MSM1G3 Foundations and Reasoning

handouts

Richard Kaye

Autumn 1999

Contents

Handout 1. Introduction 2Handout 2. The importance of providing full justification 4Handout 3. Truth values and truth tables 5Handout 4. More on truth tables 6Handout 5. Proofs 8Handout 6. The Greek alphabet, and other symbols 12Handout 7. Definitions and substitution 13Handout 8. Variables and relations 15Handout 9. Summary on equivalence relations and ordered sets 17Handout 10. Definitions of functions on quotients 19Handout 11. Construction of Z and Q 20Handout 12. Introduction to R 21Handout 13. Scope of variables 23Handout 14. Quantifiers and quantifier logic 24Handout 15. Induction 25Handout 16. More on Induction 27Handout 17. Axioms for functions 29Handout 18. Commutative Rings 30Handout 19. Groups and quotients 32Handout 20. Revision notes 33

Additional handouts

Handout 21. Ordered Structures and fields 34Handout 22. Boolean rings 35

1

MSM1G3 handout 1: Introduction

What is this module about? This handout outlines my interpretation ofthe syllabus, which has be provided elsewhere.

There are three main ‘elements’ of any mathematical theory.

1. Theorems and proofs—what is true, and why.

2. Examples—what the theory applies to.

3. Counterexamples—examples showing certain statements are not true.

This module introduces you all of these, to the way mathematics is developed,the need for precision in mathematical language, and in particular the ideaof proof.

There are four main blocks of material concerning proofs that we will needto cover. (At the moment, you should probably skip quickly over these, sincemost of the words here will be unfamiliar, but this list may well turn out to beuseful for revision purposes later.)

Mathematical Language. Precision. Common shorthands such as ‘iff’, ‘s.t.’ and‘QED’. Mathematical symbols such as =, 6, etc. Logical symbols such as ¬(‘not’), ∧ or & (‘and’), ∨ (‘or’), ⇒ (‘implies’), and⇔ (‘iff’). Variables, x, y, z . . . .Quantifiers, ∀ (‘for all’) and ∃ (‘there exists’).

Methods of reasoning and proof. Some or all of the following will be discussed:direct argument; constructive/nonconstructive reasoning; mathematical induc-tion; use of the contrapositive; reductio ad absurdum; use of ‘invariants’; howto read and write proofs.

Axiom systems. What are these and why are they needed? Examples and Coun-terexamples. Investigation of axioms systems for numbers.

Set theory. Equivalence relations, partitions, constructions using these. Families ofsets. Applications, providing examples and counterexamples.

Most of this material is ‘just’ logic, and is somewhat ‘abstract’ and ‘dry’taken on its own. So it needs to be put in context by illustrations from realmathematics. A major part of this course is to see how all the above ma-terial is put to use . Since mathematics is first and foremost about numbers,most of the illustrations I will take will be based around exploring the idea ofnumber. These illustrations will be the main mathematical part of this module,and will not always be easy, but I hope that you will find them stimulating.

What will you need to get out of this module for the exam in January andfor other courses you take during your time in Birmingham?

1. You will need to have some appreciation of the importance of precisionand rigour in mathematics.

2. You will need to have some background knowledge of foundations of theidea of number.

3. You will need to be able to read and use the precise language of mathe-matics.

4. You will need to be able to construct and write down some proofs of yourown.

These handouts are intended to supplement lectures, but are not a replace-ment for the lectures. I intend to put detailed factual information in thehandouts—the sort of stuff that you could easily copy down incorrectly in alecture. But you will need to follow the lectures to understand these handouts.

2

In particular, most of the illustrations and examples that I cover will be donein lectures and will not be on the handouts, so these handouts will most likelybe rather difficult reading on their own. You will also find ‘exercises’ in thehandouts; these are somewhat open-ended and are simply intended to get youthinking about what you have just been reading.

There will also be separate exercise sheets on which the coursework for thismodule will be based. These questions are not so open-ended and are intendedfor you to practice—indeed many of them are exam-type questions. Theseexercises will be marked ‘A’ (straightforward questions, similar in difficulty tothe compulsary Section ‘A’ of the exam), ‘B’ (more difficult questions, aboutthe sam difficulty as Section ‘B’ questions on the exam), or ‘C’ (more difficultstill).

Organization. There are two lectures every week (Tuesday 12 noon andThursday 10 am in LRA) and a slightly complicated system of ‘tutorials’ and‘supervisions’ on Tuesdays and Thurdays starting in week 2, which is the sub-ject of a different hand-out. You will be expected to attend all lectures, andalso one tutorial or supervision every week from week 2 onwards. If you haveunavoidable difficulties concerning attendance (medical problems, etc.) pleaselet me or Dr Kyle know as soon as possible so that we may make alternativearrangements and/or compensate your mark accordingly so that you are notdisadvantaged.

Reading List. The three books listed here are all fairly modern textbooksaimed at first-year undergraduates taking a course similar to this one. I won’tfollow any particular textbook, but any or all of these books may be foundhelpful.

Keith Devlin. Sets, functions and logic. Chapman and Hall Mathematics Se-ries. Chapman & Hall, London, second edition, 1992. $19.99.

D. L. Johnson. Elements of Logic via Numbers and Sets. Springer, 1998.

Daniel J. Velleman. How to prove it. Cambridge University Press, Cambridge,1994. $15.95.

The next two books are ‘classics’, with plenty of interesting mathematicsand other entertaining insights into mathematics, well-worth dipping into, andprovided here for background reading and further entertainment.

Littlewood’s miscellany. Edited by Bela Bollobas. Cambridge University Press,1986.

R. Courant and H. Robbins. What is Mathematics? Oxford University Press,1996.

3

MSM1G3 handout 2: The importance of provid-ing full justification

I will give several examples in lectures where a bald statement without reasoningis of little or no value at all, but the same statements backed up by reasoning ismuch more useful—even if the statement, or reasoning, or both, are technicallyincorrect.

The most important part of mathematics is the reasoning behindthe answers, not the answers themselves.

Reasoning is important because:

• An answer to a problem may be correct, but only applies to that partic-ular problem. The same reasoning may give answers to a lot of similarproblems.

• An answer to a problem may be wrong. If the reasoning is given, it maybe that a small mistake can easily be rectified to produce a correct answer.

• The reasoning highlights the assumptions made about a real-life problem,or how the problem was modelled mathematically. If the outcome of thereasoning is clearly incorrect, you learn why the assumptions or modelwas incorrect. In some cases, you can exploit the real-life situation andthe ways in which it is different to the perceived model.

• A properly argued reason can often give much more than was originallyintended. For example, most proofs that there are infinitely many primestell you how to compute a prime number p greater than any given n. Somemethods are better than others!

• A well thought-out reason or proof can suggest other avenues of investi-gation or research.

In this module, you will only get credit for answers provided withappropriate reasoning. You will get zero marks for the answer alone,whether or not it is correct! This applies even if the question you areanswering doesn’t appear to ask you for your reasons. I will encourageall other lecturers to take the same position (though in some modules this isnot always feasible).

4

MSM1G3 handout 3: Truth values and truth ta-bles

In mathematics, a statement is eitherTRUE (T) or FALSE (F). There is nothird alternative.

(3.1)

This assertion (called ‘the law of excluded middle’) comes from analysis ofall the proofs and arguments that have been made in the past which mathe-maticians readily accept as being correct. All these proofs contain assertions ofvarious kinds, and the proofs can all be understood as saying these assertionsare true, or false, as the case may be. For example, ‘2+2 = 4’ is true, not ‘prob-ably true’ or ‘sometimes true’, or ‘usually true’, or ‘somewhat true’. Likewise,‘0 < −3’ is false. (Later on, we will need to qualify (3.1) slightly, and say thata statement may depend on its context. For example, ‘x > 5’ depends on whatx is: if x = 8 then it is true, and if x = −1 then it is false. But whatever x is,‘x > 5’ is either true or false.)

The ‘values’ T and F, for True and False, are called truth values.Mathematical language is (fortunately) not at all like ordinary language

where the truth of a statement is a much more complicated matter. In fact,mathematical language is much simpler than normal language. An alternativeway of understanding (3.1) is to see it as saying that none of the ‘fuzzy’ state-ments you get in English is precise enough for mathematics and is therefore notallowed.

Exercise 3.1 What is the truth value, if any, of the following statements ofEnglish?

(a) It is raining. [Suppose it is only spitting lightly.]

(b) I am tall.

(c) This sentence is false.

(d) This sentence is true.

Since there are only two truth values, T and F, and every statement is eithertrue or false, it is an easy matter to explain words like not, and and or. Thestatement ‘not S’ is true when S is false, and false when S is true. Similarly for‘and’ and ‘or’.

So the mathematical meaning or ‘not’, ‘and’, ‘or’, can be conveniently definedby a table (called a truth table), as follows.

P Q ¬P ¬Q P ∧Q P ∨QT T F F T TT F F T F TF T T F F TF F T T F F

The symbol ¬ is used for ‘not’ because it looks like a modified minus sign. ∧ and∨ look a bit like the intersection and union signs, ∩ and ∪. (Recall, x ∈ A ∩Bis true if and only if x ∈ A and x ∈ B. Similarly x ∈ A ∪B is true if and onlyif x ∈ A or x ∈ B.) If you like, you can use & instead of ∧. There isn’t anygood alternative symbol for ∨ though.

5

MSM1G3 handout 4: More on truth tables

We now come on to words such as ‘implies’ and ‘if . . . then’ which will berepresented in this course by⇒. (Some people use→; I won’t here, and stronglyrecommend that you don’t either.) The truth-table definitions of these can causesome difficulties unless you remember that all we are doing is defininga precise mathematical meaning for the symbol ⇒ and not giving anexplanation of ‘implication’ in the English language.

We want the simplest meaning for P ⇒Q (read: ‘if P then Q’, or ‘P impliesQ’, or even ‘P arrow Q’) which depends on the truth values of P and Q only(and not on what particular statements they actually are). Some examples willmotivate this.

1 + 1 = 2 implies 2 + 2 = 4 (T )1 + 1 = 2 implies 1 + 1 = 3 (F )1 + 1 = 3 implies 2 + 2 = 6 (T )1 + 1 = 3 implies 0 + 0 = 0 (T )

To see the last two implications are true, try multiplying the first equation by2, and then by 0. Of course, the truth of an implication is a completely differentthing to the truth of its conclusion.

From this we see that the only possible truth table for ⇒ is this one:

P Q P ⇒QT T TT F FF T TF F T

You should learn this table, as it is essential for most of this courseand the exam.

According to the truth table, any false statement P implies any other state-ment at all. This is contrary to many people’s feelings about ‘natural’ (i.e.,nonmathematical) language, but really is correct for mathematical language.To illustrate the differences, have a go at the following exercise—first do it asa non-mathematician, then as a mathematician, and see if your answers aredifferent!

Exercise 4.1 Are the following true or false?

(a) If there are little green men on Mars, then the Earth is visited once a dayby UFOs.

(b) If you are hopping up and down while reading this, then it is raining.

(c) If 1 + 1 = 3 then I am a Dutchman.

(d) If 1 + 1 = 3 then Dr Hermans is Flemish.

WARNING. The symbol ⇒ is perhaps the most misused symbolin all of mathematics! Be careful!

The words ‘if and only if’ or ‘iff’ are symbolized by⇔. By ‘P iff Q’, or P⇔Q,we mean ‘P implies Q and Q implies P ’. Using the truth table definition for‘and’, we get

P Q P ⇒Q Q⇒ P (P ⇒Q) ∧ (Q⇒ P )T T T T TT F F T FF T T F FF F T T T

6

so ⇔ is defined by the truth table

P Q P ⇔QT T TT F FF T FF F T

Notice that P ⇔Q is true exactly when P and Q have the same truth value, so⇔ is like ‘equals’ for truth values.

The symbols ⇒ and ⇔ are related to ⊆ and = for sets. Notice that for twosets A,B, we have:

• A ⊆ B iff ‘x ∈ A⇒ x ∈ B’ is true for all x; and

• A = B iff ‘x ∈ A⇔ x ∈ B’ is true for all x.

Precidence. Just as for usual arithmetic, where we need some conventionsto understand things like 1 + 2 · 3− 4, we need rules to read some of the morecomplicated propositions built from ¬, ∧, ∨, ⇒, and ⇔.

The rules of precidence I shall use are

1. ¬ is more binding than ∧, ∨, ⇒, and ⇔.

2. ∧, ∨ are both more binding than ⇒, and ⇔.

3. Any other ambiguities will be made clear with brackets in the usual way.

WARNING. It is all too common to see statements like A⇔B⇔C. What doesthis mean? According to the conventions above this ought to be (A⇔B)⇔C orA⇔ (B⇔C), and the brackets should have been given to clarify the ambiguity.But unfortunately it is often used to mean (A⇔B) ∧ (B⇔ C).

Exercise 4.2 Show (by means of truth tables, for example) that (A⇔B)⇔Cor A⇔ (B⇔ C) are equivalent. Find an example that shows (A⇔ B)⇔ C isnot equivalent to (A⇔B) ∧ (B⇔ C).

Necessary and sufficent conditions. The two words ‘necessary’ and ‘suf-ficient’ are very important in mathematics and have very precise mathematicalmeanings.

If an implication A⇒B holds, we say that the condition or statement A issufficient for B to hold. (It is as if we are looking for something that ‘forces’ Bto be true. If A⇒B holds then A does this so is said to be sufficient.) Similarly,if A⇒B, then B is said to be a necessary condition for A to hold. (If we arelooking around for evidence that A is true, then if we discover B is true, it maywell be that A is true—but this isn’t a proof! What we do know, however, isthat if A⇒ B holds and B is false then A would have to be false, i.e., B is anecessary consequence of A being true.)

Often, in a problem we need to find ‘necessary and sufficient conditions for astatement A to hold.’ In other words, we need to find conditions B1, B2, . . . , Bn

such that A⇒Bi for all i (they are all necessary) and (B1 ∧B2 ∧ · · · ∧Bn)⇒A(taken together, they are sufficient). Alternatively, we need to find a statementB1 ∧B2 ∧ · · · ∧Bn which is equivalent to A. However, the equivalence is rarelyproved in one go. It is much more usual to prove the equivalence in two parts:first proving necessity and then sufficiency.

7

MSM1G3 handout 5: Proofs

Before we introduce some methods of proof to our system, I want to introduceone new symbol, ⊥. This means ‘contradiction’, and is always false. (If youprefer you can use F for it. Some people use other symbols; symbols that looksomething like ×× or ×÷ are popular.)

I am now going to introduce some rules that you can always use when youare proving statements. It is straightforward to see why they are all valid.

PL1 From the statement A ∧B, you may deduce A.From the statement A ∧B, you may deduce B.

PL2 From the two statements A and B, you may deduce A ∧B.

PL3 From the statement A, you may deduce A ∨B.From the statement B, you may deduce A ∨B.

PL4 From the two statements A ∨B and ¬A, you may deduce B.From the two statements A ∨B and ¬B, you may deduce A.

PL5 From the two statements A⇒B and ¬A⇒B, you may deduce B.

PL6 From the statement ¬¬A, you may deduce A.

PL7 The two statements A and ¬A together form a contradiction and you maydeduce ⊥ from them.

PL8 From the two statements A and A⇒B, you may deduce B.

PL9 From the statement A⇔B, you may deduce A⇒B.From the statement A⇔B, you may deduce B⇒A.

PL10 From the two statements A⇒B and B⇒A, you may deduce A⇔B.

The last two rules, PL11 and PL12, concern assumptions. We may introducean assumption A—which may be any statement at all—at any time during aproof, but the fact that an assumption has been introduced must be indicated, inthis system by a vertical line on the left. The assumption can only be eliminatedby one of the following rules.

PL11 If you can deduce B from assumption A, you can deduce A⇒B withoutthe assumption:

A (Assumption)...B

A⇒B

PL12 If you can deduce a contradiction ⊥ from assumption A, you can deduce¬A without the assumption:

A (Assumption)...⊥¬A

PL1–12 are the rules for Propositional Logic. Some of these rules have names:PL8 is called modus ponens, and PL12 is called reductio ad absurdum.

I will provide several examples of proofs using these rules, both in lecturesand here.

8

Proposition 5.1 For all statements A,B,C, the following is true:

(A⇒B)⇒ ((C⇒A)⇒ (C⇒B)).

Proof

A⇒B (1) Assumption

C⇒A (2) Assumption

C (3) AssumptionA (4) PL8 (2,3)B (5) PL8 (1,4)

C⇒B (6) PL11 (3,5)

(C⇒A)⇒ (C⇒B) (7) PL11 (2,6)

(A⇒B)⇒ ((C⇒A)⇒ (C⇒B)) (8) PL11 (1,7)

2

The box symbol 2 is used to mark the end of a proof. Some people use‘QED’ (from the latin phrase, Quod Erat Demonstrandum, meaning ‘what wasto be proved’) instead.

In rules PL1–12, you are not allowed to use any previous statement thatdepended on any assumptions that have been eliminated, but you can use anystatement that has previously been obtained using the current assumptions. Inother words, you can use any statement either with no line to the left (no extraassumptions were needed to get these) or any statement immediately to theright of one of the current vertical lines. Consider the following example.

A (1)B (2)

C (3)D (4)

E (5)F (6)

G (7)

H (8)I (9)

J (10)K (11)? (12)

To deduce the statement marked ?, you could use any of the precedingstatements A–K except E and F .

Here are some more important examples for you to consider. The first showsthat an implication and its contrapositive are equivalent.

9

Proposition 5.2 (Contrapositive) For all statements A and B, A⇒B and¬B⇒¬A are equivalent.

Proof

A⇒B (1) Assumption

¬B (2) Assumption

A (3) AssumptionB (4) PL8 (1,3)⊥ (5) PL7 (2,4)

¬A (6) PL12 (3,5)

¬B⇒¬A (7) PL11 (2,6)

(A⇒B)⇒ (¬B⇒¬A) (8) PL11 (1,7)

¬B⇒¬A (9) Assumption

A (10) Assumption

¬B (11) Assumption¬A (12) PL8 (9,11)⊥ (13) PL7 (10,12)

¬¬B (14) PL12 (11,13)B (15) PL6 (14)

A⇒B (16) PL11 (10,15)

(¬B⇒¬A)⇒ (A⇒B) (17) PL11 (9,16)(A⇒B)⇔ (¬B⇒¬A) (18) PL10 (8,17)

2

You should annotate the next proofs yourself, explaining which rule was usedand from which previous line(s).

Proposition 5.3 (Duality) The statements ¬(A∨B) and ¬A∧¬B are equiv-alent.

Proof For one direction we may use,

¬(A ∨B) (1) Assumption

A (2)A ∨B (3)⊥ (4)

¬A (5)

B (6)A ∨B (7)⊥ (8)

¬B (9)¬A ∧ ¬B (10)

and for the other,

10

¬A ∧ ¬B (1) Assumption¬A (2)¬B (3)

A ∨B (4)B (5)⊥ (6)

¬(A ∨B) (7)

2

The other duality law says that the statements ¬(A ∧B) and ¬A ∨ ¬B areequivalent. The proof of this is left as an exercise.

Writing your own proofs. The next level of difficulty arises when you areasked to make up your own proofs rather than just follow them. However, thisisn’t as scary as you might think.

Firstly, remember that in this course I won’t be asking you to write downreally complicated proofs; most of the exercises are quite simple, especially ifyou know the proof rules PL1–12 well. (Often these proof rules give plenty ofclues on how to proceed.) Secondly, there are some very basic strategies forproving statements that if you follow carefully should stand you in good stead.Examples in lectures explain these in detail, but the most important point is toread the statement or formula you are proving carefully first. You willdiscover that there is a natural way to read statements (looking at the ‘mostimportant’ or ‘outermost’ connective first) that almost tells you how to proceedwith the proof.

For example, the most important connective in

(A⇒¬B)⇒ ((A⇒B ∨ C)⇒ (A⇒ C))

is the second implication sign, and this tells you to try to prove

(A⇒B ∨ C)⇒ (A⇒ C)

from the assumption

A⇒¬B.

The second implication in

(A⇒B ∨ C)⇒ (A⇒ C)

is the most important so you need to indroduce

A⇒B ∨ C

as an assumption and prove

A⇒ C.

Finally, we see that we need to introduce a further assumption, A and want toprove C. At this point the problem should be quite easy to finish off by means ofthe PL rules. (Try it out now!) So this question (which came from last year’sexam) is really only a matter of reading and counting brackets and knowing thePL rules.

11

MSM1G3 handout 6: The Greek alphabet, andother symbols

The table following lists the Greek alphabet. Not all of the symbols here arecommonly used in mathematics, but most are and the list is well-worth learning!Note some ‘alternative forms’ of the lower case letters are added here too. (Inthe case of π and σ, the alternative forms are rarely used in mathematics.)

Capital Small Name Equivalent

A α Alpha AB β Beta BΓ γ Gamma G∆ δ Delta DE ε ε Epsilon EZ ζ Zeta ZH η Eta EΘ θ ϑ Theta ThI ι Iota IK κ Kappa KΛ λ Lambda LM µ Mu MN ν Nu NΞ ξ Xi XO o Omicron OΠ π $ Pi PR ρ % Rho RΣ σ ς Sigma ST τ Tau TΥ υ Upsilon UΦ φ ϕ Phi PhX χ Chi ChΨ ψ Psi PsΩ ω Omega O

(Note: ‘Ch’ is as in Scottish ‘Loch’)

The next table lists some more symbols you might come across in your threeor four years here.

Symbol Name

∇ nabla or del& ampersand or and∂ partial-d∞ infinityℵ aleph† dagger‡ double dagger

12

MSM1G3 handout 7: Definitions and substitu-tion

Sometimes a particular sentence occurs so often it makes sense to introduce ashorthand or definition. For example, consider the following proof.

Let A be the statement P ⇒ P (1) DefinitionA⇔ (P ⇒ P ) (2) From Definition

A (3) AssumptionA ∧A (4) PL2

A⇒ (A ∧A) (5) PL11(P ⇒ P )⇒ ((P ⇒ P ) ∧ (P ⇒ P )) (6) Substitution

Notice that the definition is equivalent to an if-and-only-if statement.This is true of ALL definitions, even though it is unfortunately rather com-

mon to see just the word ‘if’ (and not ‘and only if’) in many definitions, as in“If the relation ∼ satisfies x ∼ x for all x, then it is said to be reflexive,” whichreally means “∼ is relexive if and only if x ∼ x holds for all x.”

Definiton rule You can introduce new symbols or terminology by giving adefinition whenever you like. A definition is an iff (⇔) statement.

Note also the new ‘substitution rule’, which says that

Substitution If statements ψ(A) and A⇔B hold then we may deduce ψ(B).

In fact, this holds for any formulas A and B, not just ‘letters’, but you cansee what is going on better with letters.

There are some subtleties here, which you won’t be used to from A-level, butare very important throughout mathematics. A major lesson to take away fromall this is that the main idea of algebra, that letters can refer to other objects,can be applied even when the objects in question are not numbers. Thus a‘formula’ such as A⇒ (A ∧ A) contains a variable A for a sentence, and A canbe substituted for any sentence you like. Similarly, if the letter f is regarded asa function, we can substitute for f in the formula 2f(x)+f(x+1). For example,if we substitute the function sin for f we get 2 sin(x) + sin(x+ 1).

I won’t give the substitution rule a number, since as it turns out everyinstance of it can be justified by the other rules, but you are free to use it atany time. There is another form of the substitution rule which is also useful,saying that

Substitution if A⇔B holds then we may deduce ψ(A)⇔ ψ(B).

Keen students might like to prove that these two substitution rules are in factequivalent in the sense that one rule follows easily from the other, and vice-versa,but I won’t be so mean as to set this as an exercise!

SUMMARY. Definitions introduce new symbols or terminology and are alwaysif-and-only-if statments (however they happen to be worded!) If-and-only-ifstatements can be usefully used in the substitution rules.

13

Summary of propositional logic

This course is not about developing a rigid framework for proofs, but to showyou how proofs are typically constructed and written down. The point of therules given above is not that they form a sufficient set of rules for the logic of‘true’ and ‘false’ propositions (though they do, and you may see something likethis proved in the third year logic course) but that each of the rules is genuinelyuseful. They are all worth remembering as useful rules for working out andwriting a proof!

My experience is that many people (and not just students—some lecturerstoo!) remember only one or two of the proof rules and sometimes get stuckbecause they have forgotten a particularly useful rule. The one everyone findseasy to remember is reductio ad absurdum (PL12). This is a very powerful ruleand used a lot—perhaps too much! The one most people forget is PL5 (fromA⇒B and ¬A⇒B deduce B), which is a pity as this is one of the most powerfulrules of all, especially if you make the right choice of statement A.

Just as important as the rules themselves is the way proofs are set out. I havedone this in a fairly rigid way, and proofs are not normally presented in sucha style—for instance they are usually given in English or some other naturallanguage—but all correct well-presented proofs will be presented in a way thatshows the structure, where the assumptions are, and where they are discharged.You should always do the same.

Students often find the process of finding a proof daunting. It need notbe so bad if one carefully examines the statement one is trying to prove andany relevant definitions. These almost always contain sufficient clues of how toproceed, and in many cases the proof almost seems to write itself!

We also discussed various extra principles that can be used in proof writing.These were,

1. Definitions. Introducing shorthands and other terminology.

2. Substitution rules.

3. Contrapositive and duality.

However, although the idea of proof rules is important, you should not worryunduely about whether your proof uses only the ‘allowed rules’. For the rest ofthis course, and for all the other courses in your undergraduate studies (with thepossible exception of some parts of the third year ‘logic’ course), you should feelfree to use your own rules, provided you do this sensibly. By ‘sensibly’, I meanthat: (1) you should be convinced about the rule’s validity in this situation;(2) if there is any possible ambiguity, you should explain the rule you are using;and (3) the rule should be relevant, bearing in mind in a proof we are usuallytrying to justify a complicated statement by means of simpler ones we knowto be right. There’s no credit in using complicated and difficult rules to provesomething simple!

Above all, you should remember the way a proof is structured, and it isalways useful to think of proof rules as somehow predetermined and already‘out there’ ready to be used when needed, even if you don’t actually make a fulllist of these rules before you start.

WARNING. Quite often, I see students mixing up numbers, statements, and,or, addition, multiplication, and goodness knows what else. Things like

1 ∧ 2 = 1 + 1 ∨ −1⇒ 2

seems to be quite common in students’ work. What on earth does it mean? Ihaven’t the faintest idea, and I certaily can’t give it any marks. Please remem-ber: you can only connect separate mathematical statements with ∧,∨,⇒,⇔,¬,not numbers or anything else.

14

MSM1G3 handout 8: Variables and relations

The statements we consider when we are applying the logical rules of the lastfew lectures are statements about numbers or sets, such as x = π, x2 > 1 + x,‘X does not have any elements’, 2 ∈ Y and ‘X and Y have the same elements’.We see straight away that we need some further principles before we can workwith such statements.

First, we need variables representing sets and numbers (for example), and weneed other constant symbols, representing specific numbers such as 0, 1, 2, e, π, . . . .These can be compared using relations, such as =, ∈, <, ∼ and 6. Our variableswill be thought of as ranging over unknown values in a domain D . A statementsuch as x ∼ y has a definite truth value for all values for x, y in D , i.e., is to beconsidered as a mathematical statement.

Definition 8.1 A relation ∼ on a domain D is transitive if whenever we haver ∼ s and s ∼ t for r, s, t in D we can deduce r ∼ t.

For example, =, < and 6 are all transitive.

Definition 8.2 A relation ∼ on a domain D is symmetric if whenever we haver ∼ s for r, s in D we can deduce s ∼ r.

For example, = is symmetric, but < and 6 are not.

Definition 8.3 (a) A relation ∼ on domain D is reflexive if we have r ∼ rfor all r in D .

(b) A relation ∼ is irreflexive if we have ¬(r ∼ r) for all r in D .

(c) A relation ∼ is not reflexive if we have ¬(r ∼ r) for some r in D .

For example, = and 6 are reflexive, and ∈ and < are irreflexive. Noteparticularly that ‘irreflexive’ is NOT the same as ‘not reflexive’. For example,the relation x ∼ y defined on N by:

x ∼ y iff x and y are both divisible by two

is neither reflexive (since 1 ∼ 1 is false) nor irreflexive (since 2 ∼ 2 is true).You can define a relation ∼ on a finite domain D by a table—juist like a

multiplication table. But for infinite domains some other definition is required.Many more examples will be given in lectures.Please note that the definitions above are essential for this module

and must be learnt. (Many students last year in their exam threw marksaway unnecessarily by not bothering to do so.)

Equivalence relations

Just about the most important examples of these relations are equivalence re-lations.

Definition 8.4 A relation ∼ is an equivalence relation if it is transitive, reflex-ive and symmetric.

For example, = is an equivalence relation. Equivalence relations and, witha little care, transitive relations in general can be used rather easily in proofs.The following trivial example shows how this may be set out.

15

5 = 1 + 4 (1)= 1 + (2 + 2) (2) as 2 + 2 = 4= 1 + ((1 + 1) + 2) (3) as 1 + 1 = 2= 1 + ((1 + 1) + (1 + 1)) (4) as 1 + 1 = 2

) 5 = 1 + ((1 + 1) + (1 + 1)) (5) transitivity of =

Similarly,

5 > 4 (1)> 1 + 2 (2) as 4 > 1 + 2> 1 + 1 (3) as 2 > 1

) 5 > 1 + 1 (4) transitivity of >

You can even use transitivity with ⇔:

A⇔ (¬A⇒A) (1) Propositional Logic⇔ (¬A⇒ (¬A⇒A)) (2) Substitution (1)⇔ (¬A⇒ (¬A⇒ (¬A⇒A))) (3) Substitution (1,2)

) A⇔ (¬A⇒ (¬A⇒ (¬A⇒A))) (4) transitivity of ⇔

This works since the relation⇔ on the domain D of mathematical sentencesis transitive too! (Note also the use of the substitution rule in lines 2 and 3above.)

The subject of equivalence classes is an important one and one will we returnto several times. As a taster, you might like to think about the following exercise.

Exercise 8.5 Let A = (n,m) : n,m ∈ Z, m 6= 0 and define ∼ on A by

(n,m) ∼ (r, s)⇔ ns = mr.

Show that ∼ is an equivalence relation. Say in your own words what it meansfor two pairs (n,m) and (r, s) to be equivalent under ∼.

Special rules for equality

The equivalence relation = (on the domain of all numbers, sets, etc.) is sospecial, we write down the formal rules for it here.

Reflexivity For any t, the statement t = t is valid.

Symmetry Given t = s you may deduce s = t.

Transitivity Given r = s and s = t you may deduce r = t.

There is a further special substitution rule for = that doesn’t apply to any otherequivalence relations.

Substitution From φ(s) and either s = t or t = s we may deduce φ(t).

You can think of this as saying that = is the very ‘finest’ equivalence relation ofall, or that all other equivalence relations are coarser or the same as =. I meanthis in the sense that, for any other equivalence relation ∼, we have,

x = y⇒ x ∼ y

or equivalently (by contrapositive!)

¬x ∼ y⇒ x 6= y.

16

MSM1G3 handout 9: Summary on equivalencerelations and ordered sets

Definitions and proofs by cases

It is possible to define an equivalence relation (or any other kind of relation) ona finite set X by giving all the cases. (There is really no difficulty here, eitherfrom a logical point of view or from the point of view of precision. The onlyquestion concerns whether this is the quickest or ‘most efficient’ way to achievethe end.)

For example, the relation x ∼ y⇔def |x − y| 6 1 on the set X = 0, 1, 2may be defined in an alternative way as

x ∼ y ⇔def (x = 0 ∧ y = 0) ∨ (x = 0 ∧ y = 1) ∨ (x = 1 ∧ y = 0) ∨(x = 1 ∧ y = 1) ∨ (x = 1 ∧ y = 2) ∨ (x = 2 ∧ y = 1) ∨(x = 2 ∧ y = 2).

This relation is symmetric and reflexive. You can see this by drawing up a table,with the first entry, x, in x ∼ y down the left and the second entry y along thetop:

∼ 0 1 20 T T F1 T T T0 F T T

The relation ∼ is reflexive since the truth values down the diagonal are ALLtrue, and is symmetric because the table is symmetric about the main diagonal.To check transitivity, you have to check all the cases—in this case it is NOTtransitive.

Equivalence relations and partitions

We have seen how an equivalence relation on a set A breaks A up into equivalenceclasses,

x/∼ = y ∈ A : x ∼ y.

These classes are disjoint:

x/∼ 6= y/∼⇒ x/∼ ∩ y/∼ = ∅

and x ∈ x/∼ so every element of A is in some equivalence class.The set A/∼ = x/∼ : x ∈ A is called the quotient of A by ∼. You can

think of this as being just like A except that we identify (i.e., make the same)any two elements of A that happen to be equivalent. This works because

x ∼ y⇒ x/∼ = y/∼

for all x, y ∈ A, as proved in lectures.Sometimes this is described by saying that x/∼ : x ∈ A is a partition of

A. A partition of A is a set B such that

1. every X ∈ B is a nonempty subset of A

2. if X,Y ∈ B and X 6= Y then X ∩ Y = ∅

3. every x ∈ A is a member of some X ∈ B

17

Notice that in a partition B of A there is, for each x ∈ A, a unique X ∈ Bcontaining x. So you can easily go back from partitions to the equivalencerelation that formed it by saying that

x ∼ y⇔def x and y are members of the same set X ∈ B.

The details here are left as an exercise for the industrious student!

Ordered sets

The remarks above concerning definitions and proofs by cases apply to orderedsets as much as they do to equivalence relations.

The axioms for a linear order relation (also called total order, or justorder) on a set A are:

reflexivity x x, for all x ∈ A;

transitivity x y and y z implies x z, for all x, y, z ∈ A;

antisymmetry x y and y x implies x = y, for all x, y ∈ A;

linearity x y or y x, for all x, y ∈ A;

Note too that it is common to say that ‘(X,) is an ordered set’ instead of‘X is a set and is an order on X.’

Sometimes relations somewhat like orders can give rise to equivalences. Sup-pose is a relation on a set A satisfying the reflexivity law, the transitivity law,and the linearity law, but not the antisymmetry law. Then gives rise to anequivalence relation, defined by

x ∼ y⇔def (x y ∧ y x).

The point of this is that we may define on equivalence classes by

x/∼ y/∼⇔def x y

and then is a linear order on A/∼.As an example of this, suppose is the relation on R defined by x y⇔def

[x] 6 [y]. Then x ∼ y ⇔def (x y ∧ y x) is just the equivalence relationstating that x and y have the same integer parts. Then R/∼ is just like anothercopy of Z, and is the usual relation on R/∼.

So, a moral to take away from this (which will be taken up later in lecturesand will certainly be important in other courses you will follow) is that byfactoring out by a suitable equivalence relation we can make a somewhat badlybehaved relation or function (like the relation defined above on R which isn’treally an order) into a better-behaved one on a factor set (like the corresponding on R/∼).

18

MSM1G3 handout 10: Definitions of functions onquotients

Suppose D is a set and ∼ is an equivalence relation on D. It happens ratheroften that we want to define a function F : D/∼ → A by the ‘recipe’

F (x/∼) = f(x) (∗)

where f : D → A is some function we are already given.For example, let D be the set Z of integers, x ∼ y⇔def 6|(x − y), and for

x ∈ Z let f(x) be the remainder on dividing x by 3. Is then the recipe (∗) avalid definition for a function F : D/∼ → Z? In this case, the answer is ‘yes’,but there are plenty of similar cases when (∗) does not define a function.

The point is that a function F must determine a single value F (X) for everypossible X in its domain. But in the case we are considering

· · · = −6/∼ = 0/∼ = 6/∼ = 12/∼ · · ·

and

· · · = −5/∼ = 1/∼ = 7/∼ = 13/∼ · · ·

and so on. So for the recipe (∗) to define a function we must check that state-ments such as

· · · = f(−6) = f(0) = f(6) = f(12) = · · ·

and

· · · = f(−5) = f(1) = f(7) = f(13) = · · ·

all hold.

Proposition 10.1 For a function F : D/∼ → A given by F (x/∼) = f(x) forsome known function f : D → A to be well-defined we need f(x) = f(y) to betrue for all x, y ∈ D such that x/∼ = y/∼, i.e., for all x, y ∈ D such that x ∼ y.

Plenty of examples have been given in lectures. An example that takes thisslightly further is Q, which was defined to be D/∼ where D = (n,m) : n,m ∈Z, m 6= 0 and the equivalence relation ∼ is defined by (n,m) ∼ (r, s)⇔def ns =rm. We define addition on Q by

(n,m)/∼⊕ (r, s)/∼ = (ns+ sm,ms)/∼

and multiplication on Q by

(n,m)/∼ (r, s)/∼ = (nr,ms)/∼.

To show these definitions to be well-defined we need to prove

Theorem 10.2 For any (n,m), (r, s), (n′,m′), (r′, s′) ∈ D with (n,m) ∼ (n′,m′)and (r, s) ∼ (r′, s′) we have (ns+ sm,ms) ∼ (n′s′ + s′m′,m′s′) and (nr,ms) ∼(n′r′,m′s′), and hence, for such elements of D, we have (ns + sm,ms)/∼ =(n′s′ + s′m′,m′s′)/∼ and (nr,ms)/∼ = (n′r′,m′s′)/∼.

This will be proved in lectures.

19

MSM1G3 handout 11: Construction of Z and Q

This summary sheet outlines the construction of these important number sys-tems from N. Proofs will be omitted—most are straightforward, one or two area little tricky.

We assume N and addition, multiplication and < on N is understood.

Construction of Z. Z is defined to be Z /∼ where Z = N × N and (r, s) ∼(u, v)⇔def r + v = u+ s. Addition and multiplication are defined by

(r, s)/∼⊕ (u, v)/∼ =def (r + u, v + s)/∼(r, s)/∼ (u, v)/∼ =def (ru+ sv, rv + su)/∼(r, s)/∼ (u, v)/∼⇔def r + v 6 s+ u

One must prove that: ∼ is an equivalence on Z ; ⊕, and are well-defined;and certain expected properties of Z hold.

N embeds into Z, by the map

θ : n 7→ (n, 0)/∼One must prove that: θ is an injection, i.e., we can’t have θ(x) = θ(y) for somex 6= y; and that θ respects the operations just given, i.e.,

θ(n)⊕ θ(m) = θ(n+m)θ(n) θ(m) = θ(n ·m)θ(n) θ(m)⇔ n 6 m

for all n,m ∈ N.When this is done, we choose to identify n with θ(n) and regard N ⊆ Z. We

also write ⊕ and and normally as +, · and 6.

Construction of Q. Q is defined to be Q/∼ where Q = Z × (Z \ 0) and(r, s) ∼ (u, v)⇔def rv = us. Addition and multiplication are defined by

(r, s)/∼⊕ (u, v)/∼ =def (rv + us, rs)/∼(r, s)/∼ (u, v)/∼ =def (ru, sv)/∼(r, s)/∼ (u, v)/∼⇔def rv 6 su if s, v > 0 or s, v < 0(r, s)/∼ (u, v)/∼⇔def rv > su if s > 0 > v or s < 0 < v

One must prove that: ∼ is an equivalence on Q; ⊕, and are well-defined;and certain expected properties of Q hold.

Z embeds into Q by the map

ψ : n 7→ (n, 1)/∼One must prove that: ψ is an injection, i.e., we can’t have ψ(x) = ψ(y) for somex 6= y; and that ψ respects the operations just given, i.e.,

ψ(n)⊕ ψ(m) = ψ(n+m)ψ(n) ψ(m) = ψ(n ·m)ψ(n) ψ(m)⇔ n 6 m

for all n,m ∈ N.When this is done, we choose to identify n with ψ(n) and regard Z ⊆ Q.

We also write ⊕ and and normally as +, · and 6.

Construction of R. The real numbers are constructed from Q is a similarway. This is the subject of the next handout.

20

MSM1G3 handout 12: Introduction to R

We have talked about the real numbers in this course quite a lot, but up to nownot actually defined them. In fact, there are several popular ways of definingthe reals. The following one is particularly beautiful.

A sequence of rationals is an infinite list a0, a1, a2, . . . , an, . . . of rationalnumbers an ∈ Q, where the indices n range over the natural numbers. We oftenwrite (an) or (an)n∈N for such a sequence. A sequence (an) of rationals has theCauchy property or is a Cauchy sequence if

∀ε=p

q> 0 ∃n0 ∈ N ∀i, j ∈ N (i > n0 ∧ j > n0⇒ |ai − aj | < ε).

(Here, the variable ε ranges over rationals since we haven’t defined the realsyet!)

For example, the sequence 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, . . . hasthe Cauchy property as |ai − aj | 6 1/10max(i,j).

Two Cauchy sequences (an), (bn) are equivalent, (an) ∼ (bn), if

∀ε=p

q> 0 ∃n0 ∈ N ∀i ∈ N (i > n0⇒ |ai − bi| < ε).

You have probably guessed the next bit. We define

R = (an) : (an) is a Cauchy sequence of rationals.

Then it can be proved that ∼ is an equivalence relation on R, and we defineR = R/∼. Many of our favourite real numbers have rather nice and easydefinitions; for example,

π = (3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, . . . )/∼.

We can also embed our rationals into R, by sending the rational number q to(q, q, q, q, . . . )/∼, the equivalence class of the constant sequence.

The order relation 6 on R and the operations of addition and multiplicationcome next.

Order. (an)/∼ 6 (bn)/∼⇔def ∀ε=pq > 0 ∃n0 ∈ N ∀i ∈ N (i > n0⇒ai 6 bi+ε).

Addition. (an)/∼+ (bn)/∼ =def (cn)/∼, where cn = an + bn for all n ∈ N.

Multiplication. (an)/∼ · (bn)/∼ =def (cn)/∼, where cn = anbn for all n ∈ N.

Of course, it is necessary to prove these are all well-defined. Due to lackof time, I cannot do this here, but very similar things will be proved in theMSMQP2 course next semester, and when you have done that course you willbe able to come back here and fill in the missing details.

The most important properties about R are as follows. Once again these allneed proof, but sadly I don’t have time.

Archimedean property. For all x ∈ R there is n ∈ N such that n > x.Equivalently, For all x > 0 in R there is n > 0 in N such that 1/n < x. Itfollows from this that Q is dense in R or in other words any real numbercan be approximated by a rational number. (See a later lecture.)

Completeness axiom. Having got the reals, we may consider sequences ofreals, (rn), and in particular Cauchy sequences of reals. We can repeat thewhole construction, defining a new equivalence relation (rn) ≡ (sn)⇔def

∀ε > 0 ∃n0 ∈ N ∀i ∈ N (i > n0⇒ |ri − si| < ε).

21

Let S be the set of Cauchy sequences of reals, and S = S /≡. So S isa set of ‘super real numbers’. Will the process ever stop? Well in factit already has done, and every Cauchy sequence in S turns out to beequivalent to a constant sequence (r, r, r, r, . . . ) of reals r. In other words,the construction of S has added nothing new that wasn’t already in R.We refer to this property of R by saying that R is complete. Anotherway of saying this is that every Cauchy sequence of reals has a real limit.(For the meanings of some of these words, see the MSMQP2 module.)

Dedekind’s form of the completeness axiom. It turns out the the com-pleteness just discussed is equivalent to a statement that says

every bounded nonempty set A ⊆ R has a least upper bound

Here, a given A ⊆ R, an upper bound for A is some x ∈ R such that y 6 xfor all y ∈ A. The set A is bounded if it has an upper bound. Some x ∈ Ris a least upper bound for A if it is an upper bound and also is least withthis property, i.e., for all y < x, y is not an upper bound, i.e., there isa ∈ A with a > y. (Note that this least upper bound need not be in theset A.) As an exercise, prove that the least upper bound of the interval(0, 1) is 1.

Uncountability. The set of reals is not countable, that is there is no sequence(rn) of real numbers such that every real number x appears in the listas rn for some n ∈ N. Thus the ‘number of real numbers’ is an infinitenumber, and is bigger than the number of natural numbers, or the numberof rational numbers.

22

MSM1G3 handout 13: Scope of variables

The scope of variable is the piece of mathematics where that variable has a clearmeaning. This idea of scope is very important and occurs in many things youare already familiar with. It is also a source of potentially quite serious errorsand confusion, so it’s worth getting right!

We start with some examples.In the formulas∫ 1

−1

(x2 + 1) dx100∑i=1

(i3 + 2)100∏j=1

(j4 + 3)

the scope of x is the formula (x2 + 1), the scope of i is the formula (i3 + 2) andthe scope of j is the formula (j4 + 3). The notations

∫ 1

−1. . . dx,

∑100i=1 . . . and∏100

j=1 . . . bind these variables—in other words it is these bits of the notationthat tell you what variable is having its scope restricted.

As a general rule of thumb, you can tell when a variable has restrictedscope when you can rename that variable and not change the overall meaningof the piece. Another way of seeing this is that if a formula involves a variable xbut its meaning doesn’t depend on the name of the variable x (for example theformula is a true-false statement, or gives a specific number), then this variablex is most likely bound in the formula in some way or other, and doesn’t reallyhave any meaning outside the formula (outside its scope).

Variables can also be bound by the quantifiers ∀ (for ‘for all’) and ∃ (for ‘thereexists’). For example, the statement that there are infinitely many primes is,

∀n ∈ N ∃p ∈ N (p > n ∧ p is prime).

The scope of the variable p is (p > n ∧ p is prime), and it is bound by ∃p ∈ N.Similarly, the scope of the variable n is ∃p ∈ N (p > n ∧ p is prime), and it isbound by ∀n ∈ N.

A subtle sort of mistake, which is unfortunately very common, is to refer to avariable outside its scope where it really doesn’t have any meaning. A typical ex-ample is, ‘what is p in the statement ∀n ∈ N ∃p ∈ N p > n∧p is prime’? Such aquestion simply cannot be answered. One might as well have asked, ‘what valuehas x got in

∫ 2

1x2 dx? or even ‘what is p in the statement ∀n ∈ N ∃q ∈ N q >

n ∧ q is prime’?

23

MSM1G3 handout 14: Quantifiers and quantifierlogic

The quantifiers are

“for all x . . . ” or ∀x . . . ,

and

“there exists x [such that] . . . ” or ∃x . . .

These bind the scope of the variable x, as discussed last time.Before we give the formal rules for quantifier logic (QL), you should note

the following general laws about quantifiers.

Duality laws. ∀ and ∃ are dual to each other in the sense that

¬∀x θ(x) ⇔ ∃x¬θ(x)¬∃x θ(x) ⇔ ∀x¬θ(x).

The rules of quantifier logic. The main rules of quantifier logic are thoseof propositional logic (PL) together with four new rules, QL1–4.

QL1 (∀ introduction) You can introduce a new variable x in a new proof block.If the conclusion of the block is θ(x), then the block can be closed and∀x θ(x) deduced.

... (1)

Let x be arbitrary (2)... (3)θ(x) (4)

∀x θ(x) (5)... (6)

The variable x must not already have meaning at the point of the proofwhere it is introduced.

QL2 (∀ elimination) From the statement ∀x θ(x) you may deduce θ(t), providedthe substitution of the term t for x in θ is valid.

QL3 (∃ introduction) From the statement θ(t) you may deduce ∃x θ(x), pro-vided the substitution of the variable x for t in θ is valid.

QL4 (∃ elimination) From the statement ∃x θ(x) you may deduce θ(a), pro-vided the substitution is valid and the variable a is a new variable notalready meaningful in this part of the proof.

24

MSM1G3 handout 15: Induction

The idea behind the principle of mathematical induction is as follows: if (1) astatement φ(n) is true for the natural number n = 0, and (2) if φ(n+ 1) is truewhenever φ(n) is true, then φ(0) is true, so we may deduce φ(1) is true, andφ(2) is true, since φ(1) is, and so on; therefore ∀n ∈ N φ(n) is true.

We can formalize this as a rule:

Induction rule. From the statements φ(0) and ∀x ∈ N (φ(x)⇒φ(x+ 1)), youmay deduce ∀x ∈ N φ(x).

There are other versions of the rule of induction. For example, suppose wehave φ(3) and ∀x ∈ N (x > 3 ∧ φ(x)⇒ φ(x+ 1)). Then by a similar argunmentwe should have ∀x ∈ N (x > 3⇒ φ(x)). In fact this follows from the rule ofinduction just given, for letting ψ(n) be the statement ‘φ(n) ∨ n < 3’ we haveψ(0), ψ(1), and ψ(2) all hold as 0, 1, 2 < 3, and ψ(3) holds since φ(3) holds.But also ∀x ∈ N (ψ(x)⇒ ψ(x + 1)) holds (just check the cases of x = 0, 1, 2separately, using the truth table for ⇒). Therefore by the induction rule wemay deduce ∀x ∈ N ψ(x), and hence ∀x ∈ N (x > 3⇒ φ(x)).

Still more versions of the induction rule can be deduced from the originalinduction rule. Here are three particularly important examples.

Total induction. From the statements φ(0) and

∀x ∈ N (∀y ∈ N (y 6 x⇒ φ(y))⇒ φ(x+ 1))

you may deduce ∀x ∈ N φ(x).

Another form of total induction. From the single statement

∀x ∈ N (∀y ∈ N (y < x⇒ φ(y))⇒ φ(x))

you may deduce ∀x ∈ N φ(x).

Least number principle. From the statement ∃x ∈ N φ(x) you may deducethat there is a least x ∈ N satisfying φ(x), i.e.,

∃x ∈ N (φ(x) ∧ ∀y ∈ N (y < x⇒¬φ(y))).

(The second form of total induction looks more elegant, since it only re-quires the one assumption, but you have to keep a clear head when readingit—remembering the truth table for ⇒! In practice, the first form is just asuseful.)

It may not be entirely obvious, but the new rules above can all be deducedfrom the original rule of induction. But total induction is a much more powerfulrule to use, since to do the induction step, i.e., to deduce φ(n + 1) in totalinduction you may assume φ(k) for all k 6 n, and not just φ(n), so you have astronger assumption to prove the same statement! The least number principlecan be seen as a useful way of stating the contrapositive of the rule of totalinduction.

Plenty of examples will be given in lectures.

Remarks on induction. (1) There are no prizes for ‘half a proof’ usinginduction. If you managed to prove ∀n ∈ N (φ(n)⇒ φ(n+ 1)) but have not yetchecked φ(0) then your proof is incomplete and you haven’t proved anythingat all (however much effort you have expended). It often seems very cruel, butonly complete proofs have any value!

(2) Induction can be a difficult rule to use—especially in more advanced orin research mathematics. The reason is the if you want to prove φ(n) holds

25

for all n ∈ N, it may not be possible to do this by induction on φ(n). To getinduction proofs to work you must make the statement φ(n) stronger. Psycho-logically, this feels completely wrong: ‘I only want to prove ∀n ∈ N φ(n),’ youmay say, ‘surely that should be easier than proving by induction on ψ the muchstronger statement ∀n ∈ N ψ(n) first?’ But unfortunately this is how inductionworks—strong statements are often easier to prove than weak ones! If your in-duction step doesn’t work out, try to make your induction hypothesisstronger.

Examples I will illustrate the ideas of induction here by two proofs of aclassical (in the correct sense of the word!) theorem due to Pythagoras.

Theorem 15.1√

2 is irrational.

Proof 1 We shall assume that there are positive x, y ∈ N such that 2x2 = y2

and using this assumption will obtain a contradiction. (PL12!)The positive number x ∈ N satisfies ∃y ∈ N 2x2 = y2, so by the least number

principle (which is a form of induction) there is a least x > 0 in N such that∃y ∈ N 2x2 = y2. Consider the y satisfying 2x2 = y2 for this least such x. As2x2 = y2 we deduce y2 is even. But the product of two odd numbers is odd,so therefore y itself must be even. Thus y = 2x′ for some x′ ∈ N. This gives2x2 = (2x′)2, so x2 = 2(x′)2. This last equation tells us that x2, and hence x,is even, so x = 2y′ for some y′ ∈ N. But then (2y′)2 = 2(x′)2 so 2(y′)2 = (x′)2.Hence x′ = y/2 < x satisfies ∃y ∈ N 2y2 = (x′)2. But 0 < x′ < x and x wastaken to be the least positive element of N such that ∃y ∈ N 2y2 = x2. This isan obvious contradiction. (We have shown that x is the least natural numbersuch that . . . , but x′ < x also satisfies . . . ). So our assumption that there arepositive x, y ∈ N such that 2x2 = y2 is false and hence

√2 is irrational.

The proof just given is the usual one quoted in text books etc., and is ratherbeautiful. But there is a second, equally elegant, and somewhat different proofof the same theorem.

Proof 2 We prove by induction on x that ∀a, b ∈ N (0 < b 6 x⇒ a2 6= 2b2)holds for all x ∈ N.

For x = 1 this is clear, as a2 6= 2 for a ∈ N. (In fact by the truth table for‘⇒’, the statement is also true for x = 0.)

Assume inductively that ∀a, b ∈ N (0 < b 6 x⇒ a2 6= 2b2) holds for somex > 1. We wish to show ∀a, b ∈ N (0 < b 6 x + 1⇒ a2 6= 2b2) is true. Ifotherwise there are a, b ∈ N with 0 < b 6 x + 1 ∧ a2 = 2b2 (by the truth tablefor ⇒). Now consider a′ = 2b − a and b′ = a − b. As a2 = 2b2 and b > 0 thenb < a < 2b, so 0 < b′ = a − b < b, and hence b′ 6 x as b′ < b 6 x + 1. Sincea2 = 2b2, we also have

(a′)2 = (2b− a)2 = 4b2 − 4ab+ a2 = 2a2 − 4ab+ 2b2 = 2(a− b)2.

So a′, b′ ∈ N are such that ¬(0 < b′ 6 x ⇒ (a′)2 6= 2(b′)2), contradictingour inductive assumption that ∀a, b ∈ N (0 < b 6 x ⇒ a2 6= 2b2). Hence∀a, b ∈ N (0 < b 6 x + 1⇒ a2 6= 2b2), as we required, and it follows from theprinciple of induction that ∀x ∈ N ∀a, b ∈ N (0 < b 6 x⇒ a2 6= 2b2).

In particular, if a, b 6= 0 are natural numbers, then b 6 x for some x ∈ N(e.g., x = b), so this implies that a2 6= 2b2, as we require.

26

MSM1G3 handout 16: More on Induction

Examples of total induction and least number principle.

Proposition 16.1 Every natural number n can be written as∑

i<k βi2i forsome k ∈ N and βi ∈ 1, 2. (If k = 0 the sum above is the empty sum and isconventionally zero.)

Proof By total induction on n in the statement ∃k ∃βi n =∑

i<k βi2i. Thebase step is the case of the empty sum, 0 =

∑i<0 βi2i. For the induction step

assume θ(l) for all l 6 n, and consider n + 1 ∈ N. Then either n + 1 = 2l + 1for some l (and n is even) in which case l =

∑i<k βi2i for some k, βi, hence

n+ 1 = 2l + 1 =∑i<k

βi2i+1 + 1 =∑

0<i<k+1

βi−12i + 1 · 20.

Alternatively, n is odd and n+ 1 = 2l + 2 for some l so

n+ 1 = 2l + 2 = 2∑i<k

βi2i + 2∑

0<i<k+1

βi−12i + 2 · 20

as required. Thus the statement is true for all n. 2

One of my favourite applications of total induction or the least number

principle concerns Pascal’s triangle—the familiar triangle of coefficients(nm

)of coefficients in (x+ 1)n satisfying(

n0

)=(nn

)= 1 and

(n+ 1m+ 1

)=(

nm+ 1

)+(nm

).

We might want to know when one of these coefficients is odd and when it iseven. To this end, define m ≺ n to mean, when m,n are written in binary, thenfor all i, the ith digit of m is 1 implies the ith digit of n is 1. E.g., 3 ≺ 11 but3 6≺ 10. Then (

nm

)is odd iff m ≺ n.

Two reasonably easy cases are obtained by noting that(2n2m

)≡(nm

)and

(2n

2m+ 1

)≡ 0

where x ≡ y means either both are odd or both are even. (These can be seen byexpanding (x + 1)2n = (x2 + 2x+ 1)n and reducing all the coefficients modulo2.) The other cases can be done as an application of the least number principle.

Fallacious proofs by induction

Proposition 16.2 All billiard balls have the same colour.Any finite set of numbers a1, a2, . . . , an has exactly one member.

Proof We prove the two paradoxical statements in the proposition by induc-tion on the variable n in the following statement:

P (n) : any n numbers a1, a2, . . . , an are all equal.

P (1) clearly holds since a1 = a1. Suppose P (n) holds and we are given n + 1numbers a1, a2, . . . , an+1. Then by P (n) we have

a1 = a2 = · · · = an

27

and by P (n) again

a2 = a3 = · · · = an+1

hence

a1 = a2 = a3 = · · · = an+1

as required, showing P (n+ 1) holds. 2

The three red noses. Three women, A,B,C, are in a closed room. Eachone has her ears painted green but doesn’t realise this, and they are all laughingat the others. A suddenly realises that she is just as laughable to the others asthey are to her.

Her argument goes as follows: ‘Suppose my ears are not funny. Then Bwill see that I am not funny and therefore realise that her ears must be funny,else what is C laughing about? But B hasn’t worked this out, so I must belaughable too.’

Next, we shall prove by induction that from n men A1, A2, . . . , An, each withpink spots painted on his forehead, A1 will eventually be able to work out thetruth about himself by observing the reaction of his colleagues. He argues thus:‘I shall assume statement P (n− 1), that of n− 1 laughable people all laughing,they will all eventually discover the truth about themselves. (P (2) is obvious.)Now, if I am not laughable, the others will all see this, so they form a set ofn− 1 people, all laughable and laughing at each other. But by P (n− 1) each ofthem will be able to see that he looks silly, so will stop laughing. However, I cansee that they are all laughing, so this is incorrect. Therefore I am laughable.’

Of course, there is an implicit assumption that the n individuals all think atthe same speed!

A game of cards. A variation is a special game of cards between two people.They play with an infinite pack of cards. Each card has two sides, for each card,one side has a natural number n on it, the other side having n + 1. There is acard like this for each n = 0, 1, 2, 3, . . . . An independent referee chooses a cardat random and holds it so that each player can only see one side of the card, soone player sees n and the other player sees n + 1. The winner of the game isthe player who sees the largest number. But each player is allowed to veto thecurrent game and pass to the next one should he or she wish.

Now, an argument similar to the last one shows that both players shouldveto every game. For if one player saw a 0 he would be sure he has lost so wouldveto. If he saw 1 and his opponent hadn’t vetoed, then his opponent would belooking at a 2 not a 0, so again he should veto. Now suppose statement P (n),that you should veto when looking at n. Then if you see n + 1 it follows thatyour opponent sees n or n+2. But your opponent would veto when seeing n, byP (n) so he sees n+2 and would win. Therefore you should veto. Thus P (n+1).

There is an obvious problem with this apparently perfectly good argument,since it appears that every player that vetos does so sure that he or she has lost,which is clearly not the case, so what went wrong?

The surprise execution. A prisoner is told he will be executed in the next 7days, and on the morning of the execution he will not know that the executionwill be on that day. ‘Good,’ says the prisoner. ‘Then I will not be executedsince it will not be the last day, since by then I will know that that is the onlyday left. Nor can it be the previous day, since the last day is impossible, and soon by induction!’

28

MSM1G3 handout 17: Axioms for functions

Corrections to handout 15. Please replace the sentence in Proof 2 of hand-out 15 that says ‘As a2 = 2b2 and b > 0 then a, b > 0 and a < b/2, so0 < b′ = a− b < b′, and hence b′ 6 x and b′ < b 6 x+ 1.’ with ‘As a2 = 2b2 andb > 0 then b < a < 2b, so 0 < b′ = a− b < b, and hence b′ 6 x as b′ < b 6 x+1.’

Binary functions. Binary functions combine two numbers in a domain D togive another number in D. Examples include +, which is a function that returnsthe sum, x+y, of two numbers x and y, and · which returns the product x ·y ofx and y. These two functions are said to be binary since they take two ‘inputs’,or arguments, x and y.

Other functions may take different numbers of arguments. An importantcase is of unary functions (i.e., taking one argument only) such as the functionx 7→ −x for the ‘negative’ of x, or x 7→ x−1 for the reciprocal of x.

Just as we had some important axioms for relations, such as reflexivity,transitivity, etc., there are important axioms for functions too. Strictly speak-ing, none of the material here is part of the syllabus for MSM1G3, but you willcertainly find it useful for later modules you take. However, the MSM1G3 syl-labus emphasises the use of mathematical language and rigour, and questions onbinary functions can (and do) turn up in exams as a test of your appreciationof mathematical language.

Definition 17.1 The binary function is associative if (x y) z = x (y z)holds for all x, y, z.

For example, + and · are associative. The associativity rule is important becauseit means one need not worry too much about brackets. For example, x+ ((y +z) + ((y+ z) +w)) is the same as (x+ y) + (z+ (y+ (z+w))), so we can notatethem both as x+y+ z+y+ z+w. Binary functions that are not associative dosometimes come up, but fortunately only very rarely. They are rather difficultto work with!

Definition 17.2 The binary function is commutative if x y = y x holdsfor all x, y.

Again, + and · are commutative. Noncommutative functions do appear fairlyregularly, and are very important, but life is much easier when all the functionsyou are working with are commutative!

Sometimes, a binary function has a zero or an identity.

Definition 17.3 A number e is an identity for if x e = e x = x for all x.A number z is a zero for if x z = z x = z for all z.

The main examples are obvious: 0 is an identity for +, 1 is an identity for ·, and0 is a zero for ·. If you find learning this last definition difficult, just rememberthat the key example is the multiplication example.

Definition 17.4 If has an identity e, then an inverse for a number x is somenumber y such that x y = y x = e.

For example, every real number x has an additive inverse, −x. Every nonzeroreal number x has a multiplicative inverse 1/x.

Definition 17.5 A binary function is distributive over another binary func-tion ∗ if x (y ∗ z) = (x y) ∗ (x z) and (y ∗ z) x = (y x) ∗ (z x) for allx, y, z.

Once again, the most obvious example is that · is distributive over + (but notthe other way round!) Note too that we showed that ∧ is distributive over ∨,and also that ∨ is distributive over ∧.

29

MSM1G3 handout 18: Commutative Rings

This handout shows how the axioms in the last handout can be used to provesome well-known properties of numbers. None of the material here is requiredfor the exam.

The rules in the previous handout are sufficient to prove some useful factsabout the real numbers, including the following theorem—one that you probablyhave taken for granted up to now but which can be proved in quite a satisfactoryway.

Theorem 18.1 (−x) · (−y) = x · y for all real numbers x, y.

The rules concerning the real numbers that we shall require are the following:+ is associative and commutative; 0 is an identity for +; · is associative andcommutative; 0 is an zero for ·; · is distributive over +; and that every realnumber has an additive inverse.

The proof is split into parts. We prove in a proposition that there is onlyone additive inverse of a given number x. The definition of −x is that it is thisinverse. Then we prove an intermediate result (a lemma) which gives the resultrequired.

Proposition 18.2 For all real numbers x, y, z, we have:

x+ y = x+ z = 0 ⇒ y = z.

Proof

x+ y = x+ z = 0 (1) assumptionx+ y = 0 (2) from (1)x+ z = 0 (3) from (1)y = 0 + y (4) zero of +

= (x+ z) + y (5) from (3)= x+ (z + y) (6) associativity= x+ (y + z) (7) commutativity= (x+ y) + z (8) associativity= 0 + z (9) from (2)= z (10) zero of +

y = z (11) transitivity of =

and we have proved the proposition. 2

Definition 18.3 The unique w such that x+ w = 0 is denoted −x.

This defines a unary function x 7→ −x and gives us the iff statement,

Definition of the − function: w = −x⇔ x+ w = 0.

Lemma 18.4 For all real numbers x, y, we have (−x) · y = −(x · y).

Proof Assume x, y are arbitrary real numbers. Then

(x · y) + ((−x) · y) = (x+ (−x)) · y (1) distibutivity= 0 · y (2) definition of −= 0 (3) 0 is zero for ·

so,

(x · y) + ((−x) · y) = 0 (4) transitivity of =(−x) · y = −(x · y) (5) definition of −

and we have proved the lemma. 2

30

Proof of theorem. Again, take arbitrary real numbers x, y. We have,

(−x) · (−y) = −(x · (−y)) (1) lemma= −((−y) · x) (2) commutativity= (−(−(y · x))) (3) lemma= (−(−(x · y))) (4) commutativity

(−x) · (−y) = (−(−(x · y))) (5) transitivity of =.

So we only need to show that (−(−(x · y))) = (x · y). But,

(x · y) + (−(x · y)) = 0 (6) definition of −(−(x · y)) + (x · y) = 0 (7) commutativity of +(x · y) = −(−(x · y)) (8) definition of −(x · y) = (−x) · (−y) (9) from lines (5,8) by

transitivity andsymmetry of =.

Definition 18.5 A [commutative] ring, (R,+, ·) or simply R, is a number sys-tem (i.e., a set R, with binary operations of + and · under which it is closed)satisfying the axioms axioms: + is associative and commutative; 0 is an identityfor +; · is associative and commutative; 0 is an zero for ·; · is distributive over+; and every number in R has an additive inverse in R.

Examples of commutative rings include the integers, Z, the rationals, Q,the reals, R, and the integers modulo n (where n is any integer greater than orequal to 2).

Nonexamples include the natural numbers, N. This is ‘almost’ a commu-tative ring, but fails to satisfy the axiom that every number should have anadditive inverse (for example there is no n ∈ N such that 1 + n = 0, since forn ∈ N we have 1 + n > 0).

Exercise 18.6 Sam Slow, a student, once provided the following reason for whyN is not a commutative ring.

‘According to the axiom, every number x should have an additiveinverse, denoted −x. But if we take the natural number 1, we seethat −1 is not a natural number so 1 doesn’t have an inverse, so Nis not a ring.’

Criticise this argument. [Hint: there is a confusion concerning the notation −1:it means both the additive inverse of 1 and also the number one less than 0. Seealso the next exercise.]

Exercise 18.7 Show that the following new definitions of ⊕ and makes Ninto a commutative ring. First define f(x) = x/2 if x is even and f(x) =−(x+ 1)/2 if f is odd, so f : N→ Z is a bijection. (Prove this!) Second, definex⊕ y = f−1(f(x) + f(y)) and x y = f−1(f(x) · f(y)).

Write out addition and multiplication tables for the first few natural numbers.What natural number is the inverse (for ⊕) of 1? If you didn’t get the point ofthe previous exercise, try it again now.

31

MSM1G3 handout 19: Groups and quotients

This handout combines the ideas of equivalence relations and algebra and intro-duces you to the idea of a group. Groups will be studied in MSM2P2, and itmay be useful to see how MSM1G3 and later modules you will study are related.None of the material here is required for the exam.

This handout takes the idea of quotient structures (for example the con-struction of Q from Z) a little further. The object is to show you how quotientgroups may be constructed, including the cases where the group is not com-mutative. This will give you a useful taste of some material from next year’sNumber theory and Group theory modules.

Quotient groups Throughout this section, let (G, ·) be a group with identitye. That is, · is associative, x · e = x = e · x for all x ∈ G, and each x ∈ G has aninverse x−1. By exercises in sheet 6, there is only one identity e and, for eachx ∈ G, only one the inverse x−1 of x.

A subgroup of G is a set H ⊆ G containing e, such that H is closed underthe operation · (i.e., for all x, y ∈ H we have x · y ∈ H) and under inverses (i.e.,for all x ∈ H we have x−1 ∈ H). A normal subgroup of G is a subgroup H suchthat for all x ∈ H and all y ∈ G, y−1 · (x · y) ∈ H.

For example, the subsets e and G itself of G are always subgroups of G.

Exercise 19.1 If G is the group Z considered with the addition operation, thenfor each positive n ∈ Z, H = x : n divides x is a subgroup of Z.

Exercise 19.2 If (G, ·) is an abelian group (i.e., · is commutative) and H ⊆ Gis a subgroup of G then H is a normal subgroup. (So the extra property ‘normal’on a subgroup is only important when the group in question is not abelian.)

We will prove the following theorem in lectures.

Theorem 19.3 Suppose that ∼ is an equivalence relation on the group (G, ·)and

x/∼ y/∼ = (x · y)/∼

defines a binary operation on G/∼ making G/∼ into a group (G/∼,). ThenN = e/∼ is a normal subgroup of G and x ∼ y⇔ x−1 · y ∈ N .

The proof of this result is made easier by defining a map

θ : G→ G/∼

by θ(x) = x/∼. This map is a homomorphism of groups, i.e., sends one group,(G, ·), to another, (G/∼,), such that θ(x · y) = θ(x) θ(y) for all x, y ∈ G.

The theorem just given has an important converse.

Theorem 19.4 If N is a normal subgroup of a group (G, ·) then

x ∼ y⇔def x−1 · y ∈ N

defines an equivalence relation on G such that

x/∼ y/∼ = (x · y)/∼

defines a binary operation of G/∼ making G/∼ into a group.

32

MSM1G3 handout 20: Revision notes

The exam will be two hours long and will be the same format as last year’s.It will consist of two sections, Section A and Section B. Section A contains anumber of shorter questions, and you should attempt all of the questions in thissection. Section B consists of three longer questions, and you should attempttwo of them. (If you attempt all three, don’t cross anything out: the marks foryour best two questions will count.) Each section carries 50 marks and you areexpected to take about an hour on each section (and about half an hour on eachSection B question).

As for all exams, I suggest you get to the exam room in good time andread all the questions carefully first. If the exam is difficult, it is the same foreveryone else, so don’t give up—keep going. If you really can’t do one part ofa question, skip it and go on to the next part. Often longer questions are inseveral parts, with the later parts relying on the answer or result in an earlierpart. If you skipped an earlier part, it sometimes makes sense to assume theresult referred to.

Aim to revise well, think carefully, make the most of the time you have inthe exam, get the maximum number of marks possible.

In this course, the exam mark will be rescaled to a mark out of 80, and thisis added to the coursework, consisting 8 tutorials marked out of 6 each whichis then scaled to give a mark out of 20. The official module description andsyllabus for the course is given here. This may be a useful guide to planningyour revision timetable.

Module description: Clear notation and precise unambiguous arguments arethe basis of modern mathematics. This course gives a sound foundation forthe correct use of mathematical language, and develops the concept of arigorous proof. These ideas are studied in relation to various fundamentalareas of mathematics—in particular the construction of the importantnumber systems in mathematics.

Objectives: That the student be able to: understand and use mathematicallanguage; understand what a proof is, in particular, why rigorous proofsare necessary, how to construct and write down a proof, and which typesof arguments are valid, and which are not.

Syllabus: The course has (or will) cover the following syllabus. You shouldalso pay attention to what aspects of this syllabus have been emphasisedparticularly in lectures etc.

1. Introduction to the basic symbols and notation of mathematics; moredetailed consideration of the use of⇒,⇐,⇔, ∀, ∃, etc.; good practicein writing mathematics; compound statements and negation of state-ments involving the universal and existential quantifiers; translatingwords into symbols and back again.

2. Applications to basic inequalities in R; simple number theory, etc.3. Introduction to the formal processes of reasoning; primitive terms;

rules of inference; direct and indirect methods of proof; fallacious ar-guments; distinguishing between finding a proof and writing a proof;axiom systems; the natural numbers, N.

4. Basic form of mathematical induction; other forms of mathematicalinduction, with proofs that they are equivalent; examples. Appli-cations of mathematical induction: inequalities in N; sums of finiteseries, etc.; analysing false “proofs by induction”.

5. Equivalence relations; construction of Z from N and Q from Z. In-formal introduction to R and completeness.

33

MSM1G3 handout 21: Ordered Structures andfields

This handout will not be lectured. It covers material that you will probablysee in another module next year, which is not in the MSM1G3 syllabus. It isimportant background material though, and you may find it useful, as a summaryof properties of number systems covered in the course perhaps.

This section introduces two new axioms to this course relating orders toaddition and multiplication, the compatibility axioms below. While these arenew to this course, you will certainly be familiar with them from MSM1G1 orfrom A-level.

Ordered rings

Sometimes, as in the case of R, Q and Z, a number system has an order relationon it, and the addition, multiplication, and order relations relate in some niceway.

Definition 21.1 An ordered ring is a commutative ring (R,+, ·) (i.e., a num-ber system in which + and · are associative and commutative, + has an identity0, · has an identity 1, every x ∈ R has an additive inverse, and the distributivitylaw for · over + holds) with a linear order 6 on R such that +, ·, and 6 satisfythe following

(+ and 6 are compatible) x 6 y⇒ x+ z 6 y + z, for all x, y, z ∈ R

(· and 6 are compatible) x 6 y ∧ 0 6 z⇒ x · z 6 y · z, for all x, y, z ∈ R

Notice the extra clause 0 6 z in the second of these axioms. This is quiteessential, as you will see when you think about the key example of the realnumbers. (E.g., 1 6 2 but −1 · 1 66 −1 · 2.)

Fields

The number systems of R and Q have some extra properties concerning multi-plication.

Definition 21.2 A field is a commutative ring (R,+, ·) such that the followingaxiom for multiplicative inverses holds.

For all x 6= 0 in R there is y ∈ R with x · y = y · x = 1.

In other words, in a field there is an identity for multiplication by nonzeroelements, and you can take the reciprocal (or multiplicative inverse) of anynonzero element.

An ordered field is an ordered ring (as above) that is also a field. Again,R and Q are key examples, but there are examples of fields which cannot beordered, such as the two element field 0, 1 of arithmetic modulo 2, and thefield of complex numbers C.

34

MSM1G3 handout 22: Boolean rings

This handout will not be lectured. It covers material that you may or may notsee in your years as a student here, but which is at the level of good and/orambitious students just completing MSM1G3. It covers a topic in the algebraof somewhat unusual-looking number systems, and gives a good flavour of themore modern kind of university-level algebra.

One of my all-time heroes is George Boole, who proposed looking at commu-tative rings R (i.e., number systems with +, · and elements 0 and 1 such that+ and · are associative and commutative; 0 is an identity for +; 0 is an zero for·; · is distributive over +; and every real number has an additive inverse) withthe extra property that

x · x = x

for all x. Such rings are called Boolean rings in his honour.Your first objection is that this looks silly: after all, real numbers don’t

obey this rule! But Boole wanted to consider other kinds of numbers too andhe had plenty of good reasons for thinking these new kinds of numbers wouldbe interesting and useful.

Example 22.1 Let R be the ring of two elements 0, 1 with 0 + 0 = 1 + 1 = 0,0 + 1 = 1 + 0 = 1, 0 · 0 = 0 · 1 = 1 · 0 = 0, 1 · 1 = 1. (You can think of this asaddition and mutiplication modulo 2.) Then R satisfies Boole’s axiom. (Noticethat 02 = 0 and 12 = 1.)

Example 22.2 Let R be the ring of all subsets X of a given nonempty setA, with 0 denoting the empty set and 1 denoting A itself. Define X + Y =(X ∪ Y ) \ (X ∩ Y ) and X · Y = X ∩ Y . Then R is a Boolean ring. (Admittedly,it takes quite some time to check all the axioms are true. Ambitious studentsshould attempt at least some of the axioms.)

For the rest of this handout, let R be a Boolean ring. Since it is a commuta-tive ring, we know that the additive inverse operation −x is well-defined. Thefollowing exercise are intended for adventurous students who have some timeavailable during the reading week. A bottle of wine or a box of chocolates isoffered as a prize for the best solution handed in to me by Monday of week 8.

Exercise 22.3 Expand the right-hand-side of x + x = (x + x)2 to deduce thatx+ x = 0 in R. Deduce that x = −x for all x in R.

Define the binary operation ⊕ on R by x⊕ y = x+ y + x · y, and the unaryoperation ′ by x′ = 1 + x.

Exercise 22.4 The operation ⊕ is associative and commutative.

Exercise 22.5 BOTH distributivity laws hold for ⊕ and · in R, i.e.,

x · (y ⊕ z) = (x · y)⊕ (x · z)

and

x⊕ (y · z) = (x⊕ y) · (x⊕ z).

for all x, y, z ∈ R.

Exercise 22.6 In R, 1 is a zero for ⊕ and 0 is an identity for ⊕.

Exercise 22.7 For all x ∈ R, x · x′ = 0 and x⊕ x′ = 1.

35