 3, 4, . . . } and E = { 2, 4, 6, 8, . . . }.

Figure 4: A one-to-one correspondence from N to E

Of course, E � N. If you think that E and N are not the same “size,”then you are not thinking of the cardinality of the sets! In fact, it is easy tosee that f : N → E given by f (n) = 2n is a one-to-one correspondence.Hence E and N have the same cardinality.

Let us turn our attention to finite sets now.

Section 2. Finite Sets 105

Exercise 5. Let A be a set. Prove that A and ∅ have the same cardinality

if and only if A = ∅.

The next three exercises are easy and important consequences of factsabout bijective functions. These properties are called, respectively, reflex-ivity, symmetry, and transitivity; these are the properties of equivalencerelations, which we studied in Section 3 of Chapter 6.

Exercise 6. Let A be a set. Prove that A and A have the same cardinality.

Exercise 7. Let A and B be sets. Prove that A and B have the same

cardinality if and only if B and A have the same cardinality.

Exercise 8. Let A, B and C be sets. Prove that if A and B have the samecardinality and B and C have the same cardinality, then A and C have thesame cardinality.

For n ∈ N, consider the sets Zn = { 1, 2, 3, . . . , n }. We will use theseto further our understanding of finite sets. For example, if A and Zn havethe same cardinality, then we can think of Zn as an index for A, that is, Amust look like { a1, a2, a3, . . . , an }.

So how do we know if { a1, a2, a3, . . . , am } and { b1, b2, b3, . . . , bn }have the same cardinality? Clearly, the answer should be whenever m = n.You may try to prove this by taking one element at a time away from eachset until one is reduced to the empty set. Then you check if the other is alsoreduced to the empty set. We will turn this into an induction argument. Wedo this first for the following special case.

Theorem 9. Zm and Zn have the same cardinality if and only if m = n.

Proof. That m = n implies Zm and Zn have the same cardinality followsfrom the fact that the identity function on Zm is a bijection.

The proof in the other direction is by induction on n; that is, n is fixedwhile m is variable. Consider the step n = 1. Then there exists a one-to-one correspondence f : Z1 → Zm . Assume m �= n. Hence, m ≥ 2 and1, 2 ∈ Zm . Since f is onto, f (1) = 1 and f (1) = 2. This contradictionshows that m = n.

To prove the inductive step, we suppose the induction hypothesis thatZm and Zn have the same cardinality implies m = n. Suppose that Zm

and Zn+1 have the same cardinality. Thus, there exists a one-to-one cor-respondence f : Zn+1 → Zm . Since 1 �= n + 1, f (1) and f (n + 1) aredistinct elements of Zm , m − 1 ∈ N. Let g be the permutation of Zm which


exchanges f (n + 1) and m while leaving everything else fixed; that is,g : Zm → Zm is defined by

g(x) =⎧⎨⎩

f (n + 1), if x = m

m, if x = f (n + 1)

x, otherwise

.

(If f (n + 1) = m, then g = IZm .) Since f and g are bijections, g ◦ f :Zn+1 → Zm is a one-to-one correspondence. Since

g( f (n + 1)) = m,

we can define h : Zn → Zm−1 by h(x) = g( f (x)). This is a one-to-one correspondence. By the induction hypothesis, m − 1 = n. Hence,m = n + 1. �

We are now in a position to define what a finite set is.

Definition 10. A set S is finite iff either S = ∅ or there exists some n ∈ Nsuch that S and Zn have the same cardinality. We say that S has cardinality0 or n, respectively, and denote this by |S| = 0 or |S| = n.

Exercise 11. Suppose A and B are finite sets with cardinalities n and m,respectively. Prove that A and B have the same cardinality if and only ifn = m.

Exercise 12. Let A be a set with |A| = n and suppose x /∈ A and y ∈ A.(a) Prove that |A ∪ { x }| = n + 1.(b) Prove that |A − { y }| = n − 1.

Exercise 13. Let B be a finite set. Prove that if A ⊂ B, then A is finite and

|A| ≤ |B|.The results of this section now enable us to prove the following

theorem. Before looking at the proof, you should read the statement andtry to prove it for yourself. While this elementary statement may appearobvious, it is not so easy to prove; in fact, the “counting elements”-typeargument you might envision has been isolated in Exercise 13.

Theorem 14. Suppose A � B. If B is finite, then A and B do not have thesame cardinality.

Proof. Pick x ∈ B − A. Now A ⊂ B − { x }. By Exercise 12(b),

|B − { x }| = |B| − 1.

Section 3. Infinite Sets 107

By Exercise 13,

|A| ≤ |B − { x }| = |B| − 1 < |B|.

Hence, |A| �= |B|. By Exercise 11, A and B do not have the same cardi-nality. �

This theorem and its proof reflect one of the most important ideaswhich we presented in Chapter 3. It is an example of breaking a proof intosmaller, easier pieces. These pieces, usually called lemmas, are easier towork on in isolation from the remainder of the theorem. Here the lemmasare the earlier exercises in this section: Exercises 12, 13 and 11.

Now that we understand finite sets, we can prove things like the fol-lowing, which seem “obvious.” For the moment, you can suppose that thefamiliar infinite sets, such as N and R, really are infinite.

Exercise 15. Let f : A → B be a one-to-one function.(a) Prove that if B is finite, then A is finite.(b) Give a counterexample to: if A is finite, then B is finite.

Exercise 16. Let f : A → B be an onto function.(a) Prove that if A is finite, then B is finite.(b) Give a counterexample to: if B is finite, then A is finite.

3. Infinite Sets.The definition of infinite sets is obvious.

Definition 17. A set S is infinite iff S is not finite.

The next exercise establishes the existence of infinite sets.

Proposition 18. N is an infinite set.

We will consider three different proofs of this statement. The first twoare similar, while the third is quite different and will come later (Exercise27).

Proof #1. Assume that N is a finite set. Then we can write N in the form

{ n1, n2, n3, . . . , nk }.

Hence, N has a largest element, nj for some j . Now nj + 1 > nj and,therefore, nj + 1 /∈ N. But, since nj is a natural number, nj + 1 is also anatural number. Contradiction! �


You may object to the proof above, asking how we know that the set

{ n1, n2, n3, . . . , nk }

has a largest element or how we know that nj + 1 > nj or, for that matter,how we know that nj +1 is a natural number. These questions are answeredby the construction of the natural numbers in Chapter 10. If you find thisproof reasonable now, that is OK. Here is a different, but similar proof.

Proof #2. Assume that N is a finite set. Then we can write N in the form

{ n1, n2, n3, . . . , nk }.

We know that 1, 2 ∈ N. Consider the number N = n1+n2+n3+· · ·+nk . Nis a natural number. However, N > nj for all j since nj + 2 > nj + 1 > nj

and 1, 2 and nj are at least two different elements of N. So N /∈ N.Contradiction! �

You may have similar objections to Proof #2 as to Proof #1. A differentproof is contemplated in Exercise 27 based on the following importantcharacterization theorem for finite and infinite sets.

Theorem 19. Let S be a set.(a) S is infinite if and only if it has a proper subset of the same cardinality.(b) S is finite if and only if it has no proper subset of the same cardinality.

As in the proof of Theorem 14, the proof of Theorem 19 is also com-pleted in steps. Theorem 14 provided half of part (b); half of part (a) is tobe proved in the following exercise.

Exercise 20. Prove that if a set B has a proper subset with the same

cardinality, then B is infinite.

Our next goal is to prove the converses of Theorem 14 and Exercise20, thereby completing the proof of Theorem 19.

Definition 21. A set S is denumerable iff S and N have the same cardinality.Sets which are denumerable are also called countably infinite.

Definition 22. A set S is countable iff S is either finite or denumerable.

Since a denumerable set is in one-to-one correspondence with N, wecan can think of it as an infinite list indexed by the natural numbers such as

{ a1, a2, a3, . . . }.

Section 3. Infinite Sets 109

Theorem 23. Every infinite set contains a denumerable subset.

Proof. Let A be an infinite set. We wish to construct a denumerable subset

D = { a1, a2, a3, . . . }by identifying elements a1, a2, etc., in that order. The proof is by inductionon the index of the elements. Choose a1 ∈ A; this is the first step, n = 1,of the induction. Next we wish to prove the inductive step: if Dk ={ a1, a2, a3, . . . , ak } ⊂ A, then there exists ak+1 ∈ A − Dk . The inductionhypothesis is that this is true for k = n; we need to show that it is then truefor k = n + 1. Assume that it is false for k = n + 1. That means thatA − Dn = ∅. Hence A = Dn+1 is finite. This is a contradiction.

That is, we have constructed a sequence of distinct elements of A:

{ a1, a2, a3, . . . },as desired. �

We should pause here to consider the last sentence of the proof ofTheorem 23. It looks deceptively simple. The induction actually providesa sequence of finite sets D1, D2, D3, . . . and not a denumerable set. In fact,mathematicians isolated an axiom in the eighteenth century that is requiredin circumstances exactly like this one. It is called the Axiom of Choiceand most, but not all, mathematicians use and accept this axiom. We donot wish to delve into the intricacies of the Axiom of Choice. Suffice itto say that whenever an infinite number of choices are made, the Axiomof Choice is being implemented, unless a universal rule exists to make ourinfinitely many choices clear in one fell swoop. For example, to choose onesock from each of an infinite number of pairs of socks requires the Axiomof Choice; to choose one shoe from each of an infinite number of pairs ofshoes does not require the Axiom of Choice since we can choose all of theleft shoes.

Having said all of this, we now present a new proof of Theorem 23,written as most mathematicians might write it.

Second Proof of Theorem 23. Let A be an infinite set. Since A is infinite,A is nonempty. Choose a1 ∈ A. Now choose a2 ∈ A different froma1; this can be done since A − { a1 } is nonempty. Continue in this way:having chosen a1, a2, a3, . . . , ak we choose ak+1 ∈ A−{ a1, a2, a3, . . . , ak }which can be done since the difference of a finite set from an infinite setis always nonempty. Continuing this process indefinitely, we construct adenumerable subset { a1, a2, a3, . . . } of A. �

The next exercise requests a proof of a useful lemma. You will thenuse it to prove the converses of Exercise 20 and Theorem 14.


Exercise 24. Suppose D is denumerable and x ∈ D. Prove that D − { x }is denumerable.

Exercise 25. Prove that every infinite set has a proper subset of the same

cardinality.

Exercise 26. Let S be a set. Prove that if no proper subset of S has the

same cardinality as S, then S is finite.

This completes the proof of Theorem 19.If it interests you, you can keep in mind that the equivalent definitions

of finite and infinite sets, as given in Theorem 19, depend on the Axiom ofChoice.

We now return to the proof that N is infinite, Proposition 18. Theequivalent definition of infinite in Exercise 25 gives us a new proof of thisfact.

Exercise 27. Use Theorem 19 to prove that N is infinite.

The fact that R is infinite now follows.

Exercise 28. Prove that R is infinite.

4. Countable Sets.As we noted, there are two kinds of infinite sets: those which are

countable and those which are not countable, that is, uncountable. Weconsider countable sets next, remembering that countable sets are eitherfinite or denumerable.

Theorem 29. Every subset of a countable set is countable.

Proof. Let A be a countable set. We have already settled the case whereA is finite in Exercise 13. So suppose that A is denumerable. Then we canlist the elements of A as, say,

A = { a1, a2, a3, . . . }.

Let B be a subset of A. Then, using the indices of the list of the elementsof A, we let

S = { n ∈ N | an ∈ B } .

Then k &→ ak defines a one-to-one correspondence from S to B. Either Sis finite or infinite. If S is finite, then B is finite and, therefore, countable.

Section 4. Countable Sets 111

Suppose S is infinite. Since we can order the elements of S starting withthe smallest, we can call the elements of S

n1, n2, n3, . . .

in increasing order. Now k &→ nk defines a one-to-one correspondencefrom N to S. Hence S and, therefore, B are denumerable. �

We next ask the question whether all infinite sets are countable? Theanswer is negative. However, we will see that the sets N, Z and Q are allcountable.

Exercise 30. Prove that all denumerable sets have the same cardinality.

Despite the facts that N is a proper subset of Z and that Z appears tohave, roughly, twice as many elements as N, they are shown to have thesame cardinality by the following exercise.

Exercise 31. Prove that Z is denumerable.

Now consider Q, the set of rational numbers. An amazing theoremproved by Georg Cantor in 1874 shows that Q also has the same cardinalityas N. The proof depends on a diagonal argument that we present next.

Theorem 32. The set of positive rational numbers, Q ∗+, is denumerable.

Proof. Consider the following table of the elements of Q ∗+:

11

21

31

41

51

61 . . .

12

22

32

42

52

62 . . .

13

23

33

43

53

63 . . .

14

24

34

44

54

64 . . .

15

25

35

45

55

65 . . .

16

26

36

46

56

66 . . .

......

......

......

. . .

We use the pattern in Figure 33 on the table above to create a list of thepositive rationals, skipping over duplicate rationals in the table as we goalong.

...

...

...

...

...

...

.................. ............... ...

...

...

...

...

...


Figure 33: Cantor’s scheme

That list is

Q ∗+ = {

1, 2, 12 , 1

3 , 3, 4, 32 , 2

3 , 14 , 1

5 , 5, 6, 52 , . . .

}.

Hence, Q ∗+ is denumerable. �

Exercise 34. Prove that Q is denumerable.

Exercise 35. Prove that a countable union of countable sets is countable.

Exercise 36. Prove that a finite product of countable sets is countable.

It turns out that countable products of finite sets need not be countable;this will be shown in Exercises 54 and 55.

Definition 37. The set of algebraic numbers is the union of the real solutionsets of all polynomial equations (in one variable) with integer coefficients.

Exercise 38. Prove that the set of algebraic numbers is denumerable.

5. Uncountable Sets.To complete this chapter we will consider uncountable sets. Yes, they

do exist; that is, there are different orders of infinity!

Definition 39. A set S is uncountable iff S is not countable.

So, every set is exactly one of the following properties: finite ordenumerable or uncountable. The existence of uncountable sets is shownnext. The proof below is again due to Georg Cantor in 1891.

Section 5. Uncountable Sets 113

Theorem 40. The subset [0, 1] ⊂ R is uncountable.

Proof. Assume that [0, 1] is countable. Then we can list the elements of[0, 1] as

[0, 1] = { r1, r2, r3, . . . }.Every real number can be written as an infinite decimal; those in [0, 1] canbe written with 0 to the left of the decimal point. Therefore, the elementsof [0, 1] can be written in the form

r1 = 0.a1,1a1,2a1,3a1,4a1,5 . . .

r2 = 0.a2,1a2,2a2,3a2,4a2,5 . . .

r3 = 0.a3,1a3,2a3,3a3,4a3,5 . . .

r4 = 0.a4,1a4,2a4,3a4,4a4,5 . . .

r5 = 0.a5,1a5,2a5,3a5,4a5,5 . . .

...

where the ai, j ’s are the j th decimal places of ri . Define the number

r = 0.b1b2b3b4b5 . . . where bn ={

3, if an,n �= 3

4, if an,n = 3.

Clearly, r �= rn for all n ∈ N since they differ in the nth decimal place.Therefore, r ∈ [0, 1], but r /∈ { r1, r2, r3, . . . }. This is a contradiction.

�In the sense of cardinality, there are more real numbers between any

two given rational numbers than there are rational numbers!

Exercise 41. Prove that R is uncountable.

There are lots of examples of uncountable sets. With the followingsix exercises, we will show that all nonempty open intervals (including R)have the same cardinality, and are therefore uncountable.

Exercise 42. Prove that R and (0, 1) have the same cardinality.

Exercise 43. Prove that R and (0, +∞) have the same cardinality.

Exercise 44. Prove that, for all real numbers a, (a, +∞) and (0, +∞)

have the same cardinality.


Exercise 45. Prove that, for all real numbers a, (−∞, −a) and (a, +∞)

have the same cardinality.

Exercise 46. Prove that, for all real numbers a and b with a < b, (a, b)

and (0, 1) have the same cardinality.

Exercise 47. Prove that all nonempty open intervals have the same cardi-

nality.

In fact, all nonempty and non-singleton intervals are uncountable andhave the same cardinality.

Exercise 48. Prove that (0, 1), [0, 1), (0, 1] and [0, 1] have the same

cardinality.

Exercise 49. Prove that all nonempty and non-singleton intervals have the

same cardinality.

Recall that a number is irrational if it is not rational. That is, the setof irrational numbers is R − Q.

Exercise 50. Prove that the set of irrational numbers is uncountable.

Recall that a number is transcendental if it is not algebraic.

Exercise 51. Prove that the set of transcendental numbers is uncountable.

The next three exercises show that uncountable sets naturally arisefrom countable processes. First, we must define products of infinite collec-tions of sets. In Section 2 of Chapter 5, we defined finite products. Actually,we only considered products of two sets and R3, a product of three sets;this is easily generalized to finite products.

A collection, S, of sets is indexed by a set K if

S = { Sk | k ∈ K } .

In an indexed collection, the elements need not be distinct; that is, anindexed collection may not be a set. For example,

{ Z | k ∈ N }

is an indexed collection consisting of denumerably many copies of Z.

Section 5. Uncountable Sets 115

Definition 52. Suppose S = { Sk | k ∈ K } is a collection of sets indexed bya set K . The product of the sets in S is the set of all functions f : K → ∪S

with f (k) ∈ Sk for all k ∈ K .

Suppose S is a set of sets; every set in S appears exactly once. In thiscase, we can think of the product of the sets in S as the set of all functionsf : S → ∪S with f (S) ∈ S for all S ∈ S. You may think of this as S

being indexed by S since S is a set. As long as you distinguish multipleoccurrences of a set S in S, you can do this.

As for notation, we use

∏k∈K

Sk or∏S∈S

S

to denote the product of the sets in S. Some mathematicians call this adirect product and some use the notation

×k∈K

Sk or ×S∈S

S.

It is one of the most amazing facts in mathematics that the productof infinitely many non-empty sets cannot generally be proven non-emptyexcept by using the Axiom of Choice. The notation

Y X =∏x∈X

Y

is also commonly used for the set of all functions from X to Y .

Example 53. The collection S = { S1, S2 } is indexed by Z2. By definition,∏k∈Z2

Sk is the set of all functions f : { 1, 2 } → S1 ∪ S2 with f (k) ∈ Sk

for both k ∈ Z2. That is, for k = 1, 2, an element of the product dependson elements s1 = f (1) ∈ S1 and s2 = f (2) ∈ S2. We can think of thisas the ordered pair (s1, s2). Of course, this is a generic element of S1 × S2

according to Definition 1 of Chapter 5.

In light of Example 53, though this is not technically correct, when Kis a finite set such as K = Zn = { 1, 2, . . . , n }, then we write

S1 × S2 × · · · × Sn =∏k∈Zn

Sk .

We ask you to believe that this is not circular: products of collections ofsets are defined in terms of functions that are defined in terms of productsof two sets.


Exercise 54. Let S = { 0, 1 } and let P = ∏n∈N S = S × S × S × . . . . (P

is a denumerable product of copies of S.) Prove that P is uncountable.

Exercise 55. Prove that the denumerable product of finite sets, each with

two or more elements, is uncountable.

Exercise 56. Prove that the power set of N is uncountable.

Moreover, P(N) and R have the same cardinality. The proof of thisdepends on showing that every element of (0, 1) has a (unique) binary rep-resentation, corresponding to a sequence of 0’s and 1’s (which correspondsto an element of P(N) as suggested by the hint for Exercise 56).

The following result is often called Cantor’s Theorem.

Exercise 57. Let S be a nonempty set. Prove that the power set P(S) and

S do not have the same cardinality.

Remark. It is interesting to note that it is not clear, from what we havedone here, whether there exists an uncountable subset of R that does nothave the same cardinality as R. The rejection of this possibility is calledthe Continuum Hypothesis. It turns out that, as is true for the Axiom ofChoice, the Continuum Hypothesis is an independent axiom; that is, bothits acceptance and its rejection are consistent with our usual set of axioms.

117

PART III

CONSTRUCTION OFNUMBER SYSTEMS

118

Chapter 8Algebra of Number Systems

1. Primary Properties of Number Systems.The goal of Part III is to construct the familiar sets of numbers, together

with the usual arithmetic operations and the usual ordering. We do this bystarting with known concepts from set theory. The empty set will give usour starting number, zero. From zero, we use set theoretic tools to constructthe natural numbers. Zero and the natural numbers lead us to the integers,then to the rational numbers, then to the real numbers and, finally, to thecomplex numbers. This succession of constructions creates each new setsuch that it extends the previous set.

In each construction, except the complex numbers, a total orderingwill be given. Indeed, we will see that there is no ordering of the complexnumbers that extends the properties of the ordering of the real numbers.

In each construction, we will also have two operations, addition andmultiplication. We have already made precise what it means for ≤ to bea total ordering (or, equivalently, what it means for < to be a strict totalordering) in Section 2 of Chapter 6. The following definition makes precisewhat it means to have an operation like addition or multiplication.

Definition 1. A binary operation on a set X is a function � : X × X → X .For x1, x2 ∈ X , we denote �(x1, x2) by x1 � x2.

We will call a binary operation, more simply, an operation. Somemathematicians prefer binary operator or operator. You should recognizeaddition and multiplication as operations. Notice the comment on notationin the definition of operation: the functional notation for addition, +(x, y),looks strange compared with the more natural x + y.

Remark 2. We always use the symbol + to denote addition. We also willfollow the usual conventions by writing multiplication in numerous ways:

x · y = x × y = (x)(y) = xy.

The usual order of operations applies: multiplications have priority overadditions. Hence, x + yz = x + (yz) and (usually!) x + yz �= (x + y)z.

Section 1. Primary Properties of Number Systems 119

Remark 3. Consider the following properties on a set X with addition andmultiplication operations:

(A0) Closure of Addition: for all x, y ∈ X , x + y ∈ X(M0) Closure of Multiplication: for all x, y ∈ X , xy ∈ X

The closure properties are often listed as special properties of addition andmultiplication. Since we have defined operations as functions from X × Xto X , the closure properties are redundant. That is,

+ : X × X → X and · : X × X → X

are functions. Therefore, if x, y ∈ X , then (x, y) ∈ X × X and, hence,

+ (x, y) ∈ X and · (x, y) ∈ X.

This notation is clumsy and we revert to the usual notation for addition andmultiplication:

x + y = +(x, y) and xy = ·(x, y).

In this chapter and throughout Part III of this text, we will concernourselves with the algebraic properties of addition and multiplication op-erations on totally ordered sets of numbers.

Definitions 4 and 8, given below, mimic the familiar properties of thereal numbers. You should pay careful attention to the properties given inthese definitions. We will not give the most general definitions for theterms being defined, which are field and ordered field, since we are onlyinterested in whether our constructions satisfy the individual properties.

Definition 4. A set X with two operations, + and ·, is a field iff the followingproperties are satisfied, for all x, y, z ∈ X :

(AC) Commutativity of Addition:x + y = y + x

(AA) Associativity of Addition:(x + y) + z = x + (y + z)

(AZ) Existence of a Unique Additive Identity:there exists a unique 0 ∈ X such that 0 + x = x + 0 = x

(AI) Existence of Unique Additive Inverses:there exists a unique −x ∈ X such that x + (−x) = (−x)+ x = 0

(MC) Commutativity of Multiplication:xy = yx

(MA) Associativity of Multiplication:(xy)z = x(yz)

(MU) Existence of a Unique Multiplicative Identity:

120 Chapter 8. Algebra of Number Systems

there exists a unique 1 ∈ X , with 1 �= 0, such that 1x = x1 = x(MI) Existence of Unique Multiplicative Inverses:

if x �= 0, there exists a unique x−1 ∈ X such that xx−1 = x−1x = 1(DP) Distributive Properties:

(x + y)z = xz + yz and x(y + z) = xy + xz

The two letters between parentheses before each property name shouldbe suggestive of the name of the property. For example, (AC) stands forAddition is Commutative. The additive identity of a field is often called azero and the multiplicative identity of a field is often called a unity: (AZ)refers to Zero and (MU) refers to Unity.

Remark 5. According to Definition 4, 0 �= 1 in a field. Suppose all ofthe other field properties hold and 0 = 1; it will follow from Exercise 20,which shows that multiplication by 0 yields 0, that X = { 0 } since, for allx ∈ X ,

x = 1x = 0x = 0.

To avoid this trivial case, most authors include the property that 0 �= 1 intheir definitions of field, as we have done in (MU).

Remark 6. Notice that the two Distributive Properties, if written as follows,might be called Factoring Properties:

xz + yz = (x + y)z and xy + xz = x(y + z).

Remark 7. Notice that if X has a commutative multiplication, then thetwo Distributive Properties are equivalent. These are often called the leftand right distributive properties. Such choices are arbitrary and are notmade consistently. In our constructions of N, Z, etc., we will always havea commutative multiplication. Similar remarks are in order for identityelements and inverses. One can define things like left identity and rightinverse; for a commutative operation, lefts and rights are equal and wesimply call them identity and inverse, respectively.

Next, we add an ordering to the mix.

Definition 8. A field X with a total ordering < is an ordered field iff thefollowing properties are satisfied, for all x, y, z ∈ X :

(PP) Product of Positives:if x > 0 and y > 0, then xy > 0

(IX) Additive Cancellation in Inequalities:if x + y < x + z, then y < z

Remark 9. Consider property (IX). That y < z implies x + y < x + zshould strike you as obvious. But how do we know this? Perhaps you

Section 1. Primary Properties of Number Systems 121

would say that this is just “substitution.” We need not appeal to an Axiomof Substitution; in Exercise 25, we prove this type of substitution is impliedby two kinds of cancellation properties.

Remark 10. Any element z that satisfies z + x = x + z = x for allx ∈ X is called an additive identity of X . Any element y that satisfiesy + x = x + y = 0 for all x ∈ X is called an additive inverse of x inX . Property (AZ) of a field guarantees the existence of a unique additiveidentity and property (AI) guarantees the existence of a unique additiveinverse for every element of the field. The uniqueness actually followsfrom the existence according to the following two propositions.

Proposition 11. If a set X with addition has an additive identity, then thisidentity is unique.

Proof. Assume that there exist two distinct additive identities 0 and 0′.Then

0′ = 0′ + 0 = 0.

Hence, 0′ = 0. Contradiction! Therefore, 0 is the unique additive identity.�

The proof of Proposition 11 is a standard proof for many kinds ofidentity elements. It is sometimes described as “letting the two identitiesfight it out.” The uniqueness of inverses is similarly proved by letting twoinverses fight it out. You should compare this with Exercise 47 and 52 ofChapter 5.

Proposition 12. Suppose X is a set with addition satisfying properties(AA) and (AZ). If x ∈ X has an additive inverse, then this inverse isunique.

Proof. Assume that −x and x ′ are distinct additive inverses of x . Then

x ′ = x ′ + 0 = x ′ + (x + (−x)) = (x ′ + x) + (−x) = 0 + (−x) = −x .

So x ′ = −x . Contradiction! Therefore, −x is the unique additive inverseof x . �

The existence of multiplicative identity and multiplicative inverses (ofnonzero elements) similarly implies their uniqueness.

Exercise 13. Prove that if a set X with multiplication has an multiplicative

identity, then this identity is unique.


Exercise 14. Suppose X is a set with multiplication satisfying properties(MA) and (MU). Prove that if x ∈ X has a multiplicative inverse, then thisinverse is unique.

As we indicated above, it is not our purpose to study fields or orderedfields in their full generality. You may already have realized that N andZ are not ordered fields. So we will really study sets with addition andmultiplication operations and, possibly, a total ordering that satisfy some,but not necessarily all, of the properties listed in this section.

2. Secondary Properties Involving Addition and Multiplication.One of the most useful aspects of algebra is that the properties char-

acterize the structures, as suggested by the attempt, in the definition ofordered field, to characterize R. Moreover, the properties themselves maybe used to prove other properties, without ever using any direct knowledgeabout the sets or operations involved.

We will consider a few secondary properties. At first, we will assumethat X has only those properties—mostly primary properties—indicated.Later we will simply let X be a field or an ordered field. We could give aminimal list of conditions that imply the desired conclusion; however, theremay exist two or more distinct minimal lists. That is, a sufficient conditionmay not be necessary. (See the comments after Exercise 16.)

The following three properties depend only on addition.

Remark 15. Suppose that X has an addition operation and that x, y, z ∈ X .The following seems reasonable:

if y = z, then x + y = x + z.

You may regard this as a kind of substitution: z is substituted for y. How-ever, you can also reason as follows: x = x and y = z imply (x, y) =(x, z), which implies, since addition is a function, that x + y = x + z.Similarly, if y = z, then y + x = z + x . These properties are sometimesreferred to as Adding Equals to Equals.

The converse of Adding Equals to Equals, for example, that x + y =x + z implies y = z, is called Cancellation; compare it with property (IX).This property requires proof.

Exercise 16: (AX) Additive Cancellation in Equations.Suppose X is a set with addition satisfying properties (AA), (AZ), and (AI).(a) Prove that x + y = x + z if and only if y = z for all x, y, z ∈ X .(b) Prove that y + x = z + x if and only if y = z for all x, y, z ∈ X .

In proving Exercise 16, you used additive inverses. Consider for amoment whether additive cancellation holds in N. Of course, for y ∈ N,

Section 2. Secondary Properties Involving Addition and Multiplication 123

5 + y = 5 + z seems to imply that y = z. We will prove this is true inChapter 10. The proof will be very different from your proof of Exercise16 since there are no additive inverses in N: for example, −5 /∈ N. As wenoted at the beginning of this section, the conditions we give are not, ingeneral, necessary and sufficient.

The next exercise gives a familiar property about double negatives.

Exercise 17: Double Additive Inverses.Suppose X is a set with addition satisfying properties (AA), (AZ), and (AI).Prove that, for all x ∈ X ,

−(−x) = x .

If your proof of Exercise 17 followed the technique of making twoinverses fight it out, then you gave the expected proof. However, there isanother proof that does not require (AA)! Notice that (AI) says that −x isthe unique inverse of x in X if and only if

x + (−x) = (−x) + x = 0.

This same condition indicates that x is the unique inverse of −x in X . Whatis −(−x)?

The following two properties depend only on multiplication and aresimilar to the previous properties of addition except for the avoidance ofmultiplication by 0.

Exercise 18: (MX) Multiplicative Cancellation in Equations.Suppose X is a set with multiplication satisfying properties (MA), (MU),and (MI).(a) Prove that xy = xz if and only if y = z for all x, y, z ∈ X with x �= 0.(b) Prove that yx = zx if and only if y = z for all x, y, z ∈ X with x �= 0.

In the proof of Exercise 18, you should have noticed that one directionof the equivalences is easy; these are often referred to as MultiplyingEquals by Equals.

Exercise 19: Double Multiplicative Inverses.Suppose X is a set with multiplication satisfying properties (MA), (MU),and (MI). Prove that, for all x ∈ X with x �= 0,

(x−1)−1 = x .

Try to prove this without using (MA).


The remaining properties in this section use both addition and multi-plication. The next property is the last one that we list depending on onlysome of the properties of a field.

In (MI) we consider multiplicative inverses of nonzero elements. Does0 have a multiplicative inverse? Can x−1 = 0? The next exercise addressesthese issues.

Exercise 20: Multiplication by 0.Suppose that X is a set with addition and multiplication satisfying properties(AZ), (AX) and (DP). Prove that, for all x ∈ X ,

0x = x0 = 0.

Since 1 �= 0, 0 has no multiplicative inverse. Also, a multiplicativeinverse cannot equal 0. In a field X , since all of the assumed propertieshold, 0x = x0 = 0 for all x ∈ X . Consider the following statement, theconverse of Exercise 20.

Exercise 21: 0 Product Property.Suppose that X is a field. Prove that, for all x, y ∈ X ,

if xy = 0, then x = 0 or y = 0.

In a field X , the following are equivalent for all x, y ∈ X :

• if x �= 0 and y �= 0, then xy �= 0 ,• if x �= 0 and xy = 0, then y = 0 for all x, y ∈ X ,• if y �= 0 and xy = 0, then x = 0 for all x, y ∈ X .

Combining the Multiplication by 0 and the 0 Product Properties, we seethat in a field X , for all x, y ∈ X ,

xy = 0 if and only if x = 0 or y = 0.

The following two properties depend on the Uniqueness of AdditiveInverses.

Exercise 22: Factoring a Minus Property.Suppose that X is a field. Prove that, for all x, y ∈ X ,

(−x)y = x(−y) = −(xy).

Section 3. Secondary Properties Involving Order 125

Exercise 23: Product of Negatives Property.Suppose that X is a field. Prove that, for all x, y ∈ X ,

(−x)(−y) = xy.

3. Secondary Properties Involving Order.The properties in this section involve operations, addition and multi-

plication, as well as an order relation.Given an ordering, < or ≤, on a set X , we have three transitive relations

on X : <, ≤, and =. (Recall, from Exercise 14 of Chapter 6, that we canalways derive < from ≤ or vice versa.) The following proposition mayseem obvious by “substitution,” but we can prove it.

Proposition 24. Suppose X is a set with a strict partial ordering < andx, y, z ∈ X. If x < y and y = z, then x < z.

Proof. By the definition of ≤, x < y implies x ≤ y and y = z impliesy ≤ z. Hence, by transitivity of ≤, x ≤ z.

Assume that x = z. By symmetry of =, z = y. By transitivity of =,x = y. However, x < y, contradicting the trichotomy of <. Hence, x �= z.Therefore, x < z. �

Since the preceding proposition involves both an order relation (<) andan equivalence relation (=), it is not simply a transitive property. Similarproperties, involving other permutations of <, ≤, and =, are given inExercise 51.

The following property is the converse of property (L2) Additive Can-cellation in Inequalities. It is more complicated than the property of AddingEquals to Both Sides of an Equation.

Exercise 25: Adding Equals to Both Sides of an Inequality.Suppose that X is an ordered field. Prove that, for all x, y, z ∈ X ,

if y < z, then x + y < x + z.

In the following exercises, we explore the dualities between less thanand negative and between greater than and positive.

Exercise 26: Negation Reverses Order Property.Suppose that X is an ordered field. Prove that

x > 0 if and only if −x < 0 for all x ∈ X .


Exercise 27: Product by Positive Preserves Order Property.Suppose that X is an ordered field. Prove that

if x > 0 and y < z, then xy < xz for all x, y, z ∈ X .

Exercise 28: Product by Negative Reverses Order Property.Suppose that X is an ordered field. Prove that

if x < 0 and y < z, then xy > xz for all x, y, z ∈ X .

Exercise 29: Squares of Nonzero Numbers are Positive.Suppose that X is an ordered field. Prove that

if x �= 0, then x2 = xx > 0 for all x ∈ X .

Exercise 30: Multiplicative Inverses Reverse Order Property.Suppose that X is an ordered field. Prove that

if y > x > 0, then 0 < y−1 < x−1 for all x, y ∈ X .

4. Isomorphisms and Embeddings.In mathematics, one often studies structures which are the same in

all respects except for the names of the elements, relations, operations,etc. which define them. Such structures are called isomorphic structures.To better understand what kind of structure we are referring to, consider Ras the underlying set of our structure with the order relation and operationsof addition and multiplication providing the structure.

Consider the following definition.

Definition 31. Let X and Y be totally ordered sets together with additionand multiplication. We denote the strict orderings by < and the operationsby + and × for both X and Y . X and Y are isomorphic structures iff thereexists a bijective function f : X → Y , called an isomorphism, satisfyingthe following three conditions for all x1, x2 ∈ X :

x1 < x2 if and only if f (x1) < f (x2),

f (x1 + x2) = f (x1) + f (x2),

f (x1 × x2) = f (x1) × f (x2).

We call the three conditions in the definition of an isomorphismorder-preserving, addition-preserving, and multiplication-preserving, re-spectively. Taken collectively, we say that an isomorphism is structure-preserving.

Section 4. Isomorphisms and Embeddings 127

The following example and the discussion and exercises following itshould shed some light on the nature of isomorphisms.

Example 32. Let E be the set of even integers. Define a function f : Z →E by f (n) = 2n. f is clearly bijective. f is order- and addition-preservingsince

m < n if and only if f (m) = 2m < 2n = f (n)

and f (m + n) = 2(m + n) = 2m + 2n = f (m) + f (n).

However, f is not multiplication-preserving since f (1 · 1) = f (1) = 2while f (1) · f (1) = 2 · 2 = 4. Note that this shows that f is not anisomorphism; it does not show that Z is not isomorphic to E using someother function as an isomorphism.

The following exercise lists some of the many ways that isomorphismspreserve structure.

Exercise 33. Suppose that X and Y are sets together with addition and mul-tiplication (no order relations are assumed here). Suppose that f : X → Yis an isomorphism, that is, a bijection which preserves addition and multi-plication. Prove each of the following:(a) If 0 is the additive identity in X , then f (0) is the additive identity in Y .(b) If 1 is the multiplicative identity in X , then f (1) is the multiplicativeidentity in Y .(c) If −x is the additive inverse of x in X , then f (−x) is the additive inverseof f (x) in Y .(d) If x−1 is the multiplicative inverse of x in X , then f (x−1) is the multi-plicative inverse of f (x) in Y .

Results such as those of the previous exercise are useful in showingthat structures are not isomorphic.

Example 34. We know that 1 is the multiplicative identity of Z. Does Ehave a multiplicative identity? Since 1 /∈ E , the answer appears to be no.However, this does not prove it since another element of E could be themultiplicative identity.

Assume n ∈ E is the multiplicative identity in E . Therefore, 2n = 2.Since 2, n ∈ E and E ⊂ Z, 2, n ∈ Z. In Z, 2 = 2(1). Hence, 2n = 2(1)

and, by Multiplicative Cancellation, n = 1. This implies that 1 ∈ E , acontradiction. (We will check that all necessary properties of Z hold inChapter 11.)

Exercise 35. Prove that Z is not isomorphic to E .


Since we think of isomorphic structures as the same, the followingidea is natural.

Exercise 36. Consider “is isomorphic to” as a relation on a collection ofsets with addition and multiplication. Prove that this is an equivalencerelation.

The equivalence classes for the relation in Exercise 36 are called iso-morphism classes.

Consider the following definition.

Definition 37. Let X and Y be totally ordered sets together with additionand multiplication. X is embedded in Y if there exists Z ⊂ Y and anisomorphism f : X → Z . The function g : X → Y defined by g(x) =f (x) is called an embedding.

Notice that an embedding is a one-to-one function that is order-pre-serving, addition-preserving, and multiplication-preserving. We couldhave done this differently by defining an embedding this way. Then theexistence of the isomorphism f : X → g(X) is assured. (You shouldverify this.)


Supplemental ExercisesExercise 38. Give reasons for every step in the proof of Proposition 11.

Exercise 39. Give reasons for every step in the proof of Proposition 12.

Exercise 40. In Exercise 20, we essentially showed that the distributiveproperties imply 0x = x0 = 0. Consider the two distributive propertiesseparately. Which one implies that 0x = 0 and which one implies x0 = 0?

Exercise 41. Suppose that X = { 0 }. How many different operations canyou find on X? Define addition and multiplication on X in the obviousway. Verify that X with this addition and multiplication is a field.

Exercise 42. Before continuing beyond this chapter, use your knowledgeof the sets N, Z+, Z, Q, R, and C to complete the following table. Foreach set, with the usual addition, multiplication, and ordering, write Yesor No for each property of an ordered field. For every No, try to givea reason or counterexample. (For example, N does not satisfy propertyA3, the existence of an additive identity, since 0 /∈ N and Z does notsatisfy property M4, the existence of multiplicative inverses, since 2 has nomultiplicative inverse in Z.)

Property N Z+ Z Q R C(AC) Commutative Addition(AA) Associative Addition(AZ) Additive Identity 0(AI) Additive Inverses

(MC) Commutative Multiplication(MA) Associative Multiplication(MU) Multiplicative Identity 1(MI) Multiplicative Inverses(DP) Distributive Properties

(IX) Cancellation in Inequalities(PP) Product of Positives

Exercise 43. Let X be the set of all 2 × 2 matrices with real entries. Usingordinary addition and multiplication of matrices, determine which of theproperties of a field hold. (Check each property for addition, (AC), (AA),(AZ), and (AI), for multiplication, (MC), (MA), (MU), and (MI), and thedistributive properties (DP).)

Exercise 44. Suppose X has addition and multiplication operations. Provethat if multiplication is commutative, then the First Distributive Property isequivalent to the Second Distributive Property.


Exercise 45. Explain why a multiplication operation is closed. Explain

why any binary operation is closed.

Exercise 46. Prove that in a field X , (−1)x = x(−1) = −x for all x ∈ X .

Exercise 47. Suppose that Z3 = { 1, 2, 3 } and addition and multiplicationare defined by the following tables.

+ 1 2 3

1 2 3 12 3 1 23 1 2 3

· 1 2 3

1 1 2 32 2 1 33 3 3 3

Prove that Z3 is a field.

Exercise 48. Consider the field Z3 from Exercise 47. Prove that the only

field automorphism of Z3 is the identity function, IZ3 .

Exercise 49. Suppose that V = { 0, 1, a, b } and addition and multiplicationare defined by the following tables.

+ 0 1 a b

0 0 1 a b1 1 0 b aa a b 0 1b b a 1 0

× 0 1 a b

0 0 0 0 01 0 1 a ba 0 a a 0b 0 b 0 b

Is V a field? If not, indicate all properties in the definition of a field thatfail.

Exercise 50. Consider V = { 0, 1, a, b } with the addition and multiplica-tion operations given in Exercise 49. Prove that the function f : V → Vdefined by f (0) = 0, f (1) = 1, f (a) = b and f (b) = a is an automor-phism; i.e., the two operations are preserved.

Exercise 51. Suppose X is a set with a partial ordering ≤ and x, y, z ∈ X .(a) Prove that if x = y and y < z, then x < z.(b) Prove that if x < y and y ≤ z, then x < z.(c) Prove that if x ≤ y and y < z, then x < z.(d) Prove that if x ≤ y and y = z, then x ≤ z.(e) Prove that if x = y and y ≤ z, then x ≤ z.

Exercise 52. Suppose F is an ordered field. Prove that, for all x, y ∈ F ,

x < y if and only if −y < −x .

Exercise 53. Suppose X is an ordered field. Prove that if 0 �= 1, then

1 > 0.

Exercise 54. Suppose X is a field. Prove that −(x + y) = (−x) + (−y).

131

Chapter 9Archimedean Ordered Fields

1. The Main Goal and Archimedean Principle.In constructing the number systems, we will check which of the prop-

erties of an ordered field are satisfied and whether the least upper boundproperty is satisfied. As we progress towards the real numbers, the new con-structs will possess more of these properties than their predecessors. Thisculminates in the construction of the real numbers; the complex numbersform an interesting digression or continuation, depending on your personalpoint of view.

The main goal of Part III of this text is to prove the following theorem,which is proved in Section 5 of Chapter 13.

Main Theorem (Theorem 28 of Chapter 13): The Uniqeness of R.There exists, up to isomorphism, a unique ordered field with the least upperbound property.

By “unique up to isomorphism” we mean unique isomorphism class.From the constructions, we will see that R, with the usual ordering andoperations, is the ordered field with the least upper bound property.

Let us now prove two important properties of ordered fields with theleast upper bound property. Let F be an ordered field with the least upperbound property. We denote the sets of natural, integral and rational fieldelements of F , respectively, by

FN = { m ∈ F | m = 1 + 1 + · · · + 1 for finitely many 1’s } ,

FZ = { m ∈ F | m = 0 or m ∈ FN or − m ∈ FN } , and

FQ = {mn−1 | m ∈ FZ and n ∈ FN

}.

The following result is usually called the Archimedean Principle.

Theorem 1. Let F be an ordered field with the least upper bound property.If x, y ∈ F and x > 0, then there exists n ∈ FN such that nx > y.

0

2xx 3x mx

yss-x

(m+1)x

132 Chapter 9. Archimedean Ordered Fields

Figure 2: The set S in the proof of Theorem 1

Proof. Let S = { nx | n ∈ FN }. Since x ∈ S, S is nonempty. Assumethat nx ≤ y for all n ∈ FN. Then y is an upper bound of S. Since F hasthe least upper bound property, S has a least upper bound, s. Now x > 0implies s − x < s. Thus, s − x is not an upper bound of S. Hence, thereexists m ∈ FN such that s − x < mx . Therefore, s < (m + 1)x . Thiscontradicts the fact that s is an upper bound of S. �

The Archimedean Principle is intimately related to the fact that thenatural numbers are not bounded.

Exercise 3. Let F be an ordered field with the least upper bound property.

Prove that FN is not bounded.

The following corollary of the Archimedean Principle is a useful,stronger version of the Archimedean Principle.

Corollary 4. Let F be an ordered field with the least upper bound property.If x, y ∈ F and x > 0, then there exists m ∈ FZ such that

mx ≤ y < (m + 1)x .

Proof. By the Archimedean Principle, there exists q ∈ FN such that qx >

y. Next, we show that there exists p ∈ FZ such that px < y. Consider twocases: either y ≥ 0 or y < 0.Case 1. If y ≥ 0, then −x < 0 ≤ y. Set p = −1 so that px < y.Case 2. If y ≥ 0, then, by the Archimedean Principle, there exists n ∈ FN

such that nx > −y. Hence, (−n)x < y. Set p = −n so that px < y.So, there exist p, q ∈ FZ with p < 0 and q > 0 such that

px < y < qx .

Consider the set

S = { n ∈ FZ | n ≥ p and nx ≤ y } .

S is nonempty since p ∈ S. If n ∈ S, then nx < y < qx and, hence,n < q. So q is an upper bound of S and S is a finite set. Since a finite,totally ordered set has a greatest element, let m be the greatest element of

Section 1. The Main Goal and Archimedean Principle 133

S. Since m ∈ S, mx ≤ y. Since m +1 > m, m +1 /∈ S. So (m +1)x �≤ y.Therefore, mx ≤ y < (m + 1)x . �

Exercise 5. Let F be an ordered field with the least upper bound property.If x, y ∈ F and x < 0, then there exists m ∈ FN such that mx ≤ y <

(m − 1)x .

For the following theorem, we sometimes say “the rational field setFQ is dense in the field F .”

Theorem 6. Let F be an ordered field with the least upper bound property.If x, y ∈ F and x < y, then there exists r ∈ FQ such that x < r < y.

Proof. We will use the Archimedean Principle and Corollary 4 to producem, n ∈ FN so that mn−1 is the desired element r of FQ.

Since x < y, y − x > 0. By the Archimedean Principle, there existsn ∈ FN such that n(y − x) > 1. Thus,

nx + 1 < ny. (7)

By Corollary 4, there exists m ∈ FZ such that m − 1 ≤ nx < m. Hence,

m ≤ nx + 1 and nx < m. (8)

Putting together the inequalities in (7) and (8) yields

nx < m ≤ nx + 1 < ny.

Therefore nx < m < ny and multiplying all three terms by n−1 gives thedesired result:

x < mn−1 < y. �

134

Chapter 10The Natural Numbers

1. Introduction.We now begin a long journey. We have used numbers (natural, integer,

rational, and real) in the previous chapters. A fundamental question is toask if these objects that we have been using even exist. One must go tothe very foundation of our axiomatic system to answer this fundamentalquestion. In our journey we will not quite go back to the beginning; insteadwe will take a more naıve approach and assume certain constructions arealready known to be realizable in our axiomatic development. The firstsuch naıve construction is the immediate successor of a set.

The following collection of “axioms” will be helpful. Depending onwhere we start in our development of set theory, these statements can beproved.

Theorem 1(Peano Axioms). (1) ∅ exists.(2) For every n ∈ Z+, there exists Succ(n) = n ∪ { n } ∈ Z+.(3) For every n ∈ Z+, 0 �= Succ(n).(4) For every m, n ∈ Z+, if m �= n, then Succ(m) �= Succ(n).(5) If S ⊂ Z+, 0 ∈ Z+, and, for every n ∈ Z+, Succ(n) ∈ S, then S = Z+.

Definition 2. The immediate successor of a set S is the set S ∪ { S }. Theimmediate successor of S will be denoted by Succ(S).

In our construction, we will actually start with 0. While we do notregard 0 as a natural number, it will be more natural for our construction todo this. In effect, we will construct the set { 0 } ∪N which we have decidedto call Z+. In the developement of numbers, the natural numbers came intouse first; the number 0 did not come into use until much later. The notionof constructing the numbers is far more recent yet; it is due to the work inset theory initiated in the nineteenth century by Georg Cantor and others.

The second naıve notion will be the principle of induction. Here, sincewe do not yet have the set N of natural numbers, we will use induction toprove a sequence of statements P(k) where

k = n, Succ(n), Succ(Succ(n)), . . . .

Section 2. Zero, the Natural Numbers and Addition 135

That is, we prove the “first” step P(n) and prove the inductive implicationP(k) implies P(Succ(k)).

2. Zero, the Natural Numbers and Addition.First we define 0.

Definition 3. Let 0 be ∅.

Next we use the immediate successor of a set to define the naturalnumbers inductively; this is another incidence of our naıvete.

Definition 4. Let 1 be Succ(0), 2 be Succ(1), 3 be Succ(2), 4 be Succ(3)

and so on. The set of all of the numbers constructed in this way is N, theset of natural numbers.

So you may have started to compile a list that looks something likethe following:

0 = ∅1 = 0 ∪ { 0 } = { ∅ }2 = 1 ∪ { 1 } = { ∅, { ∅ } }3 = 2 ∪ { 2 } = { ∅, { ∅ }, { ∅, { ∅ } } }

. . .

Unfortunately, having done this will not help us in the development of thenatural numbers.

Definition 5. Let n ∈ N. The set of successors of n or descendency set ofn is the subset

Desc(n) = { Succ(n), Succ(Succ(n)), Succ(Succ(Succ(n))), . . . }.For p ∈ N, we will use the notation Succp(n) to denote the pth successorof n, which is

Succ(. . . (Succ(n)) . . . ),

where the Succ is taken p times.

Note that n /∈ Desc(n). Otherwise, there would exist p ∈ N such thatn = Succp(n). ............................

Definition 6. We define a relation < on N by n < m if and only if m ∈Desc(n).

Exercise 7. Prove that < is a strict total ordering on N and, in fact, N is

well-ordered.

Using < and the usual rules for a strict total ordering, we can writen ≤ m, m > n and m ≥ n.

Before defining the operation of addition, we consider a type of inverseoperation to the immediate successor.

136 Chapter 10. The Natural Numbers

Definition 8. Let n ∈ N. The number m ∈ Z+ is the immediate predeces-sor of n iff the immediate successor of m is n. We denote the immediatepredecessor of n by Pred(n). That is, if the immediate predecessor of nexists, then n = Succ(Pred(n)).

Lemma 9. For all n ∈ Z+, n = Pred(Succ(n)).

Proof. Consider Pred(Succ(n)). By definition of the immediate predeces-sor, it is the number whose immediate successor is Succ(n), that is, n.

�

Next we define addition. The definition is again inductive.

Definition 10. Let m, n ∈ Z+. Define m + n to be

m + n ={

m, if n = 0

Succ(m + Pred(n)), if n > 0

m + n is called the sum of m and n. This defines addition on Z+.

Lemma 11. For all n ∈ Z+ and p ∈ N, n + p = Succp(n). Moreover,

Desc(n) = { n + 1, n + 2, n + 3, . . . }.

Proof. We proceed by induction on p. For p = 1,

n + 1 = Succ(n + Pred(1)) = Succ(n + 0) = Succ(n).

For the inductive step, assume the inductive hypothesis n + p = Succp(n).Now

n + Succ(p) = Succ(n + Pred(Succ(p))) = Succ(n + p)

= Succ(Succp(n)) = Succp+1(n). �

Exercise 12. Prove that, for all m, n ∈ N, m < n if and only if there exists

a unique p ∈ N such that m + p = n.

Let us consider some of the properties of addition in Z+.

Section 2. Zero, the Natural Numbers and Addition 137

Theorem 13. Addition on Z+ is associative.

Proof. We wish to prove that for all m, n, p ∈ Z+, (m + n) + p = m +(n+ p). Let m and n be arbitrary elements of Z+. We proceed by inductionon p.

Consider p = 0. By the definition of addition, (m + n) + 0 = m + nand m+(n+0) = m+n. Since = is an equivalence relation, (m+n)+0 =m + (n + 0).

For the inductive step, we assume the inductive hypothesis

(m + n) + p = m + (n + p).

Now, by definition of addition, Lemma 9 and the induction hypothesis,

(m + n) + Succ(p) = Succ((m + n) + Pred

(Succ(p)

))= Succ

((m + n) + p

)= Succ

(m + (n + p)

).

Also, using the definition of addition twice and Lemma 9 twice,

m + (n + Succ(p)) = Succ(m + Pred

(n + Succ(p)

))= Succ

(m + Pred

(Succ

(n + Pred

(Succ(p)

))))= Succ

(m + (n + p)

).

Using the fact that = is an equivalence relation yields the desired result.�

We next wish to show that addition on Z+ is commutative. Recall thatm + 0 = m for all m ∈ Z+ by definition of addition. The next exercise isthe first step in the proof of commutativity.

Exercise 14. Prove that 0 + m = m for all m ∈ Z+.

The following is a corollary.

Exercise 15. Prove that n = Succn(0) for all n ∈ N.

Together with the definition m + 0 = m, Exercise 14 shows that 0 isthe additive identity element of Z+. The following exercise justifies use ofthe definite article when we say “the” additive identity.

Exercise 16. Prove that 0 is the unique additive identity element in Z+.

Moreover, N does not have an additive identity.

138 Chapter 10. The Natural Numbers

Exercise 17. Prove that addition on Z+ is commutative.

We will refer to the following as the additive cancellation property forequality.

Exercise 18. Let m, n, p ∈ Z+. Prove that m + n = m + p if and only if

n = p.

Next we consider the relationship between the total ordering < andaddition.

Exercise 19. Let m, n ∈ Z+. Prove that m + n = 0 if and only if

m = n = 0.

Exercise 20. Let m, n, p ∈ Z+. Prove that m + n < m + p implies n < p.

We will refer to the previous as the additive cancellation property forinequality. When we say the cancellation properties for addition, we willmean both for equalities and for inequalities. Of course, the operation maychange!

3. Multiplication.Next we tackle multiplication in Z+.

Definition 21. Let m, n ∈ Z+. Define mn to be

mn ={

0, if n = 0

mPred(n) + m, if n > 0

mn is called the product of m and n. This defines multiplication on Z+.

We next examine the properties of multiplication. Since the definitionof multiplication resembles the distributive property, it seems the naturalplace to start.

Exercise 22. Prove that, for m, n, p ∈ Z+, (m + n)p = mp + np.

Exercise 23. Prove that 1 is the multiplicative identity for Z+.

Exercise 24. Prove that multiplication on Z+ is commutative.

Combining the distributive and commutative properties yields the sec-ond distributive property: for all m, n, p ∈ Z+, m(n + p) = mn + mp.

Section 3. Multiplication 139

Exercise 25. Prove that multiplication on Z+ is associative.

Exercise 26. Let m, n ∈ Z+. Prove that mn = 0 if and only if m = 0 or

n = 0.

The following is known as the multiplicative cancellation property.

Theorem 27. Let m, n, p ∈ Z+. Prove that if mn = mp and m �= 0, thenn = p.

Proof. Assume n �= p. By trichotomy, either n < p or n > p. First, letus suppose that n < p. Hence, there exists q ∈ N such that p = n + q (byExercise 12). Now

mn + 0 = mn = mp = m(n + q) = mn + mq.

By additive cancellation, 0 = mq. By Exercise 26, m = 0 or q = 0. Sincem �= 0, we must have q = 0. This contradicts that q ∈ N. The case n > palso leads to a contradiction. Therefore, n = p. �

Exercise 28. Complete the proof of Theorem 27: consider the case of

n > p.

Next we have the other cancellation property for multiplication.

Exercise 29. Let m, n, p ∈ Z+ and m �= 0. Prove that n < p if and only

if mn < mp.

140

Summary of the Properties of the Non-negative Integers

The set Z+ = { 0, 1, 2, 3, 4, 5, . . . }, with addition, multiplication and anorder relation, satisfies all of the following properties, for all m, n, p ∈ Z+:

(AC): m + n = n + m(AA): (m + n) + p = m + (n + p)

(AZ): 0 is the unique element such that 0 + m = m + 0 = m(AX): m + n = m + p if and only if n = p and

n + m = p + m if and only if n = p(MC): mn = nm(MA): (mn)p = m(np)

(MU): 1 is the unique element such that 1m = m1 = m(MX): for m �= 0, mn = mp if and only if n = p and

for m �= 0, nm = pm if and only if n = p(DP): (m + n)p = mp + np and m(n + p) = mn + mp(PP): if m > 0 and n > 0 then mn > 0(IX): n + m < p + m if and only if n < p and

n + m < p + m if and only if n < p

Despite the fact that there are no additive inverses for nonzero elements ofZ+, the following Difference Property holds: for all m, n ∈ Z+,

there exists a unique d ∈ N such that m +d = n if and only if m < n.

This property can be thought of as the definition of the relation <. Recallthat m ≤ n is defined by m < n or m = n.

141

Chapter 11The Integers

1. Introduction: Integers as Equivalence Classes.In Chapter 10, we constructed the following set:

Z+ = { 0, 1, 2, 3, 4, 5, . . . }.We called this the set of non-negative integers. We constructed two oper-ations, addition and multiplication, on Z+ and a strict order relation <. Inthis chapter, you will not need to refer back to the construction in Chapter10; rather, we have listed on the previous page all of the relevant propertiesof the construction. These properties of Z+ are all that we will need in thischapter.

Addition and multiplication were the two operations defined on Z+.Notice that subtraction and division were never mentioned. In fact neitheroperation can be defined universally on Z+. The deficiency of Z+ withrespect to subtraction vis-a-vis addition necessitated its extension to the setof the integers which will be discussed in this chapter. For each pair ofnumbers m and n we wish the difference m − n to have a meaning, that is,we require the existence of a number q such that m = n + q.

A natural way to proceed would be to attempt to expand Z+ into Z.One way to do this is to assume that, for each m ∈ Z+, there exists a numberm ′ such that m + m ′ = 0; we call m ′ the (additive) inverse of m. Then theset of all integers Z would be the set of all natural numbers together withtheir inverses and 0.

Setting aside for a moment the problem of defining the new negativeelements, it turns out that this approach makes it difficult to give even acumbersome definition of addition. Such a definition, with several differentcases, makes the proofs of the basic properties of the operations in Z quitetedious.

Therefore, it is preferable to start with the following notion. We notethat, for example,

−1 = 0 − 1 = 1 − 2 = 2 − 3 = 3 − 4 = 4 − 5 = . . .

142 Chapter 11. The Integers

Similar statements can be made for all negative integers and, in fact, forall integers. We can therefore formulate a definition of the integers asequivalence classes of pairs of nonnegative integers.

This new construction has the obvious disadvantage that Z no longercontains Z+ as we had previously defined it. However, in Section 4, weshow that Z+ is embedded in Z; this allows us to think of our previouslyconstructed set Z+ as a “subset” of Z.

We start with a definition of a set which we will call Z, the set ofintegers.

Definition 1. Define a relation ∼ on Z+ × Z+ by (m, n) ∼ (p, q) if andonly if m + q = p + n.

Exercise 2. Prove that the relation ∼ on Z+×Z+ is an equivalence relation.

Definition 3. The set of integers is (Z+ × Z+)/∼. We denote this set asZ. We denote the equivalence class of (m, n) by [m, n]; this is an elementof Z.

2. A Total Ordering of the Integers.First, let us define a relation on Z that we will show is a total ordering.

Definition 4. Define the relation < on Z by [m, n] < [p, q] if and only ifm + q < p + n.

It is important to realize that the symbol < is used in two different waysin the definition above. The first is a new relation which we are defining onZ. The second is our old strict total ordering on Z+. We could use differentsymbols for each, e.g., use subscripts as in <Z+ and <Z, but we find thesetoo hideous. With care, there need be no confusion.

As with any operation on equivalence classes, we must check thatthis is well-defined. That is, we must check that all representatives of theequivalence classes of (m, n) and (p, q) satisfy the defining condition forthe operation <. Before doing this, we remind you of Proposition 24 inChapter 8: if x < y and y = z, then x < z for all x, y, z ∈ X and a strictpartial ordering < on X .

Proposition 5. The relation < on Z is well-defined.

Proof. Suppose that [m, n], [p, q] ∈ Z and [m, n] < [p, q]. Thus, m +q < n + p. We must show that if (m ′, n′) ∈ [m, n] and (p′, q ′) ∈ [p, q],then m ′ + q ′ < n′ + p′. Since (m ′, n′) ∈ [m, n] and (p′, q ′) ∈ [p, q],

m ′ + n = m + n′ and p′ + q = p + q ′,

Section 2. A Total Ordering of the Integers 143

respectively. Adding these two equations gives us the following:

(m ′ + n) + (p + q ′) = (m + n′) + (p′ + q).

Using the commutative and associative properties of addition in Z+ yields:

(m ′ + q ′) + (n + p) = (n′ + p′) + (m + q). (6)

Since m + q < n + p,

(m ′ + q ′) + (m + q) < (m ′ + q ′) + (n + p). (7)

Now, (6) and (7) imply

(m ′ + q ′) + d) + (m + q) < (n′ + p′) + (m + q).

By cancellation in Z+,

m ′ + q ′ < n′ + p′. �

Notice that we used only properties (A1), (A2) and (L3) for Z+ in theproof above.

Exercise 8. Prove that < is a strict total ordering on Z.

As for any strict total ordering <, there is an associated total ordering≤: define [m, n] ≤ [p, q] iff [m, n] < [p, q] or [m, n] = [p, q].

Recall that Z+ is well-ordered. That is, Z+ is totally ordered andevery nonempty subset of Z+ has a least element. This total ordering onZ+ extends to (or induces) a total ordering on Z, but ≤ is not a well-orderingon Z.

Exercise 9. Prove that < is not a well-ordering on Z.

Theorem 10. For all [m, n] ∈ Z, there exists a unique p ∈ Z+ such that

[m, n] ={

[p, 0], m ≥ n

[0, p], m ≤ n.

Proof. If m = n, then m + 0 = n + 0 = 0 + n. Hence, [m, n] = [0, 0]and p = 0.

If m > n, then, by the Difference Property for Z+, there exists aunique p such that m = n + p. Hence, m +0 = n + p and [m, n] = [p, 0].

If m < n, then there exists a unique p such that n = m + p. Hence,m + p = n + 0 and [m, n] = [0, p]. �

At this point in our discussion we tire of writing [m, n] for an integerand go back to the more usual looking form p ∈ Z. We will write p for[p, 0], 0 for [0, 0] and −p for [0, p]. You should be very careful here.Nothing has changed; elements of Z are still equivalence classes of orderedpairs!


Definition 11. Let m ∈ Z. m is positive iff m > 0 and m is negative iffm < 0.

Exercise 12. Let m ∈ Z. Prove that exactly one of the following holds: m

is positive, m is negative, or m = 0.

3. Addition of Integers.We next define the operation of addition on Z. The idea behind the

definition is that we want (m − n) + (p − q) = (m + p) − (n + q).

Definition 13. Define addition in Z by

[m, n] + [p, q] = [m + p, n + q]

for all [m, n], [p, q] ∈ Z.

It is important to realize that the symbol + is used in two differentways in the definition of addition. The first is addition in Z, that is, additionof equivalence classes. The second and third are our old addition in Z+.

As with any operation on equivalence classes, we must check that thisis well-defined. That is, we must check that the equivalence class of thesum (m + p, n + q) is independent of the choice of representatives of theequivalence classes [m, n] and [p, q].

Exercise 14. Prove that addition on Z is well-defined.

Recall that addition in Z+ is commutative and associative with a uniqueidentity. The corresponding properties for addition in Z follow.

Exercise 15. Prove that addition on Z is commutative.

Exercise 16. Prove that addition on Z is associative.

Exercise 17. Prove that 0 is the unique additive identity for Z.

For algebraic structures, we usually denote the additive identity ele-ment as 0. Recall that 0 = [0, 0]. Again you should see that there are twomeanings of 0 in the previous statement.

We come to the punch line of this chapter now: the purpose of definingnegative integers was to have additive inverses.

Exercise 18. Prove that every element of Z has a unique additive inverse.

Section 4. Multiplication of Integers 145

As one might expect, the additive inverse of m is denoted by −m. Wecan now define subtraction: m − n = m + (−n) for all m, n ∈ Z.

Exercise 19. Prove that, for every m ∈ Z, −(−m) = m.

The proof of the cancellation properties is now much simpler than itwas for Z+.

Exercise 20. Prove that, for all m, n, p ∈ Z, if m + n = m + p, then

n = p.

Commutativity of addition and Exercise 20 imply that, for all m, n, p ∈Z, if n + m = p + m, then n = p.

Exercise 21. Prove that, for all m, n, p ∈ Z, if m + n < m + p, then

n < p.

4. Multiplication of Integers.Next, we consider multiplication. Since we want (m − n)(p − q) =

mp − mq − np + nq , we define multiplication as follows.

Definition 22. Define multiplication in Z by

[m, n][p, q] = [mp + nq, np + mq]

for all [m, n], [p, q] ∈ Z.

Proposition 23. Multiplication on Z is well-defined.

Proof. Suppose (m ′, n′) ∈ [m, n] ∈ Z and (p′, q ′) ∈ [p, q] ∈ Z. Hence,m ′ + n = n′ + m and p′ + q = q ′ + p. We will use properties of additionin Z+. Multiplying the previous equations by equals yields the following:

m(q ′ + p) = m(p′ + q)

n(p′ + q) = n(q ′ + p)

(n′ + m)p′ = (m ′ + n)p′

(m ′ + n)q ′ = (n′ + m)q ′

By the distributive properties,

mq ′ + mp = mp′ + mq

np′ + nq = nq ′ + np

n′ p′ + mp′ = m ′ p′ + np′

m ′q ′ + nq ′ = n′q ′ + mq ′


By adding these equations, we obtain

(mq ′ + mp) + (np′ + nq) + (n′ p′ + mp′) + (m ′q ′ + nq ′)

= (mp′ + mq) + (nq ′ + np) + (m ′ p′ + np′) + (n′q ′ + mq ′).

By the associative and commutative properties, we rewrite this as

(mp + nq) + (n′ p′ + m ′q ′) + (mq ′ + np′ + mp′ + nq ′)

= (np + mq) + (m ′ p′ + n′q ′) + (mq ′ + np′ + mp′ + nq ′).

By the cancellation property,

(mp + nq) + (n′ p′ + m ′q ′) = (np + mq) + (m ′ p′ + n′q ′),

which implies

[m, n][p, q] = [mp + nq, np + mq]

= [m ′ p′ + n′q ′, n′ p′ + m ′q ′]

= [m ′, n′][p′, q ′]

Therefore, multiplication is well-defined. �

Exercise 24. Prove that multiplication on Z is commutative.

Exercise 25. Prove that multiplication on Z is associative.

Exercise 26. Prove that 1 is the unique multiplicative identity for Z.

Notice that we again have labeled a multiplicative identity 1. However,Z is lacking multiplicative inverses. To prove this, we will use the followingexercises.

Exercise 27. Prove that the distributive property holds in Z.

Exercise 28. Prove that mn > 0 if and only if m > 0 and n > 0, or m < 0

and n < 0.

Alas, Z still has deficiencies, as the following indicates.

Section 5. Embedding the Natural Numbers in the Integers 147

Proposition 29. 2 has no multiplicative inverse in Z.

Proof. Assume that there is an m ∈ Z such that 2m = 1. If m ≤ 0, then2m ≤ 0 and, hence, 2m �= 1. If m > 0, then 2m = (1 + 1)m = m + m >

m ≥ 1. In any case, 2m �= 1 and hence 2 has no multiplicative inverse inZ. �

We end this section with two more standard algebraic properties.

Exercise 30. Prove that, for all m ∈ Z, 0m = 0.

Exercise 31. Prove that, for all m, n ∈ Z, m(−n) = (−m)n = −(mn).

5. Embedding the Natural Numbers in the Integers.It is important to realize that, contrary to what you might have ex-

pected, N and Z+, as constructed in Chapter 10, are not subsets of Z =(Z+ × Z+)/∼. This is obviously true since an element of Z is defined asthe equivalence class of a pair of elements of Z+.

However, this poses no problem, since we can think of Z+ as a subsetof Z in a very natural way. Recall the discussion of isomorphic structuresin Section 4 of Chapter 8. In our case, we have Z+ and Z as the underlyingsets of our structures and the order relations and operations of addition andmultiplication providing the organization of these structures.

Do you see an isomorphic copy of Z+ in Z? If so, you are now in aposition to prove the following exercise.

Exercise 32. Prove that there exists an embedding of Z+ into Z.

Since isomorphic structures can be thought of as identical, we willsimply say, from now on, that Z+ is a subset of Z. To be more precise, wecould have called our original Z+ something else and called its isomorphicimage Z+! That would be OK until we define the rational numbers, whenwe will have to repeat the process of finding an embedding yet again (andagain for real numbers and yet again for complex numbers).


Supplemental ExercisesExercise 33. Explain what is wrong with the notation [−p, 0] ∈ Z. (The

statement [p, 0] + [−p, 0] = 0 is nonsense!)

Exercise 34. Explain what is wrong with the following proof of the ex-istence of additive inverses in Z (Exercise 18): For every [p, 0] ∈ Z,[p, 0] + [0, p] = [p, p] = [0, 0] = 0. Similarly, [0, p] + [p, 0] =[p, p] = [0, 0] = 0.

149

Summary of the Properties of the Integers

The set of integers, Z, with addition (+), multiplication (·) and a strict totalordering (<), satisfies all of the following properties, for all m, n, p ∈ Z:

(AC): m + n = n + m(AA): (m + n) + p = m + (n + p)

(AZ): 0 is the unique element such that 0 + m = m + 0 = m(AI): −m is the unique element such that −m + m = m + (−m) = 0

(AX): m + n = m + p if and only if n = p andn + m = p + m if and only if n = p

(MC): mn = nm(MA): (mn)p = m(np)

(MU): 1 is the unique element such that 1m = m1 = m(MX): for m �= 0, mn = mp if and only if n = p and

for m �= 0, nm = pm if and only if n = p(DP): (m + n)p = mp + np and m(n + p) = mn + mp(PP): if m > 0 and n > 0 then mn > 0(IX): n + m < p + m if and only if n < p and

n + m < p + m if and only if n < p

The following properties also hold: for all m, n, p ∈ Z+,

Multiplication by 0: 0m = m0 = 0Factoring a Minus: (-m) n = m(-n) = -(mn)Negation Reverses Order: m > 0 if and only if −m < 0

150

Chapter 12The Rational Numbers

1. Introduction: Rationals as Equivalence Classes.We now turn to the construction of the rationals which, from a historical

point of view, were actually identified before the integers; it seems easier tounderstand what one half of a melon means than to understand zero melonor minus one melon.

As we commented in the introduction of the previous chapter, to makedivision between any two integers as universally possible as multiplication,we need an extension of our number system. The construction of Q fromZ is quite similar to the construction of Z from Z+ in the previous chapter.

For each pair of integers m and n, we require the existence of a uniquerational number q (to be viewed as the quotient m/n), such that

m = nq.

Division by n = 0 is not allowed for the following reasons. Suppose n = 0.Since any product with 0 in Z equals 0 we would get

m = 0q = 0.

By the transitivity of equality on Z, m would have to be 0. This is no goodeither since

0 = 0q

has infinitely many solutions in Z.*Given a rational number q, the numbers m and n satisfying the equality

m = nq , i.e., mn = q, are not uniquely determined. For example

1

2= 2

4= 3

6= 4

8= 5

10= . . . .

Therefore, we define the rationals to be equivalence classes of pairs ofintegers.

Recall that Z∗ = Z − { 0 }, the set of all nonzero integers.

*So if anyone asks you “What is 0 divided by 0?” you can answer with any number youlike! More seriously, this may remind you of limits like lim

x→0

kxx = k for every k ∈ R.

Section 2. A Total Ordering of the Rationals 151

Definition 1. Define a relation ∼ on Z×Z∗ by (m, n) ∼ (p, q) if and onlyif mq = pn.

Exercise 2. Prove that ∼ is an equivalence relation on Z × Z∗.

Definition 3. The set of rational numbers is the set (Z×Z∗)/∼. We denotethis set by Q. The equivalence class of (m, n) is denoted by [m, n]; that is,[m, n] is an element of Q.

We could have chosen to define rational numbers differently, onlyallowing positive n (for the “denominator”). The following two exercisesshow that this is possible.

Exercise 4. Prove that, for all (m, n) ∈ Z × Z∗, [m, n] = [−m, −n].

Exercise 5. Prove that, for all [m, n] ∈ Q, there exists (p, q) ∈ [m, n] with

q > 0.

Remark 6: Important! For the remainder of this chapter, when we con-sider a rational number [m, n], we will assume that n > 0. It does notmuch matter whether you think that Q = (Z × N)/∼ or you think we onlychoose representatives (m, n) of rational numbers with n ∈ N.

2. A Total Ordering of the Rationals.We now define a total ordering on Q. The definition is suggested by

the following statement:

m

n<

p

qimplies mq < pn.

Notice that Remark 6 is crucial here: since n and q are positive, this works.On the contrary,

−1

2<

1

−3does not imply (−1)(−3) < (2)(1).

Definition 7. Define the relation < on Q by [m, n] < [p, q] if and only ifmq < pn.

Again, note that the symbol < is used in two different ways in Defi-nition 7.

Exercise 8. Prove that the relation < on Q is well-defined.

152 Chapter 12. The Rational Numbers

Next, we check that < is a strict total ordering on Q.

Exercise 9. Prove that < is a strict total ordering on Q.

Remember that a strict total ordering is trichotomous. Of course, wecan define a total ordering ≤ in terms of < and =. As was the case for thetotal ordering on Z, < is not a well-ordering.

Exercise 10. Prove that < is not a well-ordering on Q. Moreover, Q has

no least element.

Let 0 = [0, 1]; we will see in Exercise 17 that 0 is the unique additiveidentity. Since n > 0, we wish that

m

n>

0

1implies m = (m)(1) > (n)(0) = 0.

Definition 11. Suppose [m, n] ∈ Q. [m, n] is positive if and only if m > 0and [m, n] is negative if and only if m < 0.

Instead of always writing an element in Q as the equivalence class ofa pair of integers, we can simply write q ∈ Q. Nothing has changed: thereis a pair (m, n) such that q = [m, n].

Exercise 12. Let q ∈ Q. Prove that exactly one of the following holds: q

is positive, q is negative, q = 0.

3. Addition of Rationals.You remember that addition of rationals (fractions) is a bit more dif-

ficult than multiplication of rationals (fractions). Nevertheless, let us startwith addition. Following the idea that

m

n+ p

q= mq + np

nq,

we next define addition.

Definition 13. Define addition in Q by

[m, n] + [p, q] = [mq + np, nq]

for all [m, n], [p, q] ∈ Q.

In the definition above, the + sign on the first side is understood todenote the operation of addition in Q, which differs from the operation of

Section 4. Multiplication of Rationals 153

addition in Z that appears on the second side and is denoted by the samesymbol.

Since Definition 13 depends on the choice of equivalence class repre-sentatives, we must check that addition is well-defined.

Exercise 14. Prove that addition on Q is well-defined.

The following four exercises establish properties (AC), (AA), (AZ)and (AI) for addition in Q.

Exercise 15. Prove that addition on Q is commutative.

Exercise 16. Prove that addition on Q is associative.

Exercise 17. Prove that 0 = [0, 1] is the unique additive identity for Q.

Exercise 18. Prove that every element of Q has a unique additive inverse.

Again, we denote the additive inverse of a number x ∈ Q by −x .Subtraction is defined as with the integers: for all x, y ∈ Q,

x − y = x + (−y).

Property (AX) is established by the following exercise.

Exercise 19. Prove that the cancellation properties hold for addition in Q:(a) for all x, y, z ∈ Q, x + y = x + z if and only if y = z and(b) for all x, y, z ∈ Q, y + x = z + x if and only if y = z.

4. Multiplication of Rationals.Since we want (

m

n

) (p

q

)= mp

nq,

we next define multiplication as follows.

Definition 20. Define multiplication in Q by

[m, n][p, q] = [mp, nq]

for all [m, n], [p, q] ∈ Q.

Again, since Definition 20 depends on the choice of equivalence classrepresentatives, we must check that multiplication is well-defined.


Exercise 21. Prove that multiplication on Q is well-defined.

The following four exercises establish properties (MC), (MA) and(MU) for multiplication in Q.

Exercise 22. Prove that multiplication on Q is commutative.

Exercise 23. Prove that multiplication on Q is associative.

Exercise 24. Prove that [1, 1] is the unique multiplicative identity for Q.

We set 1 = [1, 1]. By now you should realize the special nature of therationals of the form [m, 1].

Set Q∗ = Q −{ 0 }. The following exercise establishes property (MI)for multiplication in Q.

Exercise 25. Prove that, for each x ∈ Q∗, there exists a unique multiplica-

tive inverse.

Again, we denote the multiplicative inverse of a number x ∈ Q byx−1. Division is defined, for all x ∈ Q and y ∈ Q∗, by

x ÷ y = x

y= x/y = xy−1.

The following exercise establishes property (MX) for multiplication in Q.

Exercise 26. Prove that the cancellation properties hold for multiplicationin Q:(a) for all x ∈ Q and y, z ∈ Q, xy = xz if and only if y = z and(b) for all x ∈ Q and y, z ∈ Q, yx = zx if and only if y = z.

5. An Ordered Field Containing the Integers.To show that Q is a field, all that remains is to verify the distributive

property (DP). First, we give an exercise that will be needed in the proofof the distributive property.

Exercise 27. Prove that [mp, np] = [m, n] in Q for all m ∈ Z and

n, p ∈ Z∗.

While you probably proved Exercise 27 by using the arithmetic of Z,we could argue as follows. We can show that [mp, np] = [m, n][p, p] andthat [p, p] = 1 (see Exercise S12). The result follows from a property of the

Section 5. An Ordered Field Containing the Integers 155

multiplicative identity. That is, Exercise 27 simply states that (x)(1) = xfor all x ∈ Q.

Exercise 28. Prove that the distributive property holds in Q.

Exercise 28 completes the proof that Q is a field.We next show that Q is an ordered field. We need to verify that

properties (PP) and (IX) hold.

Exercise 29. Suppose x, y ∈ Q. Prove that xy > 0 if and only if either

x > 0 and y > 0, or x < 0 and y < 0.

Exercise 30. Prove that cancellation in inequalities holds in Q:(a) for all x, y, z ∈ Q, x + y < x + z if and only if y < z and(b) for all x, y, z ∈ Q, y + x < z + x if and only if y < z.

This shows that Q is an ordered field.Of course, we “think” of Z as a subset of Q even though it is not.

Exercise 31. Prove that there exists an embedding of Z into Q.

The next two exercises show that Q satisfies the Archimedean Principleand as well as a stronger version of the Archimedean Principle.

Exercise 32. Prove that if x, y ∈ Q and x > 0, then there exists n ∈ Nsuch that nx > y.

Exercise 33. Prove that if x, y ∈ Q and x > 0, then there exists n ∈ Zsuch that nx ≤ y < (n + 1)x .

The next exercise shows that there are many rational numbers using a“denseness” type of property.

Exercise 34. Prove that if x, z ∈ Q and x < z, then there exists y ∈ Qsuch that x < y < z.

As discussed in Chapter 6, Q does not have the least upper boundproperty. We postpone a proof of this fact until we have constructed thereal numbers. At that point we will be able to show that

√2 ∈ R and use

the denseness of Q in R to derive a contradiction from the assumption thatthe least upper bound property holds for Q.

It is interesting to note that Q is an ordered field without the least upperbound property that nevertheless satisfies the Archimedean Principle. Youshould compare this with Theorem 1 of Chapter 9.


Supplemental Exercises

Exercise S1. Explain why [1, 0] is not an element of Q.

Exercise S2. Verify the following using Definition 3 of elements of Q:

(a) [1, 2] = [2, 4] in Q,(b) [−11, −6] = [11, 6] in Q,(c) [−3, 5] = [3, −5] in Q.

Exercise S3. Use Definition 3 to do the following:

(a) find (m, n) ∈ [5, −7] in Q such that n > 0,(b) find (p, q) ∈ [−16, 3] in Q such that q < 0,(c) find (r, s) ∈ [−6, −13] in Q such that s > 0.

Exercise S4. Show that [0, n] = [0, 1] in Q for all n ∈ Z∗.

Exercise S5. Show that [2, 3] = [2n, 3n] in Q for all n ∈ Z∗.

Exercise S6. Verify the following using Definition 7 of order on Q:

(a) [1, 2] < [2, 3] in Q,(b) [−10, 3] > [−7, 2] in Q,(c) [−10, 3] > [7, −2] in Q. [Be careful!]

Exercise S7. Prove that (m, n) ∈ Z × Z∗ represents a positive rationalnumber if and only if mn > 0 and (m, n) represents a negative rationalnumber if and only if mn < 0.

Exercise S8. Suppose that m, n ∈ N. For each of the following pairs ofelements of Q, determine which is greater:

(a) [m, n] and [m, n + 1](b) [m, n] and [m + 1, n]

Exercise S9. Repeat Exercise S8 with m ∈ Z∗− (i.e., m < 0) and n ∈ N.

Exercise S10. Suppose that m, n ∈ N. Under what condition on m and n

is it true that [m, n + 1] ≤ [m − 1, n]?

Exercise S11. Compute the following using Definition 13 of addition onQ:

(a) [2, 3] + [3, 4] in Q,(b) [−2, 5] + [−7, 3] in Q,(c) [−4, 8] + [9, 18] in Q.

Exercise S12. Show that [n, n] = [1, 1] in Q for all n ∈ Z∗.

Exercise S13. Show that 0 �= 1 in Q.


Exercise S14. Compute the following using Definition 20 of multiplicationon Q:

(a) [2, 3][3, 4] in Q,(b) [−2, 5][−7, 3] in Q,(c) [−4, 8][9, 18] in Q.

158

Summary of the Properties of the Rationals

The set of rationals, Q, with addition, multiplication and an order relationis an ordered field. That is, all of the following properties are satisfied, forall x, y, z ∈ Q:

(AC): x + y = y + x(AA): (x + y) + z = x + (y + z)(AZ): 0 is the unique element such that 0 + x = x + 0 = x(AI): −x is the unique element such that −x + x = x + (−x) = 0

(AX): x + y = x + z if and only if y = z andy + x = z + x if and only if y = z

(MC): xy = yx(MA): (xy)z = x(yz)(MU): 1 is the unique element such that 1x = x1 = x(MI): x−1 is the unique element such that x−1x = xx−1 = 1

(MX): for x �= 0, xy = xz if and only if y = z andfor x �= 0, yx = zx if and only if y = z

(DP): (x + y)z = xz + yz and x(y + z) = xy + xz(PP): if x > 0 and y > 0, then xy > 0(IX): y + x < z + x if and only if y < z and

y + x < z + x if and only if y < z

159

Chapter 13

The Real Numbers

1. Dedekind Cuts.The set of rationals, Q, has the defect that it does not have the least

upper bound property. Another way of looking at this is to say that thepolynomial x2 − 2 has no rational roots. In this chapter we construct thefamiliar real numbers. The first constructions were given, independently,by Richard Dedekind and Georg Cantor in 1872. In this chapter, we willpresent the construction due to Dedekind.

We saw, in Chapter 6, that{

x ∈ Q∣∣ 0 < x2 < 2

}does not have a

least upper bound in Q. We want a set like this to represent the real number√2 for us. In this spirit, what set should represent −√

2? We modify ouroriginal idea somewhat to let

{x ∈ Q

∣∣ x < 0 or x2 < 2}

and{

x ∈ Q∣∣ x < 0 and x2 > 2

}

to represent√

2 and −√2, respectively. That is, a real number is to be

represented by the set of all rational numbers that we wish to be less thanthe given real number. Then it makes sense to let

{ x ∈ Q | x < q }

represent the rational number q. Let us formalize this.

Definition 1. A Dedekind cut is a set D ⊂ Q satisfying

(1) D �= ∅,(2) D �= Q,(3) if x ∈ D, y ∈ Q and y < x , then y ∈ D,(4) if x ∈ D, then there exists y ∈ D such that x < y.

A Dedekind cut is also called a cut or a real number. The set of all cuts isdenoted R.

160 Chapter 13. The Real Numbers

We have defined a cut D to be a nonempty, proper subset of Q withno greatest element containing all rational numbers less than any numberin D. Note that another reasonable candidate to represent q is

D′ = { x ∈ Q | x ≤ q } .

Since we are trying to connect real numbers with the least upper bounds ofsets of rational numbers that are bounded above, sets like D′ are not useful.Note that the least upper bound of D′ is the greatest element, q. Property(4) precludes cuts from having a greatest element.

Throughout this chapter you should keep in mind that real numbersare defined as sets of rational numbers; two real numbers are equal if andonly if they are the same subset of Q.

Important Remark. It will be useful in this chapter to denote all rationalnumbers by lower case letters and all cuts by upper case letters. We willassume this to be the case unless otherwise specified.

2. Order and Addition of Real Numbers.Before continuing with the construction, let us consider property (3)

of a cut D that x ∈ D and y < x implies y ∈ D. This means that Dcontains all rational numbers which are less than some rational number inD. Two statements equivalent to this are given in the following exercise.

Exercise 2. Suppose D ∈ R and x, y ∈ Q. Prove that the following areequivalent.(a) If x ∈ D and y < x , then y ∈ D.(b) If x ∈ D and y /∈ D, then y > x .(c) If x /∈ D and x < y, then y /∈ D.

First, we define a total ordering on R in the natural way.

Definition 3. Define the relation < on R by A < B iff A � B.

You may find it more appealing to rewrite the condition as A ≤ B ifand only if A ⊂ B.

Exercise 4. Prove that < is a total ordering on R.

Exercise 5. Prove that R has the least upper bound property.

Next we define addition, whose definition is also quite natural.

Section 2. Order and Addition of Real Numbers 161

Definition 6. Define addition on R by

X + Y = { x + y | x ∈ X and y ∈ Y } .

Unlike our previous constructions, we must prove that X + Y ∈ R,that is, the closure property of addition.

Exercise 7. Prove that addition on R is closed.

Exercise 8. Prove that addition on R is commutative.

Exercise 9. Prove that addition on R is associative.

Exercise 10. Prove that { x ∈ Q | x < 0 } is the unique additive identity on

R.

As always, we denote the additive identity as 0.The existence of additive inverses is more difficult. Intuitively, we

might let the additive inverse of X be the set of all x such that −x /∈ X .This will not quite work since this set may have a greatest element. Forexample, consider

X = { x ∈ Q | x < 0 } .

The set

{ x ∈ Q | −x /∈ X } = { x ∈ Q | −x �< 0 } = { x ∈ Q | x ≤ 0 }has 0 as its greatest element and hence is not a cut.

We define the following set:

−X = { x ∈ Q | −x /∈ X and −x is not the least element of Q − X }= { x ∈ Q | there exists y > 0 such that −x − y /∈ X } .

We must show that −X ∈ R, and X + (−X) = 0 or −X + X = 0.

Lemma 11. For all X ∈ R, −X ∈ R.

Proof. Recall the 4 properties in Definition 1 of a cut. By property (2),X �= Q. So there exists y ∈ Q such that y /∈ X . Set x = −y − 1. Hence,y = −x − 1 = −(x + 1) /∈ X . Since −(x + 1) < −x , −x /∈ X andx ∈ −X . This proves property (1), −X �= ∅.

If y ∈ X , then −y /∈ −X . This proves property (2), −X �= Q.Choose x ∈ −X and w > 0 such that −x − w /∈ X . Suppose y < x .

Then −y − w > −x − w and hence −y − w /∈ X . Therefore, y ∈ −Xand property (3) is proved.

For x and w as in the previous paragraph, z = x + w2 > x and

−z − w2 = −x − w /∈ X . Therefore, z ∈ −X and property (4) is proved.

�


Lemma 12. For all X ∈ R, X + (−X) = 0.

Proof. Let x ∈ X and y ∈ −X . Thus, −y /∈ X . By Exercise 2, −y > x .Therefore, x + y < 0 and X + (−X) ≤ 0.

On the other hand, choose v ∈ 0. Let w = − v2 . Since v < 0, w > 0.

By Exercise 33 of Chapter 12, a strong version of the Archimedean Propertyof Q, there exists n ∈ Z such that nw ∈ X but (n + 1)w /∈ X . Set x = nw

and y = −(n + 2)w. Since −y − w = (n + 2)w − w = (n + 1)w /∈ X ,y ∈ −X . Now x + y = nw − (n + 2)w = −2w = v. Therefore,0 ≤ X + (−X).

Combining the above results gives us X + (−X) = 0. �

Exercise 13. Complete the proof that every element of R has a unique

additive inverse.

Having proved the properties for addition, we know from Chapter 8,that several secondary properties hold.

Exercise 14. Prove that cancellation in inequalities holds: X +Y < X + Z

if and only if Y < Z for all X, Y, Z ∈ R.

3. Multiplication of Real Numbers.Defining multiplication in R is a little less natural. First we consider

the product of positive real numbers. We use the desired rules of signs todefine products involving negative real numbers.

Definition 15. Let X, Y ∈ R. If X > 0 and Y > 0, then

XY = { xy | x ∈ X and y ∈ Y } .

Otherwise, define the product as follows:

XY =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

0, if X = 0 or Y = 0

−[(−X)Y ], if X < 0 and Y > 0

−[X (−Y )], if X > 0 and Y < 0

(−X)(−Y ), if X < 0 and Y < 0

We will prove the primary properties for multiplication on the set

R ∗+ = { X ∈ R | X > 0 }

of positive real numbers first.

Section 4. Embedding the Rationals in the Reals 163

Exercise 16. Prove that multiplication on R ∗+ is closed.

This also proves the product of positives property: if X > 0 andY > 0, then XY > 0 for all X, Y ∈ R.

Exercise 17. Prove that multiplication on R ∗+ is commutative.

Exercise 18. Prove that multiplication on R ∗+ is associative.

Exercise 19. Prove that { x ∈ Q | x < 1 } is the multiplicative identity on

R ∗+.

As always, we denote the multiplicative identity as 1.

Exercise 20. Prove that every X ∈ R ∗+ has a multiplicative inverse.

Second, prove the primary properties for multiplication on R.

Exercise 21. Prove that multiplication on R is closed.

Exercise 22. Prove that multiplication on R is commutative.

Exercise 23. Prove that multiplication on R is associative.

Exercise 24. Prove that 1 is the multiplicative identity on R.

Exercise 25. Prove that every X ∈ R with X �= 0 has a multiplicative

inverse.

As always, we denote the multiplicative inverse of X as X−1. We havenow shown that R is an ordered field.

4. Embedding the Rationals in the Reals.Once again, we have not defined R to extend Q. However Q is naturally

embedded in R. By renaming things, we can think of Q as a subset of R.

Exercise 26. Prove that there exists an embedding of Q into R.

Collecting the results of this chapter thus far, we have proved thefollowing theorem. [A subfield of a field is a subset which is also a fieldunder the operations on the ambient set.]


Theorem 27. R is an ordered field with the least upper bound propertyand with Q as a subfield of R.

From Theorem 1 of Chapter 9, we know that R satisfies the Archime-dean Principle and from Theorem 6 of Chapter 9, we know that Q is densein R. That is:

• if x, y ∈ R and x > 0, then there exists n ∈ N such that nx > yand

• if x, y ∈ R and x < y, then there exists q ∈ Q such that x < q < y.

5. Uniqueness of the set of Real Numbers.So far we have shown that R has certain properties. Now we will

show that those properties completely characterize R; that is, every X withthese properties is isomorphic to R. This is the Main Theorem mentionedin Section 1 of Chapter 9.

Theorem 28 (Main Theorem). There exists, up to isomorphism, a uniqueordered field with the least upper bound property.

The proof of the Main Theorem is long; we will break it into eightsteps! Suppose F is an ordered field with the least upper bound property.We will write everything having to do with F —its elements, order relationand operations—in bold. We will prove Theorem 28 by constructing anisomorphism f : R → F . Here is an outline of the construction:

Step 1: construct a function fZ : Z → FStep 2: extend fZ to a function fQ : Q → FStep 3: extend fQ to a function f : R → FStep 4: show that f preserves orderStep 5: show that f is one-to-oneStep 6: show that f is ontoStep 7: show that f preserves additionStep 8: show that f preserves multiplication

Step 1. Define fZ inductively by fZ(0) = 0 and

fZ(n) =

⎧⎪⎪⎨⎪⎪⎩

n times︷︸︸︷1 + . . . + 1, if n > 0

−(1 + . . . + 1)︸︷︷︸|n| times

, if n < 0

Let us give some obvious notation for the n-fold sums and its negation:

n = 1 + . . . + 1 and − n = −(1 + . . . + 1).

Section 5. Uniqueness of the set of Real Numbers 165

Exercise 29. Prove that fZ : Z → F preserves operations.

Step 2. We will write all rational numbers in the form m/n where m ∈ Zand n ∈ N. (Of course, n = n/1.) We extend the function fZ of Step 1 tofQ : Q → F by defining

fQ(m/n) = m n−1.

By the following exercise, the right-hand side is defined.

Exercise 30. Prove that if n ∈ N, then fQ(n) = fZ(n) = n �= 0.

The function fQ is well-defined since

m

n= p

q�⇒ mq = np �⇒ mq = np �⇒ mn−1 = pq−1,

by field properties of Q and F and Exercise 29. Note that fQ extends fZsince m ∈ Z implies

fQ(m/1) = m1−1 = m1 = m = fZ(m).

Exercise 31. Prove that fQ : Q → F preserves operations.

Step 3. Let x ∈ R. Thinking of x as a cut, x = { q ∈ Q | q < x }. Extendthe function fQ of Step 2 to f : R → F by

f (x) = sup fQ({ q ∈ Q | q < x }) = sup{

fQ(q) | q ∈ Q and q < X}.

If the least upper bound exists, it is unique and f is well-defined.

Exercise 32. Prove that sup fQ({ q ∈ Q | q < x }) exists.

Recall from Section 1 of Chapter 9, the definitions of the sets FN, FZ

and FQ of natural, integral and rational field elements of F , respectively.

Exercise 33. Prove that fQ(N) = FN, fQ(Z) = FZ and fQ(Q) = FQ.

We next prove that f extends fQ , that is, for all q ∈ Q, f (q) = fQ(q).Since, by Exercise 31, fQ(p) < fQ(q) for all p < q, we know that

f (q) = sup{

fQ(p) | p ∈ Q and p < q} ≤ fQ(q).


Assume that f (q) < fQ(q). Since FQ is dense in F (Theorem 6 of Chapter9) and by Exercise 33, there exists r ∈ Q such that

f (q) < f (r) < fQ(q).

Exercise 34. Complete the proof that f extends fQ by deriving a contra-

diction.

Step 4.

Exercise 35. Prove that f preserves order.

Step 5.

Exercise 36. Prove that f is one-to-one.

Step 6.

Exercise 37. Prove that f is onto.

Step 7.

Exercise 38. Prove that f preserves addition.

Step 8.

Exercise 39. Prove that f preserves multiplication.

Proof of Theorem 28. Since we have shown that every ordered field withthe least upper bound property is isomorphic to R, Theorem 28 followsfrom the fact that isomorphism is an equivalence relation. �

The construction of the isomorphism above may seem a bit contrivedto you at first. Nothing could be further from the truth. In fact, it is naturaland unique, as the following exercises demonstrate.

Exercise 40. Prove that the only isomorphism of R onto itself is the identity

function.

Section 5. Uniqueness of the set of Real Numbers 167

Definition 41. Suppose X is a field (respectively, ordered field). An iso-morphism from X onto X is called an automorphism of X .

The result of Exercise 40 can be stated that the identity function IR isthe unique (ordered) field automorphism of R.

Exercise 42. Let F be an ordered field with the least upper bound property.Prove that the function f , constructed above to prove the Main Theorem,is the unique isomorphism from R onto F .

168

Chapter 14

The Complex Numbers

1. Introduction.In this chapter, we will consider a rigorous construction of the field C

of complex numbers. The addition and multiplication operations on C willbe defined to preserve the operations on R. We will show that C is a fieldcontaining R as a subfield. Let us start informally.

Complex numbers arise when we consider solutions of equations likex2 = −1. We declare that there is a solution i to this equation. Havingdefined i , we allow the standard algebraic operations on the real numbers towork on (the closure of) the set containing the real numbers together withthe new number i , always obeying the property that i2 = −1. The complexnumbers, defined in this way, are of the form a + bi , where a and b are realnumbers. Of course, we should be careful to define the multiplication biand the addition a + bi .

For a moment, let us suppose that we have defined the complex num-bers. Using familiar algebraic properties, we see that

(−i)2 = (−1)2(i)2 = (1)(−1) = −1.

So, our original equation, x2 = −1, has two solutions! Actually, this agreeswith the notion that a generic quadratic polynomial should have two roots.The choice in the definition of i appears in Definition 14 below. However,this choice is inconsequential as Exercise 20 will show.

While addition and multiplication of complex numbers is quite natural,ordering the complex numbers is a different story. Suppose we have a totalordering on C which agrees with the usual ordering on the subset R. Sincea total ordering is trichotomous, either i = 0 or i > 0 or i < 0. Of course,i = 0 is just absurd. Suppose for a moment that i > 0. If C is to be anordered field, then multiplying positives yields a positive. That is, i2 > 0.But −1 �> 0 in R. The other case is no better, as you can check. This alsowill be made precise below, in Exercises 17 and 18.

Section 2. Algebra of Complex Numbers 169

2. Algebra of Complex Numbers.We must first define the set, C, of complex numbers as well as the

operations of addition and multiplication on C.

Definition 1. The set of complex numbers is the set C = R2.

Definition 2. Addition of complex numbers (a, b) and (c, d) is defined by

(a, b) + (c, d) = (a + c, b + d).

Definition 3. Multiplication of complex numbers (a, b) and (c, d) is de-fined by

(a, b)(c, d) = (ac − bd, ad + bc).

These operations are clearly well-defined; unlike the constructions ofthe integers and the rationals, this construction does not use equivalenceclasses. We next verify that C is a field.

Let us start with the properties of addition.

Exercise 4. Prove that addition on C is commutative.

Exercise 5. Prove that addition on C is associative.

Exercise 6. Prove that there exists a unique additive identity for C.

As before, we let 0 denote the additive identity; that is, 0 = (0, 0).

Exercise 7. Prove that every element of C has a unique additive inverse.

Denote the additive inverse of a number z ∈ C by −z.Now, let us continue with the properties of multiplication.

Exercise 8. Prove that multiplication on C is commutative.

Exercise 9. Prove that multiplication on C is associative.

Exercise 10. Prove that there exists a unique multiplicative identity 1 for

C.

Once again, we let 1 denote the multiplicative identity (1, 0).

Exercise 11. Prove that every nonzero z ∈ C has a unique multiplicative

inverse.

170 Chapter 14. The Complex Numbers

Denote the multiplicative inverse of a number z ∈ C by z−1.It remains to prove the distributive property.

Exercise 12. Prove the distributive property for C.

This completes the proof that C is a field. We will come back to thequestion of order on C and whether C is an ordered field or not in the nextsection.

Consider the following computation of a product of real numbers:

(a, 0)(b, 0) = (ab, 0).

Since this looks a lot like the equation a · b = ab for multiplication ofreal numbers, we will write a for (a, 0) and b for (b, 0). This suggests acandidate for an embedding of the field R in the field C, which is postponeduntil Exercise 19.

Next, we will identify the imaginary number i .

Exercise 13. Show that (0, 1)2 = −1. Is there any other complex number

whose square equals −1?

Definition 14. The element (0, 1) ∈ C is called i .

Of course, we have made a choice in Definition 14 since (0, −1)2 =−1 also. The nature of this choice will be explained by Exercise 20.

Since

(a, b) = (a, 0) + (0, b) = (a, 0)(1, 0) + (b, 0)(0, 1)

= (a, 0) · 1 + (b, 0) · i = a · 1 + b · i,

we will write a + bi for (a, b) ∈ C. We call a the real part of (a, b) andb the imaginary part of (a, b). For all b ∈ R, the numbers (0, b) ∈ C arecalled imaginary.

3. Order on the Complex Field.Let us look at order relations on C. Whenever we take products of

ordered sets, we can define an order relation on the product called thelexicographic or dictionary order. We will do this specifically for C ratherthan the general situation.

Definition 15. Define the lexicographic ordering ≺ by (a, b) ≺ (c, d) iffa < c, or a = c and b < d.

Exercise 16. Prove that ≺ is a total ordering on C and that (a, 0) ≺ (b, 0)

if and only if a < b.

Section 4. Embedding the Real Numbers in the Complex Numbers 171

Exercise 17. Prove that ≺ does not make C an ordered field.

Unfortunately, the lexicographic ordering is not a bad choice sinceno other ordering makes C an ordered field either. To prove this you canconsider the both of the following possibilities: −1 < 0 or −1 > 0.

Exercise 18. Prove that there is no total ordering on C which makes it an

ordered field.

4. Embedding the Real Numbers in the Complex Numbers.Of course, R is not a subset of C as we have defined C. Again, we

can find an embedding of R into C which preserves the operations so thatwe may consider R as a subfield of C.

Exercise 19. Prove that there exists an embedding of R into C.

Perhaps, we were a bit terse in Exercise 19; let us elaborate. So far,our embeddings have all had the property of preserving order as well asoperations. In fact, if you thought of C with the lexicographic ordering, thenthe ordering on R is quite compatible. However, since C does not have anatural ordering (i.e., one which makes it an ordered field), we really havein mind a weaker kind of embedding which does not include the order-preserving property, an embedding of a field rather than an ordered field.

We would be done with complex numbers now except for one smalldetail. The definition of i left us some wiggle room as the following exerciseexplains.

Exercise 20. Prove that there are exactly two automorphisms f : C → Csuch that f (x) = x for all x ∈ R.

Since the automorphisms in Exercise 20 induce the identity IR whenrestricted to R, we say that these automorphisms of C extend the identityon R.

By Exercise 40 of Chapter 13, IR is the unique automorphism on R.Suppose f : C → C is an automorphism such that f (R) = R. Theinduced function on R preserves operations (and order) and is one-to-onesince f has these properties, making it an isomorphism onto its range R; theinduced function is an automorphism of R, hence, equal to IR. Therefore,the statement in Exercise 20 can be strengthened: there are exactly twoautomorphisms f : C → C such that f (R) = R.

In fact, this result can be strengthened further. This is done in thefollowing exercise.

172 Chapter 14. The Complex Numbers

Exercise 21. Prove that if f : C → C is an isomorphism, then f (R) = R.

Hence, there are exactly two automorphisms on C.

173

Chapter 15

The Real Numbers Accordingto Cantor

1. Convergence of Sequences of Rational Numbers.In Chapter 13, we constructed the set of real numbers from the set of

rational numbers using the method of Dedekind cuts. We saw that thesecuts formed a type of “completion” of the rationals with respect to the leastupper bound property. In this chapter, we will give a second construction ofthe reals from the rationals using the method of Cauchy sequences devisedby Cantor. This construction will emphasize another way in which thereals “complete” the rationals. If the construction by cuts can be regardedas essentially algebraic, then the construction by Cauchy sequences can beregarded as analytic.

Recall that a sequence can be regarded as a denumerable list of num-bers, not necessarily distinct. Since we will only be interested in rationalnumbers and how they define real numbers, we restrict our attention tosequences of rational or real numbers, that is, sequences in Q or R.

Definition 1. Let X be a set. A sequence in X is a function from N to X .

Of course, you will very rarely see a sequence denoted as, for instance,f : N → Q. Usually a sequence will be denoted as either

{ qn } or { q1, q2, q3, q4, . . . }.

The connection between f and this notation is that qn = f (n) for all n ∈ N.You should note that the notation for sequences looks like a set. This ismisleading since a term in the sequence may repeat. On the other hand, weuse the same notation to represent the range of the sequence, which is a set.

Example 2. A = { n } and B = { 1, 1, 1, 1, . . . } denote two sequences inQ. The range of A is N and the range of B is { 1 }. Of course, we couldalso write the sequences as A = { 1, 2, 3, 4, . . . } and B = { 1 }.

174 Chapter 15. The Real Numbers According to Cantor

If you have seen sequences before, you will recall the major issuewas the convergence of sequences. This will be our primary concern also,especially since they do not converge in general. To explain, we need adefinition and some examples.

Definition 3. A sequence { qn } in Q (respectively, R) converges to q ifffor every ε > 0 there exists a natural number N such that |qn − q| < ε

whenever n ≥ N . A sequence { qn } in Q is convergent if there is someq ∈ Q such that { qn } converges to q. A sequence { qn } in R is convergentif there is some q ∈ R such that { qn } converges to q.

The idea is that a sequence { qn } in Q converges to the limit q ifthe terms qn get closer and closer to q as n gets larger, that is, as you gofurther into the endless tail of the sequence. Note, a sequence in Q is alsoa sequence in R. However, a sequence in Q may converge in R but not inQ.

Example 4. (a) The sequence { 0 } converges to 0 since the difference from0 is always 0.(b) The sequence { 1

n } also converges to 0 since 1n can be made as small

as we wish, simply by choosing n large enough; this is the ArchimedeanProperty of Q: for any positive ε ∈ Q there exists n ∈ N such that nε > 1(that is, 0 < 1

n < ε).(c) The sequence

{ 2, 1.5, 1.42, 1.415, 1.4143, 1.41422, 1.414214, . . . }

(in which the successive terms are derived by reducing the last decimalplace of its predecessor by one and appending, in the next decimal place,one plus the corresponding decimal place in the decimal representation of√

2) does not converge in Q.

The goal of the construction in this chapter is to make sequences likethe one in the last example convergent in the larger set; that is, convergentto

√2 ∈ R.Before continuing with sequences, we must prove an important in-

equality known as the Triangle Inequality. It will be very useful for ussince it deals precisely with expressions like |qn − q| in the definition ofconvergent sequence.

Exercise 5. Prove that, for all p, q ∈ R, |p + q| ≤ |p| + |q|. Moreover,

|p + q| ≥ |p| − |q|.Properties like the following about convergent sequences can be useful.

Section 2. Cauchy Sequences of Rational Numbers 175

Exercise 6. Suppose { xn } is a sequence in R which converges to x and { yn }is a sequence in R which converges to y. Prove that { xn + yn } convergesto x + y.

Exercise 7. Suppose { xn } is a sequence in R which converges to x and { yn }is a sequence in R which converges to y. Prove that { xn − yn } convergesto x − y.

Exercise 8. Suppose { xn } is a sequence in R which converges to x and{ yn } is a sequence in R which converges to y. Prove that { xn yn } convergesto xy.

Exercise 9. Suppose { xn } is a sequence in R which converges to x and { yn }is a sequence in R which converges to y. Prove that { xn/yn } converges tox/y provided that yn �= 0 for all n ∈ N and y �= 0.

2. Cauchy Sequences of Rational Numbers.What is it about convergent sequences which makes them convergent?

In some sense, a convergent sequence requires two things. First, it requiresthe existence of a limit q, in our set, Q or R, for the sequence to convergeto. Second, it requires the behavior of the sequence which makes q thenumber to which it converges. Notice that if the terms qn get closer andcloser to q, then they also get closer and closer to each other. The followingexercise should help to explain this better.

Exercise 10. Suppose that { qn } is a convergent sequence in either Q orR. Prove that for every ε > 0 there exists a natural number N such that|qm − qn| < ε whenever both m ≥ N and n ≥ N .

Sequences with the property in the exercise above were first consideredby A. Cauchy.

Definition 11. A sequence { qn } in Q or in R is called a Cauchy sequenceiff for every ε > 0 there exists a natural number N such that |qm − qn| < ε

whenever both m ≥ N and n ≥ N .

The converse of Exercise 10 is, of course, false. Consider again thefollowing sequence.

{ 2, 1.5, 1.42, 1.415, 1.4143, 1.41422, 1.414214, . . . }given in Example 4(c).This sequence is Cauchy since terms past the k th termdiffer in, at worst, the k th decimal place, that is, by at most 1

10k . However,

this sequence fails to converge in Q; of course, it converges to√

2 in R.


By Exercise 10, every sequence in Q or R which converges is Cauchy.By the previous example the converse is false in Q. In the remainder ofthis section, we will prove that the converse is true in R.

The sequence { n } is clearly not Cauchy. In fact, Cauchy sequencesnever have this kind of unbounded character, as we show in the next exercise.First, we give some definitions.

Definition 12. A sequence { qn } in Q or in R is bounded iff there existsq ∈ Q such that |qn| ≤ q for all n ∈ N. { qn } is bounded above iff thereexists q ∈ Q such that qn ≤ q for all n ∈ N. { qn } is bounded below iffthere exists q ∈ Q such that qn ≥ q for all n ∈ N.

Clearly, a sequence is bounded if and only if it is both bounded aboveand bounded below. On the other hand, a sequence can be bounded aboveand not bounded, or bounded below and not bounded as the sequences{ −n } and { n } demonstrate.

Exercise 13. Suppose { qn } is a Cauchy sequence in Q (or R). Prove that

{ qn } is bounded.

Definition 14. A sequence { qn } in Q or in R is increasing iff qn+1 ≥ qn

for all n ∈ N. { qn } is decreasing iff qn+1 ≤ qn for all n ∈ N. { qn } ismonotone iff it is either increasing or decreasing.

Note that a constant sequence { q } is both increasing and decreasing!Of course, you can imagine how a strictly increasing or strictly decreasingsequence should be defined.

A subsequence is much like a subset: a subsequence of a sequence{ a1, a2, a3, a4, . . . } is a sequence

{ an1 , an2 , an3 , an4 , . . . }

where n1 < n2 < n3 < n4 < . . . . The following is usually called theBolzano-Weierstrass Theorem.

Theorem 15. Every bounded monotone sequence in R is convergent.

Proof. Suppose { xn } is an increasing sequence in R which is boundedabove. By the least upper bound property, the range of the sequence has aleast upper bound x . We wish to show that the sequence converges to x .Since x is an upper bound, xn ≤ x for all n ∈ N. Fix an ε > 0. Thereexists an N ∈ N such that x − ε < xN ≤ x , since otherwise x − ε wouldbe a smaller upper bound than the least upper bound x . Since the sequenceis increasing,

x − ε < xN ≤ xn ≤ x

Section 3. Cantor’s Set of Real Numbers 177

for all n ≥ N . This proves the convergence of { xn } to x .The other case is left as an exercise. �

Exercise 16. Complete the proof of Theorem 15.

Lemma 17. Every sequence in R has a monotone subsequence.

Proof. Suppose { xn } is a sequence in R. Consider the set

P = { n ∈ N | xn ≥ xm for all m > n } .

(If you graph the sequence, the n’s in P indicate “peaks” in the graph whichare higher than all later points.) Either P is infinite or finite.

In the case when P is infinite, write P = { n1, n2, n3, n4, . . . }, wherenk+1 > nk for all k ∈ N. Then the subsequence { xn1 , xn2 , xn3 , xn4 , . . . } isdecreasing.

In the case when P is finite, set n1 greater than any n ∈ P , for example,1 + max P . Since n1 /∈ P , there exists n2 > n1 such that xn1 < xn2 .Continue in this way to produce a subsequence which is strictly increasing.

�The following is also sometimes called the Bolzano-Weierstrass The-

orem.

Exercise 18. Prove that every bounded sequence in R has a convergent

subsequence.

We can now prove the converse of Exercise 10 in R.

Theorem 19. Every Cauchy sequence in R is convergent.

Proof. Suppose { xn } is a Cauchy sequence in R. By Exercise 13, { xn }is bounded. By Lemma 17, there is a monotone subsequence { xnk }. ByTheorem 15, { xnk } converges; let x be the limit of the subsequence.

Let us fix ε > 0. Since { xn } is Cauchy, there exists a natural numberN such that |xm −xn| < 1

2ε whenever both m ≥ N and n ≥ N . Since { xnk }converges to x , there exists a natural number M such that |xnk − x | < 1

2ε

whenever k ≥ M . Pick k such that k ≥ M and nk ≥ N . By the TriangleInequality (Exercise 5), for all n ≥ N ,

|xn −x | = |(xn −xnk )+(xnk −x)| ≤ |xn −xnk |+|xnk −x | <1

2ε+ 1

2ε = ε.

Therefore, the Cauchy sequence { xn } also converges to x . �


3. Cantor’s Set of Real Numbers.For Dedekind cuts, we defined a real number x as the set of all rational

numbers less than x . Now, we wish to define a real number x using Cauchysequences via those sequences which converge to x .

Definition 20. Suppose that x = { xn } and y = { yn } are Cauchy sequencesin Q. Define a relation ∼ on the set of all Cauchy sequences in Q by x ∼ yif and only if the sequence { xn − yn } converges to 0.

To rephrase the above definition, x ∼ y if and only if for every rationalnumber ε > 0 there exists a natural number N such that |xn − yn| < ε

whenever n ≥ N .

Exercise 21. Prove that the relation ∼ is an equivalence relation on the set

of all Cauchy sequences in Q.

Definition 22. The set of equivalence classes of ∼ is the set of Cantor’sreal numbers which we denote by RC .

Of course, we wish to show that R and RC are isomorphic, both beingthe set of all real numbers. First, we need to define order and operationson RC .

It is natural to think that [x] < [y] should mean that [y] − [x] > 0and, hence, [y] − [x] > ε for some positive rational ε.

Definition 23. Define a relation < on RC by [x] < [y], for Cauchy se-quences x = { xn } and y = { yn } of rational numbers, if and only if, thereexists a rational number ε > 0 and there exists a natural number N suchthat xn + ε < yn whenever n ≥ N .

Notice that Definition 23 of [x] < [y] seems to depend on the choicesof representatives of the equivalence classes [x] and [y].

Exercise 24. Prove that the relation < on RC is well-defined.

From here on we will assume that { xn } is a representative sequencein [x] and so on for other elements of RC .

As always, we define [x] ≤ [y] to mean [x] < [y] or [x] = [y].

Example 25. Let xn = − 1n and yn = 1

n for all n ∈ N. Then [x] = [y] = 0

since |xn − yn| = 2n converges to 0. Note that xn < yn for all n ∈ N. That

is, there does not exist an N ∈ N such that such that yn ≤ xn whenevern ≥ N . Therefore, we could not define ≤ by comparing the terms xn andyn directly.

Exercise 26. Prove that < is a (strict) total ordering on RC .

Next we define the operations in the obvious way.

Section 4. The Isomorphism from Cantor’s to Dedekind’s Reals 179

Definition 27. Define addition on RC by [x] + [y] = [{ xn + yn }].

Exercise 28. Prove that addition on RC is well-defined.

Definition 29. Define multiplication on RC by [x][y] = [{ xn yn }].

Exercise 30. Prove that multiplication on RC is well-defined.

By the way, RC completes Q in the way desired in Section 1. Ifq ∈ Q, then it is represented by the constant sequence in RC , that is,q = [{ q }] ∈ RC is the embedded image of q ∈ Q. So a sequence { xn } inQ can be viewed as a sequence { xn } in RC .

Exercise 31. If x = { xn } is a Cauchy sequence in Q, then { xn } converges

to [x] in RC .

Next we could check that RC with this ordering and these operationshas all of the properties of an ordered field with the least upper boundproperty and that Q embedded as a subfield. Only the least upper boundproperty would give us any serious work. We know all such structures areisomorphic and, in particular, RC is isomorphic to R.

However, we will construct the isomorphism directly.

4. The Isomorphism from Cantor’s to Dedekind’s Reals.We next wish to give an isomorphism f : R → RC . Take a cut

representing some x in R. If we can extract a sequence from the cut whichconverges to x , this gives us a way to define a candidate for our isomorphism.By the way, it is much harder to define the isomorphism in the oppositedirection. This makes sense since we know all about R but we know almostnothing about RC . You will make frequent use of the properties of R inproving the existence of the isomorphism.

Recall that the cut x is the set { q ∈ Q | q < x in R }. Pick q1 ∈ x .Set q2 = q1 +1 if q1 +1 < x or q2 = q1 + 1

2 if q1 + 12 < x < q1 +1 or etc.

Continue in the same way to define q3, q4, etc. This defines an increasingsequence { qn } in Q. We will check that this sequence is Cauchy and thendefine f : R → RC by f (x) = [{ qn }].

Exercise 32. Prove that the process in the previous paragraph actually

defines a sequence and that it is Cauchy in Q and converges to x in R.

We define f : R → RC by f (x) = [{ qn }]. We must check that thisis well-defined, that is, independent of the initial choice q1.


Exercise 33. Prove that f is well-defined.

It remains to show that f is an isomorphism, that is, a structure-preserving bijection.

Exercise 34. Prove that f is order-preserving and, hence, one-to-one.

Exercise 35. Prove that f is addition-preserving.

Exercise 36. Prove that f is multiplication-preserving.

Exercise 37. Prove that f is onto.

This completes the proof that the structure RC we constructed in thischapter is really the same set of real numbers R with its order relation andoperations.

181

PART IV

HINTS

182

Chapter 16

Hints for (and Comments on)the Exercises

Hints for Chapter 1.

Exercises 3, 4 and 5. These are meant to inspire thought rather thanspecific answers.

Exercise 10. Think trilateral rather than triangle; that is, use linesegments, not angles.

Exercise 12. Use scale and length of line measure for betweennessand congruence of segments, respectively. Use angle measure similarly.


Exercise 18. That is, if p and q is true, then p is true and if p istrue, then p or q is true. Of course, p ∧ q ⇒ q and q ⇒ p ∨ q are alsotautologies.

Exercise 19. That is, if p implies q is true, then p is false or q is true.Exercise 20. These are known as the reflexive, symmetric, and tran-

sitive properties of equivalence, respectively.Exercise 21. This is why the phrases ”if and only if” and ”necessary

and sufficient” make sense.Exercise 22. The point of this exercise is to make sure that you un-

derstand the value of truth tables! Every row in a truth table represents anentire paragraph in words.

Exercise 24. Be careful! (d) is impossible!Exercise 27. You can do this in words or with truth tables. This is a

very important tool for constructing proofs.Exercise 29. (a) is exactly Exercise 27 after a change in variables.

You can disprove with a counterexample or with truth tables.Exercise 30. This is the same kind of algebra exercise as you have

often seen before.

Chapter 16. Hints for (and Comments on) the Exercises 183


Exercise 3. Mimic the first part of the proof of Proposition 1.Exercise 5. Mimic the proof of Proposition 4.Exercise 6. Use proof by brute force.Exercise 7. Use proof by brute force.Exercise 8. Use proof by brute force.Exercise 11. Consider when 2x + 1 is nonnegative and when it is

positive.Exercise 12. You may use the facts that products of positive real

numbers are positive and (−1)2 = 1.Exercise 14. Try proof by contraposition. You may assume that, for

integers p and q, q �= 0 implies that p2 +q2 �= 0; this is proved in Exercise23.

Exercise 15. Since this is the converse of Exercise 12, that does nothelp to prove it. Try proof by contraposition.

Exercise 16. Break the equivalence into two implications: use directproof for one and contraposition for the other.

Exercise 22. For each, mimic the proof of Theorem 18. The phrase“is even” can be replaced by “is divisible by 2;” this becomes “is divisibleby 3” Similarly, the condition m = 2k becomes m = 3k.

Exercise 23. Use proof by brute force for one direction and proof bycontradiction for the other. You can use Exercise 12.

Exercise 24. Multiply both sides of an inequality by x .Exercise 25. Suppose there are n ≥ 2 people. Use proof by contra-

diction. Consider the set of numbers of people that each person knows.Show that someone must know no one else while someone else must knowevery one.

Exercise 33. Where is the first step in the induction?Exercise 34. Examine how the situation is different when n + 1 = 2

from when n + 1 = 3.Exercise 35. Consider

[1 + 3 + 5 + · · · + (2n − 1)

] + (2(n + 1) − 1)

in a proof by induction.

Exercise 36. Usen+1∑k=1

1k(k+1)

=(

n∑k=1

1k(k+1)

)+ 1(

n+1)(

(n+1)+1) in a

proof by induction.Exercise 37. Use proof by induction.Exercise 38. Use proof by induction (on n).Exercise 41. Examine the first application of the inductive step, going

from 100 = 1 to 100+1 = 1.Exercise 42. Given n, n is either prime or the product of two integers

strictly between 1 and n.

184 Chapter 16. Hints for (and Comments on) the Exercises

Exercise 43. Consider 0.Exercise 44. Consider f (x) = |x |.


Remember Venn diagrams and logic!Exercise 6. Is it true that if x ∈ ∅, then x ∈ X? Recall that statements

such as ∅ ⊂ X are sometimes called vacuously true.Exercise 7. Remember that ⊂ does not mean proper subset.Exercise 8. You could use (p ⇒ q) ∧ (q ⇒ r) ⇒ (p ⇒ r) (Supple-

mental Exercise S5 of Chapter 2), but this is actually straightforward.Exercise 11. Use (p ⇔ q) ⇔ (p ⇒ q) ∧ (q ⇒ p) (Exercise 21 of

Chapter 2).Exercise 13. Use Exercises 7, 8, and 11.Exercise 16. Use Exercises 6 and 11; use the tautology p ∧ F ⇔ F ,

where F is always false (Supplemental Exercise S33 of Chapter 2).Exercise 17. Use the tautology p ∧ p ⇔ p (Supplemental Exercise

S22 of Chapter 2).Exercise 21. Use the tautology p ∨ F ⇔ p (Supplemental Exercise

S32 of Chapter 2).Exercise 22. Use the tautology p ∨ p ⇔ p (Supplemental Exercise

S21 of Chapter 2).Exercise 23. Break the compound statement into A ∩ B ⊂ A and

A ⊂ A ∪ B. Use Exercise 8 for the second part.Exercise 24. You could use Exercise 23 (A ∩ B ⊂ B) to show that

A ∩ B ⊃ A implies A ⊂ B.Exercise 25. You could use Exercise 23 (A ⊂ A ∪ B) to show that

A ∪ B ⊂ B implies A ⊂ B.Exercise 26. Use both parts of Exercises 25 and 24.Exercise 27. Use p ∧ q ⇔ q ∧ p and p ∨ q ⇔ q ∨ p (Supplemental

Exercises S15 and S16 of Chapter 2).Exercise 28. Use (p∧q)∧r ⇔ p∧(q∧r) and (p∨q)∨r ⇔ p∨(q∨r)

(Supplemental Exercises S17 and S18 of Chapter 2).x ∈ A ∩ (B ∪ C) means x ∈ A and x ∈ (B ∪ C), which means x ∈ A,

and x ∈ B or x ∈ C . Notice how the comma plays the role of parentheses.Be careful! It is unclear what the following means: x ∈ A and x ∈ B orx ∈ C .

Exercise 29. Use p ∧ (q ∨ r) ⇔ (p ∧ q)∨ (p ∧ r) and p ∨ (q ∧ r) ⇔(p ∨ q) ∧ (p ∨ r) (Supplemental Exercises S19 and S20 of Chapter 2).

Exercise 36. Recall that [a, b) = { x ∈ R | x ≥ a and x < b } and usethe definition of intersection.

Exercise 37. See Figure 31 and Example 34.Exercise 39. Recall the definitions of [a, b), (a, b), and union.


Exercise 40. Compare with Exercise 29.Exercise 43. Recall the definition of A − B.Exercise 44. Consider B = ∅ and A = B.

You have been using Venn diagrams, right?Exercise 45. Recall the definition of A−B; logical and is commutative

and associative.Exercise 46. Try ∅, { 1 } and { 1, 2 }Exercise 47. Follow the Venn diagram.Exercise 48. This is similar to Exercise 47.Exercise 50. Recall the definitions of A − B and B ′.Exercise 51. (B ′)′ = B is analogous to ¬(¬p) ⇔ p. Use Exercise

50.Exercise 52. Follow the Venn diagram.Exercise 54. Remember that ∅ and X are always subsets of X . You

must prove that ∅ is the only subset of ∅!Exercise 55. We have seen that |P(∅)| = 1 = 20, |P(Z1)| = 2 = 21,

and |P(Z1)| = 4 = 22. Power, get it?Exercise 56. Use Exercise 8.Exercise 57. Both are true: P(A) ∩ P(B) = P(A ∩ B).Exercise 58. Only one is true. X ⊂ A ∪ B does not imply that X ⊂ A

or X ⊂ B.Exercise 59. Neither one holds! Check ∅; compare with Supplemental

Exercise S35.


Exercise 3. They are, respectively, 3-dimensional space, a 2-dimen-sional lattice of points, 2 horizontal parallel lines, and 2 vertical parallellines (given that the plane has the standard orientation).

Exercise 4. Assume that A × B �= ∅; assume that A �= ∅ and B �= ∅.Exercise 5. Assume that (A × B) ∩ (B × A) �= ∅; assume that

A ∩ B �= ∅.Exercise 15. This is probably the more familiar definition; it is just a

matter of the meaning of f (x).Exercise 18. Graph the half-parabola y = √

x .Exercise 19. Use your knoweldge of precalculus and/or calculus.Exercise 23. If y ∈ f (A), then there exists x ∈ A such that y = f (x).Exercise 24. Suppose y ∈ f (A ∪ B); suppose y ∈ f (A) ∪ f (B).Exercise 26. Suppose y ∈ f (A ∩ B).Exercise 28. Use contraposition.Exercise 29. Compare with Example 25 and Exercise 26.Exercise 31. This is probably the more familiar definition.Exercise 33. Use the same rule and domain with a “smaller” codomain.


Exercise 34. Compare with Exercise 31; uniqueness comes from one-to-one.

Exercise 35. Think about functions you know: polynomial, trigono-metric, exponental, etc.

Exercise 37. Unravel g( f (x)).Exercise 40. That is, show (( f ◦ g) ◦ h)(x) = ( f ◦ (g ◦ h))(x).Exercise 41. Start with f : X → Y , g : W → Z , Y ⊂ W , and

x1 �= x2 in X .Exercise 42. Notice that the codomain and range of f equals the

domain of g. Determine g( f (X)).Exercise 44. By definition of function, g(y1) �= g(y2) implies y1 �= y2.

Consider functions on small finite sets for your counterexample.Exercise 45. By definition of function, f (X) ⊂ Y .Exercise 47. Assume there is another identity function i : X → X .

Show that i ◦ IX = i and i ◦ IX = IX .Exercise 49. What are the domains and codomains of f ◦ g and g ◦ f ?Exercise 51. Compare with Exercises 44 and 45.Exercise 52. Assume f has another inverse function g. Consider

g ◦ f ◦ f −1 in two ways using the associative property.Exercise 55. If x1, x2 ∈ f −1(y), then f (x1) = y = f (x2).Exercise 56. x ∈ f −1(A ∩ B) is equivalent to f (x) ∈ A ∩ B is

equivalent to f (x) ∈ A and f (x) ∈ B, etc.Exercise 57. By (c), for (b), f cannot be one-to-one; use small finite

sets for your counterexample. (d) is false!Exercise 58. By (c), for (b), f cannot be onto. (d) is false!Exercise 62. No! State the domains and codomains explicitly.Exercise 63. You must show that the domains are equal.Exercise 64. Be sure that the range of f is not a subset of the domain

of g.


Exercise 4. This is different from Exercise 3; ⊂ is different from �and the structure of R is different from that of P(U ).

Exercise 5. Consider the relation =.Exercise 6. You may try to consider relations on small finite sets.Exercise 7. Since there are eight properties, this means to give 28 =

256 examples! Since relations cannot be both reflexive and nonreflexive,cannot be both symmetric and asymmetric, and trichotomous implies con-nected, this eliminates more than half of the cases. Give at least a few, ifnot all.

Exercise 10. Check reflexive and nonreflexive.


Exercise 11. Compare the symbols ≺ and " with < and ≤, respec-tively; recall that our ⊂ is denoted ⊆ by some authors.

Exercise 14. x " y and y " x imply x " x , which implies x = x ;x ≺ x implies x �= x .

Exercise 18. Connected but not trichotomous contradicts nonreflex-iveness.

Exercise 23. Assume there are two candidates for greatest element.Use antisymmetry to show that they must be equal.

Exercise 24. Assume that a greatest element is not maximal. Usenonreflexivity of ≺.

Exercise 25. Assume there exists a greatest element and show maximalelements are equal.

Exercise 26. Use connectedness.Exercise 31. Assume there are two least upper bounds. Use antisym-

metry to show that they must be equal.Exercise 32. Consider intervals in R.Exercise 33. Check the properties.Exercise 40. See Chapter 7, Section 2.Exercise 42. Use s ≈ t to show that [s] ⊂ [t] and [t] ⊂ [s].Exercise 43. Show that if x ∈ [s] ∩ [t], then [s] = [t].Exercise 46. Check the properties.Exercise 48. Check the properties.Exercise 50.Exercise 52.


Exercise 5. Show that any function f : ∅ → ∅ satisfies the conditionsof a bijection vacuously.

Exercise 6. Consider IA.Exercise 7. Consider the inverse of a one-to-one correspondence.Exercise 8. Consider the composition of two one-to-one correspon-

dences.Exercise 11. Use Exercise 7, Exercise 8 and Theorem 9.Exercise 12. For (a), extend a one-to-one correspondence Zn → A

by n + 1 &→ x . (b) follows from (a).Exercise 13. Proceed by induction on |B|.Exercise 15. f induces a one-to-one correspondence onto a subset of

B.Exercise 16. There exists a one-to-one correspondence between B

and a subset of A.


Exercise 18. If f : Zn → N (n ≥ 2) is a one-to-one correspon-dence, show that the sum of the elements of the range is not in the range,contradicting that f is onto.

Exercise 20. Compare with Theorem 14.Exercise 24. For a one-to-one correspondence f : N → D, set

k = f −1(x). Map n &→ f (n + 1) for n ≥ k.Exercise 25. Suppose X is an infinite set. By Theorem 23, there is a

denumerable subset D of X . Choose x ∈ D. By the proof of Exercise 24,D and D − { x } have the same cardinality. Use the identity function onX − D to complete the construction of a bijection from X onto X − { x }.

Exercise 26. Compare with Exercise 25.Exercise 30. By transitivity.Exercise 31. Define a one-to-one correspondence by, for example,

alternating positive and negative.Exercise 34. Use the one-to-one correspondence of Theorem 32 to

define a one-to-one correspondence by alternating positive and negative.Exercise 35. Make an array of lists of the countable sets and use the

“zig-zag” method of the proof of Theorem 32.Exercise 36. Use induction.Exercise 38. For polynomials of a fixed degree there are countably

many algebraic numbers.Exercise 41. See Theorem 29.Exercise 42. Consider a tangent function.Exercise 43. Consider “half of” a tangent function.Exercise 44. Consider a translation function (that is, of the form

f (x) = x + t).Exercise 45. Consider the negation function.Exercise 46. Consider an affine function (that is, of the form f (x) =

mx + b).Exercise 47. Use Exercises 42–46.Exercise 48. For (0, 1) and [0, 1) and for (0, 1] and [0, 1], use the

technique of the proof of Exercise 25.Exercise 49. Generalize Exercise 48, considering different cases.

(This is similar to the method of Exercises 42–47.)Exercise 50. Assume R − Q is countable. Then Q ∪ (R − Q) is

countable.Exercise 51. Similar to Exercise 50.Exercise 54. Assume it is countable. Use the diagonal technique used

in the proof that (0, 1) is uncountable (Theorem 40).Exercise 55. Use Exercise 54.Exercise 56. Assume it is countable. A subset S of N, that is, an

element of P(N), can be represented by a sequence of 0’s and 1’s where the


n-th entry, sn , in the sequence indicates n /∈ S if sn = 0 or n ∈ S if sn = 1.Use the diagonal technique on the list of such sequences.

Exercise 57. Assume that there is a one-to-one correspondence f :S → P(S). Let A = { s ∈ S | s /∈ f (s) }. Show that A ∈ P(S) butA /∈ f (S).


Exercise 13. This is similar to Exercise 11.Exercise 14. This is similar to Exercise 12.Exercise 16. Add −x .Exercise 17. Use Exercise 12.Exercise 18. This is similar to Exercise 16.Exercise 19. This is similar to Exercise 17.Exercise 20.Exercise 21.Exercise 22.Exercise 23.Exercise 25.Exercise 26.Exercise 27.Exercise 28.Exercise 29.Exercise 30.Exercise 33.Exercise 35. Show that E has no multiplicative identity element.

[Compare with the proof of Proposition 29 in Chapter 11.]Exercise 36.


Exercise 3.Exercise 5.


Exercise 7. Use proof by contradiction to show that N is well ordered:assume S is a nonempty subset with no least element and show by inductionthat 1 and its successors SuccP(1) are all in the complement of S.

Exercise 9.Exercise 11.Exercise 12.Exercise 13.


Exercise 14. Use induction.Exercise 15.Exercise 16. Assume there is another additive identity z. Consider

0 + z.Exercise 17. For the inductive step, it is easiest to convert to powers

of Succ.Exercise 18.Exercise 19.Exercise 20.Exercise 22. Use induction on p.Exercise 23. Show 1m = m by induction. Remember to show unique-

ness!Exercise 24. Show 0m = 0 by induction. For the inductive step, use

Exercise 23.Exercise 25.Exercise 26.Exercise 27.Exercise 29.


Exercise 2. Use (AC) for Z+ to show that ∼ is reflexive and symmetricon Z+ × Z+; use (AC) and (AA) for Z+ to show that ∼ is transitive onZ+ × Z+.

Exercise 8.Exercise 9. Show that, for any [m, n] ∈ Z, [m, n + 1] < [m, n].Exercise 12.Exercise 14.Exercise 15. Use (AC) for Z+.Exercise 16. Use (AA) for Z+.Exercise 17. Remember to show both existence and uniqueness. Also,

cf., Proposition 11 of Chapter 8.Exercise 18. It is easier to start with [m, n] than p or −p. Remember

to show both existence and uniqueness. Also, cf., Proposition 12 of Chapter8.

Exercise 19. Cf., Exercise 17 of Chapter 8.Exercise 20. Cf., Exercise 16 of Chapter 8.Exercise 21. Use (PP) for Z+. ?????Exercise 24. Use (MC) and (MA) for Z+.Exercise 25. Use (MC) and (MA) for Z+.Exercise 26. Remember to show both existence and uniqueness. Also,

cf., Exercise 13 of Chapter 8.Exercise 27. Cf., Exercise 14 of Chapter 8.


Exercise 28.Exercise 30.Exercise 31.Exercise 32. Given p ∈ Z+, what element of Z resembles p?


Exercise 2. Use (MC) for Z to show that ∼ is reflexive and symmetricon Z × Z∗; use (MC), (MA) and (MX) for Z to show that ∼ is transitive onZ × Z∗.

Exercise 4. Use the Negation Reverses Order and Factoring a MinusProperties for Z.

Exercise 5. For n < 0, use Exercise 4.Exercise 8. Use (MC), (MA) and (IX) for Z.Exercise 9. Use (MC), (MA) and (MX) for Z to show that < is

transitive; use the trichotomy of < on Z to show that < is trichotomous.Exercise 10. Show that [m − 1, n] < [m, n] for any [m, n] ∈ Z.Exercise 12. Use the trichotomy of <.Exercise 14. Use (MC), (MA), (MX) and (DP) for Z.Exercise 15. Use (AC) and (MC) for Z.Exercise 16. Use (AA), (MA) and (DP) for Z.Exercise 17. Remember to show both existence and uniqueness. Also,

cf., Proposition 11 of Chapter 8.Exercise 18. Remember to show both existence and uniqueness. Also,

cf., Proposition 12 of Chapter 8.Exercise 19. Cf., Exercise 16 of Chapter 8.Exercise 21. Use (MC) and (MA) for Z.Exercise 22. Use (MC) for Z.Exercise 23. Use (MA) for Z.Exercise 24. Use (MC) and (MA) for Z. Also, cf., Exercise 13 of

Chapter 8.Exercise 25. Use (MC) and (MA) for Z. Also, cf., Exercise 14 of

Chapter 8.Exercise 26. Cf., Exercise 18 of Chapter 8.Exercise 27. Use (MC) and (MA) for Z.Exercise 28. Use (MC), (MA), (MX) and (DP) for Z.Exercise 29. Use (PP) for Z.Exercise 30. Cf., Exercise 25 of Chapter 8.Exercise 31. You already know that 0 &→ [0, 1] and 1 &→ [1, 1].Exercise 32. Use the Archimedean Principle on Z.Exercise 33. Cf., Exercise 4 of Chapter 9.Exercise 34. Take the average of p and q.



Exercise 2.Exercise 4.Exercise 5. Suppose S ⊂ R is nonempty and bounded above by U ;

that is, S ⊂ U for all S ∈ S. Set L = ∪S. Show that S ∈ R and S is theleast upper bound of S.

Exercise 7.Exercise 8.Exercise 9.Exercise 10. Show that X + 0 ⊂ X and X ⊂ X + 0. Similarly for

0 + X .Exercise 11.Exercise 12.Exercise 13.Exercise 14. Use the Cancellation Property for Addition (Exercise 16)

to show strict inequality.Exercise 16.Exercise 17.Exercise 18.Exercise 19.Exercise 20.Exercise 21.Exercise 22.Exercise 23.Exercise 24.Exercise 25.Exercise 26.Exercise 27.Exercise 28.Exercise 29.Exercise 30.Exercise 31.Exercise 32.Exercise 33.Exercise 34.Exercise 35. Thinking of cuts, x < y implies f (x) ≤ f (y). By

density of rationals, there exists q, r ∈ Q such that x < q < r < y.Exercise 36. Use Exercise 35.Exercise 37. For q ∈ F , set S = { q ∈ Q | f (q) < a } and x = sup S.

Show that f (x) = q.


Exercise 38. If x, y ∈ R and q ∈ Q such that f (x + y) < f (q) <

f (x) + f (y), then there exist q1, q2 ∈ Q such that q1 + q2 = q, x < q1,and y < q2.

Exercise 39. Start with positive real numbers.Exercise 40. Use the properties of isomorphisms to show that f (x) =

x for all x ∈ Z, then for all x ∈ Q, and finally for all x ∈ R.Exercise 42. Consider g−1 ◦ f for isomorphisms f, g : R → F .


Exercise 4.Exercise 5.Exercise 6.Exercise 7.Exercise 8.Exercise 9.Exercise 10.Exercise 11.Exercise 12.Exercise 13.Exercise 16. Use Definition 3 to prove the properties of a total order-

ing. To see this is not an ordered field, compare i with 0 and 0 with −1 andconsider the product of positives property of ordered fields.

Exercise 17.Exercise 18. Either i > 0 or i < 0. (Why?) Use Exercise 26 of

Chapter 8.Exercise 19.Exercise 20. Consider the image of i and the preservation of multipli-

cation.Exercise 21.


Exercise 5.Exercise 6.Exercise 7.Exercise 8.Exercise 9.Exercise 10.Exercise 13.Exercise 15.Exercise 16.Exercise 17.


Exercise 18.Exercise 19.Exercise 21.Exercise 24.Exercise 26.Exercise 28.Exercise 30.Exercise 31.Exercise 32.Exercise 33.Exercise 34. Use the denseness of Q in R.Exercise 35.Exercise 36.Exercise 37.

195

Postscript and Selected Ref-erences

Completion of Part II of these notes, together with a sequence incalculus, should provide a student with more than adequate preparation for acourse in linear algebra or abstract algebra or number theory. Actually, PartII alone should suffice; however, calculus may provide for some examplesin all such courses.

Completion of Parts II and III of these notes, together with a sequencein calculus, should provide a student with more than adequate prepara-tion for a course in analysis (or advanced calculus) or general (point-set)topology or axiomatic set theory or logic.

Conway, Lakatos, Knuth

References

[Bell] Bell, E. T., The Development of Mathematics, Dover, New York, 1945, 1972.[Halmos] Halmos, Paul, Finite Dimensional Vector Spaces, Springer-Verlag, New York,

19??.[Herstein] Herstein, I. N., Topics in Algebra, Ginn, Waltham, MA, 1964.[Rudin] Rudin, Walter, Principles of Mathematical Analysis, Third Edition, McGraw-Hill,

New York, 1976.[Spivak] Spivak, Michael, Calculus, Second Edition, Publish or Perish, Wilmington, DE,

1980.

196

Index

The numbers after the keywords refer to item numbers; for example, 3.2refers to Definition 2 (or Example 2, or Remark 2, etc.) in Chapter 3. Thenotation “iv” refers to page iv, concerning notation, which appears beforethe Preface; there are no other page numbers yet—sorry!

Symbols�, iv⊂, iv, 4.3�, iv∅, 4.2∈, 4.1{ }, 4.2(a, b), iv[a, b), iv(a, b], iv[a, b], ivC, iv, 14.1i , 14.14N, iv, 10.4Q, iv, 12.3Q+, ivQ∗

+, ivR, iv, 13.1, 15.22R+, ivR∗

+, ivZ, iv, 11.3Zn , ivZ+, iv, 10.4Z∗, iv

AAddition on C, 14.1Addition on Q, 12.13Addition on R, 13.6Addition on Z+, 10.10

Index 197

Addition on Z, 11.13Algebraic Nnumbers, 7.37And, 2.3Antisymmetric relation, 6.1(d)Archimedean Principle, 8.1

BBijective function, 5.32, 5.55Birkhoff, G.D., 1.6Bolzano-Weierstrass property, 15.15Bounded sequence, 15.12

CCardinality of sets, same, 7.2Cartesian product of sets, 5.1, 7.52Cauchy sequence, 15.11Codomain of a function, 5.10Complex numbers, set of, iv, 14.1Composition of two functions, 5.36, 5.60Contradiction, 2.ContradictionDefContraposition, proof by, 3.13Connected relation, 6.1(f)Convergent sequence, 15.3Countable set, 7.22Cut, 13.1

DDecreasing sequence, 15.14Dedekind cut, 13.1Denumerable set, 7.21Dense, 8.6Descendency set, 10.5Difference of sets, 4.42Disjoint sets, 4.18, 4.33Domain of a function, 5.10

EElement, 4.1Embedding, 11.37Empty set, 4.2Equal functions, 5.20Equal sets, 4.9Equivalence class, 6.41Equivalence relation, 6.37

F

198 Index

Field, 8.4Finite set, 7.10Function, 5.10

GGreatest element, 6.20

H

IIdentity function, 5.46Iff, 3.10Image of a set under a function, 5.13Inclusive or, 2.4Increasing sequence, 15.14Induction, 3.26Infinite set, 7.17Injective function, 5.27, 5.55Integers, set of, iv, 11.3Intersection of two sets, 4.14Intersection of a collection of sets, 4.30Inverse image of a set under a function, 5.53Invertible function, 5.48Isomorphism, 8.31

J

K

LLeast upper bound, 6.28Least upper bound property, 6.34Less than on C, 14.15Less than on Q, 12.7Less than on R, 13.3, 15.23Less than on Z+, 10.6, 10.12Less than on Z, 11.4Lexicographic ordering, 14.15

MMathematical or, 2.4Maximal element, 6.19Mod, for an equivalence relation, 1.6, 6.45Mod, for a partition, 6.47Monotone sequence, 15.14Multiplication on C, 14.1Multiplication on Q, 12.20Multiplication on R, 13.15

Index 199

Multiplication on Z+, 10.21Multiplication on Z , 11.22

NNatural numbers, set of, iv, 10.4Negative integer, 11.11Nonnegative integers, set of, ivNonnegative rational numbers, set of, ivNonnegative real numbers, set of, ivNonreflexive relation, 6.1(b)Nonzero integers, set of, ivNot, 2.2

OOne-to-one function, 5.27, 5.55Onto function, 5.30, 5.55Or, 2.4Ordered field, 8.8Ordering on C, 14.15Ordering on Q, 12.7Ordering on R, 13.3, 15.23Ordering on Z+, 10.6Ordering on Z, 11.4

PPairwise disjoint collection of sets, 4.32Partial ordering, 6.9Partition, 6.44Positive integer, 11.11Positive rational numbers, set of, ivPositive real numbers, set of, ivPower set, 4.53Predecessor, 10.8Principle of Induction, 3.26Product of sets (cartesian), 5.1, 7.52Proper subset, 4.4

QQuantifier, Hidden, 3.33

RRange of a function, 5.14Rational numbers, set of, iv, 12.3Real numbers, set of, iv, 13.1, 15.22Reflexive relation, 6.1(a)Relation, 6.6Relation, properties of a, 6.1

200 Index

SSame cardinality, 7.2Sequence, 15.1Set, 4.1Set difference, 4.42Strict partial ordering, 6.8Strict total ordering, 6.16Subset, 4.3Subset, proper, 4.4Successor, 10.2Surjective function, 5.30, 5.55Symmetric relation, 6.1(c)

TTautology, 2.TautologyDefTotal ordering, 6.17Transitive relation, 6.1(e)Trichotomous relation, 6.1(g)

UUncountable set, 7.39Union of two sets, 4.19Union of a collection of sets, 4.38Upper bound, 6.27

V

WWell ordered set, 6.36

X

Y

ZZero, 10.3

Bridge to Abstract Mathematics

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