psm - c1 - general course information

Proof and Structure in MathematicsLecturers: Sarah Hart and Ben Fairbairn

Module Code: EMMS095S5 Level: 5 Credit Value: 30 Credits

Web: http://www.ems.bbk.ac.uk/for students/bsc maths/coursematerials/PSM

Aims This module aims to introduce the essential logical and theoretical concepts necessaryfor the degree-level study of mathematics. At the end of the module you will have obtaineda thorough grounding in the basic mathematical concepts which will be used throughout thedegree programme. What makes mathematics different from the sciences is the emphasis placedon proof. In this module you will be introduced to various common methods of proof, such asproof by contradiction and proof by induction, as well as the basic components of propositionallogic: connectives, quantifiers and truth tables. The main number sets (natural numbers, inte-gers, rational, real and complex numbers) will be defined and their properties studied. Binaryrelations and binary operations and the concept of a group will be introduced as well.

Teaching and Assessment Teaching for this module will take place throughout the year,with eight evenings of lectures on Monday nights in each of the Autumn and Spring Terms andtwo evenings of revision and consolidation in the Summer Term (dates for all these go onlineas soon as they are available). Of the final mark for the module, 80% is based on a three-hourexam in June and the other 20% is from assessed coursework. Coursework consists of a shortproblem set (worth 5%) for each four-week teaching block; you will have at least three weeksto complete each one. The examination in May/June has two sections. Section A (worth 40%)consists of eight compulsory short questions. Section B (also worth 40%) contains four longerquestions of which you must answer two.

Resources for the Module Typed lecture notes will be provided. These will cover all youneed to know. However in lectures there may be additional examples for extra practice whichyou will find useful to write down. In the notes there are worked examples and exercises foryou to try, including some exam-style questions at the end of each chapter. Full solutions willbe given to these exercises in due course. Two past examination papers will be provided in theSummer Term. All resources including coursework will be made available online.

I will be on maternity leave from the beginning of December; I will teach the first six lectures,finishing November 22nd. Ben Fairbairn will teach the rest of the module starting with lectureson the 6th and 13th December. If you have any difficulties, feel free to ask either in or outsideof lectures. Until the end of November, you can email me at [email protected], phone on 0207631 6437 or come to my office, room 753 of the main Malet Street building. I have an ‘officehour’ during teaching weeks when I am available without appointment and you can drop inwith any queries. This year it is on Mondays from 4:30 to 5:45 pm. At other times it is best tomake an appointment by email or phone. From December, Ben Fairbairn will be your contact.He will have the same office and phone number.

2 Proof & Structure

Syllabus

1. The Language of Mathematics: Statements; theorems; definitions; logical connectives(not, and, or, implies); truth tables; tautologies and equivalent statements; the universaland existential quantifiers; negating if/then statements and statements involving quan-tifiers; some elementary proofs; contrapositive proofs; converses and counterexamples;proof by contradiction; Euclid’s proof of the infinity of primes; proof by induction; stronginduction. 4 weeks

2. The Number Sets: Integers, rationals and real numbers: The well ordering property;the division theorem; the fundamental theorem of arithmetic; the Euclidean division al-gorithm; algebraic properties of the natural numbers and integers; congruence and mod-ular arithmetic; solution of linear congruences; The rational numbers, the real numbers;boundedness; the completeness axiom. 4 weeks

3. Complex numbers and Infinite Sequences: the complex numbers: arithmetic ofcomplex numbers, polar form, De Moivre’s Theorem and applications. Definition of asequence; limits and other terminology; null sequences; combining null sequences; con-vergent sequences; combining convergent sequences; the ‘sandwich theorem’; divergentsequences; the monotone convergence theorem; infinite series. 4 weeks

4. Binary relations, Binary operations and Groups: binary relations and equivalencerelations; binary operations; the properties of commutativity and associativity; identityelements and inverses; idempotents; operation multiplication tables; definition of a group;examples from geometry, permutations, matrices, number sets; cyclic groups and abeliangroups; orders of elements and groups; subgroups; Lagrange’s Theorem. 4 weeks

In addition there will be 2 evenings of consolidation and revision lectures in the Summer Term.

Learning Outcomes Learning outcomes for the module are provided in your handbook andare also available on the web. At the end of each chapter of the notes you will find more specificlearning outcomes for that chapter, such as particular theorems that you need to know.

Recommended Books I will cover in lectures everything you need to know to pass themodule; however you may find the following books of interest; they contain many additionalexamples and exercises which you may wish to try. They are all in the library.

• Numbers and Proofs, R.J.B.T. Allenby, Arnold, Butterworth-Heinemann (1997).

• Sets, Functions and Logic, Keith Devlin, Chapman & Hall (2003).

• An introduction to mathematical reasoning, P.J. Eccles, CUP, Cambridge (1997).

• A concise introduction to Pure Mathematics, Martin Liebeck, Chapman & Hall (2000).

• How to read and do proofs, D. Solow, John Wiley and Sons, New York (2004).

• Introduction to Group Theory (Second Edition), W. Ledermann, A.J.Weir, Addison Wes-ley Longman, (1996).

• Numbers, groups and codes, J.F. Humphreys, M.Y. Prest, Cambridge University Press(1989).

Chapter 1

Mathematical Language and Proof

We begin this chapter looking at some sequences and an example showing why proofs are sonecessary, and follow up with an example of a proof – of one of the most famous theorems ofall time. We move on to a discussion of how to make our statements more precise, looking atmathematical notation for and, or, not and implies. You will learn what is meant by thecontrapositive, converse and negation of a statement, and how to write these down. Wealso define the symbols ∀ (for all) and ∃ (there exists). Equipped with these new concepts,we then begin to look at how proofs can be constructed, and introduce some methods of proof.

1.1 1,2,3 . . .

Sequences of numbers can arise in all sorts of ways – sometimes the same sequence crops up inmany different areas of mathematics. In each of the following questions, you are asked to workout the first few terms of a sequence and then make a conjecture about what you think the nth

term is. (Here n will always be a positive integer.)

Exercise 1.1 Look at the grid below. Imagine you have to travel from point O to one of thepoints P1, P2, . . . on the top line. You are only allowed to travel along grid lines and you areonly interested in ‘shortest paths’.

P5P4P3P2P1

��

��

��

��

��

��O

For example, in the diagram below the left-hand shows a correct shortest path to P4, and theright hand picture shows a path that is not as short as possible, so we do not allow it.

P4��

��

��

��O

P4��

��O

yes no

Let an be the number of different shortest paths to Pn. Then for example a1 = 1 and a2 = 2.Find a3 and a4. What do you think an is, in general?

Exercise 1.2 Imagine you are in charge of painting a block of flats. Town planning regulations(bizarrely!) state that each level must be painted either black or white, and you can only painta level white if it has a black level immediately below it.

4 Proof & Structure Chapter 1: Mathematical Language and Proof

Below are two legal ways and one illegal way of painting a 4-storey block.

legal legal illegal

Let bn be the number of legal ways of painting an n-storey block of flats. Find b1, b2 and b3.How would you find bn in general?

Exercise 1.3 You have a chocolate bar with n+1 cubes. You wish to break it into individualcubes. You may only break along the lines, and one ‘move’ consists of breaking one piece alongone straight line. The example below shows one way of breaking up a four-cube bar.

move 3move 2move 1

Let cn be the smallest number of moves required to break up an (n + 1)-cube chocolate bar.(So the illustration above shows that c3 ≤ 3.) Make a conjecture for cn.

Exercise 1.4 This question is about partitions. A partition of a positive integer is a way ofsplitting it up into a sum of positive integers. So one partition of 7 is just 7, but you could alsohave 4 + 3 or 5 + 2. Note that we count, say, 4 + 3 and 3 + 4 as the same partition, it is onlythe numbers that matter, not the order. Let pn be the number of partitions of n. Go forth andconjecture!

Exercise 1.5 Finally in this section, coins. Imagine it’s the old days and we have such thingsas threepenny bits. You have a stock of 1p, 2p and 3p coins. Let qn be the number of waysyou can make n pence. So for example 4p could be made using a 3p and a 1p coin, or two 2pcoins, and there are other ways. Investigate.

1.2 Why bother with proof?

Example 1.2.1 Let n be an integer (with n ≥ 1). Draw n points on the circumference of acircle and join every pair of points with a straight line. What is the greatest number of regionsinto which the circle can be divided? The answer isn’t obvious, so let’s try some examples.

n 1 2 3 4 5Regions 1 2 4 8 16

What do you think the answer is for n = 6? What do you think the answer is in general? Makea conjecture before you read on!

1.2 Why bother with proof? Proof & Structure 5

In fact, however you place 6 points on the circle, you can never create more than 31 regions.(For 7 and 8 the answers are 57 and 99.) So your conjecture was (probably) wrong. The actualanswer, and a proof of it, are in Appendix 1 (available to download from the module website).The moral of the story is that you should never believe something until it has been proved.

A word of warning/comfort: this question and Exercises 1.1 – 1.5 are more open ended thanthe typical exam question. Do play around with these questions though; I’ll discuss them inthe worked solutions. But don’t worry if you can’t completely solve all of them – for one ofthese sequences there is no known formula!

So what is a proof? It is a sequence of deductions in which each statement follows logically fromthe last. The first statement is what we assume to be true; it is called the hypothesis. Thefinal statement is what we want to deduce, and is called the conclusion. The steps in-betweenshould be detailed enough to convince a reader that the argument is valid. As an illustrationof this, let’s consider the following well known result.

Theorem 1.2.2 (Pythagoras) In a right-angled triangle, the square of the hypotenuse is thesum of the squares of the other two sides.

Alternatively: If ABC is a triangle with ∠ABC = 90◦, then AC2 = AB2 +BC2. Here thehypothesis is ABC is a triangle with ∠ABC = 90◦ and the conclusion is AC2 = AB2 +BC2.

In general, when a theorem is given in the form ‘if p, then q’, then p is the hypothesis and q isthe conclusion.

Proof Let a = AB, b = BC and c = AC. We have to show that c2 = a2 + b2.

Consider a square with sides of length a + b. We can fit four copies of the triangle ABC intothe square, one at each corner, as in the diagram:

a b

xa c c b

c a

b a

a

cbb c

B A

C

The quadrilateral inside has all its sides of length c. In addition, its angles are all right-angles.To see this, remember that angles in a triangle add up to 180◦ and angles on a straight lineadd up to 180◦. Therefore

x+ ∠BCA+ ∠BAC = 180◦ = 90◦ + ∠BCA+ ∠BAC

giving x = 90◦.


Hence the inner quadrilateral is a square of side c. We can now work out the area S of theouter square in two different ways. Firstly the area is just the square of the length of its sides.Thus

S = (a+ b)2 = a2 + 2ab+ b2.

Alternatively, the area of the square is the area of the inner square (with side c) plus four timesthe area of ABC. Thus

S = c2 + 4× (12ab) = c2 + 2ab.

Combining the two expressions for S gives c2+2ab = a2+2ab+b2 or, equivalently, c2 = a2+b2.This is what we wanted to prove, and so we are finished. We write the symbol � to denote theend of a proof. �

Commentary That all seems very neat! But how did anyone come up with it? It seems tocome from nowhere. That is because we are presented with a fait accompli – we don’t see thevarious attempts at proof, the ones that don’t work, the ones that work but are untidy, and soon. However, when the final proof is presented most of this work ends up being thrown out.In this module we will present many proofs; often we will also discuss where the idea for theproof came from. That is, we will spend some time looking through the mathematician’s wastepaper bin.

1.3 Statements and Theorems

In the mathematical sense, a “statement” (or “proposition”) is a sentence which makes an as-sertion: a given statement is true or false, but not both. Here are ten examples of statements.

S1 2 + 2 = 4;

S2 51 is prime;

S3 The square of an even integer is divisible by 4;

S4 The sum of two odd integers is even;

S5 x2 − 3x > 0;

S6 If x > 1, then x2 − 12x+ 35 > 0;

S7 If 2n − 1 is prime, then n is prime;

S8 Any even integer greater than 2 can be expressed as the sum of two primes;

S9√2 is irrational;

S10 The number of shortest paths to Pn (see Exercise 1.1) is n.

Something like ‘3 + 7’ is not a statement – no assertion is made, so it can’t be said to be trueor false.

Note that the truth or falsity of S5 depends on the value of x. As soon as we give a value to xthe statement becomes true or false. Statements for which the truth depends on the value ofa variable or variables are called predicates. Of the statements given only S5 is a predicate.Of the others S1, S3, S4, S7, S9 and S10 are true, the truth of S8 is not known (but it iscertainly either true or false) and S2 and S6 are false.

Statements which are always true are called theorems (of course we can’t know they are alwaystrue until we’ve proved it). Mathematicians also use other words for certain types of theorems:

1.4 Definitions Proof & Structure 7

a proposition is a result which is not considered significant enough to be called a theorem;a lemma is usually a rather technical result often required in the proof of a more importanttheorem; and a corollary is a result that follows as a consequence of another theorem. Arelated issue is that of a conjecture. This is a statement that is thought to be true butfor which there is no known proof. For example, statement S8 is a conjecture and is calledGoldbach’s conjecture.

1.4 Definitions

Consider statements S1 – S10 given in Section 1.3. Before we can decide whether any ofthem is true we need to know the meanings of certain terms. In particular, for S3 we needto know what it means for an integer to be even. There are various ways to give the meaningof ‘even’. We might say that an integer is even if there is no remainder when we divide by 2.Alternatively we could say that an integer is even if it is a multiple of 2. These are both validdefinitions. However, when defining a term it is worthwhile taking a little care to make surethat the definition is easy to work with. In this case the two definitions given for ‘even’ do notlend themselves to easy algebraic manipulation. The following definition is easier to work with.

Definition 1.4.1 Let m be an integer. Then m is even if there is an integer n such thatm = 2n.

Technical note: There is a problem with this definition. To see this observe that the definitioncan be rephrased as: if m is an integer and there is an integer n such that m = 2n, then m iseven. From this it is clear that 0,±2,±4, . . . are all even, but it does not tell us anything aboutthe other integers like 3, -7, 21 and so on. The problem can be resolved if we use the phrase‘if and only if’ instead of ‘if’ in the original definition. In practice mathematicians do not dothis – instead we have the convention that we always treat the if in a definition as an ifand only if. With this in mind we now conclude this section with some further definitionsrequired in the rest of this chapter.

Definition 1.4.2 Let m be an integer. Then m is odd if there is an integer n such thatm = 2n+ 1.

Definition 1.4.3 Let m and n be integers. Then m divides n if there is an integer q such thatn = mq. If m divides n then we say m is a factor or a divisor of n and that n is divisibleby, or a multiple of, m.

Definition 1.4.4 A positive integer p is prime if p ̸= 1 and the only positive divisors of p are1 and p. A positive integer greater than 1 which is not prime is called composite.

Example 1.4.5 The first ten primes are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.

Exercise 1.6 Show that if p is a prime number with p ≥ 5, then either there exists an integern such that p = 6n + 1, or there exists an integer n such that p = 6n + 5. [Hint: any integeris of the form 6n, 6n+ 1, 6n+ 2, 6n+ 3, 6n+ 4, or 6n+ 5 for some n. But some of these formscan be factorised.]

Definition 1.4.6 A real number x is rational if there are integers m and n with n ̸= 0 suchthat x = m

n.


Note that all integers are rational since for any integer n we have n = n1.

The previous definition relies on the notion of a real number. Although we have not definedsuch numbers, we can think of a real number as an infinite decimal r0.r1r2r3 . . . where eachri is an integer and 0 ≤ ri ≤ 9 for i ≥ 1. A real number which is not rational is called irrational.

We have now introduced several sets of numbers, each of which has a mathematical symbol.These symbols are listed below.

N: the natural numbers 0, 1, 2, 3, . . .;

Z: the integers . . . ,−2,−1, 0, 1, 2, 3, . . .;

Z+: the positive integers 1, 2, 3, 4, . . .;

Q: the rational numbers;

R: the real numbers.

1.5 Some Grammar

In spoken and written English the meaning of particular words can vary depending on context.In mathematics we have to be more precise about what certain words mean. In this section weconsider some such words.

In what follows, we denote statements by letters p, q and so forth.

not If p is a statement, then not p, denoted ¬p, is a statement which is false when p is trueand true when p is false. So if p is the statement ‘2+2=5’, then ¬p is the statement ‘2+2 ̸= 5’.If we represent true with a T and false with an F then we can express this information in aso-called truth table, as follows:

p ¬pT FF T

The statement ¬p is also called the negation of p. Sometimes the negation of statements iseasy to spot. For example, the negation of statement S9 (

√2 is irrational) given earlier is ‘

√2

is not irrational’ or more simply ‘√2 is rational’. However, the negation of other statements

can be difficult and we require more grammar before we can do this with confidence.

Exercise 1.7 Write out the negation of each of the statements S1–S10 given earlier (don’tworry if you find some of them difficult – we’ll acquire tools to make this easier in comingsections).

Note that the statement ¬(¬p) (‘not not p’) has the same truth values as p. We say that¬(¬p) is equivalent to p. In this sense mathematical grammar is much stricter than linguisticgrammar. For example compare ‘I am happy’ with ‘I am not unhappy’, and ‘they are friendly’with ‘they are not unfriendly’. They do not have quite the same meanings.

1.5 Some Grammar Proof & Structure 9

and If p and q are statements, then the statement p and q, denoted by p ∧ q, is true if bothp and q are true, and false otherwise. The truth table for p ∧ q is:

p q p ∧ qT T TT F FF T FF F F

or The word ‘or’ in English can have two uses. Consider the following two situations:

• You have just won first prize in a TV game show. The host of the show says that youcan take the car or £20,000. Given this choice he would not expect you to take both!

• The manager of a company says that the ideal candidate for a job would be either amathematician or a Birkbeck student. Presumably she would not object to a candidatewho was a Birkbeck mathematics student.

In the first instance the word or is being used exclusively (you cannot take both prizes) and inthe second instance the word is being used inclusively (having both of the required attributesdoes not stop you being a candidate for the job). When used linguistically it is usually clearfrom the context whether the word or is inclusive or exclusive. In mathematics such ambiguityis not desirable, so we have the convention that the word or is always used inclusively.

Therefore, if p and q are statements, then the statement p or q, denoted by p ∨ q, is truewhenever p is true or q is true (or both), and is false only when both p and q are false. Thetruth table for p ∨ q is as follows.

p q p ∨ qT T TT F TF T TF F F

if . . ., then Consider the following statements of the form if p, then q.

• (At the local supermarket) ‘If you’re trying to eat healthily, then we’ve hundreds of low-fatproducts.’

• If you leave your umbrella at home, then it will rain.

In the first statement, the shop has hundreds of low-fat products whether or not you’re tryingto eat healthily. So, assuming the second part is true, the truth value of the first part ‘you’retrying to eat healthily’ is irrelevant – the whole if/then statement is still true. The only timethe if/then statement is false is if you are trying to eat healthily, and you discover that theshop doesn’t have hundreds of low-fat products. That is, when the first part is true and thesecond part is false.

Let’s look at the second statement, ‘if you leave your umbrella at home, then it will rain’. Whenis this false? In other words, when have I lied? There are four possibilities to consider. First


suppose that you leave your umbrella at home, and it rains. Then there’s no reason to call mea liar; what I said would happen, happened. What if you leave your umbrella at home and itdoesn’t rain? Then my statement would definitely be false.

Now suppose you take your umbrella with you. Whether it rains or not, my statement makes nopredictions about this event, so you cannot say that the statement is false in this case. Again,the only time the statement is false is when the first part is true and the second part is false.

In a mathematical context, if p and q are statements, then the statement if p, then q, denotedby p ⇒ q, is false when p is true and q is false, and true otherwise. The truth table is givenbelow.

p q p ⇒ qT T TT F FF T TF F T

There are other ways of expressing ‘if p, then q’. The ones you are most likely to come acrossare: p implies q; p only if q; p is sufficient for q; q is necessary for p; q ⇐ p.

Given the mathematical definition of ‘if . . ., then’, try the following exercise:

Exercise 1.8 Which, if any, of the following statements (a) – (d) are true?

(a) If 17 is prime, then 5 does not divide 17.

(b) If 17 is prime, then 5 does not divide 15.

(c) If 15 is prime, then 5 does not divide 17.

(d) If 15 is prime, then 5 does not divide 15.

The symbols ¬,∧,∨ and ⇒ are called connectives. We can use them to combine statementsto make more complicated statements. The truth values of these new statements can be workedout using truth tables.

Example 1.5.1 The truth table for (p ∨ q) ∧ p is:

p q p ∨ q (p ∨ q) ∧ pT T T TT F T TF T T FF F F F

Exercise 1.9 Let p and q be statements. Consider the statement (p ∧ q) ⇒ p. When wouldyou expect this statement to be true? Construct the truth table for (p ∧ q) ⇒ p. Do theresulting truth values agree with your expectation.

1.6 Tautologies and Equivalent Statements Proof & Structure 11

if and only if If p and q are statements, then p if and only if q, denoted by p ⇐⇒ q, istrue when p and q have the same truth values and false otherwise. The truth table for p ⇐⇒ qis given below. Note that often ‘if and only if’ is abbreviated to iff.

p q p ⇐⇒ qT T TT F FF T FF F T

1.6 Tautologies and Equivalent Statements

A tautology is a statement which is always true. For example, for any statement p thestatement p ∨ ¬p is a tautology. For more involved statements we can check whether they aretautologies using truth tables.

Example 1.6.1 The statement (p ∧ q) ⇒ (p ∨ q) is a tautology. We can verify this from thefollowing truth table.

p q p ∧ q p ∨ q (p ∧ q) ⇒ (p ∨ q)T T T T TT F F T TF T F T TF F F F T

Since the truth value in the last column is always T , it follows that (p ∧ q) ⇒ (p ∨ q) is atautology.

Exercise 1.10 Let p and q be statements. Use truth tables to show that the statementsp ⇒ (p ∨ q) and ((p ⇒ q) ∧ ¬q) ⇒ ¬p are tautologies.

Two statements p and q are equivalent if they always have the same truth values. As wenoted earlier p and ¬(¬p) are equivalent for any statement p. If p and q are equivalent, thenp ⇐⇒ q is a tautology and vice versa.

Again we can use truth tables to verify that two statements are equivalent, as the followingtwo examples illustrate.

Example 1.6.2 Let p and q be statements. We will show that p ⇒ q is equivalent to ¬p ∨ q.To do this we construct the following truth table.

p q p ⇒ q ¬p ¬p ∨ qT T T F TT F F F FF T T T TF F T T T

Since the truth values in the third and fifth columns are the same, the two statements areequivalent.


Example 1.6.3 Let p and q be statements. We will show that ¬(p∧q) is equivalent to ¬p∨¬q.To do this we construct the following truth table.

p q p ∧ q ¬(p ∧ q) ¬p ¬q ¬p ∨ ¬qT T T F F F FT F F T F T TF T F T T F TF F F T T T T

Since the truth values in the fourth and seventh columns are the same, the two statements areequivalent.

Exercise 1.11 Let p and q be statements. Use a truth table to show that ¬(p ∨ q) and¬p ∧ ¬q are equivalent. (The equivalence of the statements given here and the equivalence ofthe statements given in Example 1.6.3 are collectively known as de Morgan’s Laws).

Exercise 1.12 Let p and q be statements. Use a truth table to show that p ⇒ q is equivalentto ¬q ⇒ ¬p.

The statement ¬q ⇒ ¬p is called the contrapositive of the statement p ⇒ q. Their equivalenceprovides us with a useful method of proof. Suppose we want to prove if p, then q. One possibleapproach would be to prove if ¬q, then ¬p. Having completed the proof it follows, sincep ⇒ q and ¬q ⇒ ¬p are equivalent, that if p, then q is true. This type of proof is called acontrapositive proof.

Exercise 1.13 Let p and q be statements. Use Example 1.6.2 along with de Morgan’s lawsand the equivalence of p and ¬(¬p) to show that ¬(p ⇒ q) is equivalent to p ∧ ¬q. With theequivalence of these two statements in mind, reconsider your answers to Exercise 1.7.

Exercise 1.14 So far we have only given examples of truth tables for statements constructedfrom one or two statements. How many rows would be needed for the truth table of a statementconstructed from (a) three statements? (b) four statements?Use a truth table to show that, for statements p, q and r, the statements p ∨ (q ∧ r) and(p ∨ q) ∧ (p ∨ r) are equivalent.

We will now show how we can use the theoretical work we have done so far to restate somemathematical statements in a different, but equivalent form. This sometimes helps as it can beeasier to see whether a particular statement is true when it is in one of these restated forms.

Example 1.6.4 Consider the statement S7: ‘if 2n − 1 is prime, then n is prime’. This isdifficult to get a handle on when trying to prove, because we really don’t know much aboutprime numbers, except that they don’t have many factors. Composite numbers (non-primes)are easier; we can straight away say ‘OK, if n is composite, then it has some factorisationn = ab with neither of a or b equal to 1’. So we’d ideally like to prove an equivalent statementin terms of numbers not being prime. Now, the statement is in the form p ⇒ q, where p is thestatement ‘2n − 1 is prime’ and q is the statement ‘n is prime’. By Example 1.6.2, p ⇒ q isequivalent to ¬p ∨ q. Here ¬p is the statement ‘2n − 1 is not prime’, or ‘2n − 1 is composite orequal to 1’. Thus statement S7 is equivalent to: ‘either 2n − 1 is not prime or n is prime’.

Alternatively, by Exercise 1.12, p ⇒ q is also equivalent to its contrapositive, ¬q ⇒ ¬p. Here¬q is the statement ‘n is not prime’. Thus statement S7 is equivalent to: ‘if n is not prime, then2n − 1 is not prime’. In this form we at least have some hope that we could use a factorisationof n to produce a factorisation of 2n − 1. We’ll do this in the next section.

1.7 Some Proofs Proof & Structure 13

Example 1.6.5 Consider the statement ‘if x > 0 and x < 1 then x2 < x’. This statementis of the form p ⇒ q where p is the statement ‘x > 0 and x < 1’ and q is the statement‘x2 < x’. The statement is equivalent to two other statements, the either/or form ¬p ∨ q andthe contrapositive form ¬q ⇒ ¬p. To work with these we need to know ¬p and ¬q. Herethe statement p is of the form p1 ∧ p2 where p1 is ‘x > 0’ and p2 is x < 1. By de Morgan’sLaws, ¬p is equivalent to ¬p1 ∨¬p2. Hence ¬p is the statement ‘either x ≤ 0 or x ≥ 1’. So thegiven statement is equivalent to ¬p∨q, which is the statement: ‘either x ≤ 0 or x ≥ 1, or x2 < x’.

Alternatively, the given statement is equivalent to its contrapositive ¬q ⇒ ¬p, which is thestatement: ‘if x2 ≥ x, then either x ≤ 0 or x ≥ 1’.

Exercise 1.15 For each of the following statements (a) – (d), find equivalent statements in

• the either/or form;

• the contrapositive form.

(a) If n2 is odd, then n is odd.

(b) If x2 = 2, then x is irrational.

(c) If x > y and y > z, then y2 > xz.

(d) If x+ y is irrational, then either x is irrational or y is irrational.

Which, if any, of the statements do you think are true?

1.7 Some Proofs

Example 1.7.1 The square of an even integer is even.

Commentary This is the same as saying: if n is even, then n2 is even. So we have to showthat n2 = 2m for some integer m. All we know is that n is even, which means that n = 2k forsome integer k. So n2 = (2k)2 = 4k2 = 2(2k2), and this provides our m. This is the basis forthe following proof.

Proof Let n be an even integer. Then there exists an integer k such that n = 2k. Thus

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

Now, since k is an integer, clearly 2k2 is an integer. Thus n2 = 2m for some integer m. Thatis, n2 is even. �

Remember, if in doubt, write down all the appropriate definitions and see if you can work fromthere.

Exercise 1.16 Prove that the square of an odd integer is odd.

Example 1.7.2 The sum of two odd integers is even.


Proof Let m and n be odd integers. Then there are integers k and l such that m = 2k + 1and n = 2l + 1. Thus

m+ n = 2k + 1 + 2l + 1 = 2k + 2l + 2 = 2(k + l + 1).

Now, since k and l are integers, it follows that k+ l+1 is an integer. That is, m+n is even. �

Note Do not make the mistake of using the same letter for k and l.

Exercise 1.17 Prove that if m and n are multiples of r then m+ n is a multiple of r and mnis a multiple of r2.

Example 1.7.3 Let n be a positive integer. If 2n − 1 is prime, then n is prime.

Commentary As we said in the last section, our attack on this proof will be to try and provethe contrapositive statement, that if n is not prime, then 2n− 1 is not prime. If n is not prime,then either n = 1 or n is composite. If n = 1, then 2n − 1 = 1 is not prime, so the statement istrue for n = 1. So assume n is composite. We first write down the definition: n is compositemeans that n = ab for some integers a, b with neither a nor b equal to 1. Thus 2n− 1 = 2ab− 1.We have to show somehow that 2ab−1 is composite. In other words we have to factorize 2ab−1.It is time to list three very useful factorisations which it is worth remembering.For any real numbers x, y and any positive integer k we have

1. x2 − y2 = (x+ y)(x− y);

2. xk − 1 = (x− 1)(xk−1 + xk−2 + · · · x2 + x+ 1); and in general

3. xk − yk = (x− y)(xk−1 + xk−2y + xk−3y2 + · · ·xyk−2 + yk−1).

We will be using the second of these in our proof.

Proof (Contrapositive) Suppose n is not prime. If n = 1, then 2n − 1 = 1 is not prime. Sosuppose n > 1 is not prime. Then n is composite. So n = ab for some integers a, b with neithera nor b equal to 1. Let x = 2a. Then

2n − 1 = 2ab − 1

= xb − 1

= (x− 1)(xb−1 + · · ·+ x+ 1).

Now, since a, b > 1, both 2a − 1 = x− 1 > 1 and (xb−1 + · · · + x + 1) > 1. Thus 2n − 1 is theproduct of two integers both of which are greater than 1. Thus 2n − 1 is composite, hence notprime. �If a number of the form 2n−1 is prime, it is called a Mersenne prime, after the French mathemati-cian Marin Mersenne, who found many primes of this form, and claimed that, for primes lessthan or equal to 257, the integer 2n−1 is prime if and only if n = 2, 3, 5, 7, 13, 17, 19, 31, 67, 127or 257. And this was in the 17th century! Incredibly this is very nearly true, and it wasn’tuntil 1903 that the claim was refuted. It turns out that 267 − 1 is composite, but it’s hardlysurprising that poor old Mersenne didn’t spot the factors 193 707 721 and 761 838 257 287.

1.7 Some Proofs Proof & Structure 15

Exercise 1.18 Following on from Example 1.7.3, you might be wondering when expressionslike 3n − 1 and 8n − 1 are prime. We can deal with all of these cases in one go. Let n and r bepositive integers. Show that if n > 1 and r > 2, then rn − 1 is composite.

Example 1.7.4 75 311 357 is not a perfect square.

Commentary The first thing to do is put down your calculator! There are better ways todo this, and it would be no help at all to you if the number was

75 311 222 333 444 555 666 777 888 999 357

which is also not a perfect square. It would be nice to find a proof that covered as general acase as possible.

If we investigate the squares of a few smallish integers such as 172 = 289 and 252 = 625 wenote that the units digit of the square is just the units digit of the square of the units digit ofthe original number. For example, the units digit of 172 is just the units digit of 72. How dowe show this in general?

An integer in base 10 is of the form arar−1 . . . a1a0 where, for each i, 0 ≤ ai ≤ 9 and ar ̸= 0.This representation is just shorthand for

ar10r + ar−110

r−1 + · · ·+ a110 + a0.

For example 289 = 2 · 100 + 8 · 10 + 9, so r = 2, a2 = 2, a1 = 8 and a0 = 9.

Now, since 10 is a factor of every term of this sum except for a0 we can write any positiveinteger as 10n + u where n and u are non-negative integers and u is the units digit when theinteger is expressed in base 10. Thus

(10n+ u)2 = 100n2 + 20nu+ u2 = 10(10n2 + 2nu) + u2

and so, in base 10, the units digit of (10n+ u)2 is just the units digit of u2.

We now have enough to write out a proof of the following result.

Proposition 1.7.5 The units digit of a perfect square when expressed in base 10 is either 0,1, 4, 5, 6 or 9.

Proof Any positive integer can be expressed in the form 10n + u where n and u are non-negative integers and u is the units digit when the integer is expressed in base 10. Now

(10n+ u)2 = 100n2 + 20nu+ u2 = 10(10n2 + 2nu) + u2

and so, in base 10, the units digit of (10n + u)2 is just the units digit of u2. Since we knowu = 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9, clearly u2 = 0, 1, 4, 9, 16, 25, 36, 49, 64 or 81. Thus the units digitof u2, and hence the units digit of a perfect square, is one of 0, 1, 4, 5, 6 or 9. �

Corollary 1.7.6 75 311 357 is not a perfect square.


Proof The units digit of 75 311 357 is 7. Thus by Proposition 1.7.5, 75 311 357 is not aperfect square. �

Exercise 1.19 Prove that for any integer n, n2 + 4 is not divisible by 7. (Hint: every integercan be expressed in the form 7q + r where r is one of 0, 1, 2, 3, 4, 5 or 6.)

Definition 1.7.7 For a real number x the modulus of x, written |x|, is equal to x if x ≥ 0and −x if x < 0. So |x| is giving us the size of x.

Theorem 1.7.8 The triangle inequality Let x and y be real numbers. Then |x + y| ≤|x|+ |y|.

Proof We have to consider the possibilities for |x|, |y| and |x+ y|, and in each case show that|x|+ |y| − |x+ y| ≥ 0. Some of these are filled in in the table below; as an exercise you shouldfill in the rest.

x y x+ y |x|+ |y| − |x+ y| ≥ 0?≥ 0 ≥ 0 ≥ 0 x+ y − (x+ y) = 0 X≥ 0 < 0 ≥ 0 x+ (−y)− (x+ y) = −2y = 2|y| X≥ 0 < 0 < 0 x+ (−y)− [−(x+ y)] = 2x X< 0 ≥ 0 ≥ 0< 0 ≥ 0 < 0< 0 < 0 < 0

The completed table will show that in every case, |x + y| ≤ |x| + |y|, and so the triangleinequality always holds. �

1.8 Converses and Counterexamples

The converse of the statement p ⇒ q is the statement q ⇒ p. It is important to note that thetruth of p ⇒ q tells us nothing about the truth of q ⇒ p. To see this consider the followingexamples.

Example 1.8.1 The statement: ‘if x = 3 and y = 5, then x+ y = 8’ is true. The converse ofthis statement is: ‘if x+ y = 8, then x = 3 and y = 5’, which is false (for example x = 2, y = 6is another possibility).

Example 1.8.2 The converse of the true statement ‘if n is even, then n2 is even’ is ‘if n2 iseven, then n is even’. In this case the converse is true. You can prove it using a contrapositiveproof. (See Exercise 1.16.)

Note that in the previous example we have assumed that n is an integer. We could (possiblyshould!) have written the statement as ‘if n is an integer and n is even, then n2 is even’. Ifwe did this, then it is important to note that the ‘n is an integer’ part carries through to theconclusion. In other words, the converse would be ‘if n is an integer and n2 is even, then n iseven’; it would not be ‘if n2 is even, then n is an integer and n is even’. To avoid this possibleconfusion it is often better to define first any symbols used in the statement. So we could write:‘Let n be an integer. If n is even, then n2 is even’. This approach makes it clear that the ‘n isan integer’ part belongs to both the hypothesis and the conclusion.

1.8 Converses and Counterexamples Proof & Structure 17

Exercise 1.20 Write out the converse of each of the statements (a) – (d) given in Exercise1.15.

Exercise 1.21 Criticise the following piece of deduction. Geniuses always behave oddly. Ibehave oddly, so I must be a genius!

Example 1.8.3 Knowing the word converse helps to describe what is wrong with the following‘proof’ that 1 = −1:

‘1 = −1 implies (1)2 = (−1)2 implies 1 = 1. Since the last statement is true, the first statementmust have been true. Therefore 1 = −1.’

It is true that 1 = −1 does imply 1 = 1. The error comes when the ‘proof’ goes on to claimthat this means the converse statement (1 = 1 implies 1 = −1) is true. As we know, the truthor otherwise of a statement tells us nothing about the truth of the converse. In this case theconverse is false. We can spot the error if we try to work the equations backwards. x2 = y2

only tells us that x = ±1. So we can’t deduce that 1 = −1 (thankfully!). Of course none ofyou would write a ‘proof’ like this. But do take care to avoid ‘proofs’ that start with equatingtwo objects you are trying to show are equal. More subtle errors than 1 = −1 can creep in ifyou risk this!

Exercise 1.22 Find the errors in the following ‘proofs’.

(a) A ‘proof’ that all numbers are equal to zero. Let x ∈ R. Suppose for a contradiction thatx = y for some y ̸= 0. Then y = x

y2 = xy

y2 − x2 = xy − x2

(y + x)(y − x) = x(y − x)

y + x = x

y = 0

This contradicts our earlier assertion that y ̸= 0. Hence x = 0.

(b) A ‘proof’ that the smallest positive number is 1. Let x be the smallest positive number.Clearly x ≤ 1. Now x2 is also positive, so by minimality of x, x2 ≥ x. Divide both sides by(the positive number) x to get x ≥ 1. Hence x = 1. [Nice to know we never have to worryabout fractions again, isn’t it?]

The truth, or otherwise, of the converse of a statement raises the question of how we can showthat a statement is false. Now, if p ⇒ q is false, then ¬(p ⇒ q) must be true. From Exercise1.13, ¬(p ⇒ q) is equivalent to p∧¬q. Hence, to show that the statement p ⇒ q is false we justneed to give an example where p is true and q is false; that is, an example where the hypothesisholds but the conclusion fails. Such an example is called a counterexample to the statement.

Example 1.8.4 The statement: ‘if x+ y = 8, then x = 3 and y = 5’ is false since x = 2, y = 6is a counterexample.

Example 1.8.5 The statement: ‘if n is prime, then 2n − 1 is prime’ is false. To see this putn = 11. Then n is prime, but 2n − 1 = 211 − 1 = 2047 = 23× 89 and so is composite.


Exercise 1.23 Find another counterexample to each of the statements in Examples 1.8.4 and1.8.5. (Hint: 213 − 1, 217 − 1 and 219 − 1 are all prime.)

Example 1.8.6 The statement: ‘if n is prime, then n is odd’ is false. For when n = 2, then nis prime but n is even.

Note that in Example 1.8.6, n = 2 is the only counterexample. However, one counterexampleis all that is required to show that a given statement is false.

Exercise 1.24 Give counterexamples to each of the following statements.

(a) If x2 = 9, then x = 3.

(b) If xy < 0, then x+ y < 0.

(c) If n is a positive integer, then 5n2 + 5n+ 1 is prime.

(d) If 6 divides mn, then 6 divides m or 6 divides n.

1.9 Quantifiers

The Existential Quantifier (there exists)

Consider the following four statements.

E1 There exists an integer n such that n2 is even.

E2 There is an integer n such that n2 = 4.

E3 For some integer n, n3 = 27.

E4 There exists an integer n such that n2 = 18.

Each of E1 – E4 has the form: There exists some integer satisfying a certain property. Inother words, if the statement is true, we can find an integer satisfying the given property. Themathematical symbol used to represent ‘there exists’ is ∃. So E1 – E4 can be rewritten usingthis symbol as follows.

E1 ∃n ∈ Z such that n2 is even.

E2 ∃n ∈ Z such that n2 = 4.



To prove an existence statement it is sufficient to find one object (in this case an integer) whichsatisfies the given property. So E1 is true since putting n equal to any even number verifies thestatement, E2 is true since n = 2 or n = −2 verifies the statement, and E3 is true since n = 3verifies the statement. Note that the number of examples available to illustrate an existencestatement is irrelevant; all we need to do is find one such example. Thus E1 and E3 are equallytrue even though there are infinitely many integers which verify E1 and only one integer whichverifies E3.It is not always necessary to give a specific example to prove an existence statement, as thefollowing example illustrates.

Example 1.9.1 Prove that ∃ a ∈ R such that a3 + 3a2 − 2a+ 11 = 0.

1.9 Quantifiers Proof & Structure 19

Proof Let f(x) = x3 + 3x2 − 2x+ 11. Then f(0) = 11 and f(−10) = −669. Thus the graphy = f(x) must cross the x-axis at some point between −10 and 0. Let a be such a point. Thenf(a) = 0, that is ∃ a ∈ R such that a3 + 3a2 − 2a+ 11 = 0. �

Note There is a direct proof of the previous example, for if a = − b2+6b+606b

, where b =3√

1620 + 60√669, then a3 + 3a2 − 2a + 11 = 0. However, this is more difficult to show than

the above proof!

Exercise 1.25 Write each of the following statements using, where possible, the symbols ∃,∈, R and Q and verify that they are true. [Note, for (c), that angles are measured in radiansrather than degrees.]

(a) There is a real number x such that x+ π is rational.

(b) For some real number y, y2 − y − 1 = 0.

(c) There exists a real number θ such that cos θ = θ.

The Universal Quantifier (for all)

Consider the following three statements.

A1 For all integers n, n2 ≥ 0.

A2 For every integer n, n2 is even.

A3 Any integer is divisible by 1.

Each of A1 – A3 has the form: For all integers a certain property is true. In other words, ifthe statement is true, then every integer must satisfy the given property. The mathematicalsymbol used to represent ‘for all’ is ∀. So A1 – A3 can be rewritten using this symbol asfollows.

A1 ∀n ∈ Z, n2 ≥ 0.

A2 ∀n ∈ Z, n2 is even.

A3 ∀n ∈ Z, n is divisible by 1.

To show that a ‘for all’ statement is true we need to give a general argument; to show it is falsewe need only give a single counterexample. Of the statements given here A1 and A3 are trueand A2 is false.

Exercise 1.26 Write each of the following statements using, where possible, the symbols ∀,∈, R and Q.

(a) For all rational numbers a, a2 is rational.

(b) For each rational number b, b3+ 2

5is rational.

(c) The cube of any real number is positive.

Which, if any, of the statements are true? Can you justify your answers?


Statements involving more than one quantifier

Many mathematical statements involve more than one quantifier. For example, the followingstatements involve more than one quantifier.

M1 For all integers m there exists an integer n such that m + n = 0. (Using symbols: ∀m ∈Z, ∃n ∈ Z such that m+ n = 0.)

M2 Given integers l and m, there is an integer n such that l2 + m2 = n2. (Using symbols,∀l ∈ Z,∀m ∈ Z, ∃n ∈ Z such that l2 +m2 = n2.

Exercise 1.27 Verify that M1 is true and that M2 is false.

Exercise 1.28 The order of quantifiers is important. The statement

∃n ∈ Z such that ∀m ∈ Z,m+ n = 0

looks similar to M1. Verify, however, that this statement is false. Explain the differencebetween M1 and the statement given here.

Negating statements involving quantifiers

Consider the statement E2, ‘∃n ∈ Z such that n2 = 4’, once more. Suppose we want to negatethis statement, that is we want to write the statement ¬E2. One possibility is: ‘there does notexist an integer n such that n2 = 4’. However, if this is true then if we take any integer itssquare cannot be 4. Thus, a more succinct way of writing ¬E2 is: ’for all integers n, n2 ̸= 4’or, using symbols: ∀n ∈ Z, n2 ̸= 4.

Now consider the statement A2, ‘∀n ∈ Z, n2 is even’, once more. One possibility for ¬A2 is:‘not all integers have an even square’. However, if this is true, then there must be some integerwhose square is not even. Thus, if we use the term ‘odd’ for ‘not even’, a more succinct way ofwriting ¬A2 is: ‘there exists an integer n such that n2 is odd, or, using symbols: ∃n ∈ Z suchthat n2 is odd.

In summary, when negating a statement involving quantifiers, we change there exists (∃) to forall (∀), we change for all (∀) to there exists (∃) and negate the conclusion.

Example 1.9.2 Consider the statement M2, ‘∀l ∈ Z, ∀m ∈ Z,∃n ∈ Z such that l2+m2 = n2’.Using the above observation we get that ¬M2 is

∃l ∈ Z, ∃m ∈ Z such that ∀n ∈ Z, l2 +m2 ̸= n2.

That is: There are integers l and m such that for all integers n, l2 +m2 ̸= n2.

Exercise 1.29 Negate the statement M1.

Quantifiers often occur in definitions. As an example we give the following important definitionsfor certain types of mappings.

Definition 1.9.3 Let f : X → Y be a function from X to Y . Then f is injective (or one-to-one) if for all x, y ∈ X with x ̸= y, f(x) ̸= f(y); f is surjective (or onto) if for all y ∈ Y ,there exists x ∈ X such that f(x) = y; f is bijective if it is both injective and surjective.

1.10 Proof by Contradiction Proof & Structure 21

Exercise 1.30 Write out the definition for a mapping to be (a) injective (b) surjective, usingthe symbols ∀ and ∃.

Exercise 1.31 Write out the definition for a mapping not to be (a) injective (b) surjective,using the symbols ∀ and ∃.

Exercise 1.32 For i = 1, 2, 3, 4, 5 let fi : R → R be defined as follows: f1(x) = 3x + 2,f2(x) = x2, f3(x) = x3, f4(x) = 2 and f5(x) = 2x. Which, if any, of these mappings is (a)injective? (b) surjective? (c) bijective?

An injective function f : X → Y in some sense preserves the set X, because every x ∈ X hasa unique image f(x) – so in particular the set {f(x) : x ∈ X} is a subset of Y which has thesame size as X. We’ll talk more about this in Chapter 2.

1.10 Proof by Contradiction

Suppose you want to prove a statement s. One approach is assume that the negation ¬s istrue. The aim is to produce a mathematical argument from this starting point which leads toa statement r that is clearly false, something like 1 = 0. Thus we have shown that (¬s) ⇒ ris true. But r is false. By the truth table for ⇒, this forces ¬s to be false too (we can’t havea true statement implying a false one). Therefore s was true all along. This is called a proofby contradiction, because we are aiming to produce a false statement (r), which is known asa contradiction. This can be a very powerful technique. In the case where s is an if. . ., thenstatement p ⇒ q, the negation of s is p ∧ ¬q. So the starting point is to assume that p is trueand q is false.

Example 1.10.1 Prove that if 10 001 is composite, then it has a prime factor less than 98.

Commentary As in Example 1.7.4, there is an obvious proof, namely that 10 001 = 73×137and, since 73 < 98, then 10 001 has a prime factor less than 98. However, it is worth lookingfor an alternative, not least since finding the given factorisation of 10 001 is not that easy.

Note If m > 1 is a factor of any positive integer a, then there is a prime factor p of a withp ≤ m. This is because either m is prime (so m = p), or m is composite and can be writtenm = m1m2 where both 1 < m1 < m and 1 < m2 < m. Either m1 is prime (so m1 = p) or m1

is composite and we can factorise m1. We just continue like this until we get a prime number.We will try a contradiction proof for Example 1.10.1. The statement is of the form p ⇒ q,so we will suppose that 10, 001 is composite (p is true), and all of the prime factors of 10 001are greater than 97 (q is false). Now if 10 001 is composite, then there exist positive integersm,n both greater than 1 such that 10 001 = mn. Now m and n are factors of 10 001 which aregreater than 1, and so a(by the Note) must be at least as large as the smallest prime factor of10 001. Thus m, n > 97 and 10 001 = mn > 97× 97 = 9409.

Bother. We haven’t got the required contradiction. A little more care is needed. Note thatnone of 98, 99 or 100 is prime. Thus if all the prime factors of 10 001 are greater than 97,then the smallest prime factor of 10 001 must be at least 101. If we try this, then we get thefollowing proof.


Proof Suppose that all of the prime factors of 10 001 are greater than 97. Then the smallestprime factor of 10 001 is at least 101. Since 10 001 is composite, there are positive integers mand n both greater than 1 such that 10 001 = mn. Now m and n are factors of 10 001 whichare greater than 1, and so (by the Note) must be at least as large as the smallest prime factorof 10 001. Thus m, n ≥ 101 and

10 001 = mn ≥ 101× 101 = 10 201.

Hence we have a contradiction. Therefore our assumption that all the prime factors of 10 001are greater than 97 was false. Hence 10 001 has a prime factor no greater than 97. �Exercise 1.33 Let n be a positive integer. Prove that if n is composite, then it has a primefactor no greater than

√n.

We conclude this section by giving contradiction proofs of two important results.

Theorem 1.10.2√2 is irrational.

Proof Suppose that√2 is rational. Then there are integers p and q, with q ̸= 0, such that√

2 = pq. Furthermore, we may suppose (cancelling if necessary) that p and q have no common

factors. Squaring both sides of√2 = p

qgives 2 = p2

q2, and hence p2 = 2q2. Thus, using Exercise

1.16, p is even, and so there is an integer k such that p = 2k. Putting p = 2k in p2 = 2q2 gives4k2 = 2q2, and so q2 = 2k2. Thus q is also even. That is, both p and q have 2 as a factor,contradicting the fact that they have no common factors. Thus

√2 is irrational. �

Exercise 1.34 Try to adapt the proof of Theorem 1.10.2 to show that√3 is irrational. Where,

if at all, are the potential difficulties?

Exercise 1.35 Follow the proof of Theorem 1.10.2, but replace√2 with

√4. At what point

does the proof break down?

The following proof is perhaps the earliest contradiction proof. It was discovered by Euclid.

Theorem 1.10.3 There are infinitely many prime numbers.

Proof Suppose that there are only finitely many primes, p1, p2, . . . , pr, say. Let

n = p1p2 · · · pr + 1.

Let p be a prime factor of n. Now n is one more than a multiple of p1, so clearly p1 does notdivide n, hence p ̸= p1. Similarly p ̸= p2, p ̸= p3, . . . , p ̸= pr. Thus p is a prime not on the list,which is a contradiction. Thus there are infinitely many primes. �

1.11 The Principle of Induction

Let P (n) be a statement concerning the positive integers Z+. Suppose we can prove:

I1 P (1) is true;

I2 for each positive integer k, P (k) ⇒ P (k + 1).

Then P (n) is true for all positive integers n.

Example 1.11.1 Prove that, for all positive integers n,n∑

r=1

r2 =1

6n(n+ 1)(2n+ 1).

1.11 The Principle of Induction Proof & Structure 23

Proof Let P (n) be the statement:∑n

r=1 r2 = 1

6n(n+ 1)(2n+ 1).

For n = 1,∑n

r=1 r2 =

∑1r=1 r

2 = 12 = 1 and 16n(n+1)(2n+1) = 1

6× 1× 2× 3 = 1. Thus P (1)

is true.Now, suppose P (k) is true. That is,

∑kr=1 r

2 = 16k(k + 1)(2k + 1). Then

∑k+1r=1 r

2 = 12 + 22 + · · ·+ k2 + (k + 1)2

=(∑k

r=1 r2)+ (k + 1)2

= 16k(k + 1)(2k + 1) + (k + 1)2 (since P (k) is true)

= 16(k + 1)(k(2k + 1) + 6(k + 1))

= 16(k + 1)(2k2 + 7k + 6)

= 16(k + 1)(k + 2)(2k + 3)

= 16(k + 1)([k + 1] + 1)(2[k + 1] + 1).

Thus P (k + 1) is true.

Since we have shown that P (1) is true and that P (k) implies P (k + 1), it follows that P (n) istrue for all positive integers n. Thus

∑nr=1 r

2 = 16n(n+ 1)(2n+ 1). �

Example 1.11.2 Show that for all positive integers n, 7n − 6n+ 3 is divisible by 4.

Commentary It is important here that you do not confuse the statement you are trying toprove (for all positive integers n, 7n − 6n+ 3 is divisible by 4) with the expression involving n,which is 7n − 6n + 3. In fact, it is worth giving the expression a name; we denote it by f(n)here, but any name will do.

In the actual induction proof that follows you would expect f(k+1) = 7k+1−6(k+1)+3 to beexpressed in terms of f(k) = 7k − 6k+3. We do this indirectly by considering f(k+1)− f(k).This is a useful device which makes it easier to deal with the algebra.

Proof Let f(n) = 7n − 6n+ 3 and let P (n) be the statement: f(n) is divisible by 4.

For n = 1, f(1) = 71 − 6 × 1 + 3 = 7 − 6 + 3 = 4. Thus f(1) is divisible by 4 and so P (1) istrue.

Suppose P (k) is true. That is, f(k) is divisible by 4. Now

f(k + 1)− f(k) = 7k+1 − 6(k + 1) + 3− (7k − 6k + 3)

= 7k+1 − 6k − 6 + 3− 7k + 6k − 3

= 7k+1 − 7k − 6

= 7k(7− 1)− 6

= 6(7k − 1)

Since 7k is odd, 7k − 1 is even and so is divisible by 2. Thus 6(7k − 1) is divisible by 4. Now

f(k + 1) = f(k) + 6(7k − 1)


and so is the sum (by the inductive hypothesis) of two integers which are both divisible by 4.Hence f(k + 1) is divisible by 4 and so P (k + 1) is true.

Since we have shown that P (1) is true and that P (k) implies P (k + 1), it follows that P (n) istrue for all positive integers n. Thus for all positive integers n, 7n − 6n+ 3 is divisible by 4.�

Exercise 1.36 Can you modify this proof to show that f(n) is divisible by 6 for all n? If not,where does it fall down?

Exercise 1.37 Use the principle of induction to show that∑n

r=1 r =12n(n+1) and

∑nr=1 r

3 =14n2(n+ 1)2.

Exercise 1.38 Prove that, for all positive integers n, 7n − 4n is divisible by 3.

1.12 Variations on the Principle of Induction

There are several different variations on the method of proof using the principle of induction.Here we consider just two variations on the main method.

Variation 1: A different starting point

There is no reason why an induction proof should start at n = 1. In fact there are certainstatements which are not true for the first few positive integers. And sometimes we prefer tostart at n = 0, in other words to consider the set of natural numbers. In this case the followingmethod can sometimes be used.Let P (n) be a statement concerning the integers Z. Suppose that we can prove:

I1′ for some integer N , P (N) is true;

I2′ for every integer k ≥ N , P (k) ⇒ P (k + 1).

Then P (n) is true for all integers n ≥ N .

Example 1.12.1 For all integers n ≥ 5, 2n > n2.

Proof Let P (n) be the statement 2n > n2. For n = 5, 25 = 32 and 52 = 25. Thus 25 > 52

and so P (5) is true.Suppose P (k) is true. That is 2k > k2. Then

2k+1 = 2× 2k

> 2k2 (since P (k) is true)

= k2 + [(k + 1)2 − 2k − 1]

= (k + 1)2 + k2 − 2k − 1.

Now if k ≥ 3, then k2 − 2k− 1 = (k− 1)2 − 2 > 0 and so 2k+1 > (k+1)2. Thus, if k ≥ 3, P (k)implies P (k + 1).Since we have shown that P (5) is true and that, for k ≥ 3, P (k) implies P (k + 1), it followsthat P (n) is true for all integers n ≥ 5. Thus for all integers n ≥ 5, 2n > n2. �

Exercise 1.39 Prove that, for all integers n ≥ 4, n! ≥ 2n.

1.12 Variations on the Principle of Induction Proof & Structure 25

Variation 2: Strong Induction

Let P (n) be a statement concerning the positive integers. Suppose that we can prove:

S1 P (1) is true;

S2 for each positive integer k, P (1), P (2), . . . , P (k) together imply P (k + 1).

Then P (n) is true for all positive integers n.

Remarks Note that as in the first variation, we have a version of strong induction which starts:for some integer N , P (N) is true.At first sight, strong induction looks significantly different from the basic induction, because weare allowed to assume all of P (1), P (2), . . . , P (k) to prove P (k+1). However, strong inductionis actually equivalent to basic induction. To see this let Q(n) be the statement: for all m ∈ Z+

with m ≤ n, P (m). Then S1 and S2 become

S1′ Q(1) is true;

S2′ for each positive integer k, Q(k) implies Q(k + 1);

which are just the requirements of basic induction. It is usually clear which type of inductionis required once you try to deduce that P (k + 1) is true.

For the purpose of the next example we define a product of primes to be either a singleprime or an expression of the form p1p2 · · · pr where each of p1, p2, . . . , pr is a prime number.

Example 1.12.2 Every integer greater than one can be expressed as a product of primes.

Proof Let P (n) be the statement that n can be expressed as a product of primes.

Since 2 is prime, P (2) is true.

Suppose P (2), P (3), . . . , P (k) are true. That is, 2, 3, . . . , k can all be expressed as a productof primes.

Consider k+1. Either k+1 is prime or it is composite. If k+1 is prime, then k+1 is obviouslya product of primes. If k+1 is composite, then there exist integers l and m with 1 < l < k+1and 1 < m < k + 1 such that k + 1 = lm. Since 1 < l < k + 1 and 1 < m < k + 1, both P (l)and P (m) are true by the induction hypothesis. That is, there are primes p1, p2, . . . , pr andq1, q2, . . . , qs such that l = p1p2 · · · pr and m = q1q2 · · · qs. Thus

k + 1 = lm = p1 · · · prq1 · · · qscan be expressed as a product of primes. Hence P (k + 1) is true.

Since we have shown that P (2) is true and that P (2), . . . , P (k) together imply P (k + 1), itfollows by induction that P (n) is true for all integers greater than 1. Thus every integer greaterthan 1 can be expressed as a product of primes. �Exercise 1.40 Look back at Exercise 1.3, in which cn is defined to be the smallest number ofmoves required to break up an (n + 1)-cube chocolate bar into individual cubes. Use stronginduction to prove that cn = n for all positive integers n. [Hint: Your first move is always tobreak the bar into two pieces along a line. Those pieces have, say l and m cubes respectively.Use the inductive hypothesis on both pieces.]


Exam-Style Questions

Section A

Exercise 1.41 The following is a statement about a triangle ABC.

If ABC is isosceles, then AB = BC.

(a) Write out the contrapositive of the statement. [2]

(b) Write out the converse of the statement. [1]

(c) Write out the negation of the statement. [2]

Exercise 1.42 Let n be an integer with n > 1.

(a) Prove that if n is odd, then n2 + 2n− 3 is composite. [3]

(b) Give a counterexample to the following statement:‘If n is even, then n2 + 2n+ 5 is prime.’ [2]

Exercise 1.43 Let p and q be statements. Construct a truth table to show the truth valuesof ¬p∧ q and (p∨ q) ⇒ ¬p. Are ¬p∧ q and (p∨ q) ⇒ ¬p equivalent? Explain your answer. [5]

Exercise 1.44 Consider the following statement.

For each real number x there is an integer n such that n > x.

(a) Write out the statement using the symbols ∃, ∀, R, Z and ∈. [3]

(b) Write out the negation of the statement. [2]

Section B

Exercise 1.45(a) Let f(n) =n∑

r=1

(2r − 1).

(i) Evaluate f(n) for n = 1, 2, 3, 4, 5 and 6. [3]

(ii) Conjecture a formula for f(n) in terms of n. [2]

(iii) Use induction to prove that your conjecture in (ii) is valid for all positiveintegers n. [5]

(b) Let n be a positive integer. Use a contrapositive proof to show that if 2n − 1 is prime, thenn is prime. [5]

(c)(i) Suppose that a, b, c and d are positive integers such that

a2 − 3b2 = 1

c2 − 3d2 = 1.

Let x = ac+ 3bd and y = ad+ bc. Show that x2 − 3y2 = 1. [3]

(ii) Noting that 22 − 3(12) = 1 and 72 − 3(42) = 1, use part (i) to find positive integers m andn, with m > 7, such that m2 − 3n2 = 1. [2]

Note: Exam questions, particularly Section B questions, often combine elements from differentchapters.

1.12 Variations on the Principle of Induction Proof & Structure 27

Historical Notes

At the end of each chapter of these notes you will find some historical titbits relevant to thetopics that have been covered. Since we would certainly call the study of proof pure mathe-matics, you might wonder where this term first arose. Francis Bacon wrote about mathematicsin Book II of ‘The Advancement of Learning’, published in 1605. He split mathematics (thenused in the plural, hence the s) into ‘pure’ and ‘mixed’ – what we would now call applied.

In the Mathematics I can report no deficience, except it be that men do not sufficientlyunderstand the excellent use of the Pure Mathematics, in that they do remedy and cure manydefects in the wit and faculties intellectual. [. . . ...] And as for the Mixed Mathematics, I mayonly make this prediction, that there cannot fail to be more kinds of them, as nature grows

further disclosed.

We’ll mention Bacon again later in the context of the mathematician Cantor.

In the first edition of the Encyclopaedia Britannica (1768-1771), the advantage of proof isdiscussed:

Pure mathematics have one peculiar advantage, that they occasion no disputes amongwrangling disputants, as in other branches of knowledge; and the reason is, because the

definitions of the terms are premised, and every body that reads a proposition has the sameidea of every part of it.

Pythagoras (circa 569 – 475 BC)

Not a great deal is known about Pythagoras. He was probably not the first to prove ‘Pythago-ras’ Theorem’, if indeed he proved it at all (the first surviving written proof from the Greekmathematical tradition is in Euclid, Proposition 47, Book 1). Lists of Pythagorean triples, thatis sets of three integers which form the sides of a right angled triangle (3,4,5; 5,12,13; and soon) have been found on Babylonian tablets and it is even speculated that three of the six tripleswith numbers smaller than 40 were known to megalithic man. This of course isn’t the sameas proving the theorem. However there were proofs by Chinese and Indian mathematicians,possibly predating Pythagoras. The Indian Mathematician Baudhayana, in roughly 800BC,includes a special case of the theorem in his Sulbasutra: ‘The rope which is stretched acrossthe diagonal of a square produces an area double the size of the original square.’ In ChinaPythagoras’ Theorem is known as the Gougu rule. Since Pythagoras hundreds of proofs havebeen devised, one even due to a U.S. President!Pythagoras’ Theorem and the elegant approach of Euclid have impressed many over the cen-turies. The following is from a biography of the philosopher Thomas Hobbes.

He was 40 years old before he looked on Geometry; which happened accidentally. Being in aGentlemen’s Library, Euclid’s Elements lay open, and ’twas the 47 El. libri I. He read the

Proposition. By G–, sayd he (he would now and then sweare an emphaticall Oath by way ofemphasis) this is impossible! So he reads the Demonstration of it, which referred him back tosuch a Proposition; which proposition he read. That referred him back to another, which healso read. Et sic deinceps [and so on] that at last he was demonstratively convinced of that

trueth. This made him in love with Geometry.


Learning Outcomes

After studying this chapter, and the related exercises, you should be able to

• appreciate why there is a need for proof in mathematics;

• realise that there is no guaranteed method for constructing proofs of a given mathematicalstatement;

• feel more confident using mathematical notation and terminology;

• understand and define the terms even, odd, prime, composite, multiple and divides withrespect to the integers;

• be aware that a real number is either rational or irrational;

• write out the truth tables of ¬p, p ∧ q, p ∨ q, p ⇒ q and p ⇐⇒ q, and use them toconstruct the truth tables of statements involving at most three simple statements;

• explain what it means for a statement to be a tautology and what it means for twostatements to be equivalent;

• state and verify de Morgan’s laws;

• rewrite a statement of the form if p, then q as an equivalent either/or statement;

• write out the contrapositive, converse and negation of a statement of the form if p, thenq, and know that this statement is equivalent to its contrapositive;

• construct proofs for elementary statements, particularly ones about even and odd integers,primes and perfect squares;

• show elementary statements are false by means of a suitable counter-example;

• rewrite statements involving quantifiers using the symbols ∀ and ∃;

• negate statements and definitions involving quantifiers;

• determine whether a simple statement involving quantifiers is true or false;

• understand the ideas involved in a contrapositive proof and in a proof by contradiction,and use them to verify elementary results;

• prove that√2 is irrational and that there are infinitely many primes;

• understand the principle of induction, and use it to prove statements about the naturalnumbers.

psm - c1 - general course information

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