Math 13 — An Introduction to Abstract Mathematics

Neil Donaldson and Alessandra Pantano

With contributions from:

Michael Hehmann, Christopher Davis, Liam Hardiman, and Ari Rosenﬁeld

November 29, 2024

Contents

Preface: What is Math 13 and who is it for? ii

1 Introduction: What is a Proof? 1

2 Logic and the Language of Proofs 6

2.1 Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2 Propositional Functions & Quantiﬁers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.3 Methods of Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.4 Further Proofs & Strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3 Divisibility and the Euclidean Algorithm 32

3.1 Divisors, Remainders and Congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.2 Greatest Common Divisors and the Euclidean Algorithm . . . . . . . . . . . . . . . . . 38

4 Sets and Functions 43

4.1 Set Notation and Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.2 Unions, Intersections and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.3 Introduction to Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5 Mathematical Induction and Well-ordering 60

5.1 Iterative Processes & Proof by Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

5.2 Well-ordering and the Principle of Mathematical Induction . . . . . . . . . . . . . . . . 67

5.3 Strong Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

6 Set Theory, Part II 77

6.1 Cartesian Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

6.2 Power Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

6.3 Indexed Collections of Sets: Union and Intersection Revisited . . . . . . . . . . . . . . 85

7 Relations and Partitions 92

7.1 Binary Relations on Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

7.2 Functions revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

7.3 Equivalence Relations & Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

7.4 Well-deﬁnition, Rings and Congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

8 Cardinality and Inﬁnite Sets 111

8.1 Cantor’s Notion of Cardinality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

8.2 Uncountable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

Preface: What is Math 13 and who is it for?

Math 13 was created by the late Howard Tucker as a traditional discrete mathematics course (the

number was a deliberate joke). It eventually evolved to become the transition class in UCI’s under-

graduate program, introducing students to abstraction and proof, and serving as the key pre-requisite

for upper-division courses such as abstract algebra, analysis, linear algebra & number theory.

The typical student is simultaneously working through lower-division calculus and linear algebra.

Knowledge of such material is unnecessary and students are encouraged to take Math 13 early both

to ease the transition from algorithmic to abstract mathematics, and so that a proof-mentality may be

brought to other lower-division classes.

This text evolved from course notes dating back to 2008. Math 13 is something of a hydra due to the

niche it occupies in UCI’s program: part proof-writing, part discrete mathematics, and part introduc-

tion to speciﬁc upper-division topics. Logic is covered only at a basic level while set theory is spread

throughout the text, the intent being for dry ‘grammar’ topics to be absorbed through engagement

with more accessible and fun ideas. By the end of the course, interested students should be prepared

for a formal study of logic and set theory at the upper-division level.

Learning Outcomes

1. Developing the skills necessary to read and practice abstract mathematics.

2. Understanding the concept of proof and becoming acquainted with multiple proof techniques.

3. Learning what sort of questions mathematicians ask and what excites them.

4. Introducing upper-division mathematics by providing a taste of what is covered in several

courses. For instance:

Number Theory & Abstract Algebra How can we perform arithmetic with remainders? Can you

ﬁgure out what on which day of the week you were born?

Geometry and Topology How can we visualize objects such as the M

obius strip? How can we use

sequences of sets to produce objects (fractals) that appear similar at all scales?

To Inﬁnity and Beyond! Why are some inﬁnities greater than others?

Useful Texts The following texts are recommended if you want more exercises and material. The

ﬁrst two are available free online; the rest were previous textbooks for Math 13.

• Book of Proof, Richard Hammack

• Mathematical Reasoning, Ted Sundstrom

• Mathematical Proofs: A Transition to Advanced Mathematics, Chartrand/Polimeni/Zhang

• The Elements of Advanced Mathematics, Steven G. Krantz

• Foundations of Higher Mathematics, Peter Fletcher and C. Wayne Patty

1 Introduction: What is a Proof?

The essential concept in higher-level mathematics is that of proof. A basic dictionary entry might

cover two meanings:

1. A test or trial of an assertion.

2. An argument that establishes the validity (truth) of an assertion.

In science and wider culture, the ﬁrst meaning predominates: a defendant was proved guilty in court;

a skin cream is clinically proven to make you look younger; an experiment proves that the gravitational

constant is 9.81ms

−2

. A common mistake is to treat such a proved assertion as unambiguously true.

Distinct juries might disagree as to whether a defendant is guilty; indeed for many crimes the truth

is uncertain, hence the more nuanced legal expression proved beyond reasonable doubt.

In mathematics we use the second meaning: a proof establishes the incontrovertible truth of some

assertion. To see what we mean, consider a simple claim (mathematicians use the word theorem).

Theorem 1.1. The sum of any pair of even integers is even.

Hopefully you believe this statement. But how do we prove it? We can test the assertion by consider-

ing examples (4 + 6 = 10 is even, (−8) + 30 = 22 is even, etc.), but we cannot expect to verify all pairs

this way. For a mathematical proof, we somehow need to test all possible examples simultaneously.

To do so, it is essential to have a clear idea of what is meant by an even integer.

Deﬁnition 1.2. An integer is even if it may be written in the form 2k where k is an integer.

Proof. Let x and y be even. Then x = 2k and y = 2l for some integers k and l. But then

x + y = 2k + 2l = 2(k + l) (∗)

is even.

The box indicates the end of the argument. Traditionally the letters Q.E.D. were used, an acronym

for the Latin quod erat demonstrandum (which is what was to be demonstrated).

Consider how the proof depends crucially on the deﬁnition.

• The theorem did not mention any variables, though these were essential to the proof. The vari-

ables k and l come for free once you write the deﬁnition of evenness! This is very common; simple

proofs are often little more than rearranged deﬁnitions.

• According to the deﬁnition, 2k and 2l together represent all possible pairs of even integers. It is

essential that k and l be different symbols: Is is clear why? What would you be proving if k = l?

• The calculation (∗) is the easy bit; without the surrounding sentences and the direct reference

to the deﬁnition of evenness, the calculation means nothing.

Notice the sleight of hand: a mathematical proof establishes truth only by reference to one or more

deﬁnitions.

Strictly speaking, the deﬁnition and theorem also depend on the meanings of integer and sum, though to rigorously

deﬁne either would take us too far aﬁeld. In any context, some concepts will be considered too basic to merit deﬁnition.

Theorems & Conjectures

Theorems are true mathematical statements that we can prove. Some are important enough to merit

names (Pythagorean theorem, fundamental theorem of calculus, rank–nullity theorem, etc.), but most

are simple statements such as Theorem 1.1.

In practice we are often confronted with conjectures: statements we suspect to be true, but which we

don’t (yet) know how to prove. Much of the fun and creativity of mathematics lies in formulating

and attempting to prove (or disprove) conjectures.

A conjecture is the mathematician’s analogue of the scientist’s hypothesis: a statement one would

like to be true. The difference in approach takes us right back to the dual meaning of proof. The

scientist tests their hypothesis using the scientiﬁc method, conducting experiments which attempt

(and hopefully fail!) to show that the hypothesis is incorrect. The mathematician tries to prove the

validity of a conjecture by relying on logic. The job of a mathematical researcher is to formulate

conjectures, prove them, and publish the resulting theorems. Attempting to formulate your own

conjectures is an essential part of learning mathematics; many will likely be false, but you’ll learn

much by ﬁguring out why!

Here are two conjectures to give us a taste of this process.

Conjecture 1.3. If n is any odd integer, then n

−1 is a multiple of 8.

Conjecture 1.4. If n is any positive integer, then n

+ n + 41 is prime.

How can we decide if these conjectures are true or false? To get a feel for things, we start by comput-

ing examples for several small integers n. (In practice, this is likely what lead to the formulation of

the conjectures in the ﬁrst place!)

n 1 3 5 7 9 11 13

−1 0 8 24 48 80 120 168

n 1 2 3 4 5 6 7

+ n + 41 43 47 53 61 71 83 97

Since 0, 8, 24, 48, 80, 120 and 168 are all multiples of 8, and 43, 47, 53, 61, 71, 83 and 97 are all prime,

both conjectures appear to be true. Would you bet $100 that this is indeed the case? Is n

−1 a multiple

of 8 for every odd integer n? Is n

+ n + 41 prime for every positive integer n? Establishing whether

each conjecture is true or false requires one of the following:

Prove it by showing it must be true in all cases, or,

Disprove it by ﬁnding at least one instance in which the statement is false.

Let us start with Conjecture 1.3. If n is an odd integer, then, by deﬁnition, we may write n = 2k + 1

for some integer k. Now compute the object of interest:

−1 = (2k + 1)

−1 = (4k

+ 4k + 1) −1 = 4k

+ 4k = 4k(k + 1)

We need to investigate whether this is always a multiple of 8. Since k is an integer, n

−1 is plainly a

multiple of 4, so everything comes down to deciding whether k(k + 1) is always even. Do we believe

A positive integer is prime if it cannot be written as the product of two integers, both greater than one.

this? We return to testing some small values of k:

k −2 −1 0 1 2 3 4

k(k + 1) 2 0 0 2 6 12 20

Once again, the claim seems to be true for small values of k, but is it is true for all k? Again, the only

way is to prove or disprove it. Observe that k(k + 1) is the product of two consecutive integers. This is

great, because for any two consecutive integers one is even and the other odd; their product must be

even. Conjecture 1.3 is indeed a theorem!

So far, our approach has been investigative. Scratch work is an essential part of the process, but we

shouldn’t expect a reader to have to ﬁght their way through such. We therefore offer a formal proof.

This is the ﬁnal result of our deliberations; investigate, spot a pattern, conjecture, prove, and ﬁnally

present our work in as clean and convincing a manner as we can.

Theorem 1.5. If n is any odd integer, then n

−1 is a multiple of 8.

Proof. Let n be an odd integer. By deﬁnition, we may write n = 2k + 1 for some integer k. Then

−1 = (2k + 1)

−1 = (4k

+ 4k + 1) −1 = 4k

+ 4k = 4k(k + 1)

We distinguish two cases. If k is even, then k(k + 1) is even and so 4k(k + 1) is divisible by 8.

If k is odd, then k + 1 is even. Therefore k(k + 1) is again even and 4k(k + 1) divisible by 8.

In both cases n

−1 = 4k(k + 1) is divisible by 8.

All that work, just for ﬁve lines of clean argument! But wasn’t it fun?

When constructing elementary proofs it is common to be unsure over how much detail to include.

We relied on the deﬁnition of oddness, but we also used the fact that a product is even whenever

either factor is even; does this need a proof? Since the purpose of a proof is to convince the reader,

the appropriateness of an argument will depend on context and your audience: if you are trying to

convince a middle-school student, maybe you should justify this step more fully, though the cost

would be a longer argument whose totality is harder to grasp. A proof that works perfectly in all

situations is unlikely to exist! A good rule is to imagine writing for another mathematician at your

own level—if a fellow student believes your argument, that’s a good sign of both its validity and

appropriateness.

We now consider Conjecture 1.4. The question is whether n

+ n + 41 is prime for every positive

integer n. When n ≤ 7 the answer is yes, but examples do not make a proof! To investigate further,

return to the deﬁnition of prime (footnote 2): is there a positive integer n for which is n

+ n + 41 can

be factored as a product of two integers, both at least 2? A straightforward answer is staring us in

the face! When n = 41 such a factorization certainly exists:

+ n + 41 = 41

+ 41 + 41 = 41(41 + 1 + 1) = 41 ·43

We call n = 41 a counterexample; there is at least one integer n for which n

+ n + 41 is not prime.

Conjecture 1.4, being a claim about all integers n, is therefore false—it has been disproved.

Planning and Writing Proofs

Your main responsibility in this course is the construction of proofs. Their sheer variety means that,

unlike in elementary calculus, you cannot simply practice tens of similar problems until the process

becomes automatic. So how do you learn to write proofs?

The ﬁrst step is to read other arguments. Don’t just accept them, make sure you believe them: check

the calculations, verify claims, and rewrite the argument in your own words adding any clariﬁcations

you deem necessary or helpful.

As you dissect others’ work, a daunting question often arises: how did they ever come up with this? As

our work on Theorem 1.5 shows, the source of a proof is often less magical than it appears; usually

the author experimented until they found something that worked. The experimentation is hidden

in the ﬁnal proof, whose purpose is to be as clean and convincing as possible. A proof is akin to a

concert performance after hours of private practice; wrong notes don’t belong in the Carnegie Hall!

In order to bridge the gap, we recommend splitting the proof-writing process into several steps.

Interpret Make sense of the statement. What is it saying? Can you rephrase in a clearer manner?

What are you assuming? A key part of this step is identifying the logical structure of the state-

ment. We’ll discuss this at length in the next chapter.

Brainstorm Convince yourself that the statement is true. First, look up the relevant deﬁnitions. Next,

think of some instances where the conditions of the statement are met. Try out some exam-

ples, and ask yourself what makes the claim work in those instances. Examples can help build

intuition about why a claim is true and sometimes suggest a proof strategy. Review other theo-

rems that use these deﬁnitions. Do you know any theorems that relate your assumptions to the

conclusion? Have you seen a proof of a similar statement before?

Sketch Build the skeleton of your proof. Think again about what you are assuming and what you

are are you trying to prove. It is often straightforward to write down reasonable ﬁrst and last

steps (the bread slices of a proof-sandwich). Try to connect these with informal arguments. If you

get stuck, try a different approach.

This step is often the longest in the proof-writing process. It is also where you will be doing

most of your calculations. You can be as messy as you like because no-one ever has to see it! Once

you’ve learned a variety of proof methods, this is a good stage at which to experiment with

different approaches.

Prove Once you have a suitable sketch, it’s time to prove the statement to the world. Translate your

sketch into a linear story. Carefully word your explanations and avoid shorthand, though well-

understood mathematical symbols such as =⇒ are encouraged. The result should be a clear,

formal proof such as you’d ﬁnd in a mathematics textbook. Although you are providing a

mathematical argument, your proof should read like prose and be written in complete sentences.

Review Finally, review your proof. Assume the reader is meeting the problem for the ﬁrst time and

has not seen your sketch. Read your proof with skepticism; consider its readability and ﬂow.

Get rid of unnecessary claims and revise the wording if necessary. Read your proof out loud.

If you’re adding extra words that aren’t written down, include them in the proof. Share your

work with others. Do they understand it without any additional input from you?

Conjectures: True or False?

Higher-level mathematics is all about the links between proofs, deﬁnitions, theorems and conjec-

tures. We prove theorems (and solve homework problems) because they make us use, and aid our

understanding of, deﬁnitions. We state deﬁnitions to help us formulate conjectures and prove theo-

rems. One does not know mathematics, one does it. Mathematics is a practice; an art as much as it is a

science.

With this in mind, do your best to prove or disprove the following conjectures. Don’t worry if some

terms or notations are unfamiliar: ask! Everything will be covered formally soon enough. At the end

of the course, revisit these problems to realize how much your proof skills have improved.

1. The sum of any three consecutive integers is even.

2. There exist integers m and n such that 7m + 5n = 4.

3. Every common multiple of 6 and 10 is divisible by 60.

4. There are integers x and y which satisfy 6x + 9y = 10.

5. For every positive real number x, the value x +

is greater than or equal to 2.

6. If x is any real number, then x

≥ x.

7. If n is an integer, then n

+ 5n must be even.

8. If x is a real number, then

≥ −x.

9. If n is an integer greater than 2, then n

−1 is not prime.

10. An integer is divisible by 5 whenever its last digit is 5.

11. If r is a rational number, then there is a non-zero integer n such that rn is an integer.

12. There is a smallest positive real number.

13. For all real numbers x, there exists a real number y for which x < y.

14. There is a real number x such that, for all real numbers y, we have x < y.

15. The sets A = {n ∈ N : n

< 25} and B = {n

: n ∈ N and n < 5} are equal. Here N denotes

the set of natural numbers.

2 Logic and the Language of Proofs

2.1 Propositions

To read and construct proofs, we need to develop the language of logic. This is to mathematics what

grammar is to English.

Deﬁnition 2.1. A proposition or statement is a sentence that is either true or false.

Examples 2.2. 1. 17 −24 = 7. 2. 39

is an odd integer. 3. God exists.

4. The moon is made of cheese. 5. Every cloud has a silver lining.

For a proposition to make sense, readers must agree on the meaning of each concept it references. In

the real world, arguments about propositions are often disagreements over deﬁnitions. For instance,

the question of whether God exists is meaningless unless we agree on which conception (Shiva,

Yahweh, Allah, Zeus, all/any of them?) is being discussed! This also illustrates that the truth status

of a proposition need not be known when stated, a particularly common situation in mathematics.

Logical Expressions: Truth Tables and Combining Propositions

To develop basic terminology, we represent abstract propositions by letters P, Q, R, . . . (similarly to

the use of x, y, z in algebra). Combinations of such (logical expressions) are easily described in tabular

form.

Deﬁnition 2.3. Let P and Q be propositions. The following truth tables deﬁne three new propositions:

• The conjunction P ∧ Q is read “P and Q.”

• The disjunction P ∨ Q is read “P or Q.”

• The negation ¬P is read “not P.”

P Q P ∧ Q P ∨Q

T T T T

T F F T

F T F T

F F F F

P ¬P

T F

F T

A tautology is a logical expression that is always true (truth table has a column of T’s), regardless of

its component propositions. A contradiction is a logical expression that is always false.

The letters T/F stand for true/false. For instance, the second line of the ﬁrst table says that if P is true

and Q is false, then the proposition “P and Q” is false; similarly “P or ‘Q” is true.

Examples 2.4. 1. By choosing explicit propositions, we may compare the logical and/or/not with

their plain English meanings. Suppose P and Q are the propositions

P: “I like purple.” Q: “I like chartreuse.”

The new propositions from Deﬁnition 2.3 might then be written

P ∧ Q: “I like purple and chartreuse.” P ∨Q: “I like purple or chartreuse.”

¬P: “I do not like purple.”

Alternative phrasings might aid readability, though be careful: “Not, I like purple,” is terrible

English! Note also that the logical or is inclusive (ﬁrst line of the truth table): with a logical or,

“I like purple or chartreuse,” means you might like both.

2. We continue the previous example by adding a third proposition R: “It is 9am.” What logical

expression might be represented by the following sentence?

“I like purple and I like chartreuse or it is 9am.”

Is it P ∧ (Q ∨ R) or is it (P ∧ Q) ∨ R? Without brackets the sentence is unclear. In fact, as the

next truth table shows, these logical expressions mean different things.

P Q R Q ∨ R P ∧(Q ∨ R) P ∧ Q (P ∧ Q) ∨ R

T T T T T T T

T T F T T T T

T F T T T F T

T F F F F F F

F T T T F F T

F T F T F F F

F F T T F F T

F F F F F F F

The moral here is that English is terrible for logic! Clear identiﬁcation of propositions is essen-

tial if you want to avoid ambiguous sentences such as the above.

3. Let S be an abstract proposition. Then S ∨( ¬S) is a tautology.

Otherwise said, no matter the meaning of S, either it or its

negation must be true.

S ¬S

S ∨(¬S) S ∧(¬S)

T F T F

F T T F

Similarly, S ∧(¬S) is a contradiction: regardless of S, it and its negation cannot simultaneously

be true.

Conditional and Biconditional Connectives

Having one proposition lead to another is of critical importance to mathematics.

Deﬁnition 2.5. Given propositions P, Q, the conditional (⇒)

and biconditional (⇔) connectives deﬁne new propositions as de-

scribed in the truth table.

For the proposition P =⇒ Q, we call P the hypothesis and Q the

conclusion.

P Q P =⇒ Q P ⇐⇒ Q

T T T T

T F F F

F T T F

F F T T

Connective propositions can be read and written in many different ways: for instance,

P =⇒ Q P ⇐⇒ Q

P implies Q P therefore Q P if and only if Q

If P, then Q Q follows from P P iff Q

P only if Q Q if P P and Q are (logically) equivalent

P is sufﬁcient for Q Q is necessary for P P is necessary and sufﬁcient for Q

Examples 2.6. 1. Here are six English sentences expressing the same conditional P =⇒ Q:

• If you are born in Rome, then you are Italian.

• You are Italian if you are born in Rome.

• You are born in Rome only if you are Italian.

• Being born in Rome is sufﬁcient for being Italian.

• Being Italian is necessary for being born in Rome.

Are you comfortable with what the propositions P and Q are in this situation?



P ∧(P =⇒ Q)



=⇒ Q is a tautology.

P Q P =⇒ Q P ∧(P =⇒ Q)



P ∧(P =⇒ Q)



=⇒ Q

T T T T T

T F F F T

F T T F T

F F T F T

Connectives are central to mathematics for many reasons. In particular:

1. The vast majority of simple theorems may be written in the form P =⇒ Q. For instance, revisit

Theorem 1.1 and the discussion that follows:

If x and y are even integers, then x + y is even. (If P, then Q)

Identifying the hypothesis and conclusion is essential if you want to understand a theorem!

2. The simplest (“direct”) proofs typically involve chaining a sequence of connectives:

P =⇒ P

=⇒ ··· =⇒ P

=⇒ Q

We’ll revisit these ideas in Section 2.3, and repeatedly throughout the course.

While the biconditional should be easy to remember, it is harder to make sense of the conditional

connective. Short of simply memorizing the truth table, here are two examples that might help.

Examples 2.7. 1. Suppose your professor says, “If the class earns a B average on the midterm, then

I’ll bring doughnuts.” The only situation in which the teacher will have lied is if the class earns

a B average but she fails to provide doughnuts.

2. (F =⇒ T really can be true!) Let P be the proposition “7 = 3” and Q be “0 = 0.” Since

multiplication of both sides of an equation by zero is algebraically valid, we see that

7 = 3 =⇒ 0 ·7 = 0 ·3 (If 7 = 3, then 0 times 7 equals 0 times 3)

=⇒ 0 = 0 (then 0 equals 0)

This argument is perfectly correct: the implication P =⇒ Q is true. It (rightly!) makes us uncom-

fortable because the hypothesis is false.

If we instead add 1 to each side of 7 = 3, we’d obtain a example where F =⇒ F is true.

The Converse and Contrapositive

Deﬁnition 2.8. The converse of P =⇒ Q is the reversed implication Q =⇒ P.

The contrapositive of P =⇒ Q is the implication ¬Q =⇒ ¬P.

It is vital to understand the distinction between these, particularly because of a crucial result.

Theorem 2.9. The contrapositive of an implication is logically equivalent to the original.

Proof. Compute the truth table and observe that its third and sixth columns are identical:

P Q P =⇒ Q ¬Q ¬P ¬Q =⇒ ¬P

T T T F F T

T F F T F F

F T T F T T

F F T T T T

Otherwise said, (P =⇒ Q) ⇐⇒ (¬Q =⇒ ¬P) is a tautology.

By contrast, the converse is not logically equivalent to the original: the converse of a true implication

could be either true or false (though see Exercise 11 for a special situation).

Example 2.10. Let P and Q be the statements

P: “Claudia has a peach.” Q: “Claudia has a fruit.”

Since a peach is certainly a fruit, P =⇒ Q is true: “If Claudia has a peach, then she has a fruit.”

The converse of P =⇒ Q is the sentence

Q =⇒ P: “If Claudia has a fruit, then she has a peach.”

This is palpably false: Claudia might have an apple! However, in accordance with Theorem 2.9, the

contrapositive is true:

¬Q =⇒ ¬P: “If Claudia does not have a fruit, then she does not have a peach.”

Negating Logical Expressions

Mathematics often requires us to negate propositions. What would you suspect to be the negation of

a conditional P =⇒ Q? Is it enough to say “P doesn’t imply Q”? But what does this mean?

We again rely on a truth table: to get the last column, recall that

negation simply swaps T and F. Can we write this column in

another way? The single T in the ﬁnal column provides a proof

of an important result:

P Q P =⇒ Q ¬(P =⇒ Q)

T T T F

T F F T

F T T F

F F T F

Theorem 2.11. ¬(P =⇒ Q) is logically equivalent to P ∧¬Q (“P and not Q”).

Example 2.12. Consider the implication

It’s morning so I’ll have coffee.

Hopefully its negation is clear:

It’s morning and I won’t have coffee.

As in Example 2.7, it might help to think about what it means for the original statement to be false.

Warning! The negation of P =⇒ Q is not a conditional. In particular it is neither:

• The converse Q =⇒ P

• The contrapositive of the converse ¬P =⇒ ¬Q

If you are unsure about this, compare the truth tables!

Our ﬁnal results in basic logic also involve negations; they are named for Augustus de Morgan, a

British logician of the 19

century.

Theorem 2.13 (de Morgan’s laws). Let P and Q be propositions.

1. ¬(P ∧Q) is logically equivalent to ¬P ∨¬Q

2. ¬(P ∨Q) is logically equivalent to ¬P ∧¬Q

Proof. For the ﬁrst law, compute the truth table: the fourth and seventh columns are identical.

P Q

P ∧ Q ¬(P ∧Q) ¬P ¬Q ¬P ∨¬Q

T T T F F F F

T F F T F T T

F T F T T F T

F F F T T T T

The second law is an exercise.

Example 2.14. Consider the sentence

I rode the subway and I had coffee.

To negate this, we might write

I didn’t ride the subway or I didn’t have coffee.

Subway Coffee Subway and Coffee

T T T

T F F

F T F

F F F

This reads awkwardly because the negation encompasses three distinct possibilities. Note how the

logical (inclusive) use of or includes the last row of the truth table: the possibility that one neither

rode the subway nor had coffee. As with Example 2.4, this is another advert for the use of logic over

English.

Aside: Algebraic Logic We can use truth tables to establish other laws of basic logic, e.g.:

Double negation ¬(¬P) ⇐⇒ P

Commutativity P ∧ Q ⇐⇒ Q ∧P P ∨ Q ⇐⇒ Q ∨ P

Associativity (P ∧Q) ∧ R ⇐⇒ P ∧(Q ∧ R) (P ∨Q) ∨ R ⇐⇒ P ∨(Q ∨R)

Distributivity (P ∧Q) ∨ R ⇐⇒ (P ∨ R) ∧(Q ∨ R) (P ∨Q) ∧ R ⇐⇒ (P ∧ R) ∨(Q ∧ R)

To make things more algebraic, we’ve replaced “is logically equivalent to” with a biconditional.

Armed with these laws, one can often manipulate logical expressions without the laborious creation

You are welcome to try memorizing these laws, though there is typically no need: De Morgan’s laws,

together with your intuitive understanding of and, or and not mean you’ll likely perform correct

manipulations regardless.

Exercises 2.1. A reading quiz and several questions with linked video solutions can be found online.

1. Express each statement in the form, “If . . . , then . . . ” There are many possible correct answers.

(a) You must eat your dinner if you want to grow.

(b) Being a multiple of 12 is a sufﬁcient condition for a number to be even.

(d) A triangle is equilateral only if all its sides have the same length.

2. Suppose “x is an even integer” and “y is an irrational number” are true statements, and that

“z ≥ 3” is a false statement. Which of the following are true?

(Hint: Label each statement and think about each using connectives)

(a) If x is an even integer, then z ≥ 3.

(b) If z ≥ 3, then y is an irrational number.

(d) If y is an irrational number and x is an even integer, then z ≥ 3.

3. Orange County is considering two competing transport plans: widening the 405 freeway and

constructing light rail down its median. A local politician is asked, “Would you like to see the

405 widened or would you like to see light rail?” The politician wants to sound positive, but to

avoid being tied to one project. What is their response?

(Hint: Think about how the word ‘OR’ is used in logic)

4. Consider the proposition: “If the integer m is greater than 3, then 2m is not prime.”

(a) Rewrite the proposition using the word ‘necessary.’

(b) Rewrite the proposition using the word ‘sufﬁcient.’

5. Suppose the following sentence is true: “If Amy likes art, then no-one likes history.” What, if

anything, can we conclude if we discover that someone likes history.

Stating the laws in this fashion is to assert that each expression is a tautology (Deﬁnition 2.3). For instance, to claim

that “¬(¬P) is logically equivalent to P” is to assert that ¬(¬P) ⇐⇒ P is a tautology.

6. Construct the truth tables for the propositions P ∨(Q ∧R) and (P ∨Q) ∧R. Are they the same?

7. Use truth tables to establish the following laws of logic:

(a) Double negation: ¬(¬P) ⇐⇒ P

(b) Idempotent law: P ∧ P ⇐⇒ P

(d) Distributive law: (P ∧Q) ∨ R ⇐⇒ (P ∨ R) ∧(Q ∨ R)

8. (a) Decide whether (P ∧¬P) =⇒ Q is a tautology, a contradiction, or neither.

(b) Explain why ¬P ∨ ¬Q is logically equivalent to P =⇒ (P ∧¬Q).



(P ∧¬Q) =⇒ F



⇐⇒ (P =⇒ Q) is a tautology. Here F represents a contradiction.

9. (a) Prove that the expressions (P =⇒ Q) ∧ (Q =⇒ P) and P ⇐⇒ Q are logically equivalent.

(b) Prove that



(P =⇒ Q) ∧(Q =⇒ R)



=⇒



P =⇒ R



is a tautology.

Why do these make intuitive sense?

10. Use logical algebra (page 11) to show that



(P ∨Q) ∧ ¬P



∧¬Q is a contradiction.

11. Compare the truth tables for P =⇒ Q and its converse. Do there exist propositions P, Q for

which both P =⇒ Q and its converse are false? Explain.

12. A friend insists that the negation of “Mark and Mary have the same height,” is “Mark or Mary

do not have the same height.” What is the correct negation? Where did your friend go wrong?

13. Suppose that the following statements are true:

(a) Every octagon is magical.

(b) If a polygon is not a rectangle, then is it not a square.

Is it true that “Octagons are rectangles”? Explain your answer.

(Hint: try rewriting each of the statements as an implication)

14. The connective ↓ (the Quine dagger, NOR) is deﬁned by a truth table.

(a) Prove that P ↓ Q is logically equivalent to ¬(P ∨ Q).

(b) Find a logical expression built using only P and the connective ↓

which is logically equivalent to ¬P.

P Q P ↓ Q

T T F

T F F

F T F

F F T

15. (Just for fun) Augustus de Morgan satisﬁed his own problem:

I turn(ed) x years of age in the year x

(a) Given that de Morgan died in 1871, and that he wasn’t the beneﬁciary of some miraculous

anti-aging treatment, ﬁnd the year in which he was born.

(b) Suppose you have an acquaintance who satisﬁes the same problem. When were they born

and how old will they turn this year?

Do your best to give a formal proof of your correctness.

2.2 Propositional Functions & Quantiﬁers

Mathematical propositions are typically more complicated that those seen in Section 2.1. In particular,

they often involve variables: for instance, “x is an integer greater than 5.”

Deﬁnition 2.15. A propositional function is a family of propositions depending on one or more vari-

ables. The collection of permitted variables is the domain.

If P is a propositional function depending on a single variable x, then for each object a in the domain

P(a) is a proposition. Typically P(x) is true for some x and false for others.

Example 2.16. Suppose P(x) is the propositional function “x

> 4” with domain the real numbers.

Plainly P(1) is false (“1

> 4” is nonsense), while P(6) is true (”6

> 4”).

Propositional functions are often quantiﬁed. English contains various quantiﬁers (all, some, many, few,

several, etc.), but in mathematics we are primarily concerned with just two.

Deﬁnition 2.17. The universal quantiﬁer ∀ is read “for all.” The existential quantiﬁer ∃ is read “there

exists.” Given a propositional function P(x), we may deﬁne two new quantiﬁed propositions:

• “∀x, P(x)” is true if and only if P(x) is true for every x in its domain.

• “∃x, P(x)” is true if and only if P(x) is true for at least one x in its domain.

When quantifying propositions it is common to describe the domain by including a descriptor after

the quantiﬁer (bounding the quantiﬁer—see the example below).

As with connectives, there are multiple ways to express quantiﬁed propositions both mathemati-

cally and in English. Symbolic quantiﬁers involve a trade-off so consider your audience: compact

statements can improve clarity but are harder to read for the uninitiated.

Example (2.16 cont.). To gain some practice with bounded quantiﬁers, we introduce the notation

x ∈ R which simply means that x is a real number.

• “∀x ∈ R, x

> 4” might be read, “The square of every real number is greater than 4.”

The quantiﬁed proposition is false since 1

> 4 is false: we call x = 1 a counter-example.

• “∃x ∈ R, x

> 4” might be read, “There is a real number whose square is greater than 4.”

The quantiﬁed proposition is true since 6

> 4 (is true): we call x = 6 an example.

Due to their importance, it is worth deﬁning these last concepts formally.

Deﬁnition 2.18. An example of “∃x, P(x)” is an element x

in the domain of P for which P(x

) is true.

To prove an existential statement is to provide an example.

A counter-example to “∀x, P(x)” is an element x

in the domain of P for which P(x

) is false. To disprove

a universal statement is to provide an counter-example.

This notation should be familiar. Don’t worry if not, for it will be properly discussed in chapter 4.

Universal Quantiﬁers and Connectives: Hidden Quantiﬁers

Universally quantiﬁed propositions are interchangeable with implications. To see this, suppose a

propositional function Q(x) is given and let P(x) be “x lies in the domain of Q.” Then

∀x, Q(x) is logically equivalent to P(x) =⇒ Q(x)

By convention, connectives containing variables are assumed to be universal. When written as a

connective, the universal quantiﬁer is often hidden.

Examples 2.19. 1. The universal statement, “Every cat is neurotic,” may instead be written,

If x is a cat, then x is neurotic.

2. Revisiting Example 2.16, we could rewrite “∀x ∈ R, x

> 4” as a connective,

x ∈ R =⇒ x

> 4 (If x is a real number, then x

> 4)

3. The following three sentences have identical meaning:

The square of an odd integer is odd. ∀n odd, n

is odd. n odd =⇒ n

odd.

The universal quantiﬁer is explicit in only one of the sentences! For even more variety, the

third sentence could be viewed as a universal statement about all integers; including the hidden

quantiﬁer in this case results in

∀n ∈ Z, n odd =⇒ n

odd.

where the symbol Z represents the (set of) integers.

We’ve already seen that disproving a universal statement requires only that we supply a counter-

example. While such might require some effort to ﬁnd, often the resulting argument is very simple.

By contrast, proving a universal statement is the same as proving a conditional connective, typically

a more involved activity. We therefore largely postpone this to the next section. Regardless, a simple

proof of the above oddness claim should be easy to follow.

Proof of Example 2.19.3. If an integer n is odd, then it may be written in the form n = 2k + 1 for some

integer k. But then

= (2k + 1)

= 4k

+ 4k + 1 = 2(2k

+ 2k) + 1

is plainly also odd.

Similarly, proving an existential statement (by providing an example) is typically more straightfor-

ward than disproving such. To help understand this duality, we need to see how to negate quantiﬁed

propositions.

By contrast, the existential quantiﬁer is never hidden: it is always explicit either symbolically (∃) or as a phrase in

English (there is, there exists, some, at least one, etc.).

Negating Quantiﬁed Propositions

To negate a proposition is to consider what it means for the proposition to be false. We already

understand what this means for a universal proposition:

“∀x, P(x)” is false if and only if there exists a counter-example.

Otherwise said, the negation of a universal statement is existentially quantiﬁed:

The negation of “P(x) is always true” is “P(x) is sometimes false.”

Repeating P(x) with ¬P(x) results in the related observation:

The negation of “P(x) is always false” is “P(x) is sometimes true.”

In summary:

Theorem 2.20. For any propositional function P(x):

1. ¬



∀x, P(x)



is logically equivalent to ∃x, ¬P(x).

2. ¬



∃x, P(x)



is logically equivalent to ∀x, ¬P(x).

Examples 2.21. 1. “Everyone owns a bicycle,” has negation “Someone does not own a bicycle.” It is

somewhat ugly, but we could write this semi-symbolically:



∀ people x, x owns a bicycle



⇐⇒ ∃ a person x such that x does not own a bicycle

2. The quantiﬁed proposition

“∃x > 0, sin x = 4,” has the form ∃x, P(x). Its negation is therefore

∀x, ¬P(x): explicitly,

∀x > 0, sin x = 4

Since sine satisﬁes −1 ≤ sin x ≤ 1, the original proposition is false and its negation true.

Warning! Do not change a quantiﬁer’s bounds: ∀x ≤ 0, sin x = 4 is wrong!

3. Take special care negating connectives: a negated hidden quantiﬁer ∀x becomes explicit.



P(x) =⇒ Q(x)



is logically equivalent to ∃x, P(x) ∧ ¬Q(x) (Theorem 2.11)

(a) (Example 2.19.3) The negation of “n odd =⇒ n

odd,” is the (false) claim

∃n ∈ Z with n odd and n

even.

(b) (Example 2.19.2) The negation of the false claim “x ∈ R =⇒ x

> 4” is the true assertion

∃x ∈ R for which x

≤ 4 (for example x = 1)

“∃x > 0” indicates that the domain of the proposition “sin x = 4” is the positive real numbers.

Multiple Quantiﬁers

A propositional function can have several variables, each of which may be quantiﬁed.

Examples 2.22. 1. The quantiﬁed proposition

∀x > 0, ∃y > 0 such that xy = 4 (∗)

might be read, “Given any positive number, there is another such that their product is four.”

Hopefully you believe that this is true! Here is a simple argument which comes from viewing

(∗) as an implication, “x > 0 =⇒ ∃y > 0 such that xy = 4.”

Proof. Suppose we are given x > 0. Let y =

. Then xy = 4, as required.

Being clear about domains is critical. Suppose we modify the original proposition:

∀x ∈ R, ∃y ∈ R such that xy = 4 (†)

Our proof now fails! The new statement (†) is false: indeed x = 0 provides a counter-example.

Disproof. Let x = 0. Since xy = 0 for any real number y, we cannot have xy = 4.

Alternatively, we could negate (†): following Theorem 2.20, we switch the symbols ∀ ↔ ∃ and

negate the ﬁnal proposition,



∀x ∈ R, ∃y ∈ R, xy = 4



⇐⇒ ∃x ∈ R, ∀y ∈ R, xy = 4 (¬(†))

Our disproof of (†) is really a proof of the negation: we provided the example x = 0, thus demon-

strating the truth of a ∃-statement. Since the negation is true, the original (†) is false.

2. Order of quantiﬁers matters! The meaning of a sentence—and its truth state—can change if we

alter the order of quantiﬁcation.

(a) ∀x ∈ R, ∃y ∈ R, x

< y

Proof. Suppose a real number x is given. Let y = x

+ 1. Then x

< y, as required.

As above, we prove this by viewing it as an implication, “If x ∈ R, then ∃y ∈ R, x

< y.”

(b) ∃y ∈ R, ∀x ∈ R, x

< y

Disproof. We demonstrate the truth of the negation, “∀y ∈ R, ∃x ∈ R, x

≥ y.”

Suppose a real number y is given. Let x =

. Then x

= y ≥ y, as required.

Don’t worry if these arguments seem difﬁcult at the moment; much more practice is coming.

For an abstract justiﬁcation of this heuristic, consider a propositional function P(x): “∃y, Q(x, y),” then



∀x, ∃y, Q(x, y)



⇐⇒ ¬



∀x, P(x)



⇐⇒ ∃x, ¬P(x) ⇐⇒ ∃x, ¬



∃y, Q(x, y)



⇐⇒ ∃x, ∀y, ¬Q(x, y)

We ﬁnish with two harder examples you might have encountered elsewhere. For this course, you do

not have to know what these statements mean, though you do have to be able to negate them.

Examples 2.23. 1. (Linear Algebra) Vectors x, y, z in R

are said to be linearly independent if

∀a, b, c ∈ R, ax + by + cz = 0 =⇒ a = b = c = 0

Since this is a conditional proposition, the expression ∀a, b, c ∈ R would likely be hidden. The

negation of this statement, what it means for x, y, z to be linearly dependent, is

∃a, b, c ∈ R, such that ax + by + cz = 0 and a, b, c are not all zero

2. (Analysis/Calculus) A function f : R → R is said to be continuous at a ∈ R if

∀ϵ > 0, ∃δ > 0 such that |x −a| < δ =⇒ |f (x) − f (a)| < ϵ

The negation, what it means for f to be discontinuous at x = a, is

∃ϵ > 0 such that ∀δ > 0, ∃x ∈ R with |x − a| < δ and |f (x) − f (a)| ≥ ϵ

The original statement contained a hidden quantiﬁer ∀x which became explicit upon negation.

Exercises 2.2. A self-test quiz and several worked questions can be found online.

1. Rewrite each sentence using quantiﬁers. Then write the negation (use words and quantiﬁers).

(a) All mathematics exams are hard.

(b) No football players are from San Diego.

2. Let P be the proposition: “Every positive integer is divisible by thirteen.”

(a) Write P using quantiﬁers.

(b) What is the negation of P?

3. A friend claims that the sentence “x

> 0 =⇒ x > 0” has negation “x

> 0 and x ≤ 0.” Why

is this incorrect? What is the correct negation?

4. Consider the quantiﬁed statement

∀x, y, z ∈ R, (x −3)

+ (y −2)

+ (z −7)

> 0 (∗)

(a) Express (∗) in words.

(b) Is (∗) true or false? Explain.

(d) Is the negation of (∗) true or false? Explain.

5. Suppose P, Q, R are propositional functions. Compute the negations of the following:

(a) ∀x, ∃y, P(x) ∧ Q(y) (b) ∀x, ∃y, ∀z, R(x, y, z)

6. Revisit Example 2.22.2. Decide whether each of the following is true or false:

(a) ∃x ∈ R, ∀y ∈ R, x

< y (b) ∀y ∈ R, ∃x ∈ R, x

< y

7. The following are statements about positive real numbers x, y. Which is true? Explain.

(a) ∀x, ∃y such that xy < y

(b) ∃x such that ∀y, xy < y

8. Which of the following statements are true? Explain.

(a) ∃ a married person x such that ∀ married people y, x is married to y.

(b) ∀ married people x, ∃ a married person y such that x is married to y.

9. Prove or disprove:

(a) For every two points A and B in the plane, there exists a circle on which both A and B lie.

(b) There exists a circle in the plane on which lie any two points A and B.

10. Consider the following proposition (you do not have to know what is meant by a ﬁeld).

All non-zero elements x in a ﬁeld F have an inverse: some y ∈ F for which xy = 1.

(a) Restate the proposition using quantiﬁers.

(b) Find the negation of the proposition, again using quantiﬁers.

11. “A function f : R → R is decreasing” means: x ≤ y =⇒ f (x) ≥ f (y).

(a) State what it means for f not to be decreasing (where is the hidden quantiﬁer?)

(b) Give an example to show that not decreasing and increasing do not mean the same thing.

12. Prove that P(x) =⇒ Q(x) is logically equivalent to



∃x, P(x) ∧ ¬Q(x)



=⇒ F.

(This extends Exercise 2.1.8c—be careful of the quantiﬁers!)

13. Consider the proposition: ∀m, n ∈ R, m > n =⇒ m

> n

(a) State the negation of the proposition.

(b) Prove that the original proposition is false.

> n

What is the largest collection (set) of real numbers A for which the proposition is true?

14. (Hard) Let (x

) = (x

, x

, . . .) denote a sequence of real numbers.

“(x

) diverges to ∞” means: ∀M > 0, ∃N ∈ R such that n > N =⇒ x

> M

“(x

) converges to L” means: ∀ϵ > 0, ∃N ∈ R such that n > N =⇒

− L

< ϵ

(a) State what it means for a sequence (x

) not to diverge to ∞. Beware of the hidden quantiﬁer!

(b) State what it means for a sequence (x

) not to converge to L.

) not to converge at all.

(d) (Challenge: non-examinable) Use the deﬁnitions to prove that the sequence deﬁned by

= n diverges to ∞, and that the sequence deﬁned by y

converges to zero.

2.3 Methods of Proof

The previous sections covered some of the language of foundational logic. While one can study this

more deeply, we focus on putting it to work in the service of mathematics. The real work begins now.

A mathematical theorem is

a justiﬁed assertion of the truth of an implication P =⇒ Q. A proof is

any logical argument justifying the theorem. The ﬁrst step in analyzing or strategizing a proof is to

identify the hypothesis P and conclusion Q.

There are four standard methods of proof; in practice longer arguments combine several of these.

Direct Assume the hypothesis P and deduce the conclusion Q.

This structure should be intuitive,

though it may help to revisit the truth table in Deﬁnition 2.5 and the tautology of Example 2.6.2.

Contrapositive Directly prove the contrapositive ¬Q =⇒ ¬P (logically equivalent to P =⇒ Q by

Theorem 2.9).

Contradiction Assume P ∧¬Q and deduce a contradiction (directly prove (P ∧¬Q) ⇒ F). Theorem

2.11 and Exercise 2.1.8c show that P =⇒ Q is true. If P(x) =⇒ Q(x) has a hidden universal

quantiﬁer, the negation means we start by assuming ∃x, P(x) ∧ ¬Q(x) (Exercise 2.2.12).

Induction This has a completely different ﬂavor; we will consider it in Chapter 5.

Each method has advantages and disadvantages: direct proofs typically have the simplest logical

ﬂow; contrapositive/contradiction approaches are useful when the negations ¬P, ¬Q are easier to

work with than P, Q themselves. All methods are equally valid, and, as we’ll see shortly, one can

often prove a simple theorem using all three approaches!

As you work through this section, pay special attention to the logical structure—to encourage this,

the mathematical level is very low. Refer to the previous sections if the logical terminology feels

unfamiliar. Now is also a good time to re-read Planning and Writing Proofs (page 4).

Direct Proofs

We begin by generalizing Example 2.19.3.

Theorem 2.24. The product of any pair of odd integers is odd.

To make sense of this, we ﬁrst need to identify the logical structure by writing the theorem in terms

of propositions and connectives. One way is to view the Theorem in the form P =⇒ Q:

• P(x, y) is “x and y are both odd.” This is our assumption, the hypothesis.

• Q(x, y) is “The product xy is odd.” This is what we wish to demonstrate, the conclusion.

• Both propositional functions are statements about integers. The Theorem is universal (“any

pair”), and so contains a (hidden) quantiﬁer ∀x, y ∈ Z.

We also need a clear understanding of the meaning of all necessary terms. To keep things simple,

we’ll treat integer and product as understood and be explicit only as to the meaning of oddness.

It might be awkward to ﬁt a theorem into this format but it can always be done. Often all that is stated is the conclusion

Q, in which case P would be the assertion “All mathematics we already know/assume to be true.”

To assume a proposition is to suppose its truth. To suppose P is false, we “assume/suppose ¬P.”

A direct proof can be viewed as a proof sandwich whose bread slices are the hypothesis and conclu-

sion (P and Q): write these down as a ﬁrst step. Next deﬁne any useful terms in the hypothesis. All

that remains is to perform a simple calculation!

Proof. Let x and y be odd integers. (state hypothesis P)

There are integers k, l for which x = 2k + 1 and y = 2l + 1. Then, (deﬁnition of odd)

xy = (2k + 1)(2l + 1) = 4kl + 2k + 2l + 1 (computation/algebra)

= 2(2kl + k + l) + 1

Since 2kl + k + l is an integer, we conclude that xy is odd. (state conclusion Q)

Observe how we wrote xy in the form 2(integer)+1 so as to make the conclusion absolutely clear.

Insufﬁcient Generality Before leaving this example, it is worth highlighting the most common

mistake seen in such arguments.

Fake Proof 1. x = 3 and y = 5 are both odd, hence xy = 15 is odd.

This is an example of the theorem. Since the theorem is universal, a single example does not constitute

a proof (recall, however, that an example proves an existential statement: Deﬁnition 2.18).

Fake Proof 2. Let x = 2k + 1 and y = 2k + 1 be odd. Then

xy = (2k + 1)(2k + 1) = 2(2k

+ 2k) + 1 is odd.

This only veriﬁes the special case where both odd integers are equal: it proves x odd =⇒ x

odd.

There is nothing wrong with trying out examples or sketching incomplete thoughts—indeed both

are encouraged!—but you need to be aware of when your argument isn’t sufﬁciently general.

For another simple direct proof, consider the sum of two consecutive integers.

Theorem 2.25. The sum of any pair of consecutive integers is odd.

The theorem is again a universal claim (“any”) of the form P =⇒ Q about two integers x, y:

• P(x, y) is “x, y are consecutive integers.”

• Q(x, y) is “x + y is odd.”

The trick is to observe that, being consecutive, we may write one integer in terms of the other. The

proof sandwich is still visible, though it would be hard to write down the last sentence without

already having settled on the trick, which is essentially the deﬁnition of “consecutive integers.”

Proof. Suppose we are given two consecutive integers. Label the smaller of these x and the other

x + 1. Their sum is then

x + (x + 1) = 2x + 1

which is odd.

Proof by Contrapositive

Here is another straightforward result about odd and even integers.

Theorem 2.26. If the sum of two integers is odd, then they have opposite parity.

The theorem is yet another universal statement (∀x, y ∈ Z) of the form P =⇒ Q:

P(x, y): “x + y is odd.” Q(x, y): “x, y have opposite parity.”

Parity means evenness or oddness: the conclusion is that one of the integers is even and the other odd.

ıvely attempting a direct proof produces an immediate difﬁculty:

Direct Proof? Suppose x + y is odd. Then x + y = 2k + 1 for some integer k. . .

We want to conclude something about x and y separately, but the direct approach lumps them together

in the same algebraic expression.

A contrapositive approach (¬Q =⇒ ¬P) suggests itself as a remedy, since the new hypothesis ¬Q,

by treating x and y separately, gives us twice as much to start with.

¬Q(x, y): “x, y have the same parity.” ¬P(x, y): “x + y is even.”

The remaining difﬁculty is that “same parity” encompasses two possibilities: x, y are either both even,

or both odd. The proof therefore contains two cases.

Proof. Suppose x and y have the same parity. There are two cases. (state hypothesis ¬Q)

Case 1: Assume x and y are both even.

Write x = 2k and y = 2l, for some integers k, l. (deﬁnition of even)

Then x + y = 2(k + l) is even. (computation)

Case 2: Assume x and y are both odd.

Write x = 2k + 1 and y = 2l + 1 for some integers k, l. (deﬁnition of odd)

Then x + y = 2(k + l + 1) is even. (computation)

In both cases x + y is even. (state conclusion ¬P)

Again observe the proof sandwich and how the argument depends on little more than the deﬁnitions

of even and odd.

When presenting a lengthier contrapositive argument, consider orienting the reader by starting with

the phrase, “We prove the contrapositive.” For simple proofs (like the above) this is unnecessary,

since the logical structure should be clear without such assistance. It is also unnecessary to deﬁne

and spell out the propositions P and Q or include any of the bracketed commentary. However, feel

free to continue this practice if you think it aids your explanation, of if you are nervous about your

proof skills.

Warning! The contrapositive is still a universal statement: ∀x, y ∈ Z, ¬Q(x, y) =⇒ ¬P(x, y). We are not negating the

theorem so do not convert ∀ to ∃!

. . . and want to guarantee some partial credit!

For another example of a contrapositive argument, we extend the ﬁrst result of this section.

Theorem 2.27. The product of two integers is odd if and only if both integers are odd.

This is a biconditional P ⇐⇒ Q, comprising two theorems in one: P =⇒ Q and Q =⇒ P (Exercise

2.1.9a). A contrapositive argument for P =⇒ Q is again suggested because Q (both integers are odd)

treats the two integers individually.

Proof. (⇒) We prove the contrapositive. Let x, y be integers, at least one of which is even. Suppose,

without loss of generality, that x = 2k is even. Then xy = 2ky is also even.

(⇐) This is precisely Theorem 2.24, which we’ve already proved.

Note de Morgan’s law: ¬(x odd and y odd) is equivalent to “x even or y even” (“at least one”).

The common phrase without loss of generality, often abbreviated WLOG, saves us from performing

a second, almost identical, argument assuming y = 2l is even. WLOG is stated when one makes a

choice which does not materially affect the argument.

Proof by Contradiction

Here is a simple result considered in several ways.

Example 2.28. Let x be an integer. We prove that if 3x + 5 is even, then 5x + 2 is odd.

We could proceed directly according to the following sketch:

3x + 5 even =⇒ 3x odd =⇒ x odd =⇒ 5x odd =⇒ 5x + 2 odd (∗)

This isn’t wrong! You should believe each implication; indeed we’ve proved most of them. It would

be nice, however, if we didn’t have to rely on so many other results. A similar contrapositive proof

(reverse the arrows and negate the propositions) would have the same weakness.

The advantage of a contradiction approach is that we have twice as much to work with: the hypoth-

esis (3x + 5 even) and the negation of the conclusion (5x + 2 even).

Proof. Suppose x is an integer for which both 3x + 5 and 5x + 2 are even. Then their sum

is also even. However,

(3x + 5) + (5x + 2) = 8x + 7 = 2(4x + 3) + 1 (†)

is odd. Contradiction (an integer cannot be both even and odd!).

Remember to write contradiction at the end so the reader knows what you’ve done!

A nice side-effect of this approach is that it suggests an alternative direct proof.

Direct Proof. For any integer x, (†) says that 3x + 5 and 5x + 2, in summing to an odd

number, have opposite parity.

The last argument in fact proves that 3x + 5 is even if and only if 5x + 2 is odd; the converse of the

original claim comes for free! Revisiting (∗), you should believe that all the arrows are reversible.

From now on we’ll reserve ‘Theorem’ for results that are worth remembering in their own right.

Such variety is one of the things that makes proving theorems fun! While the choice of proof method

is largely a matter of personal taste, remember your audience. Our ﬁnal direct argument is very slick

but risks confusing an elementary reader rather than empowering them.

Three Proofs of the Same Result We ﬁnish this section with three proofs of the same result. All are

based on the same factorization of a polynomial

+ 4x

−2x −20 = (x −2)(x

+ 6x + 10) = (x −2)



(x + 3)

+ 1



and the well-known fact that ab = 0 ⇐⇒ a = 0 or b = 0 (see Exercise 14). Since the mathematics is

so simple, pay attention to and compare the logical structures—which do you prefer?

Example 2.29. Let x be a real number. We prove that x

+ 4x

−2x −20 = 0 =⇒ x = 2.

Direct Proof. Suppose x

+ 4x

−2x −20 = 0. By factorization, (x −2)[(x + 3)

+ 1] = 0,

so at least one of the factors must be zero. Since (x + 3)

+ 1 ≥ 1 > 0, we conclude that

x −2 = 0, from which x = 2.

Contrapositive Proof. Suppose x = 2. Since (x + 3)

+ 1 ≥ 1 > 0, we see that

+ 4x

−2x −20 = (x −2)[(x + 3)

+ 1] = 0

Contradiction Proof. Suppose x

+ 4x

−2x −20 = 0 and x = 2. Then

0 = x

+ 4x

−2x −20 = (x −2)[(x + 3)

+ 1]

Since x = 2, we have x −2 = 0. It follows that (x + 3)

+ 1 = 0. However, (x + 3)

+ 1 ≥ 1

for all real numbers x, so we have a contradiction.

Exercises 2.3. A self-test quiz and several worked questions can be found online.

1. Prove or disprove the following conjectures.

(a) There is an even integer which can be expressed as the sum of three even integers.

(b) Every even integer can be expressed as the sum of three even integers.

(d) Every odd integer can be expressed as the sum of three odd integers.

2. For any given integers a, b, c, if a is even and b is odd, prove that 7a −ab + 12c + b

+ 4 is odd.

3. Prove that if n is an integer greater than 1, then n! + 2 is even.

(n! = n(n −1)(n −2) ···1 is the factorial of the integer n)

4. (a) Let x ∈ Z. Prove that 5x + 3 is even if and only if 7x −2 is odd.

(b) Can you conclude anything about 7x −2 if 5x + 3 is odd?

The Hungarian mathematician Paul Erd

os referred to simple, elegant proofs as ‘from the Book,’ as if the Almighty kept

a tome of perfect proofs. As with all matters spiritual, one person’s Book is likely very different to another’s. . .

5. Consider the following proposition, where x is assumed to be a real number.

−3x

−2x + 6 = 0 =⇒ x = 3

(a) Is the proposition true or false? Justify your answer. Is its converse true?

(b) Repeat part (a) for the proposition x

−3x

−2x + 6 = 0 =⇒ x = 3.

6. Below is the proof of a result. What result is being proved?

Proof. Assume that x is odd. Then x = 2k + 1 for some integer k. Then

−3x −4 = 2(2k + 1)

−3(2k + 1) − 4 = 8k

+ 2k −5 = 2(4k

+ k −3) + 1

Since 4k

+ k −3 is an integer, 2x

−3x −4 is odd.

7. Here is another proof. What is the result this time?

Proof. Assume, without loss of generality, that x = 2a and y = 2b are both even. Then

xy + xz + yz = (2a)(2b) + (2a)z + (2b)z = 2(2ab + az + bz)

Since 2ab + az + bz is an integer, xy + xz + yz is even.

8. Consider the following proof of the fact that (for m an integer) if m

is even, then m is even. Can

you re-write the proof so that it doesn’t use contradiction?

Proof. Suppose that m

is even and m is odd. Write m = 2k + 1 for some integer k. Then

= (2k + 1)

= 4k

+ 4k + 1 = 2(2k

+ 2k) + 1

is odd. Contradiction.

9. Here is a ‘proof’ that every real number x equals zero. Find the mistake.

x = y =⇒ x

= xy =⇒ x

−y

= xy −y

=⇒ (x − y)(x + y) = (x −y)y

=⇒ x + y = y

=⇒ x = 0

10. Prove or disprove: An integer n is even if and only if n

is even.

11. Let n and m be positive integers. Prove n

m is even if and only if n and m are not both odd.

12. Let x and y be integers. Prove x

+ y

is even if and only if x and y have the same parity.

13. Let n be an integer. Prove n

+ n + 58 is even.

14. Suppose a, b ∈ R. Prove that ab = 0 ⇐⇒ a = 0 or b = 0.

15. Numbers of the form

k(k+1)

, where k is a positive integer, are called triangular numbers. Prove

that n is the square of an odd number if and only if

n−1

is triangular.

2.4 Further Proofs & Strategies

The arguments in this section are slightly trickier and more representative of typical mathematics.

Some of these results are indeed quite famous and worth knowing in their own right. We will also

introduce lemmas and corollaries which are used to break up the presentation of complex results.

Proving Universal Statements

Most of the results we’ve seen thus far have been universal. As discussed on page 14, any theorem

P(x) =⇒ Q(x) is implicitly universal, albeit with the quantiﬁer ∀x typically hidden. Revisit Exam-

ples 2.22 to consider how multiple quantiﬁers ﬁt into our proof framework; here is another example.

Example 2.30. We prove: ∀x ≥ 0, ∃y < 0 such that x

< y

View the claim as an implication “x ≥ 0 =⇒ (∃y < 0 such that x

< y

)” and prove directly.

Proof. Suppose x ≥ 0 is given. Deﬁne y = −

√

−1. Then y ≤ −1 < 0 and

= x

+ 1 + 2

√

≥ x

+ 1 > x

The conclusion is existentially quantiﬁed, so we deduce it by supplying an example (“Deﬁne y. . . ”).

Since y is existentially quantiﬁed after x, note that it is allowed to depend on x! An argument such as

this likely needs some scratch work to ﬁnd a suitable y: don’t expect to create such in one shot!

For a more involved example of a universal result, here is a famous inequality relating the arithmetic

and geometric means of two numbers.

Theorem 2.31 (AM–GM inequality). If x, y are non-negative real numbers, then

x + y

≥

√

with equality if and only if x = y.

This requires some unpacking! First try an example to calm the nerves (e.g.,

3+5

= 4 ≥

√

15). It

should also be clear that both sides are equal whenever x = y. Now consider the logical structure:

x, y ≥ 0 =⇒ Q(x, y); the challenge lies in making sense of Q. There are really two separate results:

1. If x, y ≥ 0, then

x+y

≥

√

2. If x, y ≥ 0, then

x+y

√

xy ⇐⇒ x = y

Concentrate on the ﬁrst since it is simpler. The hypothesis (x, y ≥ 0) doesn’t give us much to work

with, so it seems sensible to play with the inequality and try to eliminate the ugly square-root:

x + y

≥

√

xy =⇒ (x + y)

≥ 4xy =⇒ x

−2xy + y

≥ 0 =⇒ (x −y)

≥ 0

Now we have something believable! The question is whether we can reverse the arrows. Only the

ﬁrst should give you any pause; it is here that we use the non-negativity of x, y.

Proof. Suppose x, y ≥ 0. Multiply out a trivial inequality:

(x −y)

≥ 0 ⇐⇒ x

−2xy + y

≥ 0 ⇐⇒ x

+ 2xy + y

≥ 4xy

⇐⇒ (x + y)

≥ 4xy

⇐⇒

x + y

≥

√

The square-root is well-deﬁned because x, y ≥ 0, and the inequality is preserved since the square-root

function is increasing. For the second result, observe that the ﬁnal inequality is an equality precisely

when all the inequalities are equalities; this is if and only if x = y.

The scratch work really helped us ﬁgure out how and where to apply the hypothesis. Notice also

how the second result came almost for free! Result 1 only needed the ⇒ direction in the proof, but

the second result used the fact that all arrows are biconditionals.

For variety, here is a contradiction proof incorporating the same calculations in a different order.

Contradiction Proof. Let x, y ≥ 0 and suppose that

x+y

√

xy. Since x + y ≥ 0, the second inequality

holds if and only if (x + y)

< 4xy. Now multiply out and rearrange:

(x + y)

< 4xy ⇐⇒ x

+ 2xy + y

< 4xy

⇐⇒ x

−2xy + y

< 0

⇐⇒ (x − y)

< 0

Contradiction (squares of real numbers are non-negative). We conclude that

x+y

≥

√

xy.

Now suppose that

x+y

√

xy. Following the biconditionals in the above calculation, we see that

equality holds if and only if (x −y)

= 0, from which we recover x = y.

The AM–GM inequality in fact holds for any ﬁnite collection of non-negative numbers x

, . . . , x

+ x

+ ···+ x

≥

√

···x

with equality if and only if all x

are equal. Proving this is a lot harder (see Exercise 15).

Disproving Existential Statements: Non-existence Proofs

Of the four combinations “prove/disprove a universal/existential statement,” we’ve now tackled

three; it remains to consider how to prove that something does not, or cannot, exist. Recall our basic

rule of negation (Theorem 2.20):



∃x, Q(x)



⇐⇒ ∀x, ¬Q(x)

To show that ∃x, Q(x) is false is to prove a universal statement.

As before (page 14), let P(x) be the

proposition “x lies in the domain of Q.” We conclude that



∃x, Q(x)



is logically equivalent to P(x) =⇒ ¬Q(x)

The notation ∄x, Q(x) is discouraged because it obscures this essential fact.

Viewed in this way as an implication, any of our proof strategies might be applicable to a non-

existence proof. Contradiction and contrapositive arguments are particularly common however,

since the right hand side (¬Q) is already a negative statement.

Example 2.32. We prove that the equation x

+ 12x

+ 13x + 3 = 0 has no positive solutions.

Before seeing a proof, consider several ways in which this claim could be presented.

Non-existence (¬(∃x, Q(x))) There are no x > 0 for which x

+ 12x

+ 13x + 3 = 0.

Universal (∀x, ¬Q(x)) For all x > 0, we have x

+ 12x

+ 13x + 3 = 0.

Direct (P ⇒ ¬Q) If x > 0, then x

+ 12x

+ 13x + 3 = 0.

Contrapositive (Q ⇒ ¬P) If x

+ 12x

+ 13x + 3 = 0, then x ≤ 0.

Contradiction (P ∧ Q false) x > 0 and x

+ 12x

+ 13x + 3 = 0 is impossible.

We present two similar arguments based on the direct and contradiction structures.

Direct proof. Suppose that x > 0. Then x

+ 12x

+ 13x + 3 > 0 since all terms are posi-

tive. We conclude that x

+ 12x

+ 13x + 3 = 0.

Contradiction proof. Assume that x > 0 satisﬁes x

+ 12x

+ 13x + 3 = 0. Since all terms

on the left hand side are positive, we have a contradiction.

Both arguments were very easy, with all difﬁculty coming from understanding why they work: pre-

cisely the above discussion! Learning/practicing how to recognize/translate a claim into an action-

able format is the essential skill here, both for the presenter and the reader.

Subdividing Theorems: Lemmas & Corollaries

It is sometimes helpful to break a proof into pieces, akin to viewing a computer program as a collec-

tion of subroutines. Often the intention is to improve the readability of a difﬁcult/complex argument,

though you may also wish to (de-)emphasize the relative importance of certain parts of a discussion.

One way to do this is to utilise lemmas and corollaries.

Lemma: A theorem whose importance you want to downplay or which will later be used

to help prove a more signiﬁcant result.

Corollary: A theorem which follows quickly from a previous result, either as a special case

or by modifying the proof in a straightforward way.

Presentation style varies: some authors and journals reserve theorem for only the most important

results, with everything else presented as a lemma or corollary; others never use these terms or just

call everything a proposition! Regardless, lemmas and corollaries are useful to have in your toolkit if

readability is your goal.

Here is a very simple result in preparation for a much more important upcoming theorem.

Lemma 2.33. Suppose n is an integer. Then n

is even ⇐⇒ n is even.

You should be able to prove this yourself, since the lemma is just a special case of Theorem 2.27. If

you are completely unsure how to start, revisit that result and the rest of Section 2.3.

Irrational Numbers

Since their deﬁnition is inherently negative, irrational numbers provide good examples of non-

existence/contradiction arguments. They are also interesting in their own right.

Deﬁnition 2.34. A real number x is said to be rational if it may be written in the form x =

for some

integers m, n. A real number is irrational if no such integers exist.

You likely know of a few irrational numbers (

√

2, π, e), but how do we prove that a given number is

irrational? Our next result is very famous, with versions dating back at least to Aristotle (c. 340 BCE).

Theorem 2.35.

√

2 is irrational.

We must disprove the existence claim ∃m, n ∈ Z,

√

2 =

. As before, consider several restatements:

Non-existence There are no integers m, n for which

√

2 =

Universal For all integers m, n, we have

√

2 =

Direct If m, n ∈ Z, then

√

2 =

Contrapositive If

√

2 =

, then m, n are not both integers.

Contradiction m, n ∈ Z and

√

2 =

is impossible.

Are the drawbacks of a direct or contrapositive approach obvious? We prove by contradiction. To

improve readability, we outsource a repeated step to the (⇒) direction of Lemma 2.33.

Proof. Suppose m, n ∈ Z and that

√

2 =

. Without loss of generality, assume m, n have no common

factors. Cross-multiply and square:

= 2n

is even =⇒ m is even (Lemma 2.33)

whence m = 2k for some integer k. But then

= m

= 4k

=⇒ n

= 2k

is even =⇒ n is even (Lemma 2.33)

We see that m and n have a common factor of 2. Contradiction.

Just as we can simplify

, the no common factors assumption is without loss of generality: it costs

nothing once we suppose

√

2 =

is rational. This last is what we contradicted! A (wrong) belief that

the no common factors assumption was contradicted means the calculation continues forever!

= 2n

=⇒ n

= 2k

=⇒ k

= 2l

=⇒ ···

The irrationality of various surds (

√

2, etc.), can be proved similarly (π and e are much harder).

We may also apply the theorem to demonstrate the irrationality of many other numbers.

Example 2.36. Suppose

√

2 − 5

√

3 = x were rational: ∃m, n ∈ Z such that x =

. Then

75 = (5

√

= (

√

2 − x)

= 2 + x

−2

√

2x =⇒

√

2 =

−73

−73n

2mn

Otherwise said,

√

2 is rational: contradiction.

Non-constructive Existence Proofs

Every existence proof we’ve thus far seen has been constructive: we’ve exhibited/constructed an

explicit example x for which Q(x) is true. Sometimes this is asks too much. Indeed it is often far

easier to show the existence of something without explicitly stating what it is. We present two famous

examples of this situation.

Theorem 2.37. There exist irrational numbers a, b for which a

is rational.

Proof. Consider the number x = (

√

. There are two possibilities:

1. x is rational. Let a = b =

√

2 and we’re done.

2. x is irrational. Let a = x and b =

√

2. Apply the usual exponential laws to see that



(

√



√

= (

√

2·

√

= (

√

= 2

In either case, a, b are irrational and a

is rational.

The proof is very sneaky: it does not provide an explicit example and does not answer the begged

question of whether (

√

is rational or irrational! In fact this number is irrational, though demon-

strating such is massively harder.

We ﬁnish with a particularly famous example of a non-constructive existence proof. This argument

dates back to Euclid’s Elements (300 BCE), the most inﬂuential textbook in mathematical history. As

ever, we need a solid deﬁnition before trying to prove anything.

Deﬁnition 2.38. An integer ≥ 2 is prime if the only positive integers it is divisible by are itself and 1.

The ﬁrst few primes are 2, 3, 5, 7, 11, 13, 17, 19, . . . It follows, though it is not completely obvious, that

every integer ≥ 2 is either prime or a product of primes (composite). In particular, every integer ≥ 2

is divisible by at least one prime. We now state Euclid’s result, and prove it by contradiction.

Theorem 2.39 (Elements, Book IX, Prop. 20). There are inﬁnitely many prime numbers.

Proof. Assume there are exactly n primes p

, . . . , p

and deﬁne the integer

Π := p

··· p

+ 1

Certainly Π is divisible by some prime p

(in our list by assumption!), as is the product p

··· p

. But

then the difference

1 = Π − p

··· p

must be divisible by p

, contradicting the fact that p

≥ 2.

If you’re interested, look up the Gelfond–Schneider Theorem (1934), Hilbert’s Seventh Problem, and what they say

about algebraic and transcendental numbers. Such ideas are far beyond the level of this text.

Exercises 2.4. A self-test quiz and worked questions can be found online.

1. Prove or disprove:

(a) There exist integers m and n such that 2m −3n = 15.

(b) There exist integers m and n such that 6m −3n = 11.

2. Prove or disprove: There exists a line L in the plane such that, for all points A, B in the plane,

we have that A, B lie on L.

3. Prove: For every positive integer n, the integer n

+ n + 3 is odd and greater than or equal to 5.

4. Let p be an odd integer. Prove that the equation x

− x − p = 0 has no integer solutions.

5. (Example 2.30, cont.) Prove or disprove: ∃y < 0 such that ∀x ≥ 0, x

< y

6. Prove or disprove:

√

2 −

√

7 is rational.

7. Prove or disprove the following conjectures about real numbers x, y.

(a) If 3x + 5y is irrational, then at least one of x and y is irrational.

(Be careful! This isn’t a logical ‘and’: what happens when you negate?)

(b) If x and y are rational, then 3x + 4xy + 2y is rational.

8. Prove by contradiction: if x and y are positive real numbers, then

√

x + y =

√

x +

√

How would you change things to make a contrapositive argument?

9. Prove that between any two distinct rational numbers there exists another rational number.

10. Consider the proposition:

For any non-zero rational r and any irrational number t, the number rt is irrational.

(a) Translate this statement into logic using quantiﬁers and propositional functions.

(b) Prove the statement.

11. (a) An integer n is not divisible by 3 if and only if ∃k ∈ Z for which n = 3k + 1 or 3k −1.

Prove that if n

is divisible by 3, then n is divisible by 3.

(b) Prove that

√

3 is irrational.

√

2 is irrational. (Hint: revisit Exercise 2.3.10)

12. (Recall Example 2.32) You are given the following facts:

• All polynomials are continuous.

• (Intermediate Value Theorem) If f is continuous on the interval [a, b] and L lies between

f (a) and f (b), then there is some x in the interval (a, b) for which f (x) = L.

• If f

′

(x) > 0 on an interval, then f is an increasing function.

Use these facts to give formal proofs of two claims that should be familiar from calculus:

(a) x

+ 12x

+ 13x + 3 = 0 has a solution x in the interval (−1, 0).

(b) x

+ 12x

+ 13x + 3 = 0 has exactly one real number solution x.

13. The real numbers satisfy the Archimedean property:

For any x, y > 0, there exists a positive integer n such that nx > y.

(a) Use the Archimedean property to show that there are no positive real numbers which are

less than

for all positive integers n.

(b) Consider the following ‘proof’ of the fact that every real number is less than some positive

integer:

Proof. Consider a real number x. For example, x = 19.7. Then x < 20 and 20 is a positive

integer.

What is wrong with this argument? Give a correct proof.

n(y −x) > 1.

(d) (Hard) Prove: ∀x, y ∈ R with x < y, ∃m, n ∈ Z for which nx < m < ny. Hence conclude

an extension of Exercise 9: between any two real numbers there exists a rational number.

14. Suppose x, y, z ≥ 0 satisfy x + y + z = 1. Use the AM–GM inequality (two-variable or full

version) to answer the following.

(a) What is the largest possible value of xyz and when does it occur?

(b) (Hard) Prove that (1 −x)(1 − y)(1 −z) ≥ 8xyz.

15. (Hard) We prove the full AM–GM inequality.

(a) When n = 3, try mimicking our earlier approach by cubing the desired inequality. Why

does this seem unwise?

(b) Prove that x ≤ e

x−1

for all real numbers x, with equality if and only if x = 1.

(Hint: Consider f (x) = e

x−1

− x and apply a derivative test from calculus)

+···+x

be the arithmetic mean. Apply part (b) to each expression x =

conclude that x

···x

≤ µ

and hence complete the proof.

3 Divisibility and the Euclidean Algorithm

This chapter introduces congruence, which generalizes the idea of integer parity (evenness/oddness).

This is of fundamental importance to the sub-disciplines of number theory and abstract algebra,

providing some of the most straightforward examples of groups and rings. We will cover the basics

in this section, returning in Chapter 7 for more formal observations.

3.1 Divisors, Remainders and Congruence

Deﬁnition 3.1. Let m, n be integers. The proposition n | m is read “n divides m,” and means

∃k ∈ Z such that m = kn

We could also say that “n is a divisor of m,” that “m is divisible by n,” or that “m is a multiple of n.”

The negated symbol ∤ is read does not divide.

Examples 3.2. 1. We write 4 | 20 since 20 = 4 ×5. The same equation also says that 5 | 20.

2. The proposition 9 ∤ 7 is read “9 does not divide 7.” It is shorthand for ¬(9 | 7).

When integers do not divide, there is a remainder left over. Your study of remainders likely goes back

to elementary school when you ﬁrst learned division: for instance,

33 ÷ 5 = 6 r 3 (“6 remainder 3”)

An important foundational result says that unique remainders always exist.

Theorem 3.3 (Division Algorithm). Suppose m, n ∈ Z with n positive. Then there exist unique

integers q, r (the quotient and remainder) for which

m = qn + r and 0 ≤ r < n

In elementary school language, m ÷n = q r r.

Examples 3.4. 1. Given m = 23 and n = 7, we have 23 = 3 ·7 + 2; that is q = 3 and r = 2.

2. If m = −11 and n = 3, then −11 = (−4) ·3 + 1; that is q = −4 and r = 1.

3. The formula m = 6q + 4, where q ∈ Z, describes all integers with remainder 4 on division by 6.

An algorithm is typically a computational process: if m > 0 one could view this as the repeated sub-

traction of n from m until the result r = m − qn satisﬁes 0 ≤ r < n. A rigorous proof requires

foundational ideas related to induction to which we will return in Chapter 5. For our current pur-

poses, we just need to know that remainders exist. Indeed our next step is to ﬁnd a way to compare

remainders without explicitly invoking the division algorithm.

The common meaning of divide is to apportion a quantity equally. Thus to divide 33 apples between 5 people, each

person gets 6 apples and 3 are left over. In grade school mathematics, fractions come much later.

Deﬁnition 3.5. Let a, b and n be integers with n positive. The proposition

a ≡ b (mod n) “a is congruent to b modulo n”

means that a and b have the same remainder on division by n. The integer n is called the modulus.

Examples 3.6. Consider remainders modulo 3 (division by 3).

1. We write 7 ≡ 10 (mod 3), since 7 = 2 ·3 + 1 and 13 = 4 ·3 + 1 have the same remainder (r = 1).

2. We write 6 ≡ 10 (mod 3), since 6 = 2 ·3 + 0 and 17 = 5 · 3 + 2 have different remainders.

Calculating using the division algorithm is tedious. Our next result is crucial in that it permits the

direct comparison of remainders. This can be treated as an equivalent deﬁnition of congruence.

Theorem 3.7. a ≡ b (mod n) ⇐⇒ n | (b − a) ⇐⇒ b = a + kn for some integer k

Proof. The second biconditional is nothing more than an application of Deﬁnition 3.1:

n | (b − a) ⇐⇒ ∃k ∈ Z such that b − a = kn

⇐⇒ b = a + kn for some integer k

Before presenting direct arguments for each direction of the ﬁrst biconditional, it is helpful to intro-

duce notation from the division algorithm:

a = q

n + r

b = q

n + r

0 ≤ r

, r

< n

=⇒ b −a = (q

−q

) n + (r

−r

) (∗)

(⇒) If a ≡ b (mod n), then a, b have the same remainder r

= r

. But then (∗) says that n | (b −a).

(⇐) Assume that n | (b −a) so that b − a = kn for some integer k. By (∗), we see that

−r

= (b −a) − (q

−q

) n = (k −q

+ q

) n

is divisible by n. Since the remainders satisfy 0 ≤ r

, r

< n, we moreover see that

−n < r

−r

< n

The only possibility is r

−r

= 0. Otherwise said, a, b have the same remainder: a ≡ b (mod n).

If you’re having trouble with the last step, think about an example! Suppose n = 26 and write

x = r

−r

. Hopefully you believe that x = 0 is the only integer satisfying the two conditions,

x is divisible by 26 and −26 < x < 26

Since the result is abstract, it is good to recap the relationship between congruence and divisibility.

• Each a ∈ Z is congruent to exactly one of the integers 0, 1, 2, . . . , n −1 modulo n: its remainder.

• a is divisible by n if and only if a ≡ 0 (mod n).

Examples 3.8. 1. We describe all integers x which are congruent to 7 on division by 11:

x ≡ 7 (mod 11) ⇐⇒ 11 | (x −7) ⇐⇒ x −7 = 11k ⇐⇒ x = 11k + 7

for some integer k.

2. To get more of a feel for the notation, consider the following conjectures:

(a) a ≡ 8 (mod 6) =⇒ a ≡ 2 (mod 3)

(b) a ≡ 2 (mod 3) =⇒ a ≡ 8 (mod 6)

Conjecture (a) is true. If a ≡ 8 (mod 6), then a = 6k + 8 for some integer k, from which

a = 6k + 8 = 3(2k + 2) + 2 =⇒ a ≡ 2 (mod 3)

Conjecture (b) is false. The counter-example a = 5 disproves this (universal) claim:

5 ≡ 2 (mod 3) and 5 ≡ 8 (mod 6)

Modular Arithmetic

Remainders have a natural arithmetic similar to that of the real numbers. We use the same symbols,

with even the congruence symbol ≡ looking a bit like an equals sign!

Modular arithmetic has many

applications, particularly to data security, with cell-phones and computers performing countless such

calculations daily. Here are the basic rules, generalizing most of what we saw in Section 2.3.

Theorem 3.9. Suppose a, b, c, d are integers and that n is some modulus. Then,

1. If a ≡ c and b ≡ d modulo n, then

a ±b ≡ c ±d and ab ≡ cd (mod n)

2. The usual associative, commutative and distributive laws of arithmetic hold for congruences:

a + (b + c) ≡ (a + b) + c a + b ≡ b + a a(b + c) ≡ ab + ac

a(bc) ≡ (ab)c ab ≡ ba

The theorem says that the operations ‘take the remainder’ and ‘add/subtract/multiply’ can be per-

formed in any order or combination, the result will be the same.

Warning! Division does not work so nicely in modular arithmetic (see Example 3.13).

Example 3.10. Find the remainder when 29 + 14 is divided by 6. We do this in two ways:

(a) First ﬁnd the sum 43, then compute its remainder: 43 ≡ 1 (mod 6) since 6 | (43 −1).

(b) Alternatively, we could ﬁnd the remainder of each component and then add:

29 + 14 ≡ 5 + 2 ≡ 7 ≡ 1 (mod 6)

This is no accident. In Chapter 7 we’ll see that congruence is an important example of an equivalence relation: a general-

ized notion of equality. Indeed, two integers are congruent if and only if something about them is equal: their remainders!

Proof. 1. We prove the multiplication rule. Suppose that a ≡ c and b ≡ d. By Theorem 3.7, we

have c = a + kn and d = b + ln for some integers k, l. Now compute:

cd − ab = (a + kn)(b + ln) − ab = (bk + al + kln)n

is divisible by n, whence ab ≡ cd. The addition/subtraction argument is almost identical.

2. The associative, commutative and distributive laws hold because x = y =⇒ x ≡ y (mod n),

regardless of n (equal numbers have the same remainder!).

The ability to take remainders before adding and multiplying is very powerful, allowing us rapidly

to perform some surprising calculations.

Examples 3.11. 1. Find the remainder when 37

423

is divided by 10.

The sheer size of 37

423

makes this appear impossible at ﬁrst glance.

Instead we think about

the rules of arithmetic modulo 10. Since 37 ≡ 7 ≡ −3 (mod 10), we see that

37 · 37 ≡ (−3) · (−3) ≡ 9 ≡ −1 ( mod 10)

This is more promising, for we can use it to simplify the original expression:

423

≡ (−3) ·(−3) ···(−3)

| {z }

423 times

≡



( −3)



211

( −3) ≡ ( −1)

211

( −3) ≡ 3 (mod 10)

2. Here we compute modulo n = 6:

+ 14

≡ 1

+ 2

≡ 1 + 8 ≡ 9 ≡ 3

It would have been madness to compute 7

+ 14

= 40356351 before ﬁnding the remainder!

3. Find the remainder when 124

·65

is divided by 11.

This needs several steps and simpliﬁcations. Since 124 = 11

+ 3 and 65 = 11 · 6 − 1, we write

124

·65

≡ 3

·(−1)

≡ (3

)

·(−1) ≡ −(27

) ≡ −(5

)

≡ −( 25

) ≡ −(3

) ≡ 2 (mod 11)

When performing these calculations:

• Replace each integer by something small with the same remainder: 37 ≡ −3(mod 10) is more

helpful than 37 ≡ 7 (mod 10), since powers of −3 are much easier to work with.

• The base of an exponential can be reduced, but not the exponent: 17

≡ 3

(mod 7) is correct,

but 3

≡ 3

(mod 7). Exponentiation is merely shorthand for repeated multiplication.

Using logarithms, a pocket calculator will tell you that 37

423

≈ 2.2 ×10

663

has 663 digits! This is no help since what

we want is the units digit, not its largest few signiﬁcant ﬁgures.

Application: On what day were you born?

While we all know our date of birth, most of us do not know on which day of the week we were born.

You can answer this question quite easily (perhaps in your head!) using modular arithmetic.

• Since 365 ≡ 1 (mod 7), a standard year advances the calendar one weekday.

• Each leap year

advances the calendar an additional day.

Can you ﬁgure the weekday today’s date in your year of birth? Thinking about the length of each

month modulo 7, you should also be able to ﬁnd your birthday.

Example 3.12. Paul Revere was born January 1

, 1735, in Boston. Given that January 1

, 2024 was a

Monday, ﬁnd the weekday of Revere’s birth.

The dates differ by 289 years, in which time there have been

288

− 2 = 70 leap years (not 1800 and

1900). The calendar has therefore advanced 289 + 70 ≡ 2 weekdays: Revere was born on a Saturday.

Division and Congruence Equations

Division in modular arithmetic behaves unexpectedly, so take care! The next section provides a

general approach (see Exercise 3.2.15). Here are two examples to whet the appetite.

Examples 3.13. 1. Even when there is a common factor, dividing both sides is perilous. For instance

42 ≡ 12 (mod 10) =⇒ ∃k ∈ Z, 42 −12 = 10k =⇒ ∃k ∈ Z, 21 −6 = 5k

=⇒ 21 ≡ 6 (mod 5)

We also divided the modulus! If we hadn’t done so, the result would be false: 21 ≡ 6 (mod 10).

2. Congruence equations are harder to solve than standard equations. For instance, we cannot

attack 2x ≡ 7 (mod 9) by division: x ≡

is meaningless since

is not an integer!

It won’t always work, but in this case a sneaky multiplication by 5 solves the problem:

2x ≡ 7 =⇒ 10x ≡ 35 =⇒ x ≡ 8 (mod 9)

Exercises 3.1. A reading quiz and several questions with linked video solutions can be found online.

1. Prove, using the deﬁnition of “divides,” that n | a and n | b =⇒ n | (a + b).

2. Let a, b, c be integers. Prove or disprove each of the following claims:

(a) a | b and b | c =⇒ a | c (b) a = b ⇐⇒ a | b and b | a

(b) a | b and a | c =⇒ a | bc (d) a | c and b | c =⇒ ab | c

3. List all integers x for which x ≡ 3 (mod 5) and −10 ≤ x ≤ 20.

4. Use the division algorithm to prove that if p is an odd prime, then p ≡ 1 or p ≡ 3 (mod 4).

In the Gregorian calendar (the de facto worldwide standard introduced in the 1600s), leap years occur in centuries

divisible by 400 and every non-century divisible by 4: for instance, 2000 was a leap year but 1900 was not.

5. Prove the ﬁrst part of Theorem 3.9: if a ≡ c and b ≡ d, then a + b ≡ c + d (mod n).

6. Find a positive integer n and integers a, b such that a

≡ b

(mod n) but a ≡ b (mod n).

7. Check explicitly that 3

≡ 3

(mod 7). (Hint: 3

= 27 . . .)

8. Compute the following remainders—use a calculator to help!

(a) 12

+ 19

on division by 10.

(b) 30

on division by 13.

251

·23

−19

on division by 5. (Hint: 17 ≡ 2 and 2

≡ −1 (mod 5))

(d) (Hard) 12

+ 2

·18

on division by 141. (Hint: what nice number is close to 141?)

9. Prove that 3 | (4

−1) for all positive integers n.

10. Let n be an integer. Prove that exactly one of n, n + 2 and n + 4 is divisible by 3.

11. (a) Let n be a positive integer. Prove that n is congruent to the sum of its digits modulo 9.

(Hint: e.g. 345 = 3 ·10

+ ···)

(b) Is the integer 123456789 divisible by 9?

12. Describe all integers x which satisfy the congruence equations:

(a) 3x ≡ 2 (mod 8) (b) 7x ≡ 28 (mod 42).

13. Abraham Lincoln was born February 12

, 1809. On what day of the week was this?

(Start by looking up the day for February 12

this year)

14. Let n be an integer.

(a) Prove: n

≡ 0 or 1 (mod 3). (Hint: prove by cases)

(b) Prove: n

≡ 0 or 1 (mod 4).

on division by 7.

(d) Find all possible remainders of n

on division by 7.

15. Use some part(s) of Exercise 14 to prove the following.

(a)

√

4m + 6 is not an integer, for any integer m ≥ −1.

(b) Any number which is simultaneously a square and a cube must be of the form 7k or 7k + 1

for some integer k.

16. Let n be an integer ≥ 2 and consider numbers of the form 11 ···11

| {z }

n times

(a) Prove every such number can be written as 4k + 3 for some k ∈ Z.

(b) Prove that no such number is a perfect square.

17. Fermat’s Little Theorem states that if p is prime and a ≡ 0 (mod p), then a

p−1

≡ 1 (mod p).

(a) Use Fermat’s Little Theorem to prove that b

≡ b (mod p) for any integer b.

(b) Prove that if p is prime then p | (2

−2).

341

−2). What does this say about the converse to part (b)?

3.2 Greatest Common Divisors and the Euclidean Algorithm

A basic goal of number theorists is to ﬁnd integer solutions to equations. For instance:

Are there any integer points on the line with equation 9x − 21y = 6? That is, does the

equation 9x −21y = 6 have any solutions, where x, y are both integers?

You might start by sketching the line (graph paper will help). What do you observe? If there are

integer points, do they seem to follow any pattern?

In this section we will see how to solve all such linear Diophantine equations.

The method introduces

a famous procedure dating at least to Euclid’s Elements (c. 300 BCE), and an important concept.

Deﬁnition 3.14. Let a and b be integers, not both zero. Their greatest common divisor gcd(a, b) is the

largest (positive) divisor of both a and b. We say that a and b are relatively prime if gcd(a, b) = 1.

Example 3.15. The positive divisors of a = 60 and b = 90 are listed in the table. The greatest common

divisor gcd( 60, 90) = 30 is plainly the largest number common to both rows.

a 1 2 3 4 5 6 10 12 15 20 30 60

b 1 2 3 5 6 9 10 15 18 30 45 90

For large integers, listing divisors is very inefﬁcient. This is where Euclid rides to the rescue.

Theorem 3.16 (Euclidean Algorithm). Let a > b be positive integers. We construct a decreasing

sequence of integers b = r

> r

> ···

1. Apply the division algorithm (Theorem 3.3): a = q

b + r

with 0 ≤ r

< b

2. If r

> 0, apply the division algorithm again: b = q

+ r

with 0 ≤ r

< r

3. If r

> 0, apply again: r

= q

+ r

with 0 ≤ r

< r

4. While r

> 0, keep repeating the division algorithm, dividing each r

i+1

by r

The algorithm eventually terminates with a remainder of zero: some r

t+1

= 0. The greatest common

divisor is the last non-zero remainder: gcd(a, b) = r

Exercise 8 provides a proof. If either/both of a, b are negative, simply ignore the signs and compute

normally: for instance gcd(−4, 34) = gcd(34, 4) = 2.

Example 3.17. We compute gcd(1260, 750) using the Euclidean algorithm. Note how each line is a

single instance of the division algorithm a = qb + r and how remainders move diagonally ↙ at each

step. For this ﬁrst example, we also summarize the data in a table.

a q b r

1260 = 1 ×750 + 510 1260 1 750 510

750 = 1 ×510 + 240 750 1 510 240

510 = 2 ×240 + 30 510 2 240 30

240 = 8 ×30 + 0 240 8 30 0

The Euclidean algorithm says that gcd( 1260, 750) = 30, the ﬁnal non-zero remainder.

Equations with integer coefﬁcients and solutions honor Diophantus of Alexandria (3

C. CE).

To apply the Euclidean algorithm to our motivational problem, we need to run it backwards. Start

with the penultimate line of the algorithm and move upwards, substituting remainders one at a time.

The result is an expression of the form gcd(a, b) = ax + by for some integers x, y.

Example (3.17, cont). We ﬁnd integers x, y such that 1260x + 750y = 30



= gcd(1260, 750)



Start by expressing 30 using the third line of the algorithm and work upwards:

30 = 510 −2 ×240 (3

/penultimate line)

= 510 −2(750 − 510) = 3 ×510 −2 ×750 (2

line)

= 3(1260 −750) −2 ×750 (1

line)

= 3 ×1260 −5 ×750

Rearranging, we see that x = 3 and y = −5 satisfy the equation 1260x + 750y = 30. To simplify,

divide everything by 30 to see that ( 3, −5) is an integer point on the line 42x + 25y = 1.

To comfortably apply the algorithm like this will require signiﬁcant practice; we’ll see another exam-

ple momentarily. The reversed algorithm works in general, yielding a very powerful result.

Corollary 3.18 (B´ezout’s Identity). Given integers a, b, not both zero, ∃x, y ∈ Z such that

gcd(a, b) = ax + by

If either of a, b are negative, apply the algorithm to

and correct the signs afterwards.

Example 3.19. We express 3 = gcd(123, 78) = 123x + 78y: remainders are underlined for clarity.

123 = 1 ×78 + 45

78 = 1 ×45 + 33

45 = 1 ×33 + 12

33 = 2 ×12 + 9

12 = 1 ×9 + 3

9 = 3 ×3 + 0











=⇒











3 = 12 −9

= 12 −(33 − 2 × 12) = 3 ×12 −33

= 3(45 −33) −33 = 3 ×45 −4 ×33

= 3 ×45 −4(78 − 45) = 7 ×45 −4 ×78

= 7(123 −78) −4 ×78

= 7 ×123 −11 ×78 (x, y) = ( 7, −11)

As an application of the power of B

ezout, here is a simple and useful criterion.

Corollary 3.20. Suppose a, b, c are integers for which gcd(a, b) = 1 and a | bc. Then a | c.

Proof. We prove directly. By the corollary, ∃x, y ∈ Z with ax + by = 1. Multiply by c to obtain

acx + bcy = c, the left side of which is a multiple of a.

Example 3.21. The claim “a | c and b | c =⇒ ab | c” is, in general, false (e.g., a = b = c = 2).

However: Suppose c = 2m = 3n for some integers m, n. Since gcd(2, 3) = 1 and 2 | (3n), the

Corollary says that 2 | n. Thus n = 2k for some k ∈ Z and c = 6k. We conclude:

2 | c and 3 | c =⇒ 6 | c

While we could have done this using Theorem 2.27, B

ezout proves the general claim whenever

gcd(a, b) = 1: e.g., 19 | c and 21 | c =⇒ 399 | c (see the video problems if you’re unsure why).

Linear Diophantine & Congruence Equations

We return, as promised to the motivating problem of ﬁnding integer points on straight lines.

Example (3.17, mk. III). We saw that (x

, y

) = (3, −5) solves 1260x + 750y = 30 (equivalently

42x + 25y = 1). Suppose (x, y) is any other integer solution and observe that

42(x − x

) + 25(y −y

) = 1 −1 = 0 =⇒ 42(x − x

) = −25(y −y

)

Since gcd( 42, 25) = 1, Corollary 3.20 says 42 | (y −y

). Write y −y

= 42t for some t ∈ Z, then

42(x − x

) = −25 ·42t =⇒ x −x

= −25t

We have therefore found all integer solutions to the original equation:

x = 3 −25t, y = −5 + 42t where t ∈ Z

This method works for all solvable linear Diophantine equations; here is the general result.

Theorem 3.22. Let a, b, c be integers where a, b are not both zero, and let d = gcd(a, b).

1. The equation ax + by = c has an integer solution (x, y) if and only if d | c.

2. If a solution (x

, y

) exists (e.g. as supplied by B

ezout), then there are inﬁnitely many such:

x = x

−

t, y = y

t, where t ∈ Z

Example (3.19, cont). We saw that (7, −11) solves 123x + 78y = 3 = gcd(123, 78).

1. Since 3 ∤ 8, the equation 123x + 78y = 8 has no integer solutions.

2. The equation 123x −78y = 6 has integer solutions since 3 | 6. Indeed (x

, y

) = (14, 22) is such

(modify signs and scale ( 7, −11) by 2). All integer solutions are then given by

x = 14 +

t = 14 + 26t, y = 22 +

123

t = 22 + 41t, where t ∈ Z

We began to consider linear congruence equations in Example 3.13. The linear Diophantine method

also applies in this context. To see why, observe that

ax ≡ c (mod n) ⇐⇒ ∃y ∈ Z such that ax = ny + c

The Theorem tells us when we can solve the right hand side, and supplies a method for doing so. To

solve ax ≡ c (mod n), simply take the x-part!

Examples 3.23. 1. To solve the congruence equation 123x ≡ 6 (mod 78) is to solve the Diophantine

equation 123x = 78y + 6. By the last example, x = 14 + 26t. Otherwise said

123x ≡ 6 (mod 78) =⇒ x ≡ 14 ( mod 26)

2. The congruence equation 4x ≡ 6 (mod 20) has no solutions. If it did, then 4x = 20y + 6 would

have integer solutions, but it does not since gcd( 4, 20) = 4 does not divide 6.

3. Here is a slightly different approach for 7x ≡ 11 (mod 23). By applying the algorithm,

gcd( 23, 7) = 1 = 7 ·10 −23 ·3 =⇒ 7 ·10 ≡ 1 (mod 23)

The original equation may now be solved via multiplication:

7x ≡ 11 =⇒ x ≡ 10 · 7x ≡ 110 ≡ 18 (mod 23)

In this context, division by 7 really means multiplication

by 10.

Exercises 3.2. A reading quiz and several questions with linked video solutions can be found

online. Some of the later questions are particularly tricky, and are included to give you a taste of

upper-division number theory and abstract algebra.

1. Use the Euclidean algorithm to compute the greatest common divisors.

(a) gcd( 20, 12) (b) gcd(100, 36) (c) gcd(207, 496)

2. For each part of Exercise 1, ﬁnd integers x, y which satisfy B

ezout’s identity gcd(a, b) = ax + by.

3. Describe all the integer points on the line 9x −21y = 6 using Theorem 3.22.

4. (a) Use Theorem 3.22 to show that there are no integer points on the line 4x + 6y = 1.

(b) Give an elementary proof (without using the Theorem) of part (a)?

5. Find all integer points (x, y) on the following lines, or show that none exist.

(a) 16x −33y = 2 (b) 122x + 36y = 3

6. (a) Show that there exists no integer x such that 3x ≡ 5 (mod 6).

(b) Find all solutions x to the congruence equation 12x ≡ 1 (mod 17).

7. Five people each take the same number of candies from a jar. Then a group of seven people

does the same: in so doing they empty the jar. If the jar originally contained 93 candies, can

you be sure how much candy each person took?

8. We complete the proof of the Euclidean algorithm (Theorem 3.16).

(a) Suppose a > b > r

> r

··· is a sequence of non-negative integers. Why must there be

only ﬁnitely many terms? This shows that the algorithm terminates with some r

t+1

= 0.

(b) Suppose that a, b, q, r are integers satisfying a = qb + r. Prove that gcd(a, b) = gcd(b, r).

(You cannot use B´ezout’s identity to do this, since B´ezout is a corollary of the algorithm!)

9. Suppose m = 0. What is gcd(m, 0)? Why? Why is B

ezout’s identity trivial in this situation?

10. Use B

ezout’s identity to prove that if k | a and k | b, then k | gcd(a, b).

In future studies, you’ll refer to 10 as the multiplicative inverse of 7 in the ring Z

, and write 7

−1

= 10 (see Exercise 18).

11. Prove: gcd(m, n) = 1 ⇐⇒ ∃x, y ∈ Z such that mx + ny = 1.

(Hint: One direction follows from B´ezout’s identity, but the other. . . )

12. Use Example 3.21.1 to prove the following.

(a) Let n ≥ 3 be an odd number. Show that n ≡ 1 (mod 3) =⇒ n ≡ 1 (mod 6).

(b)

n(n + 1)(2n + 1) is an integer.

13. Let a, b, p ∈ Z and p ≥ 2. Recall that only the positive divisors of a prime are itself and 1.

(a) Suppose p is prime and that p | ab. Prove that p | a or p | b.

(Hint: if p ∤ a, use Corollary 3.20)

(b) Suppose p satisﬁes the property ∀a, b ∈ Z, p | ab =⇒ p | a or p | b. Prove that p is prime.

(Hint: prove the contrapositive)

14. (Hard) Show that if a is relatively prime to both b and c then it is relatively prime to bc.

(Hint: suppose d | a and d | bc and try to prove that d = ±1)

15. The general rule for congruence division is as follows:

ac ≡ bc (mod n) =⇒ a ≡ b (mod

gcd(c,n)

)

(a) Use the rule to ﬁnd all solutions to the congruence equation 22x ≡ 66 (mod 77).

(b) We prove the rule. Let d = gcd(c, n):

i. Explain why gcd(

) = 1

ii. If ac ≡ bc (mod n), prove that

| (a − b).

16. (a) Prove part 1 of Theorem 3.22 using B

ezout’s identity.

(b) Prove part 2 by mimicking the method in Example 3.17, (mk. III).

17. (Hard) Apply the discussion of the Euclidean algorithm and linear equations to the following.

(a) Complete the table.

If n is a positive integer, make a conjecture for the

value of gcd( 2n, n + 1) and prove it.

n 1 2 3 4 5 6

gcd( 2n, n + 1)

(b) Show that gcd( 5n + 2, 12n + 5) = 1 for every integer n.

18. The set of remainders Z

= {0, 1, 2, . . . , n −1} is called a ring when equipped with addition

and multiplication modulo n. We say that b ∈ Z

is an inverse of a ∈ Z

ab ≡ 1 (mod n)

(a) Show that 2 has no inverse modulo 6.

(b) Prove that a has an inverse modulo n if and only if gcd(a, n) = 1. Conclude that the only

sets Z

for which all non-zero elements have inverses are those for which n is prime.

4 Sets and Functions

Sets are the fundamental building blocks of mathematics, supplying the language used to describe

mathematical objects and to group objects according to shared characteristics. While our primary

focus is learning to understand and employ set notation, the mathematical discipline of set theory is

far more ambitious: set theorists deﬁne all basic mathematical objects—numbers, addition, functions,

etc.—purely in terms of sets!

We will only scratch the surface of set theory; indeed long before one

can accept the beneﬁt of such an approach, it is necessary to develop a signiﬁcant level of familiarity

with sets and their basic operations.

4.1 Set Notation and Subsets

Without any attempt to deﬁne the meaning of object, we offer a na

ıve deﬁnition.

Deﬁnition 4.1. A set is a collection of objects, namely its elements or

members.

The proposition “x is an element/member of the set A,” is written x ∈ A,

usually read simply “x is in A.”

If y is a member of some other set, but not of A, we instead write y /∈ A

(“y is not in A”). In essence, this is the negation ¬(y ∈ A).

As in the deﬁnition, it is typical to use upper-case letters (A, B, C, . . .) for abstract sets and lower-case

letters for elements/members.

Venn diagrams are useful for visualizing abstract sets. A set is represented by a region in the plane,

with elements depicted by dots. The diagram in the deﬁnition represents a set A comprising at least

four elements a

, a

and x. The element y does not lie in A.

Example 4.2. Let A be the set of (names of) US states. Then Michigan ∈ A and Saskatchewan /∈ A.

Deﬁnition 4.3. Let A and B be sets.

1. Sets are equal, written A = B, when they have precisely the same elements.

2. A is a subset of B, written A ⊆ B, when every element of A is also an

element of B.

3. A is a proper subset of B when both A ⊆ B and A = B. To stress this,

we could write A ⊊ B. The Venn diagram represents a proper subset.

The following observations are merely translations of the deﬁnition—do they make sense to you?

Equality: A = B ⇐⇒ A ⊆ B and B ⊆ A. (∗)

Subset: A ⊆ B ⇐⇒



x ∈ A =⇒ x ∈ B



⇐⇒



∀x ∈ A, x ∈ B



Not a subset: A ⊈ B ⇐⇒ ∃x ∈ A for which x /∈ B.

This is an impractical approach for most working mathematicians, most of the time. Within axiomatic set theory it can

take many, many pages of development to justify writing 1 + 1 = 2: the issue is that rigorous deﬁnitions—using sets—are

ﬁrst required of the notions one, two, equals and add. . .

Roster & Set-Builder Notation

Roster notation is ideal for describing small sets: simply list the elements in any order between curly

brackets {, }.

Example 4.4. A = {3,

} is the set containing the numbers 3 and

. For instance 3 ∈ A, but 7 /∈ A.

Since order doesn’t matter, we could also write A = {

, 3}. Now let Let B = {3}. Plainly,

• A ⊈ B since

∈ A and

/∈ B (∃x ∈ A for which x /∈ B).

• B ⊆ A since 3 (the only element of B) lies in A (∀x ∈ A, x ∈ B). Indeed B ⊊ A is a proper subset

since A = B.

Roster notation is rarely used in practice, due to its limited utility for larger sets. As an alternative. . .

Set-builder notation describes the elements of a set in terms of some common property. Suppose U is

some (already understood) set and P(x) a propositional function with domain U, then

A :=



x ∈ U : P(x)



(“A is the set of x in U such that P(x)”)

deﬁnes a set A as the subset of U whose elements x satisfy the property P(x). A vertical separator |

can be used instead of a colon: in some contexts the choice is essential for clarity.

Examples 4.5. 1. We continue Example 4.4. Note that 2x

−7x + 3 = (2x −1)(x −3) and recall that

R represents the set of real numbers and Z the set of integers. In set-builder notation, our sets

may be written

A =



x ∈ R : 2x

−7x + 3 = 0



, B =



x ∈ Z : 2x

−7x + 3 = 0



In this case the qualifying proposition P(x) is “2x

−7x + 3 = 0.”

We can also express the fact that B ⊆ A in this notation (this time with a vertical separator),

B =



x ∈ A



x ∈ Z} (“B is the set of elements x in A such that x is an integer”)

2. Let X = {2, 4, 6} and Y = {1, 2, 5, 6}. There are many options for how to write these in set-

builder notation. For instance:

X =



n ∈ Z :

n ∈ {1, 2, 3}



, Y =



n ∈ Z



1 ≤ n ≤ 6 and n = 3, 4



We now practice the opposite skill by converting ﬁve sets from set-builder to roster notation.



x ∈ X : x is divisible by 4



= {4} S



y ∈ Y : y is odd



= {1, 5}



x ∈ X



x ∈ Y



= {2, 6} S



x ∈ X : x /∈ Y



= {4}



y ∈ Y



y is odd and y − 1 ∈ X} = {5}

Can you ﬁnd alternative descriptions in set-builder notation for the sets S

, . . . , S

above? Take

your time getting used to this notation, since translating between various descriptions of a set

is essential to reading mathematics.

3. We use the set C = {0, 1, 2, 3, . . . , 24} to describe D = {n ∈ Z : n

−3 ∈ C} in roster notation.

Start by expanding the criterion for membership in D:

−3 ∈ C ⇐⇒ n

∈



3, 4, 5, . . . , 25, 26, 27



Since n must be an integer, it follows that D = {±2, ±3, ±4, ±5}.

4. To express E = {0, 2, 6, 12, . . .} in set-builder notation, we might spot a pattern and decide that

E =



n ∈ Z : n = m(m + 1) for some integer m ≥ 0



Unfortunately, we cannot guarantee our correctness! Perhaps the correct formula is

n = m(m + 1) + m(m −2)(m −6)(m −12) (!!)

In the ﬁrst case the next term in the sequence is 4 ·5 = 20, whereas in the second it is 20 + 128 =

148. For larger sets, the clarity afforded by set-builder notation is essential!

Common Sets of Numbers

We’ve used some of this notation already and much of the rest should be familiar.

Natural numbers N =



1, 2, 3, 4, . . .



is the set of positive integers.

Integers Z =



. . . , −3, −2, −1, 0, 1, 2, 3, . . .



Rational numbers Q =



: m ∈ Z and n ∈ N







a, b ∈ Z and b = 0



Real numbers R. Even a rudimentary deﬁnition is too involved for this text.

Complex numbers C =



x + iy : x, y ∈ R



where i

= −1. We won’t make use of these.

Examples 4.6. 1. For instance: 7 ∈ N, π ∈ R, −

/∈ Z,

√

2 /∈ Q and 3 +

√

5i ∈ C.

2. The basic symbols can be modiﬁed is natural ways. For example:

• N



0, 1, 2, 3, 4, . . .



= Z



x ∈ Z : x ≥ 0



. By some called the whole numbers (W).

• Z

≥5



5, 6, 7, 8, . . .





x ∈ Z : x ≥ 5



denotes the integers greater than or equal to 5.

• R



x ∈ R : x > 0



is the set of positive real numbers.

• 4Z =



. . . , −8, −4, 0, 4, 8, 12, . . .





x ∈ Z : 4 | x



is the set

of integer multiples of 4.

This can also be used for non-integer multiples, e.g. πZ =



. . . , −π, 0, π, 2π, . . .



• 2Z + 1 =



x ∈ Z : x ≡ 1 (mod 2)



is the set of odd integers.

3. Intervals are commonly encountered subsets of the real numbers. For instance:

• [1, π] =



x ∈ R



1 ≤ x ≤ π



is a closed interval

• [−4, 7.21) = {x ∈ R



−4 ≤ x < 7.21} is a half-open interval.

• (−∞,

√

2) = {x ∈ R



x <

√

2} is an inﬁnite (open) interval.

We assume the reader is comfortable with the real line where number corresponds to length. A rigorous development

of R is a matter for an upper-division analysis course.

Be careful with the second version—the colon is the such that separator while | denotes the property “4 divides x.”

In view of the natural subset relationships N ⊊ Z ⊊ Q ⊊ R ⊊ C, we consider a simple result.

Lemma 4.7 (Transitivity of Subset). Suppose A ⊆ B and B ⊆ C. Then A ⊆ C.

Proof. Think back to the criteria following Deﬁnition 4.3. Suppose A ⊆ B and B ⊆ C. Then

x ∈ A

(A⊆B)

=⇒ x ∈ B

(B⊆C)

=⇒ x ∈ C

We conclude that A ⊆ C.

Compare this to Exercise 2.1.9b: if we rewrite each subset relation as an implication, the proof struc-

ture becomes (x ∈ A ⇒ x ∈ B) ∧(x ∈ B ⇒ x ∈ C) =⇒ (x ∈ A ⇒ x ∈ C). This is typical of basic set

results: a translation often reduces the problem to one of the standard rules of logic.

Cardinality and the Empty Set

It is helpful to introduce some terminology to describe the size of a set.

Deﬁnition 4.8. A ﬁnite set contains a ﬁnite number of elements: this number is its cardinality

A set with inﬁnitely many elements is said to be an inﬁnite set.

The symbol ∅ denotes the empty set: a set containing no elements (cardinality zero,

∅

= 0).

Examples 4.9. 1. If A =



1, 3, π,

√

2, 103



, then

= 5.

2. Let B =



4, {1, 2}, {3}



. The elements of B are 4, {1, 2} and {3}, therefore

= 3. It doesn’t

matter that the element {1, 2} ∈ B is also a set!

3. Recall some basic trigonometry:



x ∈ [0, 4π] : cos x =





5π

7π

11π



has cardinality 4.

−1

π 2π 3π 4π

4. There are many, many representations of the empty set in set-builder notation: for example

∅ =



x ∈ R : x

= −1





x ∈ N : x

+ 3x + 2 = 0





n ∈ N : n < 0



In general, if X is any set and P(x) is false for all x ∈ X, then

∅ =



x ∈ X : P(x)



Cardinality is very simple for subsets of ﬁnite sets: if B is ﬁnite, so is any subset, and we have

A ⊆ B =⇒

≤

(Is it obvious why the converse is false?!) For inﬁnite sets, cardinality is more subtle; we’ll return to

this matter and uncover some of its bizarre and fun consequences in Chapter 8.

The existence of the empty set is sometimes considered an axiom: an assumption made without proof. Provided one

accepts that set-builder notation always deﬁnes a set (itself an axiom!) and that at least one set X exists, the empty set may

be deﬁned as in the example; a suitable property P(x) might be something like “x /∈ {x}.”

We ﬁnish with a couple of simple results regarding the empty set.

Lemma 4.10. Let A be a set.

1. If

= 0, then A = ∅. The empty set is the unique set with cardinality zero.

2. ∅ ⊆ A and A ⊆ A

Proof. Consider the claim ∅ ⊆ A. By the observations following Deﬁnition 4.3, this means

x ∈ ∅ =⇒ x ∈ A

This is true (for any set A!) since there are no elements x satisfying the hypothesis.

1. Suppose A has cardinality zero. Repeating and combining with the above observation, we see

that ∅ ⊆ A and A ⊆ ∅. We conclude that A = ∅.

2. We already know that ∅ ⊆ A. For the second part, simply observe that x ∈ A =⇒ x ∈ A.

Exercises 4.1. A reading quiz and several questions with linked video solutions can be found online.

1. Describe the following sets in roster notation: that is, list their elements.

(a)



x ∈ N : x

≤ 3x



(b)



n ∈ {0, 1, 2, 3, . . . , 19} : n + 3 ≡ 5 (mod 4)



(c)



n ∈ {−2, −1, 0, 1, . . . , 23} : 4 | n



(d)



x ∈

Z : 0 ≤ x ≤ 4 and 4x

∈ 2Z + 1



(e)



y ∈ R : y = x

for some x ∈ R with x

−3x + 2 = 0



2. Describe the following sets in set-builder notation (look for a pattern).

(a)



. . . , −3, 0, 3, 6, 9, . . .



(b)



−3, 1, 5, 9, 13, . . .



(c)



, . . .



3. Each of the following sets of real numbers is a single interval. Determine the interval.

(a)



x ∈ R : x > 3 and x ≤ 17



(b)



x ∈ R : x ≰ 3 or x ≤ 17



(c)



∈ R : x = 0



(d)



x ∈ R

−

: x

≥ 16 and x

≤ 27



4. Is the set {x ∈ Z : −1 ≤ x < 43} ﬁnite or inﬁnite? If ﬁnite, what is its cardinality?

5. What is the cardinality of the set

∅,



∅





∅, {∅}



? What are its elements?

6. Let A = ∅, B = {A}, C =



{A}



and D =



A, {0}, {0, 1}



Answer the following true or false:

(a) 0 ∈ A (b) A ∈ B (c) A ∈ C (d) B ∈ C (e) A ∈ D

(f) B ∈ D (g) 0 ∈ D (h) {0} ∈ D (i) {1} ∈ D

7. List all the proper subsets of {1, 2, 3}.

If P(x) is always false, then (∀x) P(x) =⇒ Q(x) is true (Deﬁnition 2.5). This is called a vacuous (empty) theorem.

8. Let A, B, C, D be the following sets:

A = {−4, 1, 2, 4, 10} B =



m ∈ Z :

≤ 12



(absolute value of m)

C =



n ∈ Z : n

≡ 1 (mod 3)



D =



t ∈ Z : t

+ 3 ∈ [4, 20)



Of the 12 subset relations A ⊆ B, A ⊆ C, . . . , D ⊆ C, which are true and which false?

9. Let A =



1, 2, {1, 2}, {3}



and B = {1, 2}. Answer the following true or false:

(a) B ∈ A (b) B ⊆ A (c) 3 ∈ A (d) {3} ⊆ A

(e) {3} ∈ A (f) ∅ ⊆ A (g) ∅ ∈ A

10. Let A = {0, 2, 4, 6, 8, 10}. Write the set B = {X ⊆ A : |X| = 2} in roster notation.

11. (a) Suppose A ⊆ B ⊆ C ⊆ A. Show that A = B = C.

(b) Is it possible for sets A, B, C to satisfy A ⊊ B ⊆ C ⊆ A? Why/why not?

12. Let A = {1,2,3,4}, and let B =



{x, y} : x, y ∈ A



(a) Describe B in roster notation (what happens when x = y?).

(b) Find the cardinalities of the following sets:

C =



x, {y}



: x, y ∈ A

and D =





x, {y}



: x, y ∈ A



13. Let A = {x ∈ R : x

+ x

− x − 1 = 0} and B = {x ∈ R : x

−5x

+ 4 = 0}. Are either of the

relations A ⊆ B or B ⊆ A true? Explain.

14. For which real numbers x > 0 do we have [0, x] ⊊ [0, x

]? Prove your assertion.

15. Let m, n ∈ N. Prove: mZ ⊆ nZ ⇐⇒ n | m.

16. Given A ⊆ Z and x ∈ Z, we say that x is A-mirrored if and only if −x ∈ A. Also deﬁne



x ∈ Z : x is A-mirrored



(a) What does it mean for x not to be A-mirrored?

(b) Find M

given B = {0, 1, −6, −7, 7, 100}.

that M

is closed under addition.

(d) In your own words, under which conditions is A = M

17. Deﬁne the set [1] =



x ∈ Z : x ≡ 1 (mod 5)



(a) Describe the set [1] in roster notation.

(b) Compute the set M

[1]

, as deﬁned in Exercise 16. Is M

[1]

equal to [1]?

[10]

equal?

Prove or disprove.

18. Consider the set A = {a, b, c, d}.

(a) Of each cardinality 0, 1, 2, 3 and 4, how many subsets has A? Is there a pattern?

(b) Completely expand the polynomial ( 1 + x)

. What do you notice about the coefﬁcients?

4.2 Unions, Intersections and Complements

In this section we construct new sets from old, modeled on the logical and, or, and not (Deﬁnition 2.3).

Deﬁnition 4.11. Let A, B be sets.

1. The union of A, B is the set of elements lying in A or in B (or in both):

A ∪ B =



x : x ∈ A or x ∈ B



(x ∈ A ∪ B ⇐⇒ x ∈ A or x ∈ B)

2. The intersection of A, B is the set of elements lying in both A and B:

A ∩ B =



x : x ∈ A and x ∈ B



(x ∈ A ∩ B ⇐⇒ x ∈ A and x ∈ B)

We say that A, B are disjoint if A ∩ B = ∅.

3. The complement of A is the set of elements not in A (with respect to

some universal set

U):



x ∈ U : x /∈ A



(x ∈ A

⇐⇒ x ∈ U and x /∈ A)

In the Venn diagram, the outer box represents the universal set U.

4. The complement of A relative to B is the set of elements in B which do

not lie in A:

B \ A = B ∩ A

(x ∈ B \ A ⇐⇒ x ∈ B and x /∈ A)



x ∈ B : x /∈ A



This can be read “B minus A” (some authors indeed write B − A).

Note also that A

= U \ A; the distinction is that the relative com-

plement B \ A does not require that A be a subset of B.

A \ B B \ A

A ∪ B

A ∩ B

Observe the notational similarities with logic: ∪ looks a bit like ∨ (OR); ∩ like ∧ (AND). The second

Venn diagram suggests the identities

A = (A \B) ∪ (A ∩ B) and B = (B \ A) ∪ (A ∩ B)

While these indeed hold, note that a Venn diagram isn’t a proof: set identities must rigorously be

proved in the style of the upcoming theorems.

Examples 4.12. Reading set notation is one of the most basic requirements of abstract mathematics.

Make sure you understand why the following examples are correct before moving on!

1. Let U = {1, 2, 3, 4, 5}, A = {1, 2, 3}, and B = {2, 3, 4}. Then

= {4, 5} B

= {1, 5} B \ A = {4} A \B = {1}

A ∪ B = {1, 2, 3, 4} A ∩ B = {2, 3} A ∩ B

= {1} A

∪ B

= {1, 4, 5}

The elements x must live somewhere! Without a universal set, should, say, {7}

be the set of integers except 7, real

numbers except 7, etc.? Often U is naturally assumed: e.g., R in calculus. The universal set is not needed for parts 1, 2 & 4:

that the union is a set is typically an axiom, while intersections and relative complements are subsets of pre-existing sets.

2. Using interval notation, let U = [−4, 5], A = [−3, 2], and B = [−4, 1). Then

= [−4, −3) ∪ (2, 5]

= [1, 5]

A \ B = [1, 2]

B \ A = [−4, −3)

A ∪ B = [ −4, 2]

A ∩ B = [ −3, 1)

−4 −3 −2 −1 0 1 2 3 4 5

[ ]

[ )

[ ) ( ]

[ ]

B \ A

[ )

A \ B

[ ]

While you should believe these from the picture, they also make for good logic practice. E.g.,

x ∈ A

⇐⇒ x /∈ A ⇐⇒ ¬



x ∈ A



⇐⇒ ¬



−3 ≤ x and x ≤ 2



⇐⇒ x < −3 or x > 2 (de Morgan’s law, Theorem 2.13)

⇐⇒ x ∈ [−4, −3) ∪ ( 2, 5] (remember that −4 ≤ x ≤ 5 (x ∈ U))

The argument illustrates the basic strategy for set computations & proofs: convert claims to

propositions (Deﬁnition 4.11 parentheses) and apply basic logic (Theorem 2.13, page 11, etc.).

Alternatively, you can prove each direction separately: this would be to show that each of the

sets A

, [−4, −3) ∪(2, 5] is a subset of the other (page 43, (∗)).

3. Let A = (−∞, 3) and B = [−2, ∞) in interval notation. Then A ∪ B = R and A ∩ B = [−2, 3).

We show the ﬁrst: for variety, this time we observe that each side is a subset of the other.

(⊆): A ∪ B ⊆ R is trivial, since everything in A, B is a real number (R is the universal set).

(⊇): Let x ∈ R. If x < 3, then x ∈ A. Otherwise, x ≥ 3 =⇒ x ≥ −2 =⇒ x ∈ B. Either way,

x ∈ A ∪ B.

For the remainder of this section, we summarize the basic rules of set algebra.

Theorem 4.13 (Union/intersection rules). Let A, B, C be sets. Then:

1. ∅ ∪ A = A and ∅ ∩ A = ∅ 2. A ∩ B ⊆ A ⊆ A ∪B

3. A ∪ B = B ∪ A and A ∩ B = B ∩ A 4. A ∪ A = A ∩ A = A

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

6. A ⊆ B =⇒ A ∪C ⊆ B ∪C and A ∩C ⊆ B ∩C

If you don’t believe a result, visualize it with a Venn diagram. We prove part 2 and half of 6.

Proof. 2. We offer direct proofs of the two results: A ∩ B ⊆ A and A ⊆ A ∪ B.

• Suppose x ∈ A ∩ B. (goal: want to prove x ∈ A ∩ B ⇒ x ∈ A)

Then x ∈ A and x ∈ B. (deﬁnition of intersection)

Plainly x ∈ A. We conclude that A ∩ B ⊆ A (deﬁnition of subset)

• Suppose y ∈ A. (goal: prove y ∈ A ⇒ y ∈ A ∪ B)

Then “y ∈ A or y ∈ B” (is true), whence y ∈ A ∪ B. (deﬁnition of union/or)

We conclude that A ⊆ A ∪B.

6. (ﬁrst half) Suppose A ⊆ B. We wish to prove that x ∈ A ∪ B =⇒ x ∈ A ∪C. However,

x ∈ A ∪C =⇒ x ∈ A or x ∈ C (deﬁnition of union)

=⇒ x ∈ B or x ∈ C (since A ⊆ B)

=⇒ x ∈ B ∪C

Our next batch of rules describe how complements interact with other set operations: parts 1 and 2

are de Morgan’s laws for sets; unsurprisingly, their proofs depend on the corresponding laws of logic.

Theorem 4.14 (Complement rules). Let A, B be sets. Then:

1. (A ∩B)

= A

∪ B

(pictured)

2. (A ∪B)

= A

∩ B

3. (A

)

= A

4. A ⊆ B ⇐⇒ B

⊆ A

Proof. We prove only part 1. As before, the natural approach is to restate the result using propositions.

x ∈ (A ∩ B)

⇐⇒ ¬



x ∈ A ∩ B



⇐⇒ ¬



x ∈ A and x ∈ B



⇐⇒ ¬



x ∈ A



or ¬



x ∈ B



(de Morgan’s ﬁrst law of logic)

⇐⇒ x ∈ A

or x ∈ B

⇐⇒ x ∈ A

∪ B

Our ﬁnal results describe the interaction of unions and intersections.

Theorem 4.15 (Distributive laws). For any sets A, B, C:

1. A ∩(B ∪C) = (A ∩B) ∪ (A ∩C)

2. A ∪(B ∩C) = (A ∪B) ∩ (A ∪C)

The Venn diagram illustrates the second result: think about adding the

colored regions.

Proof. We prove only the ﬁrst result.

x ∈ A ∩(B ∪ C) ⇐⇒ x ∈ A and x ∈ B ∪C

⇐⇒ x ∈ A and



x ∈ B or x ∈ C



⇐⇒



x ∈ A and x ∈ B





x ∈ A and x ∈ C



(distributive law, page 11)

⇐⇒ x ∈ A ∩ B or x ∈ A ∩C

⇐⇒ x ∈ (A ∩ B) ∪(A ∩C)

Remember, if you prefer, you can prove these equalities in two stages: S = T ⇐⇒ S ⊆ T and S ⊇ T.

Exercises 4.2. A reading quiz and several questions with linked video solutions can be found online.

1. Describe each set straightforwardly as you can: e.g.,



x ∈ R : x

< 9 and x

< 8



= (−3, 3) ∩ (−∞, 2) = (−3, 2)

(a)



x ∈ R : x

= x



(b)



x ∈ R : x

−2x

−3x ≤ 0 or x

= 4



(c)



y ∈ R : ∃x ∈ R with y = x

and x = 1



(d)



z ∈ Z : z

is even and z

is odd



(e)



y ∈ 3Z + 2 : y

≡ 1 (mod 3)



2. Let A = {1, 3, 5, 7, 9}, B = {1, 4, 7, 10} and U = {1, 2, . . . , 10}. What are the following sets?

(a) A ∩ B (b) A ∪ B (c) B \ A (d) A

(e) (A \B)

(f) A

∩ B

(g) (A ∪B) \ (A ∩ B)

3. In Example 4.12.2, use logic to formally justify the assertions B

= [1, 5], A ∩ B = [−3, 1), and

A ∪ B = [ −4, 2]. If you prefer, use the ‘subset of each side’ approach of Example 4.12.3.

4. Give formal proofs of the following parts of Theorems 4.13, 4.14 and 4.15.

(a) ∅ ∩ A = ∅ (b) A ∩(B ∩C) = (A ∩B) ∩ C

)

= A (d) A ∪(B ∩C) = (A ∪B) ∩ (A ∪C)

(e) A ⊆ B ⇐⇒ B

⊆ A

5. By showing that each side is a subset of the other, give a formal proof of the set identity

A = (A \B) ∪ (A ∩ B)

Now repeat your argument using only results from set algebra (Theorems 4.14 and 4.15).

6. Prove the identity A ∪B = A ⇐⇒ B ⊆ A for any sets A, B.

7. Prove the identities for any sets A, B, C:

(a) (A ∩B ∩C)

= A

∪ B

∪C

(b) (A ∪B) \ (A ∩ B) = (A \ B) ∪ (B \ A)

8. Prove or disprove the following conjectures (Hint: revisit Section 2.4).

(a) ∃x ∈ R \Q such that x

∈ Q (b) ∀x ∈ R \Q we have x

∈ Q

9. Let A ⊆ R, and let x ∈ R. We say that x is far away from the set A if and only if:

∃d > 0 such that A ∩[x −d, x] = ∅

If this does not happen, we say that x is close to A.

(a) Draw a picture of a set A and elements x, y such that x is far away from and y is close to A.

(b) State the meaning of “x is close to A” (negate “x is far away from A”).

i. Show that x = 4 is far away from A using the deﬁnition.

ii. Let A = {1, 2, 3}. Show that x = 1 is close to A.

(d) For general A ⊆ R, show that if x ∈ A, then x is close to A.

(e) Let A = (a, b) be a bounded interval. Is the end-point a far away from A? What about b?

4.3 Introduction to Functions

Sets become a lot more useful and interesting once you start transforming their elements! This is

accomplished using functions. In this section we introduce some basic concepts and notation, much

of which should be familiar. A formal deﬁnition will be given in Chapter 7, but for the present a

ıve notion will sufﬁce.

Deﬁnition 4.16. Let A, B be sets. A function f : A → B is a rule assigning to each input a ∈ A a

single output f (a) ∈ B. Various sets are associated to f :

Domain: dom( f ) = A is the set of inputs to the function.

Codomain: codom( f ) = B is the set of potential outputs.

Image of a subset U ⊆ A: the set of outputs given inputs in U

f (U) :=



f (u) ∈ B : u ∈ U



Range: range( f ) = f (A) = {f (a) ∈ B : a ∈ A} is the set of

realized outputs, the image of the domain A.

f (A)

Inverse image (or pre-image) of a subset V ⊆ B: the set of inputs which are mapped into V

−1

(V) :=



a ∈ A : f (a) ∈ V



The rule deﬁning a function can also be described using arrow notation f : a 7→ b.

Examples 4.17. 1. It is common to graph functions whose codomain is a subset of the real numbers:

the domain and range are found by projecting the graph onto the two axes.

For instance if f : [−3, 2) → R is the square function

f : x 7→ x

(equivalently f (x) = x

)

then dom( f ) = [−3, 2) and range( f ) = [0, 9]. We could

also calculate other images/pre-images, for example,



[−1, 2)





: −1 ≤ x < 2



= [0, 4)

−1



( −10, 2]





x ∈ [−3, 2) : −10 < x

≤ 2



= [−

√

f (x)

−3 −2 −1 0 1 2

range

domain

2. Deﬁne f : Z → {0, 1, 2} by f : n 7→ n

(mod 3). The table

shows a few examples (remember dom( f ) = Z is inﬁnite!).

n 0 1 2 3 4 5 6 7

f (n) 0 1 1 0 1 1 0 1

Exercise 3.1.14 conﬁrms what the table suggests, that range( f ) = {0, 1}. Here also is a single

inverse image (revisit the previous sections if you’re unsure of the notation):

−1



{1}





x ∈ Z : x

≡ 1 (mod 3)





x ∈ Z : x ≡ 1 or 2 (mod 3)



= (3Z + 1) ∪ (3Z + 2)

3. Let A = {0, 1, 2, . . . , 7} be the set of remainders modulo 8

and deﬁne two functions f , g : A → A:

f (n) = 3n (mod 8) g(n) = 6n (mod 8)

n 0 1 2 3 4 5 6 7

f (n) 0 3 6 1 4 7 2 5

g(n) 0 6 4 2 0 6 4 2

This time the table completely describes the functions. Observe that

range( f ) = A f



{1, 5}



= {3, 7} f

−1



{1, 2, 3, 4}



= {1, 3, 4, 6}

range(g) = {0, 2, 4, 6} g



{1, 5}



= {6} g

−1



{1, 2, 3, 4}



= {2, 3, 6, 7}

If you’re unsure, compute these as we did in the last two examples.

4. Let A = {0, 1, 2, 3, 4} and let B = {two-element subsets of A}. Deﬁne

f : A → B : a 7→ {a, a + 1 (mod 5)}

where we take the remainder modulo 5. You should be able to convince yourself that

range( f ) =



{0, 1}, {1, 2}, {2, 3}, {3, 4}, {4, 0}





{1, 4}





f (1), f ( 4)





{1, 2}, {4, 0}



and f

−1



{2, 4}, {4, 0}



= {4}

Injections, Surjections and Invertibility

We turn our attention to perhaps the most important properties a function can possess.

Deﬁnition 4.18. Let f : A → B be a function. We say that f is:

1. Injective (1–1, an injection) if distinct inputs produce distinct outputs. Otherwise said (we state

the contrapositive),

f (a

) = f (a

) =⇒ a

= a

(“∀a

, a

∈ A” is typically hidden)

2. Surjective (onto, a surjection) if every potential output is realized: B = range( f ). Equivalently,

∀b ∈ B, ∃a ∈ A such that f (a) = b

This merely expresses B ⊆ range( f ); the reverse inclusion range( f ) ⊆ B holds for any function.

3. Bijective (invertible, a bijection) if it is both injective and surjective. Equivalently,

∀b ∈ B, ∃ a unique a ∈ A such that f (a) = b

When f is bijective, its inverse function is f

−1

: B → A : b 7→ a.

These are universal statements, so counter-examples are enough to demonstrate the negations:

f not injective: ∃a

= a

∈ A such that f (a

) = f (a

)

f not surjective: ∃b ∈ B such that ∀a ∈ A, f (a) = b

Examples (4.17, cont.). We brieﬂy revisit our previous examples.

1. Let f : [−3, 2) → R : x 7→ x

• f is non-injective: f ( 1) = f (−1) provides a counter-example. (a

= 1 = −a

)

• f is non-surjective: there is no x ∈ [−3, 2) for which x

= −5. (b = −5)

We can obtain a related injective function by shrinking the domain,

for instance g : [0, 2) → R : x 7→ x

. Indeed

g(x

) = g(x

) =⇒ x

= x

=⇒ x

= x

since x

, x

∈ [0, 2) are non-negative. By also shrinking the codomain,

we get a surjective (now bijective) function: h : [0, 2) → [0, 4) : x 7→ x

Proof of surjectivity: Given y ∈ [0, 4), let x =

√

y, then y = h(x).

0 1 2x

2. f : Z → {0, 1, 2} : n 7→ n

(mod 3) is neither injective nor surjective.

• f is non-injective: for instance f (1) = f ( 2).

• f is non-surjective: range( f ) = {0, 1} = {0, 1, 2} = codom( f ).

3. Given f , g : A → A where A = {0, 1, . . . , 7} as in the table:

• f is bijective: all elements of codom( f ) appear exactly

once in the f -row.

• g is non-injective: e.g., g( 0) = g(4).

• g is non-surjective: e.g., 1 /∈ range(g) = {0, 2, 4, 6}.

n 0 1 2 3 4 5 6 7

f (n) 0 3 6 1 4 7 2 5

g(n) 0 6 4 2 0 6 4 2

4. Let A = {0, 1, 2, 3, 4} and f : A →



two-element subsets of A



: a 7→ {a, a + 1 (mod 5)}

• f is injective: suppose a

, a

∈ A, then

f (a

) = f (a

) =⇒ {a

, a

+ 1 (mod 5)} = {a

, a

+ 1 (mod 5)} =⇒ a

= a

• f is not surjective: e.g., {1, 3} /∈ range( f ).

You should have seen the approach of the next example in other classes.

Example 4.19. We show that f : (−∞, 2) → (1, ∞) : x 7→ 1 +

(x−2)

bijective by computing its inverse function. Just solve for x in terms of y:

y = 1 +

(x −2)

=⇒ (x − 2)

y −1

=⇒ f

−1

( y) = x = 2 −

y −1

The sign of the square-root was chosen so that x ∈ dom( f ) = (−∞, 2).

−1 0 1 2x

Composition of Functions

We consider how injectivity and surjectivity interact with composition of functions.

Deﬁnition 4.20. Given functions f : A → B and g : B → C,

their composition is the function

g ◦ f : A → C : a 7→ g



f (a)



Note the order: to compute (g ◦ f )(a), ﬁrst apply f , then g.

f g

g ◦ f

f (a)



f (a)



In practice, some restriction of domains might be required in order to deﬁne a composition.

Example 4.21. If f (x) = x

and g(x) =

x−1

, then

(g ◦ f )(x) =

−1

and ( f ◦ g)(x) =

(x −1)

Even though ±1 are legitimate inputs for f , dom(g ◦ f ) = R \ {±1} is implied so as to prevent

division by zero.

Theorem 4.22. Let f : A → B and g : B → C be functions. Then:

1. If f and g are injective, then g ◦ f is injective.

2. If f and g are surjective, then g ◦ f is surjective.

It follows that the composition of bijective functions is also bijective.

Proof. Suppose f and g are injective and let a

, a

∈ A. Then

(g ◦ f )(a

) = (g ◦ f )(a

) =⇒ g



f (a

)



= g



f (a

)



=⇒ f (a

) = f (a

) (since g is injective)

=⇒ a

= a

(since f is injective)

That is, g ◦ f is injective. We leave part 2 to the exercises.

Somewhat surprisingly, the converse of this theorem is false. If a composition is injective or surjective,

only one of the original functions is required also to be.

Theorem 4.23. Suppose f : A → B and g : B → C.

1. If g ◦ f is injective, then f is injective.

2. If g ◦ f is surjective, then g is surjective.

The picture illustrates what can happen: f is only injective,

g is only surjective, but g ◦ f is bijective.

Example 4.24. Here is a formulaic version of the picture in the theorem. Make sure you’re comfort-

able with the deﬁnitions and draw pictures or graphs to help make sense of what’s going on.

f : [0, 2] → [−4, 4] : x 7→ x

(injective only)

g : [−4, 4] → [0, 16] : x 7→ x

(surjective only)

g ◦ f : [0, 2] → [0, 16] : x 7→ x

(bijective!)

Proof. This time we leave part 1 for the Exercises. Let c ∈ C and assume g ◦ f is surjective. But then

∃a ∈ A such that c = (g ◦ f )(a) = g



f (a)



Otherwise said, ∃b(= f (a)) ∈ B for which c = g(b): that is, g is surjective.

Functions and Cardinality

Injectivity and surjectivity are intimately tied to the notion of cardinality. In Chapter 8, we will use

such functions to deﬁne cardinality for inﬁnite sets. For the present we stick to ﬁnite sets.

Theorem 4.25. Let A and B be ﬁnite sets. The following are equivalent:

≤

2. ∃f : A → B injective 3. ∃g : B → A surjective

Moreover,

⇐⇒ ∃f : A → B bijective.

The theorem asserts that any one of the three numbered statements is true if and only

if all are. It might appear that six arguments are required but, by proving in a circle,

we only need three: for instance



⇒



holds because



⇒



and



⇒



The proof is very abstract, but if you focus on the picture it should make sense.



=⇒

Proof. Suppose

= m,

= n and label the elements in roster notation:

A = {a

, a

, . . . , a

} B = {b

, b

, . . . , b

}



⇒



If m ≤ n, deﬁne f : A → B by f (a

) = b

as in the

picture. This is injective since the b

, . . . , b

are distinct.



, . . . , a

}



, . . . , b

z }| {

m+1

, . . . , b

}



⇒



Suppose f : A → B is injective. Without loss of generality, label the elements of B such that

= f (a

) for 1 ≤ k ≤ m. Deﬁne the surjective function g : B → A as in the picture:

g(b

) :=

(

if k ≤ m

if k > m



⇒



Suppose g : B → A is surjective. Without loss of generality, label the elements of B such

that a

= g(b

) for 1 ≤ k ≤ m. Since the b

must be distinct, we see that n ≥ m.

When m = n, the constructed functions are plainly bijections with f

−1

= g.

The elements b

m+1

, . . . , b

could be mapped anywhere; we choose a

for simplicity.

Exercises 4.3. A reading quiz and several questions with linked video solutions can be found online.

1. For each of the following functions f : A → B determine whether f is injective, surjective or

bijective. Prove your assertions.

(a) f : [0, 3] → R where f (x) = 2x.

(b) f : [3, 12) → [0, 3) where f (x) =

√

x −3.

√

+ 9.

2. Suppose that f : [−3, ∞) → [−8, ∞) and g : R → R are deﬁned by

f (x) = x

+ 6x + 1, g(x) = 2x + 3

Compute g ◦ f and show that it is injective.

3. (a) Find a set A so that the function f : A → R : x 7→ sin x is injective.

(b) Find a set B so that the function f : R → B : x 7→ sin x is surjective.

4. A function f : R → R is even if and only if ∀x ∈ R, f (−x) = f (x).

(a) Prove that f : x 7→ x

is even.

(b) By negating the deﬁnition, state what it means for a function not to be even.

(d) Prove or disprove: for every f , g : R → R even, the composition h = f ◦ g is even.

5. The picture in Deﬁnition 4.16 illustrates a function

f : {a

, a

} → {b

, b

}

State the following:

a a

f (a) b

(a) range( f ) (b) f



, a

}



−1



}



(d) f

−1



}



6. (a) Let A = {a, b, c}, B = {1, 2, 3, 4}and f : A → B be the function deﬁned by f (a) = f (c) = 1

and f (b) = 3. State the following:

i. f

−1



{1}



(b) f

−1



{3}



−1



{1, 3}



(d) f

−1



{2, 4}



(b) Let g : [−1, ∞) → R : x 7→ x

+ 2x + 1. Compute g

−1



(0, 2]



−1



{−1, 1}



7. Deﬁne f : ( −∞, 0] → R and g : [0, ∞) → R by

f (x) = x

, g(x) =

(

1−x

x < 1

1 − x x ≥ 1

Is g ◦ f : (−∞, 0] → R surjective? Justify your answer.

8. Recall Example 4.17.3. Consider the nine functions f

: A → A : x 7→ kx (mod 10), where

k = 1, 2, . . . , 9. Find the range of each f

. Can you ﬁnd a relationship between k and the

cardinality of range( f

9. Let f : R → R

be the function deﬁned by f (x) = e

. Explain why the following “proof” that

f is surjective is incorrect. Then, give a correct proof.

Proof? Let e

∈ R

be arbitrary. Then f (x) = e

. We conclude that f is surjective.

10. (a) Show there is a bijection between Z and 2Z.

(b) Let S be the set of all circles in the plane which are centered at the origin. Find a bijection

between S and R

11. Prove that the composition of two surjective functions is surjective.

12. Suppose that g ◦ f is injective. Prove that f is injective.

13. Let f : A → B. Prove the following:

(a) f is injective if and only if ∀b ∈ B, f

−1



{b}



has at most one element.

(b) f is surjective if and only if ∀b ∈ B, f

−1



{b}



has at least one element.

(Taken together: f is bijective ⇐⇒ f

−1

is a function ( f is invertible).)

14. Prove that functional composition is associative. That is, if f : A → B, g : B → C, and h : C → D

are functions, then for all a ∈ A we have



f ◦ (g ◦ h)



(a) =



( f ◦ g) ◦ h



(a)

15. Following Theorem 4.22, the composition of bijective functions f , g is itself bijective. Give a

brief explanation as to why (g ◦ f )

−1

= f

−1

◦ g

−1

16. Let f : A → B and X ⊆ A. Fill in the blanks to complete a proof of the following facts:

(a) X ⊆ f

−1



f (X)



. (b) If f is injective, then X = f

−1



f (X)



Proof. (a) x ∈ X =⇒ f (x) ∈ . Let y = f (x), then x ∈ ⊆ f

−1



f (X)



(b) a ∈ f

−1



f (X)



=⇒ , whence ∃x ∈ X with f (a) = f (x) . By injectivity, ,

whence a ∈ X. We conclude that . Combine with part (a) for the result.

17. Let f : A → B be a function and let Y ⊆ B. Prove the following facts:

(a) f



−1

(Y)



⊆ Y. (b) If f is surjective, then f



−1

(Y)



= Y.

18. (Hard) Let f : A → B be a function and X, Y ⊆ A.

(a) Prove that f (X ∩Y) ⊆ f (X) ∩ f (Y).

(b) If f is injective, prove that f (X ∩Y) = f (X) ∩ f (Y).

5 Mathematical Induction and Well-ordering

In Section 2.3 we discussed three methods of proof: direct, contrapositive, and contradiction. The

fourth standard method, induction, has a very different ﬂavor. Before discussing this formally, we

consider some contexts in which induction arguments often arise.

5.1 Iterative Processes & Proof by Induction

Recursive processes are very common in mathematics and its applications: an initial value x

deter-

mines a sequence of values (x

) via a recurrence relation x

n+1

= f (x

). A typical approach to such

problems is to hypothesize a general formula x

= g(n)—spot a pattern!—before proving the validity

of the formula. Induction is the proof method often employed in such situations. To get us started,

we investigate a famous game.

The Tower of Hanoi

Circular disks of decreasing radius are stacked on three pegs. A single move consists of removing

one disks from the top of a stack then placing it on an empty peg or on top of a larger disk. If we start

with 10 disks on the ﬁrst peg, how many moves are required to transfer all disks to another peg?

To get a feel for the problem, try playing the game with small numbers of disks. Suppose n disks

require r

moves. Then:

• r

= 1 since there is only one disk to move!

• r

= 3 is demonstrated in the picture.

Further experimentation will hopefully convince you that r

= 7, at which point you might be ready

to hypothesize a general formula—if not, experiment more!

Conjecture 5.1. The Tower of Hanoi with n disks requires r

= 2

−1 moves.

To make progress we need to think abstractly. If we have a stack of n + 1 disks, then to move the

largest disk all others must be stacked on a single peg. Moving n + 1 disks to another peg is therefore a

three-step process:

1. Move the smallest n disks to another peg (r

moves);

2. Move the largest disk (one move);

3. Move the remaining disks on top of the largest (r

moves).

moves

1 move

moves

The upshot is that r

satisﬁes a recurrence relation: r

n+1

= r

+ 1 + r

= 2r

+ 1.

We are now in a position to prove our conjecture.

Proof. Certainly the formula r

= 2

−1 holds when n = 1 disk (one disk requires r

= 1 move).

Now suppose that n disks require r

= 2

−1 moves, where n ∈ N is some ﬁxed number. Then n + 1

disks require

n+1

= 2r

+ 1 = 2(2

−1) + 1 = 2

n+1

−1 (∗)

moves. Since n was arbitrary, we see that we’ve proved an inﬁnite family of implications:

= 2

−1) =⇒ (r

= 2

−1) =⇒ (r

= 2

−1) =⇒ ··· =⇒ (r

= 2

−1) =⇒ ···

Since the ﬁrst proposition (r

= 2

−1) holds, we conclude that all do: r

= 2

−1 for all n ∈ N.

To answer the original question, 10 disks require r

= 2

−1 = 1023 moves; at one move per second

this would take 17 minutes, 3 seconds.

Proof by Induction

The above argument is an example of a proof by induction. We invoke this method when we want

to prove a sequence of propositions P(1), P(2), P(3), . . . , one for each natural number. The abstract

structure of an induction proof consists of two separate arguments:

1. Base case: Prove P(1).

2. Induction step: Prove P(n) =⇒ P(n + 1) (for each n ∈ N). During this phase, P(n) is termed

the induction hypothesis.

The result is an inﬁnite chain of implications:

P(1) =⇒ P(2) =⇒ P(3) =⇒ P(4) =⇒ P(5) =⇒ ···

Since P(1) is true (base case), all remaining propositions P(2), P(3), P(4), . . . are also true.

Re-read the Tower of Hanoi proof; can you separate the base case and the induction step? Since this

is an introduction, our presentation was informal. A few modiﬁcations should be made to produce a

formal argument.

• Set-up the proof by stating, “We prove by induction.” It might also be helpful to spell out the

propositions P(n) and to tell the reader what variable (n) controls the induction.

• Label the base case and induction step to aid the reader.

• After the induction step is complete, state your conclusion. In the above we would replace

everything after (∗) with, “By mathematical induction, r

= 2

−1 for all n ∈ N.”

Here is a straightforward and famous result, where we write the proof in our new language.

Theorem 5.2. The sum of the ﬁrst n positive integers is

∑

k=1

k =

n(n + 1)

You should be familiar with summation notation

∑

k=1

k = 1 + 2 + 3 + ···+ n from calculus: if not, ask.

Proof. We prove by induction. For each n ∈ N, let P(n) be the proposition

∑

k=1

k =

n(n + 1).

Base case (n = 1):

∑

k=1

k = 1 =

1( 1 + 1), says that P( 1) is true.

Induction Step: Fix n ∈ N and assume P(n) is true for this n

n. We compute the sum of the ﬁrst n + 1

positive integers and use the induction hypothesis P(n) to simplify:

n+1

∑

k=1

k =

∑

k=1

k + (n + 1) =

n(n + 1) + (n + 1) (induction hypothesis)



1 +



( n + 1) =

( n + 2)(n + 1)

( n + 1)



( n + 1) + 1



Therefore P(n + 1) is true.

By mathematical induction, we conclude that P(n) is true for all n ∈ N. Otherwise said,

∀n ∈ N,

∑

k=1

k =

n(n + 1)

Note how we grouped

( n + 1)



( n + 1) + 1



so that it is obviously the right hand side of P(n + 1).

We present several more examples in a similar vein, though done a little faster. As is typical, we don’t

explicitly introduce the notation P(n), though you should feel free to continue doing so if you ﬁnd it

helpful. Aim to lay out your formal arguments in a similar style.

Example 5.3. We prove by induction that, for all n ∈ N,

2 + 5 + 8 + ··· + (3n −1) =

n(3n + 1) (†)

Base case (n = 1): The proposition (†) is trivially true: 2 =

·1 ·(3 ·1 + 1).

Induction Step: Fix n ∈ N and assume (†) holds for this value of n. Then

2 + 5 + ··· + (3n −1) + [3(n + 1) −1] =

n(3n + 1) + 3n + 2

(3n

+ 7n + 4) =

( n + 1)(3n + 4)

( n + 1)



3(n + 1) + 1



which is the required proposition for n + 1.

By mathematical induction, (†) holds for all n ∈ N.

For brevity we labelled the desired proposition (what we’d might call P(n)) by (†) so it could be

referenced. The structure is similar to Theorem 5.2: since the goal is to evaluate a sum, the induction

step is little more than adding the same thing (3n + 2) to both sides of the induction hypothesis. In

fact, the example could have been proved directly as a corollary of Theorem 5.2—can you see how?

Our next two examples are a little harder, requiring more creativity to invoke the induction hypoth-

esis. Both can alternatively be proved directly using modular arithmetic (Chapter 3).

Examples 5.4. 1. We prove by induction: ∀n ∈ N, the integer 17

−4

is divisible by 13.

Base case (n = 1): Plainly 17

−4

is divisible by 13.

Induction Step: Fix n ∈ N and assume that 17

−4

= 13k for some k ∈ Z. Then

n+1

−4

n+1

= 17





−4

n+1

= 17



13k + 4



−4

n+1

(induction hypothesis)

= 17 ·13k + (17 − 4)4

= 13



17k + 4



is divisible by 13.

By mathematical induction, 17

−4

is divisible by 13 for all n ∈ N.

2. We prove by induction: if n ∈ N, then n(n + 1)(2n + 1) is divisible by 6.

Base case (n = 1): The proposition reads 1 · (1 + 1) · (2 ·1 + 1) = 6, which is divisible by 6.

Induction Step: Fix n ∈ N and assume that n(n + 1)(2n + 1) = 6k for some k ∈ Z. Then

( n + 1)(n + 2)



2(n + 1) + 1



−6k = (n + 1)



( n + 2)(2n + 3) −n(2n + 1)



= (n + 1)



+ 7n + 6 −(2n

+ n)



= 6(n + 1)

from which

( n + 1)



( n + 1) + 1



2(n + 1) + 1



= 6



( n + 1)

+ k



is divisible by 6.

By mathematical induction, n(n −1)(2n −1) is divisible by 6 for all n ∈ N.

Scratch work is your friend! Unless things are very simple, start with some scratch work for the

hard part: the induction step. Explicitly state the propositions P(n) and P(n + 1) and try to manipulate

one into the other. Here are the relevant propositions for Example 5.4.1:

P(n): ∃k ∈ Z such that 17

−4

= 13k

P(n + 1): ∃l ∈ Z such that 17

n+1

−4

n+1

= 13l

Since 17

is common to both, it is natural to try multiplying both sides of the equation in P(n) by 17;

if you re-read the example, you’ll see that this is essentially the induction step! For Example 5.4.2,

you might try multiplying out the cubic expressions

n(n + 1)(2n + 1) and (n + 1)(n + 2)(2n + 3)

and comparing coefﬁcients. Since the leading term in both is n

, the difference is quadratic and there-

fore much easier to think about. . .

Remember that scratch work isn’t a proof; while it might make perfect sense to you, it isn’t a proof

unless a reader can follow it without assistance. Once you think you understand the induction step,

lay out the entire proof cleanly: set-up, base case, induction step, conclusion. As an example of what

happens when you don’t, here is a typical attempt at Example 5.3 by someone new to induction:

P(n + 1) = 2 + 5 + ··· + (3n −1) + [3(n + 1) − 1] =

( n + 1)



3(n + 1) + 1



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

( n + 1)(3n + 4)

+ n + 6n + 4 = 3n

+ 7n + 4

Is this a good argument? While there are many issues,

the work isn’t without merit: the required

calculation is present (left side of 1

line = left side of 2

). While helpful as scratch work, a substan-

tial re-write is needed to make this convincing to a reader.

We ﬁnish this section with a trickier example of this thinking at work.

Example 5.5. An L-shaped tromino is an arrangement of three squares in an “L” shape. We claim:

If any single square is removed from a 2

×2

square gird, then

the remaining grid may be tiled by L-shaped trominos.

The claim has the form ∀n ∈ N, P(n), but note that P(n) is itself universal.

The picture shows one of the sixteen possible examples when n = 2. To get

an idea of how to structure the induction step, think how you might use

2 × 2 grids to analyse a 4 × 4 grid: the picture shows how!

• Whatever square we remove, one quarter of the 2

×2

grid is a 2

×2

grid with one square removed: this is tilable (the blue tromino).

• Place a single tromino (orange in the picture) so that one of its squares lies in each remaining

quadrant. What’s left of each quadrant is a 2

×2

grid with one missing square: again tilable.

This scratch work is really an argument P(1) =⇒ P(2)! It remains only to formalize this intuition

into a general proof. We proceed by induction on n.

Base case (n = 1): If a single square is removed from a 2 × 2 grid, the three remaining squares form

single L-shaped tromino.

Induction step: Fix n ∈ N and assume that after removing any square from any 2

× 2

grid, the

remainder is tilable. Now take any 2

n+1

×2

n+1

grid and remove a square.

• By the induction hypothesis, the 2

×2

quadrant containing the removed square is tilable.

• Place a single tromino in the center so that one of its squares lies in each remaining quad-

rant. What’s left of each quadrant is a 2

×2

grid with one missing square, each of which

is tilable by the induction hypothesis.

By induction, we conclude that every 2

×2

grid is tilable by trominos after any square is removed.

• There is no set-up, base case or conclusion, and the word induction is missing. The argument also needs some English.

• P(n) has not been deﬁned. If you don’t deﬁne it, don’t write it.

• P(n + 1) is a proposition: it cannot equal a number! Replacing “P(n + 1) =” with “P(n + 1) ⇐⇒” would correct this.

• There are no conditional connectives to indicate the logical ﬂow. Moreover, read top to bottom, the argument is

essentially P(n) ∧ P(n + 1) =⇒ T, rather than the correct induction step P(n) =⇒ P(n + 1).

Exercises 5.1. A reading quiz and several questions with linked video solutions can be found online.

1. Suppose you move one disk on the Tower of Hanoi per second.

(a) One of the oldest versions of the problem has monks transferring a tower of 64 disks.

Roughly how many years would this take?

(b) In a realistic human lifetime, how large a tower could be moved?

2. Imagine you cut a large large piece of paper in half and stack the two pieces on top of each

other. You then repeat the process, cutting all sheets in half and making a single taller stack.

Cut and stack

If a single sheet of paper has thickness 0.1 mm, how many times would you have to repeat the

cut-and-stack process until the stack of paper reached to the sun? (≈ 150 million kilometers).

Prove that you are correct.

3. A room contains n people. Everybody wants to shake everyone else’s hand (but not their own).

(a) Suppose n people require h

handshakes. If person n + 1 enters the room, how many

additional handshakes are required? Obtain a recurrence relation for h

n+1

in terms of h

(b) Hypothesize a general formula for h

, and prove it by induction.

4. (a) In Example 5.4.2, what is the proposition P(n + 1)?

(b) In the induction step of Example 5.4.2, explain why it would be incorrect to write

P(n + 1) − P(n) = (n + 1)



( n + 2)(2n + 3) −n(2n + 1)



= (n + 1)(2n

+ 7n + 6 −2n

−n)

= 6(n + 1)

∑

k=1

n(n + 1)(2n + 1).

5. Prove by induction that for each natural number n, we have

∑

k=0

= 2

n+1

−1.

6. Consider the statement: If n is a natural number, then

∑

k=1

( n + 1)

(a) What, explicitly, is

∑

k=1

(b) What would be meant by the expression

∑

k=1

, and why is it different to

∑

k=1

7. (a) Prove by induction that ∀n ∈ N we have 3 | (2

+ 2

n+1

(b) Give a direct proof that 3 | (2

+ 2

n+1

) for all integers n ≥ 1.

8. Prove by induction that for every n ∈ N we have n ≡ 5 or n ≡ 6 or n ≡ 7 (mod 3).

9. Prove by induction that, for all n ∈ N,

1 · 2 + 2 · 3 + 3 · 4 + ··· + n(n + 1) =

n(n + 1)(n + 2)

10. (a) Show, by induction, that for all n ∈ N, the number 4 divides the integer 11

−7

(b) More generally, prove by induction that (a −b) | (a

−b

) for any a, b, n ∈ N.

11. (a) Find a formula for the sum of the ﬁrst n odd natural numbers. Prove your assertion.

(b) Use Theorem 5.2 to give an alternative direct proof of your formula.

12. Find the error in the following “proof” of the statement, “All cats have the same color fur.”

Proof. Let P(n) be the proposition, “Any set of n cats have the same color fur.” We prove by

induction on n.

Base case (n = 1): Any cat has the same color fur as itself.

Induction step: Fix n ∈ N and assume P(n). Take any set S = {C

, C

, . . . , C

n+1

} of n + 1 cats.

The set S \{C

} has n cats; by the induction hypothesis all have the same color fur. Again

by the induction hypothesis, all cats in S \{C

} have the same color fur. Combining these

observations, we see that all cats in S have the same color fur. Since S was arbitrary, we

see that P(n + 1) holds.

By induction, P(n) is true for all n ∈ N, which establishes the claim.

13. Use induction, the product rule, and the fact that

x = 1 to prove the power law from calculus:

∀n ∈ N,

= nx

n−1

14. A (real) polynomial of degree n is a function p(x) = a

+ a

n−1

+ ··· + a

x + a

, whose

coefﬁcients a

are real numbers and where a

= 0.

(a) Prove: for all n ∈ N,

= p

(x)e

where p

(x) is some polynomial of degree n.

(b) (Hard) Let p(x) be a polynomial of degree n ≥ 1. Show p has at most n roots.

(Hint: induct on the degree n)

15. Consider the following scratch work. Determine what result is being proved, then convert the

scratch work into a formal proof of that result.

(1 + x)

n+1

= (1 + x)

(1 + x) ≥ (1 + nx)(1 + x)

= 1 + x + nx + nx

= 1 + (n + 1)x + nx

≥ 1 + (n + 1)x

5.2 Well-ordering and the Principle of Mathematical Induction

In this section we think more carefully about the logic behind induction, and tie it to a fundamental

property of the natural numbers.

Deﬁnition 5.6. A non-empty set of real numbers A is well-ordered if every non-empty subset of A

contains a minimum element.

To test if a set A is well-ordered, we need to check all of its non-empty subsets. The deﬁnition could

be written as equivalently as follows (in the second line we expand what is meant by a minimum):

• If B ⊆ A and B = ∅, then min B exists.

•



(B ⊆ A) ∧ (B = ∅ )



=⇒



∃b ∈ B, ∀x ∈ B, b ≤ x



To show that A is not well-ordered, we need only exhibit a non-empty subset B with no minimum.

Examples 5.7. 1. A = {4, −7, π, 19, ln 2} is a well-ordered set. There are 31(!) non-empty subsets of

A, each of which has a minimum element.

Can you justify this fact without listing the subsets? It might be easier to think about why any

ﬁnite set A = {a

, . . . , a

} ⊆ R is well-ordered. . .

2. The interval A = [3, 10) is not well-ordered. Indeed B = (3, 4) is a non-empty subset with no

minimal element. While you should believe this, let’s prove it anyway!

We need to prove that ∀b ∈ B, ∃x ∈ B with x < b. Given any b ∈ (3, 4), observe that x :=

b+3

satisﬁes

3 < x < b < 4 from which x ∈ B and x < b

You could also argue by contradiction (if b ∈ B is minimal, then. . . ).

3. The integers Z are not well-ordered. For instance, Z is a non-empty subset of itself and there is

no minimal integer.

You might suspect (wrongly!) that every well-ordered set is ﬁnite. That the natural numbers form a

well-ordered inﬁnite set is, for us, an axiom,

a foundational claim forming part of our basic concep-

tion of the natural numbers.

Axiom 5.8 (Well-ordering Principle). N is well-ordered.

Also known as the least natural number principle, well-ordering is applied widely throughout mathe-

matics. In fact we’ve already done so in this text! Consider the set of positive remainders generated

by the Euclidean algorithm (Theorem 3.16) when applied to natural numbers a > b:

{. . . , r

, r

, b, a} ⊆ N

Well-ordering guarantees that this set has a minimal element r

(which turns out to be gcd(a, b)); this

is essentially the argument for Exercise 3.2.8(a).

There are many ways to deﬁne the natural numbers. Typically well-ordering is either an axiom (essentially part of the

deﬁnition) or a theorem. Compare with Exercise 11 for an alternative approach.

Armed with the well-ordering principle, we can justify the method of proof by induction.

Theorem 5.9 (Principle of Mathematical Induction). For each n ∈ N, let P(n) be a proposition.

Additionally make the two standard assumptions:

Base case: P(1) is true

Induction step: ∀n ∈ N, P(n) =⇒ P(n + 1)

Then P(n) is true for all n ∈ N.

Before attempting a proof, consider how the theorem could be written as a pure implication:

P(1) ∧



∀n ∈ N, P(n) =⇒ P( n + 1)



=⇒



∀n ∈ N, P(n)



This helps us select a proof strategy: a direct approach seems hard since the conclusion is universal; a

contrapositive approach requires an ugly negation of the hypothesis; a proof by contradiction seems

most sensible since negation of the conclusion is straightforward.

Proof. We argue by contradiction. Assume the base case, the induction step, and that ∃n ∈ N for

which P(n) is false. The set of natural numbers

S :=



k ∈ N : P(k) is false



is therefore non-empty (n ∈ S). Well-ordering guarantees that s := min S exists. Observe:

• s ∈ S =⇒ P(s) is false.

• The base case tells us that s = 1. Thus s ≥ 2 and s −1 ∈ N.

• s −1 < min S =⇒ P(s −1) is true.

• The induction step (P(s −1) =⇒ P(s)) tells us that P( s) is true.

Contradiction: P(s) cannot be both true and false!

Now we have the proof, it is straightforward to extend the principle of induction. For any integer m

(positive, negative or zero), the set

≥m

= {n ∈ Z : n ≥ m} = {m, m + 1, m + 2, m + 3, . . .}

is also well-ordered. By changing the base case to P(m) and replacing N with Z

≥m

, we immediately

obtain the proof of a more general principle of induction.

Corollary 5.10 (Induction with base case m). Fix an integer m. For each integer n ≥ m, let P(n) be

a proposition. Suppose:

Base case: P(m) is true

Induction step: ∀n ∈ Z

≥m

, P(n) =⇒ P(n + 1)

Then P(n) is true for all n ∈ Z

≥m

The intuitive concept is exactly as before, just with a different base case!

P(m) =⇒ P(m + 1) =⇒ P(m + 2) =⇒ P(m + 3) =⇒ ···

Examples 5.11. 1. For all integers n ≥ 2, we prove that

∑

k=2

k(k −1)

= 1 −

(∗)

Base case (n = 2): When n = 2, (∗) reads

∑

i=2

i(i−1)

= 1 −

Induction step: Assume that (∗) is true for some ﬁxed n ≥ 2. Then

n+1

∑

i=2

k(k −1)

∑

i=2

k(k −1)

( n + 1)n

= 1 −

n(n + 1)

(induction hypothesis)

= 1 −

( n + 1) − 1

n(n + 1)

= 1 −

n + 1

which is precisely (∗) when n is replaced by n + 1.

By induction, (∗) holds for all integers n ≥ 2.

2. For all integers n ≥ 4, we claim that 3

> n

We really do need the given base case: when n = 3, the claim 3

> 3

is false! As is often the

case, it helps to do some scratch work. The induction hypothesis allows us to see that

n+1

= 3 ·3

> 3n

The proof of the induction step thus hinges on being able to show that 3n

≥ (n + 1)

. There

are many ways to convince yourself of this, for instance

≥ (n + 1)

⇐⇒ 3 ≥



n + 1





1 +



(†)

The right side decreases as n increases; since n ≥ 4, the right side is at most





125

< 2,

whence (†) holds for all n ≥ 4.

We now prove the original claim by induction.

Base case (n = 4): Observe that 3

= 81 > 64 = 4

Induction step: Fix n ∈ Z

≥4

and suppose that 3

> n

. By (†), we see that

n+1

= 3 ·3

> 3n

≥ (n + 1)

By induction, we conclude that 3

> n

whenever n ∈ Z

≥4

You might have encountered this example in calculus as a telescoping series:

∑

k=2

k(k −1)

2 · 1

3 · 2

+ ···+

n(n −1)



1 −





−



+ ···+



n −1

−



= 1 −

Taking the limit as n → ∞ results in

∞

∑

k=2

k(k−1)

= 1. Induction does this without the ambiguous ellipses (···).

Example 5.12. We prove an extended version of de Morgan’s law for sets (Theorem 4.14(a)): for any

collection of sets A

, . . . , A

where n ≥ 2, we have

∩···∩ A

)

= A

∪···∪ A

(‡)

Base case (n = 2): A

∩ A

= A

∪ A

is precisely the standard de Morgan identity.

Induction step: Fix n ∈ N

≥2

and suppose (‡) holds for all collections of n sets. Given a collection of

n + 1 sets, we see that

∩···∩ A

∩ A

n+1

)



∩···∩ A

) ∩ A

n+1



= (A

∩···∩ A

)

∪ A

n+1

(de Morgan again!)

= A

∪···∪ A

∪ A

n+1

(induction hypothesis)

By induction, the claim (‡) holds for any collection of n sets.

We could have approached the argument as a standard induction with base case n = 1. Instead

we deliberately chose n = 2, both to avoid confusion (the n = 1 case A

= A

isn’t helpful or

interesting) and to highlight the importance of de Morgan’s law for two sets to the entire argument.

Proof by Minimal Counter-example

Sometimes authors present induction arguments as contradiction proofs in the style of Theorem 5.9,

with s = min S being known as the minimal counter-example. Here are two variations on this idea; the

ﬁrst is a straight translation of an induction where the base case and induction step are clear.

Examples 5.13. 1. We prove: for all n ∈ N

∑

k=0

= 2

n+1

−1.

Suppose to the contrary and let s ≥ 0 be the minimal counter-example.

• Since

∑

k=0

= 2

= 1 = 2

0+1

−1, we see that s = 0. (base case)

• Since the claim holds for s −1, we see that

∑

k=0

= 2

s−1

∑

k=0

= 2

+ 2

−1 = 2

s+1

−1 (induction step)

This contradicts the fact that s is a counter-example!

2. We re-prove Theorem 2.35:

√

2 /∈ Q. Suppose

√

2 is rational and consider the set

S =



x ∈ N : ∃y ∈ N with x

= 2y



If S is non-empty, let s = min S, and let t ∈ N be such that s

= 2t

. Plainly t < s. Since s

even, s is also even, and we can write s = 2k. But then

= 2t

=⇒ t

= 2k

=⇒ t ∈ S

But this contradicts the minimality of s.

This approach is often used in number theory in the guise of Fermat’s method of inﬁnite descent.

Aside: Well-ordering more generally Deﬁnition 5.6 is a weak version of a much deeper concept.

Informally, to well-order a set means to list its elements in some order so that every non-empty subset

has an initial element with respect to that order.

For instance, the set of negative integers Z

−

= {. . . , −4, −3, −2, −1} is not well-ordered with respect

to the standard ordering of the integers, but is well-ordered with respect to the reverse ordering

−

= {−1, −2, −3, −4, . . .}

The principle of mathematical induction is easily modiﬁed to accommodate theorems of the form

∀n ∈ Z

−

, P(n): the base case is P(−1) and the induction step justiﬁes the chain

P(−1) =⇒ P(−2) =⇒ P(−3) =⇒ ···

All the inﬁnite well-ordered sets we’ve thus far seen have “looked like” the natural numbers, how-

ever more esoteric examples exist. For instance, the following well-ordered set looks like two copies

of the natural numbers, one following the other:

A =



, . . . , 1,

, . . .





1 −

: n ∈ N



∪



2 −

: n ∈ N



Every non-empty subset of A really does have a minimum! It is possible to modify induction to

apply to propositions indexed by well-ordered sets like this, though an extra step is required to deal

with limit elements (like 1 ∈ A) with no immediate predecessor. If your further studies include set

theory, you’ll likely spend much time considering well-orders and their associated ordinals.

Exercises 5.2. A reading quiz and several questions with linked video solutions can be found online.

1. Prove that the interval (−2, 5] has no minimum element.

2. Prove that every ﬁnite set of real numbers is well-ordered.

3. (a) Suppose that n ≥ 3. Prove that



n+1



< 2.

(b) Hence or otherwise, prove that n

< 2

for all natural numbers n ≥ 5.

4. Consider the following result. For every natural number n ≥ 2,



1 −



1 −



1 −



···



1 −



n + 1

(a) If the statement is written in the form ∀n ∈ N

≥2

, P(n), what is the proposition P(n)?

(b) Prove the result by induction.

5. Prove the geometric series formula: if r = 1 and n ∈ N

, then

∑

k=0

1−r

n+1

1−r

6. For all integers n ≥ 3, prove that

∑

k=3

k(k−2)

−

2n−1

2n(n−1)

7. Prove: for any n ∈ N,

∑

i=1

< 2

(Hint: prove the stronger fact that

∑

i=1

< 2 −

for all n ≥ 2)

8. The set A

= {1, 2, 3} satisﬁes the property that the sum of its elements (1 + 2 + 3 = 6) is

divisible by every element of A

(a) Use induction to prove that for any n ≥ 3, there is a set A

of n natural numbers such that

the sum of its elements is divisible by every element of A

(b) Prove by contradiction that no set of two natural numbers satisﬁes this property.

9. Suppose that x

+ 4y

= 3z

has a solution (x, y, z) where all three are positive integers.

(a) By considering remainders modulo 3, prove that 3 | z. Thus create a new solution (X, Y, Z)

in positive integers, where Z < z.

(b) Use the method of minimal counter-example to prove that x

+ 4y

= 3z

has no solutions

where x, y, z ∈ N.

10. We use the fact that N

is well-ordered to prove the division algorithm (Theorem 3.3).

If m ∈ Z and n ∈ N, then ∃ unique q, r ∈ Z such that m = qn + r and 0 ≤ r < n.

Given m, n, deﬁne

S = N

∩



m + nZ





k ∈ N

: k = m −qn for some q ∈ Z



(a) (Existence) Show that S is a non-empty subset of N

. By well-ordering, deﬁne r := min S.

Prove that 0 ≤ r < n.

(b) (Uniqueness) Suppose two pairs of integers (q

, r

) and (q

, r

) satisfy m = q

n + r

and

0 ≤ r

, r

< n. Prove that r

= r

11. (Hard) We consider a version of Peano’s axioms for the natural numbers.

i. (Initial element) 1 ∈ N

ii. (Successor function) f (n) = n + 1 is a function f : N → N

iii. (No predecessor of the initial element) 1 ∈ range( f )

iv. (Unique predecessor/order) f is injective: m + 1 = n + 1 =⇒ m = n

v. (Induction) Any subset A ⊆ N with the following properties equals N:

1 ∈ A and ∀a ∈ A, a + 1 ∈ A

(a) Replace N with Z in each axiom. Which are true and which false?

(b) Let T =



( m, n) : m, n ∈ N



be the set of all ordered pairs of natural numbers.

i. Let f : T → T be the function f (m, n) = (m + 1, n). Letting the pair (1, 1) play the role

of 1, and f the successor function, decide which of Peano’s axioms are satisﬁed by T.

ii. Repeat the question for the same initial element and

f : T → T : (m, n) 7→

(

( m −1, n + 1) if m ≥ 2

( m + n, 1) if m = 1

(Hint: let A = {1} ∪range( f ) in the induction axiom)

(d) Prove that N, as deﬁned by Peano, is well-ordered (with respect to x < x + 1, etc.).

5.3 Strong Induction

The principle of mathematical induction (Theorem 5.9) is often known as weak induction. Strong

induction differs primarily in that the induction step can assume more than one previous proposition.

Theorem 5.14 (Principle of Strong Induction). Let l ≥ m be ﬁxed integers and suppose P(n) is a

proposition, one for each n ∈ Z

≥m

. Suppose:

Base case(s): P(m), P(m + 1), . . . , P(l) are true

Induction step: ∀n ≥ l,



P(m) ∧ P(m + 1) ∧···∧ P(n)



=⇒ P(n + 1)

Then P(n) is true for all n ≥ m.

Exercise 6 provides a proof by showing that strong and weak induction are equivalent. We instead

concentrate on a few examples. The additional difﬁculty of strong induction comes from determining

how many base cases are required and in phrasing the induction hypothesis: in practice one rarely

needs to employ all the propositions P(m), . . . , P(n).

Example 5.15 (Fibonacci numbers). This famous sequence ( f

)

∞

n=1

= (1, 1, 2, 3, 5, 8, 13, 21, . . .) is

deﬁned by the 2

-order recurrence relation

(

n+1

= f

+ f

n−1

if n ≥ 2

= f

= 1

While the Fibonacci sequence seems to be increasing, it also appears to be less than doubling at each

step, suggesting the claim

∀n ∈ N, f

< 2

We prove this using strong induction. Two base cases are suggested since the sequence is deﬁned by

two initial conditions ( f

= f

= 1): in the language of the Theorem, m = 1 and l = 2. Moreover, the

fact that each term from f

onwards is the sum of its two predecessors suggests that the induction step

requires only the explicit use of two propositions.

Proof. Base cases (n = 1, 2): f

= 1 < 2

and f

= 1 < 2

Induction step: Fix n ≥ 2 and suppose

that f

n−1

< 2

n−1

and f

< 2

. Then

n+1

= f

+ f

n−1

< 2

+ 2

n−1

< 2

+ 2

= 2

n+1

By induction, f

< 2

for all n ∈ N.

The Fibonacci numbers satisfy many identities which can often be established by induction (see, for

instance, Exercises 3 & 4).

To follow Theorem 5.14 precisely, we should assume that f

< 2

for all k ≤ n. Do so if you like, though our phrasing

is more typical. Since we only make explicit use of two cases in the induction step, it is clearer to state these concretely

rather than introducing the new variable k.

It is instructive to consider why we really needed strong induction to prove our Fibonacci example.

Here are two broken attempts to prove the claim by weak induction.

Broken Proof A. Base case (n = 1): f

= 1 < 2

Induction step: Fix n ≥ 2 and suppose that f

< 2

. Then

n+1

= f

+ f

n−1

< 2

+ ????

What is the problem? The induction hypothesis assumes f

< 2

, but nothing about f

n−1

: we are

stuck! Let’s correct this ﬂaw by making the induction hypothesis as in the correct proof.

Broken Proof B. Base case (n = 1): f

= 1 < 2

Induction step: Fix n ≥ 2 and suppose that f

n−1

< 2

n−1

and f

< 2

. Then

n+1

= f

+ f

n−1

< 2

+ 2

n−1

< 2

+ 2

= 2

n+1

By induction, f

< 2

for all n ∈ N.

Where is the problem now? Consider the ﬁrst instance, n = 2, in which the induction step is invoked:

= f

+ f

< 2

+ 2

We haven’t proved enough base cases to get us started: the single base case establishes f

< 2

, but

not f

< 2

. The induction step correctly establishes the chain of implications

P(1) ∧ P(2) =⇒ P(3), P(3) ∧ P(4) =⇒ P(5), P(4) ∧ P(5) =⇒ P(6), . . .

but the process only gets started if we prove both base cases P(1) and P( 2).

The moral here is to try the induction step as scratch work. Your attempt should tell you if you need

strong induction and how many base cases are required.

Example 5.16. A sequence of integers (a

)

∞

n=0

is deﬁned by

(

n+1

= 5a

−6a

n−1

if n ≥ 1

= 0, a

= 1

We prove by induction that a

= 3

−2

for all n ∈ N

Proof. Base cases (n = 0, 1): a

= 0 = 3

−2

and a

= 1 = 3

−2

Induction step: Fix n ≥ 1 and suppose that a

= 3

−2

for all k ≤ n. Then

n+1

= 5a

−6a

n−1

= 5(3

−2

) −6(3

n−1

−2

n−1

)

= (15 −6) 3

n−1

+ (10 −6)2

n−1

= 3

n+1

−2

n+1

By induction, a

= 3

−2

for all n ∈ N

The induction step requires n ≥ 2: since f

n−1

= f

doesn’t exist, f

n+1

= f

+ f

n−1

is meaningless when n = 1.

For another sequential induction example in the same vein, see Exercise 5.3 where three base cases

are required and the induction step explicitly uses three propositions.

To see strong induction in all its glory, with the induction step making use of all previous proposi-

tions, we prove the existence part of the Fundamental Theorem of Arithmetic, which states that all

integers ≥ 2 can be (uniquely) expressed as a product of primes: e.g., 3564 = 2

×3

×11.

Theorem 5.17. Every integer n ≥ 2 is either prime or a product of primes.

This provides the missing piece in our discussion of Euclid’s Theorem (2.39) on the existence of

inﬁnitely many primes. First recall Deﬁnition 2.38: p ∈ N

≥2

is prime if and only if its only positive

divisors are itself and 1. A non-prime q ∈ N

≥2

is said to be composite: ∃a, b ∈ N

≥2

such that q = ab.

Proof. We prove by induction.

Base case (n = 2): The only positive divisors of 2 are itself and 1, hence 2 is prime.

Induction step: Fix n ≥ 2 and suppose that every natural number k satisfying 2 ≤ k ≤ n is either prime

or a product of primes. There are two possibilities:

• n + 1 is prime. Certainly n + 1 is divisible by a prime: itself!

• n + 1 is composite. Then n + 1 = ab for some natural numbers a, b ≥ 2. Plainly a, b ≤ n.

By the induction hypothesis, both a, b are prime or products of primes. Therefore n + 1 is

also the product of primes.

By induction we see that all natural numbers n ≥ 2 are either prime, or a product of primes.

Think carefully about why only one base case is required!

Exercises 5.3. A reading quiz and several questions with linked video solutions can be found online.

1. Deﬁne sequence (b

)

∞

n=1

as follows:

(

n+1

= 2b

+ b

n−1

if n ≥ 2

= 3, b

= 6

Prove by induction that b

is divisible by 3 for all n ∈ N.

2. Deﬁne a sequence (c

)

∞

n=0

(

n+1

−

225

n−2

if n ≥ 2

= 0, c

= 2, c

= 16

Prove by induction (use three base cases!) that c

= 5

−3

for all n ∈ N

3. Let f

be the n

Fibonacci number (Example 5.15). Prove the following by induction ∀n ∈ N:

(a)

∑

k=1

= f

n+1

(b) f

≥





n−2

(Hints: Weak induction is good enough for (a); why?)

4. Extending Exercise 3(b), prove Binet’s formula for the n

Fibonacci number:

√



−



where ϕ =

(1 +

√

5) and

ϕ =

(1 −

√

(ϕ is the famous golden ratio: ϕ,

ϕ are the solutions to the quadratic equation x

= x + 1)

5. Prove by induction that every n ∈ N can be written in the form

n = 2

+ 2

+ ···+ 2

ℓ

for some ℓ ∈ N and distinct integers r

, r

, . . . r

ℓ

≥ 0.

(Hints: try proving in the style of Theorem 5.17; consider the cases when n + 1 is even/odd separately)

6. Prove the principle of strong induction (Theorem 5.14) by applying weak induction to a new

family of propositions Q(n) via:

Q(n) ⇐⇒ P(m) ∧ P(m + 1) ∧ ··· ∧ P(n)

7. Consider the proof of Theorem 5.17.

(a) If the Theorem is written in the form ∀n ∈ N

≥2

, P(n), what is the proposition P(n)?

(b) Explicitly carry out the induction step for the three situations n + 1 = 9, n + 1 = 106 and

n + 1 = 45. How many different ways can you perform the calculation for n + 1 = 45?

ing 2 ≤ k ≤

n+1

are prime or products of primes.

8. In this question we use the alternative deﬁnition of prime (Exercise 3.2.13).

An integer p ≥ 2 is prime if and only if ∀a, b ∈ N, p | ab =⇒ p | a or p | b.

Let p be prime, let n ∈ N, and let a

, . . . , a

be natural numbers such that p | a

···a

. Prove

by induction that,

p | a

for some i ∈ {1, 2, . . . , n}

(Hint: n = 1 isn’t really part of the induction, but you can treat it as a base case)

9. The Fundamental Theorem of Arithmetic states that every integer n ≥ 2 can be written as a product

of prime factors in a unique way (up to reordering of the prime factors). In other words,

i. n = p

··· p

for some primes p

, p

, . . . , p

, and,

ii. If n = q

···q

ℓ

for primes q

, q

, . . . , q

ℓ

, then k = ℓ and p

= q

after possibly reordering

the prime factors.

Part i. is Theorem 5.17. Using Exercise 8, or otherwise, supply a proof of part ii.

Strictly, this is deﬁnition of prime, whereas Deﬁnition 2.38 deﬁnes a subtly different concept: irreducibility. Within the

integers, Exercise 3.2.13 says that these concepts are synonymous.

6 Set Theory, Part II

In this chapter we return to set theory and consider several more-advanced constructions.

6.1 Cartesian Products

You have been working with Cartesian products for years, referring to a point in the plane R

by its

Cartesian co-ordinates (x, y), an ordered pair where each co-ordinate (x, y) is a member of the set R. The

same approach can be used for any sets.

Deﬁnition 6.1. The Cartesian product of sets A and B is the set of ordered pairs

A × B =



(a, b) : a ∈ A and b ∈ B



Otherwise said: (a, b) ∈ A × B ⇐⇒ a ∈ A and b ∈ B.

Examples 6.2. 1. The Cartesian product of the real line R with itself is the usual xy-plane.

As you’ve seen in other classes, rather than writing R × R which

is unwieldy, we denote this set



(x, y) : x, y ∈ R



More generally, the set of n-tuples of real numbers is



, x

, . . . , x

) : x

, x

, . . . , x

∈ R



= R ×R ×···× R

| {z }

n times

−2

−2 2 4

(2, 3)

2. If A = {1, 2, 3} and B = {α, β}, then the Cartesian product of A and B is

A × B =



(1, α), (1, β ), (2, α), (2, β), (3, α), (3, β)



The order of terms in an ordered pair really matters: the Cartesian product with the roles re-

versed is

B × A =



( α, 1), (β, 1), (α, 2), (β, 2), ( α, 3), (β, 3)



While A × B = B × A, note that both Cartesian products have six (= 3 ·2 = 2 · 3) elements.

3. The menu in a restaurant might be summarized set-theoretically:

Mains = {ﬁsh, steak, eggplant, pasta}, Sides = {asparagus, salad, potatoes}

The Cartesian product Mains ×Sides is the set of all possible meals consisting of one main and

one side. It should be obvious that there are 4 ×3 = 12 possible meal choices.

The last two examples illustrate one of the simplest properties of Cartesian products, in a result which

indeed justiﬁes the very use of the word product!

Strictly this should be deﬁned inductively, e.g., R

:= R

×R =



(x, y), z



: x, y, z ∈ R

, but this is very tedious!

Theorem 6.3. If A and B are ﬁnite sets, then

A × B

Proof. Label the elements of each set and list the elements of A × B lexicographically. If

= m and

= n, then

A × B =



, b

), (a

, b

), (a

, b

), ··· (a

, b

), (a

, b

), (a

, b

), ··· (a

, b

), (a

, b

), (a

, b

), ··· (a

, b

)



Every element of A ×B is listed exactly once. There are m rows and n columns, so

A × B

= mn.

Set Identities These may be established as we’ve done previously (Section 4): convert everything

into propositions regarding elements of sets and use basic logic. If you’re feeling more conﬁdent, you

might also be able to invoke previously established rules of set algebra.

Examples 6.4. 1. Consider the complement of a Cartesian product A × B. If you had to guess an

expression for (A ×B)

, you might mistakenly think it is A

×B

. Let us think more carefully:

(x, y) ∈ (A × B)

⇐⇒ ¬((x, y) ∈ A × B) (deﬁnition of complement)

⇐⇒ ¬(x ∈ A and y ∈ B) (deﬁnition of A × B)

⇐⇒ x ∈ A or y ∈ B (de Morgan (logic))

By contrast, (x, y) ∈ A

× B

⇐⇒ x /∈ A and x /∈ B, so (A ×B)

= A

× B

. Indeed the

complement of a Cartesian product is not a Cartesian product! For more on this, see Exercise 5.

2. Let A, B, C, D be any sets. We prove that (A × B) ∪ ( C × D) ⊆ (A ∪C) ×(B ∪ D).

(x, y) ∈ (A × B) ∪(C × D) =⇒ (x, y) ∈ A × B or (x, y) ∈ C ×D

=⇒ (x ∈ A and y ∈ B) or (x ∈ C and y ∈ D)

=⇒ (x ∈ A or x ∈ C) and ( y ∈ B or y ∈ D)

=⇒ x ∈ A ∪C and y ∈ B ∪ D

=⇒ (x, y) ∈ (A ∪C) × (B ∪ D)

All implications except the red arrow are in fact biconditional.

To convince yourself of its truth, ﬁrst consider what must be

true about x, then do the same for y. The picture provides a

visualization where A, B, C, D are intervals of real numbers:

• (A ×B) ∪ (C ×D) is the solid shaded region.

• (A ∪C) × (B ∪D) is the larger dashed square.

If x ∈ C \ A and y ∈ B \ D, then (x, y) ∈ (A ∪C) ×(B ∪D) but

(x, y) ∈ (A × B) ∪ (C × D), so we do not, in general, expect

these sets to be equal.

A × B

C × D

Exercises 6.1. A reading quiz and several questions with linked video solutions can be found online.

1. (a) Suppose that A = {1, 2} and B = {3, 4, 5}. State the set A × B in roster notation.

(b) Sketch both A × B and B × A using dots in R

. What do you observe about your pictures?

A × B ×C =



(a, b, c) : a ∈ A, b ∈ B, c ∈ C



If C = {6, 7} and A, B are as above, state the set A ×B ×C in roster notation.

2. Rewrite the condition (x, y) ∈ (A

∪ B) × (C \ D) in terms of (some of) the propositions

x ∈ A, x ∈ A, x ∈ B, x ∈ B, y ∈ C, y ∈ C, y ∈ D, y ∈ D

3. Consider the intervals A = [2, 5] and B = (0, 4) of real numbers.

(a) Express the set (A \ B)

in interval notation, as a disjoint union of intervals.

(b) Draw a picture of the set (A \ B)

×(B \ A).

(Be careful: In this problem B = (0, 4) is an interval (a subset of R), not a point in R

4. A straight line in R

is a subset of the form

a,b,c



(x, y) : ax + by = c



, for some constants a, b, c, with a, b not both zero

(a) Draw the set A

1,2,3

. Is it a Cartesian product?

(b) Which straight line subsets in the plane R

are Cartesian products? Otherwise said, ﬁnd a

condition on the constants a, b, c for which the set A

a,b,c

is a Cartesian product.

5. Draw a picture, similar to that in Example 6.4.2, which illustrates the fact that

(A × B)

= (A

× B

) ∪(A

× B) ∪ (A × B

)

Now give a rigorous proof of the claim.

6. Let A = [1, 3], B = [2, 4] and C = [2, 3]. Prove or disprove:

(A × B) ∩(B × A) = C ×C

(Hint: Draw the sets A × B, B × A and C ×C in the Cartesian plane)

7. Prove that A ∩B = ∅ ⇐⇒ (A × B) ∩(B × A) = ∅.

(Hint: try the previous question ﬁrst)

8. Let A, B, C be sets. Prove:

(a) A ×(B ∪C) = (A ×B) ∪ (A ×C)

(b) A ×(B ∩C) = (A ×B) ∩ (A ×C)

9. (a) Give an explicit example of sets A, B, C, D such that

(A × B) ∪(C × D) = (A ∪C) × (B ∪ D)

(b) For any sets A, B, C, D, prove that

(A ∪C) × (B ∪ D) = (A × B) ∪(A × D) ∪ ( C × B) ∪ (C × D)

10. Prove by induction. For all n ∈ N, if A

, . . . , A

are ﬁnite sets, then

×···× A

···

11. (a) Suppose

= 3, and

= 4. What are the minimum and maximum values for the

cardinalities

(A × B) ∩(B × A)

and

(A × B) ∪(B × A)

(b) (Hard) More generally, suppose

= m,

= n and

A ∩ B

= c. What can you say

about the cardinalities

(A × B) ∩(B × A)

and

(A × B) ∪(B × A)

12. Let A and B be nonempty sets. Deﬁne functions π

: A × B → A and π

: A × B → B by

(a, b) = a and π

(a, b) = b respectively (these are called projection maps).

(a) If A = B = R and X = [1, 3], Y = (2, 4], then X × Y ⊆ A × B. Compute the images

(X ×Y) and π

(X ×Y).

(b) Let Z be any set and suppose there are functions ρ

: Z → A and ρ

: Z → B. Show that

there is a unique function h : Z → A × B such that ρ

= π

◦ h and ρ

= π

◦ h.

13. Let E ⊆ N ×N be the smallest subset satisfying the following conditions:

• Base case: ( 1, 1) ∈ E

• Generating Rule I: If (a, b) ∈ E then (a, a + b) ∈ E

• Generating Rule II: If (a, b) ∈ E then (b, a) ∈ E

(a) Show in detail that ( 4, 3) ∈ E.

(b) Show by induction that for every n ∈ N, (1, n) ∈ E.



(a, b) ∈ N × N : gcd(a, b) = 1



(Hint: what do the generating rules have to do with the Euclidean algorithm?)

14. (Hard) A strict set-theoretic deﬁnition requires that ordered pair (a, b) be deﬁned as a set for

instance (a, b) :=



a, {a, b}



. We prove that (a, b) = (c, d) ⇐⇒ a = c and b = d.

(a) The regularity axiom of set theory says there is no set a for which a ∈ a. Use this to prove

that the cardinality of



a, {a, b}



is two.

(b) Prove that (a, b) = (c, d) =⇒



a = c and b = d





a = {c, d} and c = {a, b}



regularity also says that this is illegal. Conclude that (a, b) = ( c, d) ⇐⇒ a = c and b = d.

6.2 Power Sets

Thus far we have used the operations of subset, complement, union, intersection and Cartesian prod-

uct to build new sets from old. There is essentially only one further method available.

Deﬁnition 6.5. Let A be a set. Its power set P(A) is the set of all subsets of A:

P(A) =



B : B ⊆ A



Otherwise said: B ∈ P(A) ⇐⇒ B ⊆ A.

Examples 6.6. 1. The set A = {1, 3, 7} has the following subsets:

0-element subsets: ∅

1-element subsets: {1}, {3}, {7}

2-element subsets: {1, 3}, {1, 7}, {3, 7}

3-element subsets: {1, 3, 7}

Gathering these together yields the power set:

P(A) =



∅, {1}, {3}, {7}, {1, 3}, {1, 7}, {3, 7}, {1, 3, 7}



The power set therefore has eight elements. Be absolutely certain you understand the difference

between ∈ and ⊆. Here are eight propositions; which are true and which false?

(a) 1 ∈ A (b) 1 ∈ P(A) (c) {1} ∈ A (d) {1} ∈ P(A)

(e) 1 ⊆ A (f) 1 ⊆ P(A) (g) {1} ⊆ A (h) {1} ⊆ P(A)

2. The set B =



{2}, 3



has precisely two elements, namely 1 and



{2}, 3



. As before, we

gather the subsets of B in a table:

0-element subsets: ∅

1-element subsets: {1},



{2}, 3



2-element subsets:



{2}, 3



Remember that to make a subset out of a single element you surround the element with braces:

1 ∈ B =⇒ {1} ⊆ B =⇒ {1} ∈ P(B)



{2}, 3



∈ B =⇒



{2}, 3



⊆ B =⇒



{2}, 3



∈ P(B)

Using different-sized braces is essential here! The power set of B has four elements:

P(B) =



∅, {1},



{2}, 3





{2}, 3





As a further exercise in making careful use of notation, we prove a simple theorem.

Only (a), (d), and (g) are true. Make sure you understand why!

Theorem 6.7. If A ⊆ B, then P(A) ⊆ P(B).

Proof. Suppose A ⊆ B and let C ∈ P(A). We must show that C ∈ P(B) .

C ∈ P(A) =⇒ C ⊆ A (deﬁnition of power set)

=⇒ C ⊆ B (since C ⊆ A and A ⊆ B)

=⇒ C ∈ P(B) (deﬁnition of power set)

We conclude that P(A) ⊆ P(B).

The converse to this is also true: P(A) ⊆ P(B) =⇒ A ⊆ B. Try proving it yourself.

Cardinality and Power Sets

Let’s investigate the relationship between the cardinality of a set and its power set. Consider a few

basic examples where we list all of the subsets, grouped by cardinality.

Set A 0-elements 1-element 2-elements 3-elements

P(A)

∅ ∅ 1

{a} ∅ {a} 1 + 1 = 2

{a, b} ∅ {a}, {b} {a, b} 1 + 2 + 1 = 4

{a, b, c} ∅ {a}, {b}, {c} {a, b}, {a, c}, {b, c} {a, b, c} 1 + 3 + 3 + 1 = 8

You’ve seen this pattern before: we are looking at the ﬁrst few lines of Pascal’s Triangle! It should be

no surprise that if

= 4, then

P(A)

= 1 + 4 + 6 + 4 + 1 = 16. The progression 1, 2, 4, 8, 16, . . . in

the ﬁnal column suggests a general result.

Theorem 6.8. Let A be a ﬁnite set. Then

P(A)

= 2

Conjuring a proof may seem daunting given how little we know about A; only its cardinality. By

introducing a variable n for the cardinality and rephrasing the theorem

∀n ∈ N

= n =⇒

P(A)

= 2

induction seems like a sensible approach. But what might the induction step look like? The basic idea

is to view a set with n + 1 elements as the disjoint union of a set with n elements and a single-element

set. It is instructive to see an example of the strategy before writing the proof.

Example 6.9. Let B = {1, 2, 3}. Delete 3 ∈ B to create a smaller set

A = B \ {3} = {1, 2}

In the table, the subsets of Y ⊆ B are split into two groups depending on whether

3 ∈ Y. Each subset Y ⊆ B either has the form X or X ∪{3} where X ⊆ A.

Plainly B has twice the number of subsets of A; two for each subset X ⊆ A.

Subsets of B

X X ∪{3}

∅ {3}

{1} {1, 3}

{2} {2, 3}

{1, 2} {1, 2, 3}

This method of pairing is exactly mirrored in the induction step of our formal proof.

Proof. We prove by induction on the cardinality of A. For each n ∈ N

, consider the proposition

Every set A with cardinality n has

P(A)

= 2

(∗)

Base case (n = 0): If n = 0, then A = ∅ (Lemma 4.10). But then P(A) = {∅} =⇒

P(A)

= 1 = 2

Induction step: Fix n ∈ N

and assume (∗) is true for this n. Let B be any set with n + 1 elements.

Choose an element b ∈ B and deﬁne A = B \{b}. The subsets of B may be separated into two

types:

1. Subsets X ⊆ B which do not contain b.

2. Subsets Y ⊆ B which contain b.

In the ﬁrst case, X is a subset of A.

In the second case we can write Y = X ∪{b}, where X is again a subset of A.

Each subset X ⊆ A therefore corresponds to precisely two subsets X and X ∪ {b} of B. Since

= n, the induction hypothesis (∗) tells us there are 2

subsets X ⊆ A, whence

P(B)

= 2

P(A)

= 2

n+1

By induction, (∗) holds for all n ∈ N

Exercise 7 offers an alternative proof.

Example 6.10. You might erroneously expect the sets P(A × B) and P(A) × P(B) to be the same.

Here is a simple counter-example to convince you otherwise!

Let A = {a} and B = {b, c}. Think about cardinalities:

P(A × B)

= 2

A×B

= 2

= 4

P(A) × P(B)

P(A)

P(B)

= 2

= 8

Since the cardinalities are different, the sets cannot be equal: P(A × B) = P(A) × P(B). But what

about subset? Might the smaller set be a subset of the larger? Again the answer is no, as can be seen

by computing the sets explicitly.

A × B =



(a, b), (a, c)



=⇒ P(A × B) =



∅, {(a, b)}, {(a, c)}, {(a, b), (a, c)}



The elements of P(A × B) are sets of ordered pairs. By contrast, the elements of P(A) × P(B) are

ordered pairs of sets:

P(A) × P(B) =



∅, {a}





∅, {b}, {c}, {b, c}





∅, ∅





∅, {b}





∅, {c}





∅, {b, c}





{a}, ∅





{a}, {b}





{a}, {c}





{a}, {b, c}



The elements of the two sets have completely different types, so there is no way that one could be a

subset of the other!

Exercises 6.2. A reading quiz and several questions with linked video solutions can be found online.

1. For each A, ﬁnd P(A) and

P(A)

(a) A = {1, 2} (b) A = {1, 2, 3} (c) A =



(1, 2), (2, 3)



(d) A =



∅, 1, {a}



(e) A =



1, 2



, 3,



4, {5}



(f) A =

(1, 2), 3,



4, {5}



2. Let A = {1, 3} and B = {2, 4}.

(a) Draw a picture of the set A ×B as a subset of R

(b) Compute P(A × B).

3. Determine whether the following statements are true or false. Justify your answers.

(a) If {7} ∈ P(A), then {7} /∈ A.

(b) Suppose that A ⊊ P(B) ⊊ C where

= 2. Then

can be 5, but

cannot be 4.

+ 1, then P(B) has at least two more elements than P(A).

(d) If A, B, C, D are cardinality-two subsets of {1, 2, 3}. Then at least two of them are equal.

4. Here are three incorrect proofs of Theorem 6.7: A ⊆ B =⇒ P(A) ⊆ P(B). Why does each fail?

(a) Let x ∈ P(A). Since A ⊆ B, we have x ∈ B. Therefore x ∈ P(B), and so P(A) ⊆ P(B).

(b) Let A = {1, 2} and B = {1, 2, 3}. Then

P(A) =



∅, {1}, {2}, {1, 2}



⊆



∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, B



= P(B)

5. (a) Prove that P(A) ∪ P(B) ⊆ P(A ∪ B). Provide a counter-example to show that we do not

expect equality.

(b) Does anything change if you replace ∪ with ∩ in part (a)? Justify your answer.

6. (a) For any set A, show there is an injection ι : A → P(A). (Explicitly construct a map, and

show that it is one-to-one.)

(b) Is there any set A such that A ∩P(A) = ∅?

7. If you’ve studied combinatorics, you’ll know that the binomial coefﬁcient

(

)

r!(n−r)!

denotes

the number of distinct ways one can choose r objects from a set of n objects.

(a) Prove directly:

(

n+1

)

(

)

(

r−1

)

whenever 1 ≤ r ≤ n.

(b) (Hard) Prove by induction that ∀n ∈ N

∑

r=0

(

)

= 2

, by using part (a) in the induction

step.

If you found this easy, try proving the binomial theorem: ∀n ∈ N

, (x + y)

∑

r=0

(

)

n−r

6.3 Indexed Collections of Sets: Union and Intersection Revisited

In Deﬁnition 4.11 we deﬁned the union of two sets A ∪ B = {x : x ∈ A or x ∈ B}, which inductively

extends to any ﬁnite union of sets

∪···∪ A

= {x : x ∈ A

for some k}

In this section we consider a stronger deﬁnition and compute unions and intersections of (potentially)

inﬁnite collections of sets.

Deﬁnition 6.11. Given a set of sets {A

} (each A

is a set!), we form its union and intersection:

[

= {x : x ∈ A

for some n} x ∈

[

⇐⇒ ∃n such that x ∈ A

= {x : x ∈ A

for all n} x ∈

⇐⇒ ∀n we have x ∈ A

We say that {A

} is pairwise disjoint if A

∩ A

= ∅ whenever m = n.

For all examples in this section, the sets A

are indexed: each n lies in some indexing set, typically

N, Z or R. It is typical to decorate the union/intersection symbols to indicate this: e.g., if n ∈ N we

might use the notation

n∈N

∞

n=1

Example 6.12. Here is a simple (ﬁnite) example to get us used to the notation. Let

= {1, 3, 5}, A

= {2, 3, 4, 6}, A

= {1, 2, 3, 6}

The indexed collection {A

: n ∈ I} is indexed by I = {1, 2, 3}. Then

[

n=1

= A

∪ A

= {1, 2, 3, 4, 5, 6}

n=1

= A

∩ A

= {3}

Lemma 6.13. Let {A

} be a set of sets. For any A

∈ {A

⊆

[

and

⊆ A

A proof is almost immediate from the deﬁnition: can you supply it?

Nested Collections

When a collection of sets is indexed by the natural numbers N in such a way that successive sets

satisfy a subset relation, we describe the collection as nested, for instance

⊇ A

⊇ ···

Since A

⊆ A

for all n, we see that

∞

n=1

= A

x ∈

∞

[

n=1

⇐⇒ ∃n ∈ N, x ∈ A

⇐⇒ x ∈ A

Computing the intersection in such a situation typically requires much more care. . .

Example 6.14. Consider the nested collection {A

: n ∈ N} of half-open intervals A





m ≤ n =⇒

≤

=⇒ A

⊆ A

=⇒ A

⊇ A

⊇ ···

The union is therefore

∞

n=1

[0,

) = A

= [0, 1).

Before considering the full intersection, we ﬁrst compute all ﬁnite intersections. The nesting condition

says that a ﬁnite intersection is simply the smallest of the listed sets: for any constant m ∈ N,

n=1





= A

∩···∩ A

= A





To ﬁnd the inﬁnite intersection, it is very tempting to take limits:

∞

n=1





= lim

m→∞

n=1





0, lim

m→∞



= [0, 0)

This is mathematical garbage! Nothing you know about limits justiﬁes either questionable equality.

Moreover, the ‘answer’ [0, 0) could only mean the empty set, which is incorrect: we claim that

∞

n=1





= {0}

Proof. First observe that

x ∈

∞

n=1

∞

n=1





⇐⇒ ∀n ∈ N, 0 ≤ x <

Certainly x = 0 satisﬁes these inequalities. It remains to eliminate the other possibilities.

• If x < 0, then x /∈ A

= [0, 1) and so x /∈

• Suppose that x > 0. There exists

N large enough so that

≤ x. Otherwise said,

x /∈ A

, whence x /∈

If the last part of the argument seems difﬁcult, try an example! If x = 0.13, observe that

0.1 < 0.13 =⇒ x /∈ A

=⇒ x /∈

∞

n=1

By modifying the endpoints of the sets A

we obtain slightly different results:

∞

n=1



) = ∅

∞

n=1



] = ∅

∞

n=1



] = {0}

How would the arguments differ from what we did above?

The moral is that you cannot na

ıvely apply limits to sequences of sets. If thinking about limits helps

your intuition, great, but you can’t trust it blindly!

The existence of N ≥

should be intuitive; it is in fact guaranteed by the Archimedean property (Exercise 2.4.13).

An indexed collection can also be nested the other way round, in which case the intersection is

straightforward (though unions need more work)

⊆ A

⊆ ··· =⇒

∞

n=1

= A

Examples 6.15. Here are a few more simple examples of computing unions and intersections of

indexed collections; some are nested, some not.

1. Let A

= [n, n + 1) ⊆ R, for each n ∈ Z. For example A

= [3, 4) and A

−17

= [−17, −16).

Every real number lies in precisely one such set (the sets A

are pairwise disjoint), whence

[

n∈Z

= R and

n∈Z

= ∅

To prove the former, note that x ∈ [n, n + 1) where n = ⌊x⌋ is the greatest integer less than or

equal to x: i.e., ∀x ∈ R, we have x ∈ A

⌊x⌋

, whence R ⊆

n∈Z

(the reversed subset inclusion

is trivial since each A

⊆ R).

2. For each n ∈ N, let A

= [−n, n] be a closed interval. This is a nested collection

⊆ A

⊆ ··· =⇒

∞

n=1

= A

= [−1, 1]

Similarly to the previous example, ∀x ∈ R we have x ∈ A

where n is any integer ≥

: we

conclude that

∞

n=1

= R.

3. For each n ∈ N, let A

= {x ∈ R :



−1



}. Before computing the union and intersection

of these sets, it is helpful to write each set as a pair of intervals. Note that



−1



⇐⇒ −

< x

−1 <

⇐⇒

1 −

1 +

The sets are nested: A

⊇ A

⊇ ···, where



−

1 +

, −

1 −



∪



1 −

1 +



=⇒

∞

[

n=1

= A

= (−

√

2, 0) ∪ (0,

√

For the intersection,

√

2−

√

∀n ∈ N, x ∈ A

⇐⇒ ∀n ∈ N,



−1



⇐⇒ x

−1 = 0

It follows that

∞

n=1

= {1, −1}.

4. Let A

= {0, 1}, A

= {1} is n ≥ 1 is odd, and A

= {2} if n ≥ 2 is even. Then,

∞

n=1

∞

[

k=n

∞

[

k=1

∩

∞

[

k=2

∩··· = {0, 1, 2}∩ {0, 1} ∩ {0, 1} ∩ ··· = {0, 1}

Think about why x ∈

∞

n=1

∞

k=n

⇐⇒ x lies in inﬁnitely many of the sets A

Unions: Don’t Confuse Sets and Elements

When working with large unions, it is easy to confuse two sets:

• {A

} is a set of sets, each element of which is some set A

•

is the set whose elements lie in at least one A

It is important to understand the difference! Sometimes the indexed collection itself is the object of

interest, other times we may want to use its union or intersection to deﬁne something new.

Example 6.16 (Projective space). Let A

be the line

through the ori-

gin in R

with gradient m ∈ R ∪ {∞}. Since every point in R

lies on

such a line, and distinct lines intersect at the origin, we see that

[

= R

and



(0, 0)



The indexed collection is known as projective space

−3

∞

−

P( R

) =



: m ∈ R ∪{∞}



and is interesting in its own right. Each element of projective space is a line, making P(R

) a very

different set to R

. This example also shows that indexing sets don’t have to be simple sets of integers!

In the next example we use an inﬁnite union to deﬁne an interesting set.

Example 6.17 (Finite Decimals). For each n ∈ N, let A

be the set of decimals of length n,



0.a

. . . a

: where each a

∈ {0, 1, . . . , 9}



For example 0.134 ∈ A

. Since 0.134 = 0.1340, we also have 0.134 ∈ A

. This is a nested collection

⊆ A

⊆ ··· =⇒

n∈N

= A



0, 0.1, . . . , 0.9



Now consider unions. If m ∈ N, then

[

n=1

= A



x ∈ [0, 1) : x has a decimal representation of length ≤ m



You might guess that the inﬁnite union would be the entire

interval [0, 1], but this is incorrect:

x ∈

[

n∈N

⇐⇒ ∃n ∈ N such that x ∈ A

⇐⇒ x is a decimal with some ﬁnite length n

The inﬁnite union

is precisely the set of x ∈ [0, 1) which have a ﬁnite decimal representation!

This is far from the entire interval: many rational numbers are excluded (e.g.,

= 0.3333 ··· /∈

and the union contains no irrational numbers.

The symbol ∞ is used to indicate the vertical line A

∞

with ‘inﬁnite gradient.’

We would include 1 = 0.9999 ···

Optional! We ﬁnish this section with a bit of fun, using an inﬁnite intersection to create a fractal set.

Example 6.18 (Cantor’s middle-third set). Starting with the interval C

= [0, 1], construct a sequence

of sets C

by repeatedly removing the middle third of all intervals in C

= [0, 1]

= [0,

] ∪[

, 1]

= [0,

] ∪[

, 1]

The sequence is drawn up to C

, though you’ll have to zoom in a long way to see the detail!

Cantor’s middle-third set is deﬁned to be the inﬁnite intersection C :=

∞

n=0

Cantor’s set has several interesting properties which we state non-rigorously.

Zero Measure (length) It should seem reasonable to write

len(C

) = 1, len(C

) =

, len(C

) =





, . . . , len(C

) =





, . . .

where len(C

) is the sum of the lengths of all intervals in C

. Since C ⊆ C

for all n, we conclude

that len(C) = 0: Cantor’s set contains no intervals of positive length.

Inﬁnite Cardinality Cantor’s set contains the endpoints of every interval removed at any stage of

its construction. In particular,

∈ C for all n ∈ N

, whence C is an inﬁnite set.

Self-similarity Let f , g : R → R be functions f (x) =

x and

g(x) =

x +

. Since C

n+1

= f (C

) ∪g(C

), we conclude

C = f (C) ∪ g(C)

f (C) g(C)

Cantor’s set consists of two shrunken copies of itself, a classic property of fractals.

We’ll analyze Cantor’s set a little and consider a related construction in Exercises 14 & 15. Other

fractal sets with a similar construction include the Sierpi´nski carpet and the von Koch snowﬂake.

Exercises 6.3. A reading quiz and several questions with linked video solutions can be found online.

1. Let I = {1, 3, 4}. Determine

r∈I

and

r∈I

, where each S

= [r −1, r + 3] is an interval.

2. For each integer n, consider the set B

= {n} ×R.

(a) Draw a picture (in the Cartesian plane) of

n=2

= B

∪ B

(b) Draw a picture of the set C = [1, 5] × {−2, 2}.

(Careful! [1, 5] is an interval, while {−2, 2} is a set containing two points)



n=2



∩C.

(d) Compute

n=2

(

∩C

)

. Compare with your answer to part (c).

In fact it is more than merely inﬁnite, it is uncountably so, as we’ll discuss in Chapter 8. The bizarre contrast between

this and the zero measure property was part of the reason Cantor introduced his set.

3. Give an example of four subsets A, B, C, D of {1, 2, 3, 4} such that all intersections of two subsets

are different.

4. For each of collection, deﬁne an interval A

such that the given collection is {A

: n ∈ N}.

Then ﬁnd both the union and intersection of the collection.

(a)



[1, 2 + 1), [1, 2 +

), [1, 2 +

), . . .



(b)



( −1, 2), (−

, 4), (−

, 6), (−

, 8), . . .



(c)



(

, 1), (

), (

), . . .



5. For each real number x, let A

= {3, −2} ∪ {y ∈ R : y > x}. Find

x∈R

and

x∈R

6. Use Deﬁnition 6.11 to prove that A

⊆ A

⊆ ··· =⇒

n∈N

= A

7. Let A

be the set of decimals of length n, as described in Example 6.17.

(a) Prove directly that the cardinality of A

is 10

(b) Prove by induction that

= 10

∞

n=1

⊆ Q.

(d) Explain why

/∈

∞

n=1

8. Suppose that ∀n ∈ N, A

= ∅ and m ≥ n =⇒ A

⊆ A

. Prove or disprove the following:

(a)

293

n=1

= ∅ (b)

293

n=1

= ∅ (c)

n∈N

= ∅ (d)

n∈N

= ∅

9. Compute the set

∞

n=1

∞

k=n

[

, 2 +

]. For a general family of sets {A

: n ∈ N} Explain why it

is is reasonable to write

x ∈

∞

[

n=1

∞

k=n

⇐⇒ x lies in all except ﬁnitely many A

10. Let {A

: n ∈ I} and {B

: n ∈ I} be indexed families of sets. Give explicit examples for which

the following hold:

(a)

(

n∈I

)

∩

(

n∈I

)

=

n∈I

∩ B

)

(b)

(

n∈I

)

∪

(

n∈I

)

=

n∈I

∪ B

)

11. (De Morgan’s laws) Let {A

: n ∈ I} be an indexed family of sets and B a set. Prove

(a) B \

(

n∈I

)

n∈I

(B \ A

)

(b) B \

(

n∈I

)

n∈I

(B \ A

)

12. Suppose we are working in a universal set U (so every set is considered a subset of U). Give an

explanation for why it makes sense to deﬁne

n∈I

= U when I = ∅.

13. (Hard) Let A

= {

∈ Q : 0 < m < n, m ∈ N}, for each n ∈ N.

(a) State A

, A

explicitly.

(b) Prove that A

⊆ A

for any p ∈ N.

n∈N

= Q ∩(0, 1).

(d) Argue that further

n∈N

= Q ∩(0, 1).

(e) Extend your proof to show that, for any ﬁxed p ∈ N,

n∈N

= Q ∩(0, 1).

14. (Hard) A ternary representation

of a number x ∈ [0, 1] is an expression

x =

∞

∑

n=1

+ ··· where each a

∈ {0, 1, 2}

We write x = [0.a

···]

. For example [0.12]

(a) Verify that [0.02101]

243

, [0.22222 ···]

= 1 and [0.020202020 ···]

(Hint: You’ll need the geometric series formula

∑

∞

n=1

1−r

for the latter two)

(b) Let C

be the n

set in the construction of Cantor’s middle-third set C (Example 6.18).

Prove by induction that C

is the set of all x ∈ [0, 1] with a ternary representation whose

ﬁrst n digits are only 0 or 2.

(Hints: Use C

n+1

= f (C

) ∪ g(C

); What does division by 3 do to a ternary representation?)

∈ C, but that it is not an endpoint of any of the deleted middle-thirds re-

moved during the construction of C.

15. (Hard) We construct a modiﬁed Cantor set and fractal curve. Starting with F

= [0, 1], repeat-

edly delete the third quarter segment of each interval to obtain a sequence of sets F

, F

, . . .:

= [0,

] ∪[

, 1], F

= [0,

] ∪[

, 1], . . .

(a) Prove that the sum of the lengths of all of all line segments in F





(b) Prove that

∞

n=1

contains no intervals of positive length.

ment, replace it with the other three sides of a square.

The ﬁrst three steps and the result of applying the pro-

cess inﬁnitely many times are shown in the pictures.

After step 1 the curve has length ℓ

and the area

under the curve is A

. After step 2 the length is

ℓ

and the area A

= A

i. Find the length ℓ

of the curve after n steps. What

is the ‘length’ of the resulting fractal curve?

ii. Repeat for the area under each curve F

. Prove that

the area between the fractal and the x-axis is

Analogous to a decimal representation x =

∑

∞

n=1

+ ··· where a

∈ {0, 1, 2, . . . , 9}.

7 Relations and Partitions

The mathematics of sets is rather basic until one has a notion of how to relate elements of sets to each

other. We are already familiar with several examples of this, for instance:

1. The usual order of numbers (e.g., 3 < 7) is a way of relating/comparing two elements of R.

2. A function f : A → B relates elements in a set A with those in B.

In this chapter we discuss a general framework based on the Cartesian product (Section 6.1).

7.1 Binary Relations on Sets

Deﬁnition 7.1. Let A and B be sets. A (binary) relation R from A to B is a set of ordered pairs

R ⊆ A ×B

A relation on A is a relation from A to itself. The domain and range of R are the sets

dom(R) =



a ∈ A : ∃b ∈ B, (a, b) ∈ R



range(R) =



b ∈ B : ∃a ∈ A, (a, b) ∈ R



If (a, b) ∈ R we can also write a Rb, and say ‘a is related to b.’ Similarly a Rb means (a, b) ∈ R.

Examples 7.2. 1. R =



(1, 3), (2, 2), (2, 3), (3, 2), (4, 1), (5, 2)



deﬁnes a relation on N. Various true

statements about this relation include

(2, 2) ∈ R, (4, 2) /∈ R, 2 R5, 3 R 2

In this case dom(R) = {1, 2, 3, 4, 5} and range(R) = {1, 2, 3}.

2. R =



[1, 3) ×(3, 4]



∪



(2t + 1, t

) : t ∈ [

, 2]



is a relation on R. This time

we have dom(R) = [1, 5] and range(R) = [

, 4].

3. For any set A, the set R =



(a, a) : a ∈ A



is a relation on A. The relation

R is simply ‘equals:’

(a, b) ∈ R ⇐⇒ a = b

4. On R, the relation ≤ can be graphed: we plot all points (x, y) ∈ R

such

that x ≤ y.

5. If A is the set of all humans, we may deﬁne R ⊆ A × A by

(a, b) ∈ R ⇐⇒ a, b have a parent/child or sibling relationship

In this example, the mathematical use of relation is similar to that in English:

I am related to my sister, and my mother is related to me.

6. If A is a set, then ⊆ is a relation on the power set P(A) (the set of subsets).

For instance, if A = {1, 2, 3} then {1} ⊆ {1, 3} but {1, 3} ⊈ {1}.

0 2 4

Example 1

0 2 4

Example 2

−2

−2 2

Example 4

7. If f : A → B is a function then



a, f (a)



: a ∈ A



is a relation! We’ll revisit this in Section 7.2.

With abstract relations, there are only a small number of things we can do.

Deﬁnition 7.3. Given a relation R ⊆ A × B, its inverse R

−1

⊆ B × A is the set

−1



( b, a) ∈ B × A : (a, b) ∈ R



To ﬁnd the elements of R

−1

, you simply switch the components of each ordered pair in R.

We say that R is symmetric if R = R

−1

(requires A = B): i.e., (x, y) ∈ R ⇐⇒ (y, x) ∈ R.

Theorem 7.4. For any relations R, S ⊆ A × B:

1. dom(R

−1

) = range(R) 2. range(R

−1

) = dom(R)

3. (R

−1

)

−1

= R 2. R ⊆ S ⇐⇒ R

−1

⊆ S

−1

5. (R ∪S)

−1

= R

−1

∪S

−1

4. (R ∩S)

−1

= R

−1

∩S

−1

7. If A = B, then R∪R

−1

and R ∩ R

−1

are both symmetric

Proof. Here are two of the arguments. Try the others yourself.

4. Suppose R ⊆ S. Then,

( b, a) ∈ R

−1

=⇒ (a, b) ∈ R (deﬁnition of inverse)

=⇒ (a, b) ∈ S (since R ⊆ S)

=⇒ (b, a) ∈ S

−1

(deﬁnition of inverse)

Therefore R

−1

⊆ S

−1

. Replacing R, S with R

−1

, S

−1

and applying part 1, we also see that

−1

⊆ S

−1

=⇒ (R

−1

)

−1

⊆ (S

−1

)

−1

=⇒ R ⊆ S

7. We prove the ﬁrst half. By part 3,

( R ∪ R

−1

)

−1

= R

−1

∪(R

−1

)

−1

= R

−1

∪R = R ∪ R

−1

Example (7.2.1, cont.). R =



(1, 3), (2, 2), (2, 3), (3, 2), (4, 1), (5, 2)



is not

symmetric. Indeed

−1



(3, 1), (2, 2), (3, 2), (2, 3), (1, 4), (2, 5)



= R

as should be plain: e.g., ( 1, 3) ∈ R but (3, 1) /∈ R. However,

R∩R

−1



(2, 2), (2, 3), (3, 2)



and

R∪R

−1



(1, 3), (3, 1), (2, 2), (2, 3), (3, 2), (4, 1), (1, 4), (5, 2), (2, 5)



0 2 4

are both symmetric. Observe the symmetry in the picture. One obtains R

−1

by reﬂecting

R in the

line y = x: a symmetric relation on R has reﬂection symmetry in this line.

Just as one does for inverse functions in calculus. . .

Exercises 7.1. A reading quiz and practice question can be found online.

1. Let R be the relation on {0, 1, 2} deﬁned by

0 R 0 0 R1 2 R1

(a) Write R as a set of ordered pairs. What are its domain and range?

(b) What is the inverse of R?

2. (a) Let R be the relation on R deﬁned by a Rb ⇐⇒

a −b

= 1. Is this relation symmetric?

(b) Let S be the relation on R deﬁned by

a S b ⇐⇒ ∃x ∈ Q \ {0} such that a = x

Is this relation symmetric?

3. Draw pictures of the following relations on the set of real numbers R.

(a) R =



(x, y) : y ≤ 2 and y ≥ x and y ≥ 2 − x



(b) S =



(x, y) : (x −4)

+ (y −1)

≤ 9



State the domain and range and draw the inverse of each relation.

4. A relation is deﬁned on N by a R b ⇔

∈ N. Let c, d ∈ N. Under what conditions can we

write c R

−1

5. Let R ⊆ {1, 2, 3, 4} × {1, 2, 3, 4} be the relation

R =



(1, 3), (1, 4), (2, 2), (2, 4), (3, 1), (3, 2), (4, 4)



(a) Find dom(R), range(R) and R

−1

(b) Compute the relations R∪R

−1

and R ∩ R

−1

, and check that they are symmetric.

6. For the relation R =



(x, y) : x ≤ y



on N, what is R

−1

, and what is the intersection R∩R

−1

7. Let R ⊆ Z ×Z be the relation m Rn iff m | n. Compute R∩ R

−1

8. Let A be a set with

= 4. What is the maximum number of elements that a relation R on A

can contain such that R ∩ R

−1

= ∅?

9. Give formal proofs of the remaining parts of Theorem 7.4.

10. Let R and S be two symmetric relations on a set A.

(a) Show R ∩ S is symmetric.

(b) Does R ∪ S have to be symmetric? Give a proof or counterexample.

11. Let R be a relation on a set A and deﬁne S = R ∪ R

−1

. Prove that S is the smallest symmetric

relation on A containing R in the following sense: if

T =

T ⊆ A × A : T symmetric and R ⊆ T



then

S =

T ∈T

(S is known as the symmetric closure of R)

7.2 Functions revisited

In Section 4.3, we na

ıvely deﬁned a function f : A → B as a rule associating to each element a ∈ A an

element f (a) ∈ B. But what do we mean by a rule? We address this issue by turning Example 7.2.7

on its head: a function f : A → B is precisely its graph.

Example 7.5. The function f : [0, 4] → [0, 2] : x 7→

√

x corresponds to

the relation



√



: x ∈ [0, 4]



⊆ [0, 4] × [0, 2]

0 2 4

√

(x,

√

The difﬁculty is that we cannot use the notation f (a) until we know that we have a function. . .

Deﬁnition 7.6. A function from A to B is a relation f ⊆ A × B satisfying two conditions:

1. dom( f ) = A (each a ∈ A is related to at least one b ∈ B)

2. (a, b

), (a, b

) ∈ f =⇒ b

= b

(each a ∈ A is related to at most one b ∈ B)

A relation f ⊆ A × B is a function if and only if every a ∈ A is the ﬁrst entry of exactly one ordered

pair (a, b) ∈ f . This is simply an abstraction of the vertical line test from calculus.

Example 7.7. Let A = [0, 4] and B = [0, 5]. Two relations f , R ⊆ A × B are drawn below.

The ﬁrst relation deﬁnes a function f : A → B since every a ∈ A corresponds to exactly one b ∈ B:

the vertical line through any a ∈ A intersects the graph in exactly one point (a, b).

The second relation R does not deﬁne a function. In fact it fails both parts of the deﬁnition:

1. The vertical line through

a ∈ A does not intersect the graph:

a /∈ dom(R).

2. The vertical line through a ∈ A intersects the graph in two points (a, b

) = (a, b

0 2 4

dom( f ) = A

range( f )

0 2 4

dom(R) = A

range(R)

Injectivity (Deﬁnition 4.18) can be rephrased in this new context: f : A → B injective means

( ∀a ∈ A) (a

, b), (a

, b) ∈ f =⇒ a

= a

Surjectivity has the same meaning as before: range( f ) = B.

Example 7.8. Let A = {1, 2, 3} and B = {p, q, r}. The relation

f =



(1, r), (2, p), (3, r)



satisﬁes both conditions necessary to be a function. In elementary

language, f (1) = r, f (2) = p and f (3) = r. Moreover f is:

Not injective since ( 1, r), (3, r) ∈ f and 1 = 3

Not surjective since range( f ) = {p, r} = B

The inverse relation

−1



(r, 1), (p, 2), (r, 3)



⊆ B × A

is not a function due to failing both conditions of Deﬁnition 7.6.

1. dom( f

−1

) = {p, r} = B. The vertical line through b = q does

not intersect (the graph of) f

−1

2. (r, 1) ∈ f

−1

and (r, 3) ∈ f

−1

, but 1 = 3. The vertical line

through b = r intersects f

−1

twice.

Note how the manner of these failures are identical to the failures of

injectivity and surjectivity. . .

1 2 3

The function f : A → B

p q r

−1

⊆ B × A: not a function

The example highlights one advantage of the relational approach: the inverse of a function f : A → B

always exists; it is the inverse relation f

−1

⊆ B ×A. The question is whether f

−1

is also a function. From

the example and our discussions in Section 4.3, you should strongly suspect the answer.

Theorem 7.9. Let f : A → B be a function and consider its inverse relation f

−1

⊆ B × A. Then

−1

is a function ⇐⇒ f is bijective (both injective and surjective)

Proof. Recalling Deﬁnition 7.6, we see that f

−1

is a function if and only if the two conditions hold:

1. dom( f

−1

) = B

2. (b, a

), (b, a

) ∈ f

−1

=⇒ a

= a

By Theorem 7.4, condition 1 is equivalent to range( f ) = B; that is, f is surjective.

Condition 2 is equivalent to (a

, b), (a

, b) ∈ f =⇒ a

= a

: i.e., f is injective.

Example (7.5 cont.). f =



(x,

√

x) : x ∈ [0, 4]



⊆ [0, 4] × [0, 2] is

Injective since (x

, y), (x

, y) ∈ f =⇒ x

= y

and x

= y

, whence x

= x

Surjective since range( f ) =



√

x : x ∈ [0, 4]



= [0, 2].

Since f is bijective, its inverse relation is a function: f

−1



( y, y

) : y ∈ [0, 2]



or, as we’d normally

write, f

−1

: [0, 2] → [0, 4] : y 7→ y

Exercises 7.2. A reading quiz and practice questions can be found online.

1. Suppose f is the relation

f =



(1, 1), (2, 3), (3, 5), (4, 7)



⊆ {1, 2, 3, 4} × {1, 2, 3, 4, 5, 6, 7}

(a) Show that we have a function f : {1, 2, 3, 4} → {1, 2, 3, 4, 5, 6, 7}. Can you ﬁnd a concise

formula f (x) to describe this function?

(b) Is f injective? Justify your answer.

that is, as sets,



(a, f (a)) : a ∈ {1, 2, 3, 4}





(a, g(a)) : a ∈ {1, 2, 3, 4}



If g is a bijective function, what is B?

2. Decide whether each of the following relations are functions. For those which are, decide

whether the function is injective and/or surjective.

(a) R =



(x, y) ∈ [−1, 1] ×[−1, 1] : x

+ y

= 1



(b) S =



(x, y) ∈ [−1, 1] ×[0, 1] : x

+ y

= 1





(x, y) ∈ [0, 1] ×[−1, 1] : x

+ y

= 1



(d) U =



(x, y) ∈ [0, 1] ×[0, 1] : x

+ y

= 1



3. (a) Express the function f : R → R : x 7→ x

+ 3 as a relation.

(b) What is the inverse relation f

−1

is not a function.

(d) Prove directly from Deﬁnition 4.18 that f is not injective and not surjective. Compare your

arguments with your answer to part (c).

4. Repeat the previous question for f : R → R : x 7→

√

−4x + 5.

5. (a) Express the function f : R → R : x 7→ x

as a relation on R.

(b) What is the inverse relation f

−1

? Is it a function? Justify your answer.

6. Let A be any set and deﬁne the relation id



(a, a) : a ∈ A



on A. Show that id

is a bijective

function (it is called the identity function on A). What is its inverse?

7. Consider the relation f =



(x, x) : x ∈ Q



∪



(x, 0) : x ∈ Q



⊆ R

(a) Prove that f is a function R → R.

(b) Is f injective? Surjective? Explain.

8. Suppose f ⊆ A × B is a function.

(a) If U ⊆ A, explain why it would be reasonable to write the image of U as

f (U) =



b ∈ B : ∃u ∈ U with (u, b) ∈ f



(Recall Deﬁnition 4.16 if you’re unsure what this means)

(b) If V ⊆ B, how would you write the inverse image f

−1

(V) in this language?

7.3 Equivalence Relations & Partitions

What do we mean when we say that two objects are equal? Mathematicians use the word ﬂexibly,

often in reference to non-identical objects which share some common property.

You’ve been doing

this for years, indeed our very ﬁrst result (Theorem 1.1: even + even = even) has exactly this ﬂavor.

To help develop a more ﬂexible notion of equality, consider three important concepts.

Example 7.10. Let X be a set. The following are immediate:

Reﬂexivity: ∀x ∈ X, x = x

Symmetry: (∀x, y ∈ X) x = y =⇒ y = x

Transitivity: (∀x, y, z ∈ X) x = y and y = z =⇒ x = z

We turn this simple example on its head to create an abstract, generalized notion of equality.

Deﬁnition 7.11. We deﬁne the following properties for a relation

∼ on a set X:

Reﬂexivity: ∀x ∈ X, x ∼ x (every element of X is related to itself)

Symmetry: x ∼ y =⇒ y ∼ x (if x is related to y, then y is related to x)

Transitivity: x ∼ y and y ∼ z =⇒ x ∼ z (if x is related to y and y is related to z,

then x is related to z)

An equivalence relation on X is a relation ∼ satisfying all three properties.

Example 7.10 says that ‘equals’ is indeed an equivalence relation on any set X. Things would be very

boring if this were the only example. . .

Example 7.12. Deﬁne a relation ∼ on Z by

x ∼ y ⇐⇒ x −y is even (otherwise said, x ≡ y (mod 2))

We claim that ∼ is an equivalence relation on Z.

Reﬂexivity: ∀x ∈ Z, x −x = 0 is even, hence x ∼ x.

Symmetry: Given x, y ∈ Z, x ∼ y =⇒ x −y even =⇒ y − x even =⇒ y ∼ x.

Transitivity: Given x, y, z ∈ Z,

x ∼ y and y ∼ z =⇒ x −y even and y −z even

=⇒ x −z = (x −y) + (y −z) is even (Theorem 1.1!)

=⇒ x ∼ z

As we’ll see shortly, an equivalence relation algebraically characterizes what it means for elements to

share a common property: here x ∼ y if and only if they have the same parity.

This goes back at least to Euclid, who used equal to refer to congruent triangles. To Euclid, congruent triangles were

essentially identical and therefore not worth distinguishing.

Symmetry is precisely as in Deﬁnition 7.3. As we’ve done, the universal quantiﬁers for symmetry and transitivity are

typically hidden. The symbol ∼ (‘tilde,’ or ‘twiddles’) is commonly used for an abstract equivalence relation. It is the same

symbol used to denote similar triangles: congruence and similarity are both equivalence relations on the set of triangles!

It would also be somewhat dull if every relation were an equivalence relation. In fact most are not.

Examples 7.13. 1. Consider the relation ≤ on the natural numbers N. We check each condition:

Reﬂexivity: True. ∀x ∈ R, x ≤ x.

Symmetry: False. For example, 2 ≤ 3 but 3 ≰ 2.

Transitivity: True. ∀x, y, z ∈ R, if x ≤ y and y ≤ z, then x ≤ z.

Plainly ≤ is not an equivalence relation on N.

2. Let X be the set of lines in the plane and deﬁne ℓ

Rℓ

⇐⇒ ℓ

, ℓ

intersect.

Reﬂexivity: True. Every line intersects itself, so ℓ R ℓ for all ℓ ∈ X.

Symmetry: True. For all lines ℓ

, ℓ

∈ A, if ℓ

intersects ℓ

then ℓ

intersects ℓ

Transitivity: False. As the picture illustrates, if ℓ

, ℓ

are par-

allel and ℓ

crosses both, then ℓ

Rℓ

, ℓ

Rℓ

and ℓ

Rℓ

ℓ

Again R is not an equivalence relation on X.

3. The relation R =



(1, 1), (1, 3), (3, 1), (3, 3)



is not an equivalence relation on X = {1, 2, 3}.

Reﬂexivity: False. For instance (2, 2) /∈ R (otherwise said, 2 R2).

Symmetry: True. In all four cases, (x, y) ∈ S =⇒ (y, x) ∈ R is clear.

Transitivity: True. Check this yourself; there are six valid combinations!

The usefulness of equivalence relations comes when we group together all related elements.

Deﬁnition 7.14. Given an equivalence relation ∼ on X, the equivalence class of x ∈ X is the set

[x] = {y ∈ X : y ∼ x} (y ∈ [x] ⇐⇒ y ∼ x)

If y ∈ [x], we call y a representative of the equivalence class [x]. The quotient of X by ∼ is the set of

equivalence classes:



∼



[x] : x ∈ X



(read ‘X modulo ∼’)

Examples 7.15. We start by revisiting our ﬁrst two examples.

1. (Example 7.10) For the equivalence relation x ∼ y ⇐⇒ x = y on a set X, each equivalence class

has precisely one element [x] = {x}; the quotient is the set of singleton subsets,



∼



{x} : x ∈ X



2. (Example 7.12) For the relation x ∼ y ⇐⇒ x −y is even, there are two equivalence classes:

[0] = {. . . , −4, −2, 0, 2, 4, . . .} = 2Z = {even integers}

[1] = {. . . , −3, −1, 1, 3, 5, . . .} = 1 + 2Z = {odd integers}

and the quotient has two elements:



∼



[0], [1]



Note that any even number is a representative of the ﬁrst class in the even/odd example: for instance

4 ∈ [0] since 4 − 0 = 4 is even.

Indeed [4] = [0], which leads us a simple piece of bookkeeping. . .

Lemma 7.16. Suppose ∼ is an equivalence relation. Then x ∼ y ⇐⇒ [x] = [y].

Observe how the proof uses all three deﬁning properties of an equivalence relation.

Proof. (⇐) By reﬂexivity, x ∈ [x]. Thus [x] = [ y] =⇒ x ∈ [y] =⇒ x ∼ y.

( ⇒) Suppose x ∼ y. We prove that [x] = [y] by showing that each side is a subset of the other.

( ⊆) Let z ∈ [x]. By deﬁnition, z ∼ x. By transitivity,

z ∼ x and x ∼ y =⇒ z ∼ y =⇒ z ∈ [y]

( ⊇) By symmetry, we also have y ∼ x. Repeating the previous argument yields [y] ⊆ [x].

Example 7.17. Let X = {students taking this course} and deﬁned, ∀x, y ∈ X,

x ∼ y ⇐⇒ x achieves the same letter-grade as y

It should be clear that this is an equivalence relation (try saying out loud what reﬂexivity, symme-

try & transitivity mean here!). There is one equivalence class for each letter-grade awarded, each

class being the set of all students who obtain a given grade. If we label the equivalence classes

, A, A

−

, B

, . . . , F, where, say, B = {students obtaining a B-grade}, then the quotient



∼



, A, A

−

, B

, . . . , F



is essentially the set of possible letter-grades! Suppose Laura & Jorge both achieve a B

; both are

representatives of this equivalence class and the following propositions are all true:

Laura ∈ B

, Jorge ∈ B

, Laura ∼ Jorge, [Laura] = [Jorge] = B

The example is a special case of a general construction.

Theorem 7.18. Suppose f is a function with domain X. Then

x ∼ y ⇐⇒ f (x) = f (y)

deﬁnes an equivalence relation on X. There is one equivalence class [x] = {y ∈ X : f (x) = f (y)} for

each value in range( f ).

The proof is an exercise. A suitable function in the previous example would be

f : X →



, A, A

−

, B

, . . . , F



where f (x) = “x’s letter grade”

This converse to the theorem always holds: if ∼ is an equivalence relation then the so-called canonical

map f : X →



∼

: x 7→ [x] satisﬁes x ∼ y ⇐⇒ f (x) = f (y)!

100

Examples 7.19. 1. The function

f : Z → {0, 1, 2, 3, 4} : x 7→ x

(mod 5) (take the remainder on division by 5)

deﬁnes an equivalence relation on Z:

x ∼ y ⇐⇒ x

≡ y

(mod 5)

2. The function f : R

→ R : (x, y) 7→ x

+ y

deﬁnes an equivalence relation

on R

(x, y) ∼ (v, w) ⇐⇒ x

+ y

= v

+ w

As stated in the Theorem, there is one equivalence classes for

each element r

∈ range( f ) = R

: if x

+ y

= r

, then

[(x, y)] =

( v, w) ∈ R

: v

+ w

= r

is the circle of radius r centered at the origin. The equivalence class

[(1, 0)] highlighted in the picture. Moreover, the quotient set is



∼

= {circles centered at the origin}

−1

−1 1

Partitions When you cut a cake, each crumb ends up in exactly one slice. To partition a set is to do

precisely this: split into disjoint subsets so that each element lies in exactly one subset.

Deﬁnition 7.20. Let X be a set. A collection {A

} of non-empty subsets A

⊆ X partitions X if

1. X =

(each x ∈ X lies in at least one subset A

)

2. If A

= A

, then A

∩ A

= ∅ (each x ∈ X lies in at most one

subset A

)

Partitions are intimately related to equivalence relations, as the next example illustrates.

Example 7.21. Partition X = {1, 2, 3, 4, 5} into subsets

= {1, 3}, A

= {2, 4}, A

= {5}

and consider the relation ∼ on X deﬁned by

x ∼ y ⇐⇒ x, y are in the same subset A

Run through the checklist: ∼ is reﬂexive, symmetric & transitive; it is an equivalence relation! More-

over, its equivalence classes are precisely the partitioning subsets A

, A

and A

[1] = [3] = {1, 3} = A

, [2] = [4] = {2, 4} = A

, [5] = {5} = A

Be careful: ∼ is a subset of R

×R

, whereas each equivalence class is a subset of R

Distinct sets A

are pairwise disjoint (Deﬁnition 4.11). It is sometimes useful in examples to permit A

= A

Otherwise said, ∼ is the set



(1, 1), (1, 3), (3, 1), (3, 3), (2, 2), (2, 4), ( 4, 2) , (4, 4), (5, 5)



⊆ X × X.

101

This tight relationship between partitions and equivalence relations is completely general: equiva-

lence relations provide a straightforward algebraic method of working with partitions.

Theorem 7.22. 1. If ∼ is an equivalence relation on X, then its equivalence classes partition X.

2. A partition {A

} of X deﬁnes an equivalence relation ∼ on X by

x ∼ y ⇐⇒ ∃A

such that x, y ∈ A

Its equivalence classes are the distinct subsets A

: otherwise said,



∼

= {A

In short, equivalence relations and partitions are essen-

tially the same thing!

The picture illustrates part 2

and the cake-cutting metaphor:

= [a], A

= [b] = [c],

b ∼ c, a ≁ b, a ≁ c

XXXXX

partition

Before reading the proof, look back at every example of an equivalence relation in this section and

convince yourself that the equivalence classes really do partition the ‘big set.’ In both parts of the

proof look for where the assumptions are used—reﬂexive, symmetric, transitive in part 1; the two

partition conditions in part 2—and think about why they are needed.

Proof. 1. Given an equivalence relation ∼ on X, we must establish two claims:

(a) Every x ∈ X lies in some equivalence class. This follows by reﬂexivity: for each x ∈ X,

x ∼ x =⇒ x ∈ [x]

Every element of X lies in the equivalence class deﬁned by itself!

(b) Distinct equivalence classes are pairwise disjoint: [x] = [y] =⇒ [x] ∩[y] = ∅. We establish

the contrapositive. If [x] ∩ [y] = ∅, then ∃z ∈ [x] ∩ [y]. But then

z ∼ x and z ∼ y =⇒ x ∼ z and z ∼ y (symmetry)

=⇒ x ∼ y (transitivity)

=⇒ [x] = [y] (Lemma 7.16)

2. We need only demonstrate the reﬂexivity, symmetry and transitivity of ∼.

Reﬂexivity: X =

=⇒ every x ∈ X lies in some A

. Thus x ∼ x for all x ∈ X.

Symmetry: x ∼ y =⇒ ∃A

such that x, y ∈ A

=⇒ y, x ∈ A

=⇒ y ∼ x.

Transitivity: If x ∼ y and y ∼ z, then ∃A

, A

such that x, y ∈ A

and y, z ∈ A

. Then

y ∈ A

∩ A

=⇒ A

∩ A

= ∅ =⇒ A

= A

=⇒ x, z ∈ A

=⇒ x ∼ z

Even more is true. If you think carefully, conditions 1 & 2 in the deﬁnition of partition (7.20) now form a disguised

version of the vertical line test for the canonical map (Theorem 7.18)!

102

Exercises 7.3. A reading quiz and practice questions can be found online.

1. Show that x ∼ y ⇐⇒ x ≡ y (mod 3) deﬁnes an equivalence relation on Z. What are the

equivalence classes?

2. (a) Let ∼ be the relation on Z deﬁned by a ∼ b ⇐⇒ a + b is even. Show that ∼ is an equiva-

lence relation and determine its equivalence classes.

(b) If ‘even’ is replaced by ‘odd’ in part (a), which of the properties reﬂexive, symmetric,

transitive does ∼ now possess?

3. For each equivalence relation on R

, identify the equivalence classes and draw several of them.

(a) (a, b) ∼ (c, d) ⇐⇒ ab = cd (b) (v, w) ∼ (x, y) ⇐⇒ v

w = x

4. Let X = {1, 2, 3, 4, 5, 6}. The distinct equivalence classes resulting from an equivalence relation

∼ on X are {1, 4, 5}, {2, 6}, and {3}. What is ∼? Give your answer as a subset of X × X.

5. ⊆ is a relation on any set of sets. Is ⊆ reﬂexive, symmetric, transitive? Prove your assertions.

6. Suppose X is a set with at least two elements. Which of the properties reﬂexive, symmetric,

transitive are satisﬁed by the relation =?

7. Let X = {ax

+ bx

+ cx + d : a, b, c, d ∈ R} be the set of polynomials of degree ≤ 3. Deﬁne a

relation R on X by

p Rq ⇐⇒ p and q have a common real root



∃x ∈ R : p(x) = q(x) = 0



For instance p(x) = (x −1)

and q(x) = x

−1 have the root 1 in common, so p Rq. Determine

which of the properties reﬂexive, symmetric and transitive are possessed by R.

8. Let A = {2

: m ∈ Z}. A relation ∼ is deﬁned on the set Q

of positive rational numbers by

a ∼ b ⇐⇒ ab

−1

∈ A

Show that ∼ is an equivalence relation and describe the elements in the equivalence class [3].

9. A relation is deﬁned on the set X = {a + b

√

2 : a, b ∈ Q, a + b

√

2 = 0} by x ∼ y ⇐⇒

∈ Q.

Show that ∼ is an equivalence relation and determine the distinct equivalence classes.

10. For the purposes of this question, a real number is x small if

≤ 1. Let R be the relation on

the set of real numbers deﬁned by x Ry ⇐⇒ x −y is small.

Prove or disprove: R is an equivalence relation on R.

11. Find the equivalence classes for the relation ∼ in Example 7.19.1.

12. For each relation R on Z, decide whether it is reﬂexive, symmetric, or transitive, and whether

it is an equivalence relation.

(a) a Rb ⇐⇒ a ≡ b (mod 3) or a ≡ b (mod 4)

(b) a Rb ⇐⇒ a ≡ b (mod 3) and a ≡ b (mod 4)

13. For Example 7.13.3, compute the ‘classes’ [x] = {y ∈ X : xRy}. What do you observe?

103

14. Let X = {1, 2, 3}. Deﬁne the relation S =



(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 3)



on X.

(a) Which of the properties reﬂexive, symmetric, transitive are satisﬁed by S?

(b) Let A

= {x ∈ X : x S n}. Show that {A

, A

} do not partition X.

T =



(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 3)



(Warning! Some of the sets A

, A

might be the same in these examples)

15. (Example 7.19.2) Prove directly that the circles A



(x, y) : x

+ y

= r



partition R

: i.e.,

[

r≥0

and A

∩ A

= ∅ =⇒ A

= A

16. Determine whether each collection {A

: n ∈ R} partitions R

. Sketch several of the sets A

(a) A



(x, y) ∈ R

: y = 2x + n



(b) A



(x, y) ∈ R

: y = (x −n)





(x, y) ∈ R

: xy = n



(d) A



(x, y) ∈ R

: y

−y

= x −n



17. Let X be the set of all humans. If x ∈ X, deﬁne the set

= {people who had the same breakfast or lunch as x}

(a) Does the collection {A

: x ∈ X} partition X? Explain your answer.

(b) Is your answer different if the or in the deﬁnition of A

is changed to and?

18. A relation R is antisymmetric if



(x, y) ∈ R



∧



( y, x) ∈ R



=⇒ x = y. Give examples of

relations R on X = {1, 2, 3} having the stated property.

(a) R is both symmetric and antisymmetric.

(b) R is neither symmetric nor antisymmetric.

−1

is not transitive.

19. Let R be a relation on X and deﬁne its reﬂexive closure S = R ∪



(x, x) : x ∈ X



. Prove that S

is reﬂexive and that, if T is any reﬂexive relation for which R ⊆ T , then S ⊆ T .

20. (a) Prove Theorem 7.18.

(b) If f has domain X, explain why



−1

( {b}) : b ∈ range( f )



forms a partition of X.

21. Deﬁne a relation ∼ on R

\{(0, 0)} by

(x, y) ∼ (v, w) ⇐⇒ ∃λ = 0 such that (λx, λy) = (v, w)

(a) Prove that ∼ is an equivalence relation.

(b) What does this relation have to do with projective space P(R

) (Example 6.16)?

22. (If you’ve studied linear algebra) We say that square matrices A, B are similar if there exists a

matrix M such that B = MAM

−1

(a) Prove that similarity is an equivalence relation on the set of n ×n matrices.

(b) What is the equivalence class of the identity matrix?



−11 15

−5 9



and



4 10

0 −6



are similar.

(Hint: diagonalize the matrices!)

104

7.4 Well-deﬁnition, Rings and Congruence

We return to our discussion of congruence (Section 3.1) in the context of equivalence relations and

partitions. The important observation is that congruence modulo n is an equivalence relation on Z, each

equivalence class being the set of all integers sharing a remainder modulo n.

Theorem 7.23. For a ﬁxed n ∈ N, deﬁne x ∼

y ⇐⇒ x ≡ y (mod n). Then ∼

is an equivalence

relation on Z.

The theorem is merely a generalization of Example 7.12 and Exercise 7.3.1. You should prove this

yourself. The equivalence classes are precisely the integers which are congruent modulo n:

[a] =



x ∈ Z : x ≡ a (mod n)





x ∈ Z : x and a have the same remainder as when divided by n





x ∈ Z : x −a is divisible by n



In this language, we can restate what it means for two equivalence classes to be equal.

Lemma 7.24. [a] = [b] ⇐⇒ a ≡ b (mod n) ⇐⇒ ∃k ∈ Z such that b = a + kn.

If the meaning of any of the above is unclear, re-read the previous section! The equivalence classes of

∼

partition the integers into exactly n subsets, one for each remainder. The quotient set is therefore



∼



[0], [1], . . . , [n −1]



We use this set to deﬁne an extremely important object.

Deﬁnition 7.25. Deﬁne operations +

and ·

on the quotient set



∼

as follows:

[x] +

[y] := [x + y], [x] ·

[y] := [x ·y]

The ring Z

is the set



∼

together with the operations +

and ·

It is important to appreciate that +

and ·

are new operations, deﬁning addition/multiplication of

equivalance classes in terms of standard addition/multiplication of integers.

Example 7.26. We work in the ring Z

. According to the deﬁnition,

[3] +

[6] = [3 + 6] = [9] = [1]

There is a potential difﬁculty: each equivalence class is large (e.g., [3] = {. . . , −5, 3, 11, 19, . . .}), so we

have lots of choice regarding representatives. For instance, since [3] = [11] and [6] = [22] we should

be able to observe that

[11] +

[22] = [3] +

[6]

Is this true? If not, then the operation +

would not be particularly useful. Thankfully this is not a

problem: according to the deﬁnition of +

, things turn out exactly as we’d like,

[11] +

[22] = [11 + 22] = [33] = [1] = [3] +

[6]

105

Consider things more abstractly. Given equivalence classes X and Y, here is the process for comput-

ing the equivalence class X +

1. Choose representatives x ∈ X and y ∈ Y so that X = [x] and Y = [y].

2. Sum the integers x and y to get a new integer x + y ∈ Z.

3. Take the equivalence class X +

Y = [x + y].

The concern is that there are inﬁnitely many possible choices for elements x ∈ X and y ∈ Y in step 1. If

is to make sense, we must obtain the same equivalence class [x + y] regardless of our choices of x

and y. This indeed happens, as we indicate with a special piece of terminology.

Theorem 7.27. The operations +

and ·

are well-deﬁned.

This is nothing more than a rehash of Theorem 3.9: compare it with what follows until you are

comfortable we are doing the same thing! Observe that

[x] =



z ∈ Z : z ≡ x (mod n)





x + kn : k ∈ Z



Otherwise said, all representatives (all possible choices in step 1) of the equivalence class [x] have the

form x + kn for some k ∈ Z. Thinking similarly for y, the Theorem requires only that we prove

∀k, l ∈ Z, [x + kn] +

[y + ln] = [x] +

[y] and [x + kn] ·

[y + ln] = [x] ·

[y]

Proof. We prove that +

is well-deﬁned.

[x + kn] +

[y + ln] = [(x + kn) + (y + ln)] (deﬁnition of +

)

= [x + y + (k + l)n] = [x + y] (Lemma 7.24)

= [x] +

[y] (deﬁnition of +

)

The argument for ·

is similar.

Compare our equivalence class notation with that from Section 3.1. For instance

[−3] +

[12] = [1] means the same thing as −3 + 12 ≡ 1 (mod 8) (∗)

Aside: Notation Given the wide usage of Z

, it is customary in higher-level mathematics to drop

the square brackets and all subscripts, simply writing



0, 1, 2, . . . , n − 1



In this language, (∗) would be written −3 + 12 = 1 in Z

. It is critical to appreciate that −3, 12 and 1

denote sets (equivalence classes) and not integers. Until you are 100% certain of this, you should stick

to either of the notations in (∗).

It is something of a mathematical joke to write 1 + 1 = 0 (in Z

). Even in this context, 1 + 1 = 2;

the seeming paradox is that 2 = 0 in Z

. This is really a very simple claim about sets: that the sum

of any two odd numbers is even! Indeed general modular arithmetic (in Z

) is little more than an

abstraction/generalization of this simple fact.

106

Functions and Partitions Our construction of Z

is a special case of a more general situation.

Deﬁnition 7.28. Let ∼ be an equivalence relation on a set X and let A be any set. Suppose we

construct f :



∼

→ A using a rule of the form

f ([x]) := “do something to a representative x”

We say that f is well-deﬁned if this construction deﬁnes a legitimate function. More formally:

[x] = [y] =⇒ f ([x]) = f ([y])

Examples 7.29. 1. Consider Z



∼

, that is X = Z where x ∼ y ⇐⇒ x ≡ y (mod 3). The rule

f ([x]) = 2x + 1 does not produce a well-deﬁned function f : Z

→ Z since, for instance

[0] = [9] but f ([ 0]) = 1 = 19 = f ([9])

2. The rule f ([x]) = [2x + 1] does produce a well-deﬁned function f : Z

→ Z

however:

[x] = [y] =⇒ y = x + 3k for some integer k

=⇒ f ([y]) = f ([x + 3k]) = [2x + 6k + 1] = [2x + 1] = f ([x])

since 6x ≡ 0(mod 3). If this seems abstract, note that f is easily

stated in tabular form since there are only three possible inputs!

[x] [0] [1] [ 2]

f ([x]) [1] [0] [2]

3. This last example is something of an advert for the advanced notation in the previous aside.

We ask for which integers k the rule f

(x) = kx produces a well-deﬁned function f

: Z

→ Z

If this is confusing, re-rewrite using either of the notations

(x) = kx (mod 6) or f

([x]

) = [kx]

Start with a special case to get a feel for things. A table

of values for f

(x) shows an immediate problem!

x 0 1 2 3 4 5 6 ···

(x) 0 1 2 3 4 5 0 ···

In Z

, we have 0 = 4, however f

(0) = 0 and f

(4) = 4 which are not equal in Z

! It follows that

is not well-deﬁned: it isn’t a function (and never was!).

Now proceed systematically.

x ≡ y (mod 4) =⇒ x −y = 4n =⇒ kx −ky = 4kn (for some n ∈ Z)

It follows that f

is well-deﬁned ( f

(x) = f

( y) ∈ Z

) if and only if 6 | 4kn for all n ∈ Z. This is

the case if and only if 6 | 4k. Otherwise said,

is well-deﬁned ⇐⇒ 6 | 4k ⇐⇒ 3 | 2k ⇐⇒ 3 | k

Since kx ∈ Z

, equivalent values of k modulo 6 won’t

change f

: there are only two well-deﬁned functions

(x) = 0 and f

(x) = 3x.

x 0 1 2 3 4 5 6 ···

(x) 0 0 0 0 0 0 0 ···

(x) 0 3 0 3 0 3 0 ···

You’ll have much more practice with problems like these when you study group theory.

Theorem 7.27 veriﬁes the well-deﬁnition of two functions +

, ·



∼



∼

→



∼

107

Just for Fun: Geometric Applications (optional!)

We consider how equivalence relations permit the easy construction of certain geometric objects and

of functions on such.

Examples 7.30. 1. Deﬁne an equivalence relation on R

(a, b) ∼ (c, d) ⇐⇒ a −c ∈ Z and b = d

The equivalence classes are horizontal strings of points with the same y co-ordinate:

[(a, b)] =



(a + n, b) : n ∈ Z



The set of equivalence classes



∼

may be visu-

alized as a cylinder by imagining rolling the plane

into a tube of circumference 1 so that all points in

a given equivalence class coincide.

Now consider the rule f ([(x, y])) = y sin(2πx).

wrap

around

[(x, y)] = [(v, w)] =⇒ y = w and x = v + n, for some n ∈ Z

=⇒ f



[(x, y)]



= y sin(2πx) = w sin(2πv) = f



[(v, w)]



Otherwise said, f :



∼

→ R is a well-deﬁned function whose domain is the cylinder. More

generally, and F : R

→ R for which (x, y) ∼ (v, w) =⇒ F(x, y) = F(v, w) may be viewed as a

function on the cylinder.

2. As a natural extension, see if you can visualize why the equivalence classes of the relation

(a, b) ∼ (c, d) ⇐⇒ a −c ∈ Z and b −d ∈ Z

on R

may be identiﬁed with a torus (the surface of a ring-doughnut). The function



[x, y]



= sin(2πx) cos(2πy)

is well-deﬁned and may thought of as having domain the torus. The pictures below illustrate

f , where the colors correspond.

0 1

−1.0

−0.5

0.0

+0.5

+1.0



[(x, y) ]



= sin(2πx) cos(2πy)

Alternatively, imagine piercing a roll of toilet paper and unrolling it: the single puncture becomes a row of (almost!)

equally spaced holes. Unfortunately for the analogy, toilet paper has purposeful thickness!

108

Equivalence relations and partitions are regularly used in this manner to describe objects in geometry

and topology. Here is a ﬁnal famous example written using the language of partitions.

Example 7.31 (M¨obius Strip). Partition the a rectangle X as follows:

• If P ∈ X does not lie on the left or right edges, place it in a subset {P} by itself.

• Otherwise, pair Q with a point on the other edge identiﬁed in the opposite direction





These subsets clearly partition the rectangle X and thus describe an equivalence relation ∼on X. The

quotient set



∼

may be visualized via the classic construction of a M

obius strip: give the rectangle

a half-twist and glue the two edges so that both points in each equivalance class coincide. This

construction allows us to analyse the strip while working on the rectangle: if you walk off the right

side of the rectangle (at

Q) you simply end up at the corresponding point (Q) on the left side!

Rectangle Half-twist and glue

Exercises 7.4. A reading quiz and practice questions can be found online.

1. Let X be the set of students in a (large) math class and deﬁne an equivalence relation on X via

x ∼ y ⇐⇒ x, y either both wear glasses, or both do not wear glasses

Does the rule f ([x]) = “x’s name” describe a well-deﬁned function? Explain.

2. (a) Explicitly check that [7] + [21] = [98] + [−5] in Z

(b) Suppose that [5] ·[7] = [8] ·[9] makes sense in the ring Z

. Find n.

3. Prove the second half of Theorem 7.27, that ·

is well-deﬁned.

4. Suppose that p is prime and that in Z

, we have [a] = [0] . Show [a]

= [0].

(Hint: Recall Exercise 3.2.13)

5. Give an explicit proof of Theorem 7.23.

6. Determine whether the following are well-deﬁned.

(a) f : Z

→ Z

: [x]

7→ [x

]

(b) g : Z

→ Z

: [x]

7→ [x

]

→ Z

where h(x) = 2x (equivalently h(x) = 2x (mod 16), or h([x]

) = [2x]

(d) j : Z

→ Z

where j(x) = 2x.

7. (a) Suppose ∀n ∈ Z, (x + 4n)

≡ x

(mod m). Find all integers m ≥ 2 for which this is true.

(b) For what m ∈ N

≥2

is the function f : Z

→ Z

: x 7→ x

(mod m) well-deﬁned.

8. In the sense of Example 7.30, can we view F(x, y) = (y

− 1) sin

( πx) as a function whose

domain is the cylinder? Explain.

109

9. (a) Suppose f :



∼

→ A is well-deﬁned. State what it means for f to be injective. What do

you notice?

(b) Prove that f : Z

→ Z

: [x]

7→ [15x]

is a well-deﬁned, injective function.

100

→ Z

300

: [x]

100

7→ [9x]

300

(Hint: You may ﬁnd it useful that 9 ·(−11) ≡ 1 (mod 100))

10. Deﬁne a partition of the sphere S



(x, y, z) : x

+ y

+ z

= 1



into subsets consisting of

pairs of antipodal points:



(x, y, z), (−x, −y, −z)



Let ∼ be the equivalence relation whose equivalence classes are the above subsets.

(a) f :



∼

→ R : [(x, y, z)] 7→ xyz is not well-deﬁned. Explain why.

(b) Prove that f :



∼

→ R

: [(x, y, z)] → (yz, xz, xy) is a well-deﬁned function.

The image of this function is Steiner’s Roman Surface; look it up!

11. Consider the relation (a, b) ∼ (c, d) ⇐⇒ ad = bc on Z ×N.

(a) Prove that ∼ is an equivalence relation.

(b) List several elements of the equivalence class [(2, 3)]. Repeat for [(−3, 7)]. What does ∼

have to do with the set of rational numbers Q?

Z ×N



∼

[(a, b)] + [(c, d)] = [(ad + bc, bd)] [(a, b)] × [(c, d)] = [(ac, bd)]

Prove that + and × are well-deﬁned (do this without using division!).

(d) (Hard) Prove that f



[(x, y)]



is a well-deﬁned bijection f :

Z ×N



∼

→ Q, and that f

transforms + and × into the usual addition and multiplication on Q: that is,



[(a, b)] + [(c, d)]



= f



[(a, b)]



+ f



[(c, d)]



, etc.

(This is essentially a formal deﬁnition of the ring of rational numbers!)

12. Deﬁne ∼ on R by x ∼ y if and only if x −y ∈ Z.

(a) Show ∼ is an equivalence relation on R.

(b) Find an example of a surjective function F : R → [0, 1) such that x ∼ y =⇒ F( x) = F(y).



∼

→ [0, 1).

13. Suppose ∼ is an equivalence relation on X and let γ(x) = [x]. Prove the following:

(a) If f :



∼

→ A is well-deﬁned, then F = f ◦γ : X → A satisﬁes x ∼ y =⇒ F( x) = F(y).

(b) If F : X → A satisﬁes x ∼ y =⇒ F(x) = F(y), then there is a unique function f :



∼

→ A

satisfying F = f ◦ γ.

14. (Just for fun!) Prove that you can cut a M

obius strip round the middle yet

still end up with a single loop.

Look up the description of a Klein bottle: how is the relationship between

it and the M

obius strip similar to the torus/cylinder relationship?

110

8 Cardinality and Inﬁnite Sets

During the late 1800’s, Georg Cantor assaulted the foundations of mathematics with his investiga-

tions of certain inﬁnite sets, in particular his middle-third set (Example 6.18). Cantor’s ideas about

inﬁnity met signiﬁcant resistance: some mathematicians and philosophers considered his approach

unnatural; he even inﬂamed religious scholars who thought his ideas an affront to the divine!

Despite initial skepticism, Cantor’s notion of cardinality is now universally accepted. By unearthing

several contradictions inherent in contemporary (na

ıve) set theory, mathematicians became con-

vinced that a rigorous axiomatic approach was necessary. Developing an axiomatic foundation to

mathematics became a core goal of early 20

century mathematics.

8.1 Cantor’s Notion of Cardinality

Recall that the cardinality

of a ﬁnite set A is simply the number of elements of A (Deﬁnition 4.8).

While “number of elements” is meaningless for inﬁnite sets, cardinality also gives rise to a relation ≤

which can be used to compare sets; this interpretation does extend to inﬁnite sets. . .

Example 8.1. Consider two sets

A = {ﬁsh, dog} B = {α, β, γ}

While the elements of A and B are completely different, the sets themselves may be compared using

cardinality: we write

≤

to indicate that B has at least as many elements as A. In Theorem

4.25, we saw that this mandates the existence of an injective function f : A → B: plainly

f (ﬁsh) = α, f (dog) = β

deﬁnes a suitable function.

Theorem 4.25 tells us how functions may be used to compare cardinalities of ﬁnite sets without counting

elements. Cantor’s seemingly innocuous idea was to turn this theorem for ﬁnite sets into a general

deﬁnition of cardinality.

Deﬁnition 8.2. The cardinality of a set A is denoted

. We compare cardinalities as follows:

•

≤

⇐⇒ ∃f : A → B injective.

•

⇐⇒ ∃g : A → B bijective.

means

≤

and

=

, that is ∃f : A → B injective but ∄g : A → B bijective.

Lemma 8.3. Equality of cardinality

is an equivalence relation on any collection of sets.

“Cardinality” is an abstract property whereby sets can be compared, rather than a value attaching to

a given set.

Regardless, the Lemma proves that cardinality partitions any collection of sets: every

set has a cardinality, and no set has more than one cardinality. It is moreover natural to identify the

cardinalities of ﬁnite sets with the cardinal numbers 0, 1, 2, 3, 4, . . .

We could use the Lemma to deﬁne

:= [A] as the equivalence class of A, though this likely feels confusing!

111

Countably Inﬁnite Sets

To make progress, it is helpful to introduce a symbol for the cardinality of the simplest inﬁnite set.

Deﬁnition 8.4. We say that A is a countably inﬁnite

set and write

= ℵ

(aleph-nought or aleph-

null), if its cardinality equals that of the set of natural numbers N:

= ℵ

⇐⇒ ∃g : N → A bijective

In the next section we’ll see why a new symbol is necessary; why ∞ doesn’t sufﬁce.

Examples 8.5. 1. The function g : N → N

≥2

deﬁned by g(n) = n + 1 is a bijection, whence

≥2

= {2, 3, 4, 5, . . .} is countably inﬁnite.

2. Let 2N = {2, 4, 6, 8, 10, . . .} be the set of positive even integers. The function

h : N → 2N : n 7→ 2n

is a bijection. It follows that

= ℵ

and that 2N is countably inﬁnite.

Theses examples demonstrate a perplexing property of inﬁnite sets. For instance, 2N is a proper subset

in bijective correspondence with N! It feels like we want to say two contradictory things:

• N has the ‘same number of elements’ as 2N (bijectivity of h).

• N has ‘twice the number of elements’ as 2N (‘half’ of N is even, and ‘half’ odd).

The source of our discomfort is that number of elements is meaningless for inﬁnite sets. However, if we

replace this phrase with cardinality, then both statements are true!

The existence of a proper subset

with the same cardinality can indeed be used as a deﬁnition of inﬁnite set (Exercise 13).

Rather than hunting for an explicit bijection, the simplest approach to these problems is often to list

the elements of a set A so that it ‘looks like’ the natural numbers, with an initial element a

followed

by all others in sequence:

A = {a

, a

, . . .} indicates a bijection g : N → A : n 7→ a

, whence

= ℵ

For instance, we could have approached our examples by listing elements in tabular form:

N n 1 2 3 4 5 6 7 8 9 10 ···

≥2

g(n) 2 3 4 5 6 7 8 9 10 11 ···

2N h(n) 2 4 6 8 10 12 14 16 18 20 ···

The required bijections are immediately visible with little need to state them explicitly. We apply this

technique to two important examples of countably inﬁnite sets.

Some authors say denumerable and use countable for sets with

≤ ℵ

. Aleph is the ﬁrst letter of the Hebrew alphabet.

This is a version of Hilbert’s Grand Hotel Problem. Imagine a hotel with an inﬁnite number of rooms: Room 1, Room 2,

Room 3, Room 4, etc. Even if the hotel is full, by moving everyone one room down the hall, space can always be found for

an additional guest. The second example is another version, where inﬁnitely many new guests may be accommodated!

We won’t pursue cardinal arithmetic in any detail, but it is completely legitimate to write 2ℵ

= ℵ

for the second

example. The ﬁrst example would be 1 + ℵ

= ℵ

112

List the set of integers in a non-standard order, alternating positive and negative terms:

Z = {z

, z

, . . .} = {0, 1, −1, 2, −2, 3, −3, 4, −4, . . .}

The function g : N → Z : n 7→ z

is bijective, and we conclude:

Theorem 8.6. The integers Z are a countably inﬁnite set.

You might feel that our argument was too quick! Is it really obvious what g is? Bijectivity is the

observation that every integer appears exactly once in our non-standard ordering. If you really want,

you can construct an explicit formula for g from a tabular representation

n 1 2 3 4 5 6 7 8 9 ···

g(n) 0 1 −1 2 −2 3 −3 4 −4 ···

=⇒ g(n) =

(

n if n even

−

( n −1) if n odd

and prove bijectivity using the formula. In practice, it is not worth the cost to be this explicit.

As you build up examples, you no longer have to compare directly to the natural numbers. Since

composition of bijective functions is bijective (Theorem 4.22/transitivity in Lemma 8.3),

B is countably inﬁnite ⇐⇒ ∃A countably inﬁnite and ∃g : A → B bijective

For instance, the set of even integers 2Z is denumerable via the bijection g : Z → 2Z : z 7→ 2z.

We use this approach to help prove the ﬁrst of Cantor’s truly counter-intuitive revelations.

Theorem 8.7. The rational numbers Q are countably inﬁnite.

Any sensible person should feel that there are far, far more rational numbers than integers, yet the

sets have the same cardinality. Bizarre!

Proof. For each pair of natural numbers a, b, place the fraction

in the b

row, a

column of an

inﬁnite square. List the positive rational numbers by tracing diagonals in a snake-like manner and

deleting any number that has already been traced (

, etc.).

Since

each appears in diagonal a + b −1 and repeats are

deleted, every positive rational number appears exactly once.

We therefore obtain an enumeration



, . . .



and conclude that Q

is countably inﬁnite. Plainly this

corresponds to a bijection g : N → Q

. To ﬁnish the proof,

extend g by deﬁning the bijective function

h : Z → Q : n 7→











g(n) if n > 0

0 if n = 0

−g(−n) if n < 0

···

which identiﬁes the negative rationals with Z

−

. By Theorem 8.6, we deduce that

= ℵ

113

Other countably inﬁnite sets appear to be even larger than Q! For example:

• Cartesian products such as Q ×Q.

• The algebraic numbers A =



x ∈ C : p(x) = 0 for some polynomial p with integer coefﬁcients



Every rational number is algebraic (

is a root of p(x) = bx − a), but so are many irrationals

(

√

2 is a root of p(x) = x

−2). Non-algebraic numbers (e.g., π and e) are termed transcendental.

Arguments for these examples are left to the exercises.

The Least Inﬁnite Cardinal?

We introduced ℵ

to represent the cardinality of the ‘simplest’ inﬁnite set N. Here is a compelling

reason why the natural numbers should indeed be considered the most simple inﬁnite set.

Theorem 8.8. A is ﬁnite if and only if

< ℵ

Otherwise said, every inﬁnite set has cardinality at least as large as the natural numbers: ℵ

may

therefore be considered the least inﬁnite cardinal.

Proof. (⇒) Express the set in roster notation A = {a

, . . . , a

}. We must prove two things:

(

≤ ℵ

) Deﬁne f : A → N by f (a

) = k for each k ∈ {1, 2, 3, . . . , n}. This is injective since

the distinct elements a

of A map to distinct integers.

(

= ℵ

) We show there are no bijections N → A. Suppose g : N → A and consider the set



{1, . . . , n + 1}





g(1), . . . , g(n + 1)



⊆ A

Since A has n elements, at least two of the values g(1), . . . , g(n + 1) must be equal. There-

fore g is not injective and consequently not bijective.

(⇐) See Exercise 11.

We’ll address the existence of inﬁnite sets with cardinality larger than ℵ

in the next section.

Exercises 8.1. A reading quiz and practice question can be found online.

1. Refresh your proof skills by proving explicitly that the following functions are bijections:

(a) g : N → N

≥2

: n 7→ n + 1 (b) h : N → 2N : n 7→ 2n

2. Prove that Z

≥−3

is countably inﬁnite by constructing a bijective function g : N → Z

≥−3

3. Prove that the set 3Z + 2 = {3n + 2 : n ∈ Z} is countably inﬁnite.

4. Show that the set of all triples of the form (n

, 5, n + 2) with n ∈ 3Z is countably inﬁnite by

providing an explicit bijection with a known countably inﬁnite set.

5. State a bijection g : (0, 1) → (4, 6) which shows that these intervals have the same cardinality.

The n = 0 case (A = ∅) works, though it feels strange: f = ∅ is a suitable injective (!) function f : ∅ → N, and there

are no functions g ⊆ N → ∅. It will help to think about the formal deﬁnition of function (Section 7.2)!

114

6. Prove Lemma 8.3 (revisit Theorem 4.22 on the composition of bijective functions.)

7. Prove that A ⊆ B =⇒

≤

(show there exists an injective function f : A → B).

8. (a) Prove that N ×N is countably inﬁnite by modifying the proof of Theorem 8.7.

(b) Combine part (a) with Theorem 8.7 to prove that Q ×Q is countably inﬁnite.

is countably inﬁnite for each n ∈ N and list elements as follows:

= {a

, a

, . . .}

= {a

, a

, . . .}

= {a

, a

, . . .}, etc.

Prove that

is countably inﬁnite (a countable union of countable sets is countable).

Warning! The remaining exercises are signiﬁcantly trickier.

9. Suppose A = ∅. Prove that

≤

if and only if there exists a surjective function g : B → A.

(Hint: Use g to construct an injective f : A → B, and vice versa)

10. Let A =



x ∈ C : p(x) = 0 for some polynomial p with integer coefﬁcients



be the set of

algebraic numbers. We prove that A is countably inﬁnite.

(a) Let M ∈ N. Prove that there are only ﬁnitely many choices of d ∈ N and a

, . . . , a

∈ Z

such that M = d +

+ ···+

(b) Let P



+ ···+ a

x + a

: d +

+ ···+

= M



. Explain why P

is ﬁnite.



x ∈ C : p(x) = 0 for some p ∈ P



is ﬁnite by using the fact that a

degree d polynomial has at most d roots.

(d) Prove that A =

M∈N

and conclude that A is countably inﬁnite.

11. We complete the proof of Theorem 8.8: if

< ℵ

, then A is a ﬁnite set.

We prove by contradiction. Suppose A is inﬁnite and that

< ℵ

. Then there exists an

injective function f : A → N. List the range of f in increasing order:

range( f ) = {n

, n

, . . .} n

< n

< ···

(a) Show that for all k ∈ N, there exists a unique a

∈ A satisfying f (a

) = n

(b) Deﬁne g : N → A by g(k) = a

. Prove that g is a bijection and hence obtain a contradiction.

12. Suppose C is countably inﬁnite and let c ∈ C.

(a) Show that there exists a bijection h : N → C with h(1) = c.

(b) Prove that D = C \{c} is countably inﬁnite.

(Hint: h be as in part (a) and use g from Example 8.5 to construct k : N → D)

13. (a) Suppose

≥ ℵ

. Show that there exists C ⊆ A for which

= ℵ

(b) Prove that a set A is inﬁnite if and only if it has a proper subset B with

(Hint: use part (a) and Exercise 12)

14. Describe an explicit bijection g : [0, 1] → (0, 1), and thus demonstrate that the intervals have the

same cardinality.

(Hint: {

: n ∈ N

≥2

} is a countably inﬁnite subset of (0, 1))

115

8.2 Uncountable Sets

Since Q seems so large, you might imagine that no sets could have strictly larger cardinality. But we

haven’t yet thought about the real numbers. . .

Deﬁnition 8.9. A set A is uncountable if ℵ

. Otherwise said, there exists an injection f : N → A

but no bijection g : N → A.

Theorem 8.10. The interval (0, 1) of real numbers is uncountable.

We denote the cardinality of the interval (0, 1) by c for continuum. The theorem may therefore be

written ℵ

< c. We prove by showing that ℵ

≤ c and ℵ

= c.

Proof. (ℵ

≤ c) The function f : N → (0, 1) : n 7→

n+1

is plainly injective:

f (n) = f (m) =⇒

n + 1

m + 1

=⇒ n = m

(ℵ

= c) Suppose, for contradiction, that g : N → (0, 1) is a bijection. Express the sequence of values

g(1), g(2), g(3), . . . as decimals:

g(1) = 0.b

···

g(2) = 0.b

···

g(3) = 0.b

···

g(4) = 0.b

···

g(5) = 0.b

···

where each b

∈ {0, . . . , 9}

Deﬁne a new decimal

x := 0.x

··· ∈ (0, 1) where x

(

1 if b

= 1

2 if b

= 1

Since x disagrees with g(n) at the n

decimal place, we see that x = g(n): that is, x is not in the

above list. However x ∈ (0, 1) and g is surjective, so x must be in the list: contradiction.

The second part of the proof is known as Cantor’s diagonal argument, since we compare the constructed

decimal x with the diagonal of an inﬁnite square of integers. Since the interval (0, 1) is uncountable,

and (0, 1) ⊆ R, it is immediate that the real numbers are also uncountable. Using only the ideas

developed so far (a combination of Exercises 8.1.5 and 14), we could prove directly that every interval

of ﬁnite length has cardinality c. It is easier, however, to delay this momentarily. . .

Even more amazingly, Cantor’s middle-third set (Example 6.18) also has cardinality c, despite seem-

ing vanishingly small! The details, and more, are in Exercise 12.

A number x ∈ (0, 1) has two decimal representations if and only if one of them terminates and the other ultimately

becomes an inﬁnite sequence of 9’s: e.g., 0.135 = 0.1349999 . . .. For this proof, we choose the terminating decimal whenever

it exists. We restrict to x

= 1, 2 later in the proof to keep away from these double representations.

116

Non-explict Comparison of Cardinalities

The following result is very useful for comparing cardinalities.

Theorem 8.11 (Cantor–Schr¨oder–Bernstein). If

≤

and

≤

, then

This seems like it should be obvious, but pause for a moment: it is not a result about numbers! The

theorem should be understood in the context of Deﬁnition 8.2, in which language it becomes:

If there exist injections f : A → B and g : B → A, then there exists a bijection h : A → B

The proof is beautiful, though a little long to reproduce here; if you’re interested, check out any ele-

mentary text on set theory. Its usefulness is that it allows us to equate cardinalities without explicitly

constructing bijective functions; injective functions are typically easier to conjure!

Example 8.12. The following functions are both injective (only—they are not bijective!):

f : (0, 1) → [0, 1] : x 7→ x (range( f ) = (0, 1) ⊊ [0, 1])

g : [0, 1] → (0, 1) : x 7→

x +

(range(g) = [

] ⊊ (0, 1))

By Cantor–Schr

oder–Bernstein, the sets (0, 1) and [0, 1] have the same cardinality (c). Note how fast

this was; we didn’t have to construct an explicit bijection h : (0, 1) → [0, 1], a much nastier business

(see Exercise 8.1.14)

As recently advertised, this example generalizes to cover all intervals.

Corollary 8.13. All intervals (of positive length) have cardinality c.

Proof. Let A be an interval (could be inﬁnite length) and choose a subinterval (a, b) ⊆ A. The follow-

ing functions are injective:

• f : (0, 1) → A where f (x) = a + (b − a)x; this maps

(0, 1) onto range( f ) = (a, b) ⊆ A.

• ι : A → R where ι(x) = x.

• g : R → (0, 1) where g(x) =

tan

−1

x; this is in

fact bijective with inverse g

−1

( y) = tan



πy −



−2 −1 0 1 2

g(x) =

tan

−1

Putting everything together:

(0, 1)

≤

(0, 1)

By Cantor–Schr

oder–Bernstein, all these sets have the same cardinality, namely c.

Further examples can be found in the exercises.

117

Cantor’s Paradoxical Theorem

For a ﬁnal punchline, we generalize Theorem 6.8 which, for ﬁnite sets A, asserted that

P(A)

= 2

is strictly larger than A itself. We now have the technology to attack this for inﬁnite sets.

Theorem 8.14 (Cantor). If A is any set, then

⪇

P(A)

The main implication is that there is no largest cardinality! We can always construct a set with strictly

larger cardinality just by taking the power set. For example,

P(R)

= c. Want an even larger

cardinality? Try P



P(R)



, or P



P(P(R))



! This process may be continued indeﬁnitely.

Proof. If A = ∅, the result is trivial. Otherwise, ﬁrst observe that f : a 7→ {a} deﬁnes an injective

function f : A → P(A), whence

≤

P(A)

To complete the argument we must show that no bijective function g : A → P(A) can exist. Suppose,

for a contradiction, that g : A → P(A) is bijective, and consider the set

X =



a ∈ A : a /∈ g(a)



Take stock for a moment and think about X. Since g(a) is a subset of A, the condition a ∈ g(a) is

legitimate, whence X is a genuine subset of A (X ∈ P(A)). A simple example will hopefully help.

Example 8.15. Let A = {1, 2, 3} and deﬁne a function g : A → P(A) by

g(1) = {1, 2}, g(2) = {1, 3}, g(3) = ∅

Since 1 ∈ g(1), 2 /∈ g(2), and 3 /∈ g(3), we see that X = {2, 3}. Since our goal is to prove that no

bijection A → P(A) can exist, it is important to note that this g is not bijective; indeed g isn’t surjective,

since X /∈ range(g) =



{1, 2}, {1, 3}, ∅



. This last observation is what ﬁnishes the proof. . .

Proof Continued. By assumption, g is surjective. Thus X ∈ range(g). Otherwise said, X = g(a) for

some a ∈ A. We ask whether this element a lies in the set X:

a ∈ X ⇐⇒ a /∈ g(a) (deﬁnition of X)

⇐⇒ a /∈ X (since X = g(a))

The conclusion a ∈ X ⇐⇒ a /∈ X is plainly a contradiction! No bijection g : A → P(A) can exist,

and so

⪇

P(A)

Cantor’s theorem played a key role in pushing set theory towards axiomatization, in part because of

a simple paradox. If a ‘set’ is merely a collection of objects, we may consider the ‘set of all sets’ S. Its

power set P(S) is a set of sets, which must be a subset of S. Plainly

P(S)

≤

; but this contradicts

Cantor’s theorem!

The remedy is a rigorous deﬁnition of ‘set.’ Axiomatic set theory describes a small number of legitimate

ways to build sets, of which we’ve seen several in these notes: e.g., union, power set, set-builder

notation. In particular, the ‘set of all sets’ cannot be legitimately constructed.

The critical condition for preventing Cantor’s paradox is that set-builder notation {x ∈ A : P(x)} can only produce a

subset of an already existing set A. The ‘set of all sets’ would have the form {x : P(x)} where x is unrestricted.

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Some Final Thoughts on the Limits of Proof

During this course we’ve learned some of the basic methods and concepts used by mathematicians.

In particular, we’ve learned how to use proofs to demonstrate the truth of statements about mathe-

matical objects. As we ﬁnish, it makes sense to reﬂect on the limits of our methods.

By the early 20

century, the discovery of various paradoxes and contradictions (such as Cantor’s)

caused a foundational crisis in mathematics. If a concept as basic as set is self-contradictory, how

are we to have faith in any mathematical conclusion?! The response to this crisis was an effort to

formulate a list of reasonable axioms from which all mathematics could be derived using basic logical

reasoning. Such an axiomatic foundation would ideally satisfy two conditions:

• Consistency: No contradiction can be derived from the axioms.

• Completeness: All true mathematical statements could be derived from the axioms.

Any hope for such a foundation was crushed in 1931, when Kurt G

odel published his famous Incom-

pleteness Theorems, showing that no such axiomatic system could exist. Very roughly, G

odel showed

that in any consistent axiomatic system strong enough to produce some basic arithmetic, there are

undecideable statements; neither deducible nor refutable from the axioms. Perhaps even worse, no

such system can prove its own consistency.

While the strongest aims of early 20

axiomatics cannot be accomplished, contemporary research

was able to provide a foundation that most modern mathematicians deem adequate. The most pop-

ular approach is to base all of mathematics on set theory—as your studies progress, you’ll see that

many of the objects you study can be formalized as sets together with functions and relations be-

tween them. We’ve started this work already: Chapter 7 says that functions and relations are them-

selves sets! Numbers like 0, 1, 2,

or even π = 3.14 . . . can be thought of as sets if one so desires.

In turn, set theory is often axiomatized using the ZFC axioms (short for Zermelo–Fraenkel set theory

with the Axiom of Choice).

While the ZFC system remains subject to G

odel’s limitations,

it has proven able to formalize most

of the mathematics actually used by current mathematicians, and has not (thus far!) produced any

inconsistencies. While there is plenty of fun to be had exploring set theory, its history and its quirks,

most modern mathematicians feel little need to dwell on the foundational issues of last century!

Exercises 8.2. A reading quiz and practice question can be found online.

1. Decide the cardinality of each set. No working is necessary.

(a) N

≤12

(b) Z

≤12



{R}



(f)

x∈R

[3 −

, 3 +

)

2. Find explicit bijections (thus showing that the given intervals have the same cardinality):

(a) f : [2, 3) → [1, 5) (b) g : [2, 3) → (1, 5] (c) h : (−3, 2) → R (d) j : R → (1, ∞)

(Hint: The proof of Corollary 8.13 should provide some inspiration—be creative)

3. Let B = [ 3, 5) ∪ (6, 10). Use the Cantor–Schr

oder–Bernstein Theorem to prove that

= c.

(Hint: State injective functions f : (0, 1) → B and g : B → (0, 1))

Perhaps the most famous undecidable statement in ZFC is relevant to our recent discussion: the continuum hypothesis is

the claim that no set has cardinality strictly between ℵ

and c; that intervals are the simplest (‘smallest’) uncountable sets.

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4. (a) Prove that f : N × N → N deﬁned by f (m, n) = 2

is injective.

(b) Use part (a) and the Cantor–Schr

oder–Bernstein Theorem to conclude that

N × N

= ℵ

| {z }

k times

| = ℵ

(d) Use part (b) to give an alternative proof that

= ℵ

5. Revisit the proof of Cantor’s Theorem, and Example 8.15.

(a) Suppose g : {1, 2, 3, 4} → P({1, 2, 3, 4}) is deﬁned by

g(1) = {2, 3}, g(2) = {1, 2}, g(3) = ∅, g(4) = {1, 2, 4}

Compute the set X =



a ∈ {1, 2, 3, 4} : a ∈ g(a)



(b) Repeat part (a) for g : N → P(N) : n 7→ {x ∈ 2N : x ≤ n}.

(Hint: Try some examples ﬁrst! What is g( 1)? Is 1 ∈ g(1)? What about 2 ∈ g(2)?)

6. Let I = R \Q be the set of irrational numbers.

(a) Prove that

≤ c.

(b) Prove that x ∈ N =⇒ x +

√

2 ∈ I. Hence conclude that ℵ

≤

= c is harder to show).

(d) Show that the transcendental (non-algebraic) numbers are uncountable (see Exercise 8.1.10).

7. Give an example of an uncountable set I and an indexed collection {A

: n ∈ I} for which all

the following conditions hold:

• Each A

is countably inﬁnite.

• If m = n, then A

= A

• For all m, n, either A

⊆ A

or A

⊇ A

•

n∈I

is countably inﬁnite.

8. The proof of Cantor’s Theorem makes use of a construction similar to Russell’s paradox. Let X

be the ‘set’ of all sets which are not members of themselves:

X = {A : A ∈ A}

(a) Suppose X is a set. By asking yourself if X is a member of itself, deduce a contradiction.

(The point of Russell’s paradox is that objects like X cannot be considered sets!)

(b) Russell’s paradox is one version of an ancient logical conundrum with many guises. E.g.,

A town has one barber, who cuts the hair of all townspeople, and only those

people, who do not cut their own hair: Who cuts the barber’s hair?

Can you explain the connection between this and Russell’s paradox?

9. In Exercise 6.3.14, we saw that Cantor’s middle-third set C is the set of all numbers in [0, 1] pos-

sessing a ternary expansion consisting only of 0’s and 2’s. By modifying the proof of Theorem

8.10, argue that C is uncountable.

(Exercise 12 establishes the stronger result

= c)

120

The remaining questions are more of a challenge: if these seem interesting, consider taking a

set theory course!

10. Express a real number x ∈ (0, 1) as a decimal x = 0.x

. . . where we choose the terminat-

ing decimal whenever there is a choice (footnote 56). Prove that

f : (0, 1) ×(0, 1) → (0, 1) deﬁned by f (x, y) = 0.x

. . .

is injective. Hence conclude that



= c.

11. If A and B are non-empty sets we let A

denote the set of all functions f : B → A.

(a) If A = {0, 1} and B = {a, b, c}, list all elements of A

. What is



(b) If A and B are ﬁnite sets, show that



: B → {0, 1} where

(x) =

(

1 if x ∈ Y

0 if x /∈ Y

Prove that Φ : P(B) → {0, 1}

deﬁned by Φ(Y) = χ

is bijective.

(If B is ﬁnite, this provides another proof that

P(B)

= 2

)

12. Let x ∈ [0, 1]. A binary expansion of x is a sequence (b

) of zeros and ones such that

x =

∞

∑

n=1

The binary expansion of x ∈ [0, 1] is (almost) unique;

if there is a choice, take the terminating

expansion. Deﬁne a function f : [0, 1] → P(N) (the set of subsets of N) by

f (x) = {n ∈ N : b

= 1 in the binary expansion of x}

(a) Prove that f is injective, and that, consequently, c ≤

P(N)

(b) Is f surjective?

(Hint: consider the set X = {2, 3, 4, . . .} ∈ P(N))

g(X) =

∞

∑

n∈X

(d) Use the Cantor–Schr

oder–Bernstein Theorem to conclude that

P(N)

= c.

(Together with Exercise 11, this shows why we could write c = 2

ℵ

)

Binary expansions are unique unless x has a terminating expansion, in which case the there is a second representation

with an inﬁnite string of 1’s: e.g., [0.011111 ···]

= [0.1]

. This discussion is beloved of computer scientists who, following

Exercise 11, might view P(N) ∼ {0, 1}

as the set of binary sequences/strings (x

, x

, . . .), where each x

∈ {0, 1}.

121