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Examiners commentaries 2012


MT2116 Abstract mathematics

Important note

This commentary reflects the examination and assessment arrangements for this course in theacademic year 201112. The format and structure of the examination may change in future years,and any such changes will be publicised on the virtual learning environment (VLE).

Information about the subject guide

Unless otherwise stated, all cross-references will be to the latest version of the subject guide (2011).

General remarks

Learning outcomes

At the end of this course and having completed the Essential reading and activities, you should:

have used basic mathematical concepts in discrete mathematics, algebra and real analysis tosolve mathematical problems in this subject be able to use formal notation correctly and in connection with precise statements in English be able to demonstrate an understanding of the underlying principles of the subject be able to solve unseen mathematical problems in discrete mathematics, algebra and real

analysis

be able to prove statements and formulate precise mathematical arguments.

Showing your working

We start by emphasising that you should always include your working. This means two things.First, you should not simply write down the answer in the examination script, but you shouldexplain the method by which it is obtained. Secondly, you should include rough working (even if it ismessy). The Examiners want you to get the right answers, of course, but it is more important thatyou prove you know what you are doing: that is what is really being examined.

We also stress that if you have not completely solved a problem, you may still be awarded marks fora partial, incomplete or slightly wrong solution; but, if you have written down a wrong answer andnothing else, no marks can be awarded. So it is certainly in your interests to include all yourworkings.

Knowing the definitions

In this course, precision and clarity are extremely important. It is vital that you know the keydefinitions and theorems exactly, so that you can quote them and use them. It is simply not possible

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to prove something using a formal definition if you only have a vague and incorrect recollection ofwhat that definition is. How could you use the formal definition of convergence of a sequence toprove that a sequence converges if you do not know what it is that you need to establish, becauseyou do not know the definition? Well, you cant. It is so important to know the definitions and thereare a number of marks to be picked up simply for knowing them.

Covering the syllabus and choosing questions

You should ensure that you have covered the bulk of the syllabus in order to perform well in theexamination: it is bad practice to concentrate only on a small range of major topics in theexpectation that there will be lots of marks obtainable for questions on these topics. Theexamination paper has some element of choice: you choose three of the four questions from SectionA and three of the four questions from Section B. If you have not covered the whole syllabus, thenyou will be limiting your choice. Assuming you have, however, covered the whole syllabus, it is avery good idea to take a little time to choose carefully: it could be that a question on your favouritetopic is in fact more difficult than a question on another topic.

Expectations of the examination paper

Every examination paper is different. You should not assume that your examination will be almostidentical to the previous years: for instance, just because there was a question, or a part of aquestion, on a certain topic last year, you should not assume there will be one on the same topic thisyear. Each year, the Examiners want to test that candidates can reason precisely mathematically,and that they know and understand a number of mathematical concepts and methods. In setting anexamination paper, they try to test whether the candidate does indeed know the methods,understands them and is able to use them, not merely whether he or she vaguely remembers them.Because of this, every year there are some questions which are likely to seem unfamiliar, or differentfrom previous years questions. You should expect to be surprised by some of the questions. Ofcourse, you will only be examined on material in the syllabus, so all questions can be answered usingthe course materials.

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Question spotting

Many candidates are disappointed to find that their examination performance is poorerthan they expected. This can be due to a number of different reasons and the Examiners

commentariessuggest ways of addressing common problems and improving your performance.We want to draw your attention to one particular failing question spotting, that is,confining your examination preparation to a few question topics which have come up in pastpapers for the course. This can have very serious consequences.

We recognise that candidates may not cover all topics in the syllabus in the same depth, butyou need to be aware that Examiners are free to set questions on anyaspect of the syllabus.This means that you need to study enough of the syllabus to enable you to answer the requirednumber of examination questions.

The syllabus can be found in the Course information sheet in the section of the VLE dedicatedto this course. You should read the syllabus very carefully and ensure that you cover sufficientmaterial in preparation for the examination.

Examiners will vary the topics and questions from year to year and may well set questions thathave not appeared in past papers every topic on the syllabus is a legitimate examinationtarget. So although past papers can be helpful in revision, you cannot assume that topics orspecific questions that have come up in past examinations will occur again.

If you rely on a question spotting strategy, it is likely you will find yourself indifficulties when you sit the examination paper. We strongly advise you not toadopt this strategy.

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Important note




Comments on specific questions Zone A

Candidates should answerSIXof the following EIGHTquestions: THREE from Section A andTHREE from Section B. All questions carry equal marks.

Section A

Question 1

(a) Let Sbe the following statement about natural numbersm and n:

Ifm and n are even, then m+ n is even.

(i) Prove that S is true.

(ii) Write down the converse ofS. Is the converse ofS true or false? Justify youranswer.

(iii) Write down the contrapositive ofS. Is the contrapositive ofS true or false?

Justify your answer.(iv) Write down the contrapositive of the converse ofS.

This question builds on the material of Chapter 1 of the subject guide.

Ifm, n are even then, for some integers k, l, m = 2k, n = 2l. Then, m + n= 2(k+ l), whichis even. This proves S is true.

The converse is: ifm + n is even, then m and n are even. This is false. For example, 1 + 1 iseven but 1 is not even.

The contrapositive ofS is: ifm + nis odd, then m is odd or n is odd. This is true becausethe contrapositive is logically equivalent to S, which is true. (In fact, something stronger istrue, as can easily be seen: ifm + n is odd, then precisely oneofm, nis odd, which implies

the weaker condition that m is odd or n is odd.)The contrapositive of the converse is: ifm or n is odd, then m + n is odd.

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(b) Let A and B be subsets of a set E. The symmetric difference ofA and B,denoted by AB, is the set of elements ofE that belong toA or B, but not bothA and B. That is, using set notation,

AB= (A B) \ (A B)

or, equivalently,AB= (A \ B) (B\ A).

Prove that, ifX=AB, then AX=B.

(Hint: show thatAXB and BAX.)This is quite a challenging question. SupposeX=PQ. We follow the hint (which isalways a good idea): we show first that PXQ and then that QPX. LetzPX.Then either (i) zP\ Xor (ii)zX\ P. In case (i), given that X=PQ, we must havezQ (otherwise we would have zX). In case (ii), given that X=PQ, we must havezPQ and hence zQ (sincezP). This shows PXQ.Now suppose zQ. Since X=PQ, ifzX, then zP and so zPX. IfzX,thenz

P(otherwise we would have z

PQ= X), so z

PX.

Question 2

(a) Prove by induction that, for all natural numbersn,

2nr=1

(1)rr3 =n2(4n + 3).

Proof by induction is discussed in Chapter 3 of the subject guide.

The statement is easily seen to be true when n = 1. Both sides are equal to 7. This is thebase case.

Suppose, inductively, that it is true for n = k, and consider n = k+ 1. We have

2(k+1)r=1

(1)rr3 = (2k+ 1)3 + (2k+ 2)3 +2kr=1

(1)rr3

= (2k+ 1)3 + (2k+ 2)3 + k2(4k+ 3)= k2(4k+ 3) (8k3 + 12k2 + 6k+ 1) + 8(k3 + 3k2 + 3k+ 1)= 4k3 + 15k2 + 18k+ 7.

This is the same as

(k+ 1)2(4k+ 7) = (k+ 1)2(4(k+ 1) + 3),

as can be seen by expanding this expression. So the result is true for n = k+ 1. Hence, byinduction, it is true for all n.

(b) Suppose f is the function f : N Z given by

f(n) =

n/2 ifn is even,(n 1)/2 ifn is odd.

Prove that f is a bijection and determine a formula for the inverse functionf1.

Functions and their properties are dealt with in Chapter 4 of the subject guide.

We need to show that f is injective and f is surjective.Injective: (We need to prove iff(x) =f(y), thenx = y.) Supposef(x) = f(x). Either: (i)f(x) =f(x)> 0, or (ii)f(x) = f(x)0. (This split into two cases is a crucial step.) In

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case (i)x and x are both even and x2 =f(x) = f(x) = x

2, so x = x. In case (ii) x, x are

both odd andx12 =f(x) =f(x) =x12 , so x= x

. So fis an injection.

Surjective: (We need to prove that for any y Z, there is some xNsuch that y = f(x).)Ify1, let x = 2y. Then x is even and so f(x) = x/2 =y. Ify0, let x = 1 2y. Then xis odd and so f(x) =(x 1)/2 =y.The formula for the inverse function follows immediately from the proof that the function issurjective: we have

f1(n) =

2n ifn11 2n ifn0.

(c) Let X= R \ {0} and let R be the relation on Xgiven by x R y if and only ifx/yis rational. Prove that R is an equivalence relation on X, and determine theequivalence class containing

2.

Equivalence relations are discussed in Chapter 5 of the subject guide. Recall what it meansto say that a relation R on a set Ais an equivalence relation. It means three things:

R is reflexive: for all x

A, x R x.

R is symmetric: for all x, yA, x R y implies y R x.x/x= 1 is rational, so the relation is reflexive. Ifx/y is rational, so is y /x, so it issymmetric. Suppose next that xRy and y Rz. Then x/y and y /z are rational numbers.Their product is therefore also rational, but this product isx/z . SoxRz , andR is transitive.

xR

2 if and only ifx/

2 is rational, which means that

[2] ={q

2 :q Q}=

2Q.

Question 3

(a)

(i) Use the Euclidean algorithm to prove that55 and 576 are coprime, andexpress 1 in the form 1 = 55x + 576y for some integers x, y.

(ii) Determine the inverse of55 in Z576.

(iii) Use your answer to (ii) to solve the congruence

55x3 (mod 576).

This is a standard, easy, question. We use the Euclidean algorithm (of Chapter 6 of thesubject guide).

By the Euclidean algorithm, we have

576 = 10.55 + 26

55 = 2.26 + 3

26 = 8.3 + 2

3 = 1.2 + 1,

so 576 and 55 have the greatest common divisor (gcd) equal to 1 and are therefore coprime.

Then,

1 = 3 2

= 3 (26 8.3) = 9.3 26= 9(55 2.26) 26 = 9.55 19.26= 9.55 19(576 10.55) = 199.55 19.576.

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Letc = cos ands= sin . Then we have

(c + is)4 =c4 + 4c3(is) + 6c2(is)2 + 4c(is)3 + (is)4 =c4 + 4ic3s 6c2s2 4ics3 + s4.Equating real parts in (c + is)4 = cos(4) + i sin(4), we have

cos(4) =c

4

6c2

s

2

+ s

4

,as required. Furthermore, sinces2 = 1 c2,

cos(4) = c4 6c2(1 c2) + (1 c2)2 = 8c4 8c2 + 1.

(c) Let X= C \ {0}, and let f :XXbe given by f(z) = z|z|2, forzX. Provethat f is a bijection.

First, we prove the function is injective. Supposef(z) =f(w). Then z |z|2 =w|w|2. Takingthe modulus of each side,|z|3 =|w|3, so|z|=|w|, Thus (since|z|2 =|w|2 = 0) we maycancel in z |z|2 =w|w|2 to obtain z = w.Surjectivity is a little harder. Supposez is given. We need to find w so thatw

|w

|2 =z . This

suggests takingw = z for some R. Ifw = z thenf(w) = z|z|2 =||3|z|2z.

So this will work if||3|z|2 = 1. So we may take =|z|2/3. So, z = f(w) wherew=|z|2/3z.

Section B

Question 5

(a) Define what it means for a sequence (an)nN to converge to LR.Use this definition to prove that if (an)nN is increasing and the subsequence(a3n)nN converges to L, then (an)nN also converges toL.

See Chapter 10 of the subject guide for the background material to this part of the question.

A sequence (an)n1 converges to L Rif, for every > 0, there is an Nsuch that for alln > N we have|an L|< .Sincea3nL, for > 0, there is an N such that, for n > N, we have|a3n L|< . SetN= 3N + 3. Then, suppose n > N. We have ana3n < L + . But we also haveana3n > L , wheren =n/3> N. It follows that, for such n,|an L|< . Thisshows thatanL.

(b) Define what it means for a function f: R R to be continuous at c.Let f: R R and g : R R be functions that are continuous at0, with f(0) =g(0).Prove that ifh : R R is a function such that f(x)h(x)g(x) for all xR,then h is continuous at 0.

See Chapter 11 of the subject guide for background material on limits of functions andcontinuity.

A function f : R R is continuous at a point c Rif for every >0 there exists >0 suchthat for every x satisfying|x c|< , we have|f(x) f(c)|< .We use the Sandwich Theorem, together with the alternative definition of continuity in

terms of sequences, which is thatf : R R is continuous at c if and only if, for everysequence (xn) converging to c, we have thatf(xn)f(c).First, notice that f(0)h(0)g(0) =f(0), and hence f(0) =h(0) =g(0) =L.

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Now, suppose that a sequence (xn) is converging to c. Hence, we have that f(xn)L andg(xn)L. We also have that f(xn)h(xn)g(xn) for all n. By the Sandwich Theorem,h(xn)L = h(0). So, h is continuous at 0.

(c) State the Intermediate Value Theorem.

Let f: [1, 1]R be a continuous function that satisfies1f(x)1 for allx[1, 1]. Show that there exists c[1, 1] such that f(c) =c.The Intermediate Value Theorem states the following: Letfbe a function continuous on[a, b]. Suppose that y is such that f(a)< y < f(b). Then there exists c[a, b] such thatf(c) = y.

Letg (x) =f(x) + x. Sincefand any polynomial are continuous on [1, 1], g is alsocontinuous on [1, 1]. We also have g(1) = f(1) 10 and g (1) =f(1) + 10. By theIntermediate Value Theorem, there is c[1, 1] such that g(c) = 0; hence, f(c) =c.

Question 6

(a) Let Sbe a subset of real numbers. State the definition of a lower bound ofSand of the infimum ofS.

Use the definition to show that1 is the infimum of the interval (1, 1).See Chapter 9 of the subject guide for related material.

An is a lower bound ofS if, for all xS, we have x.An is the infimum ofS if is a greatest lower bound ofS; that is, is a lower bound ofSand, for every other lower bound ofS, we have .LetS:= (1, 1) ={x R| 1< x 1. Since 0S, we have 0. Takex= ( 1)/2. Then x (1 1)/2 =1, soxS. However,this is a contradiction becausex < .

(b) State the greatest lower bound property ofR.

Let A be a nonempty set of positive real numbers, and let

B={a| aA}.

Prove that infB exists, and that infB =

infA.

The greatest lower bound property is: every non-empty set bounded below has an infimum.

For every aA, we have a0, hence 0 is a lower bound ofAand we must have infA0.Consequently, for all aA, we have ainfA0 and soa infA. Hence, B isnon-empty and bounded below, from which we deduce that infB exists and infB

infA.

If infB >

infA, then (infB)2 >infA and (infB)2 could not be a lower bound ofA.Hence, there would be an aA such that (infB)2 > a. This would mean, however, thatinfB >

a a contradiction.

So, infB =

infA.

(c) Show that the polynomialp(x) = 2x3 + 2x2 10x + 5 has two real roots in theinterval (0, 2).

Sincep(x) is a polynomial function, we know that it is continuous on [0 , 1] and [1, 2]. We

note thatp(0) = 5> 0, p(1) =1< 0 and p(2) = 9> 0, so, by the Intermediate ValueTheorem, there are c1(0, 1) and c2(1, 2) such that p(c1) = p(c2) = 0.

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Question 7

(a) Let F = R \ {0} and for each a, bF define a b=ab.(i) Show that (F, ) is a group.

(ii) Give an example of a finite subgroup of (F, ) of order larger than 1 and anexample of an infinite subgroup of (F, ) not equal to (F, ).

See Chapters 12 and 13 of the subject guide for relevant reading. Recall, from Chapter 12 ofthe subject guide, that (S, ) is a group if and only if the following properties hold: x, yS, x yS[G0, closure property] x,y,zS, (x y) z = x (y z)[G1, associativity property] eSsuch thatxS, e x= x e= x [G2, identity property] xS,x1 Ssuch that x x1 =x1 x= e [G3, inverse property]

We verify the group axioms, as follows.

(G0): for every a, bG, we have a, b= 0, hence a b=ab= 0 and a bG.(G1): for every a, b, cG, we have

(a b) c=(a b)c=(ab)c= abc =a(bc) =a(b c) = a (b c).(G2):1G is the identity because a (1) =a(1) = a and (1) a=(1)a= a.(G3): for every aG, 1/aG (becausea= 0 and 1/a= 0 has no solution) anda (1/a) =a(1/a) =1 and (1/a) a=(1/a)a=1.H={1, 1} is a finite subgroup ofG with group table

* -1 1

-1 -1 11 1 -1

H = Q \ {0} is an infinite subgroup that is not equal to G

(b) Let G be a group with identity e.

(i) State the definition of the order of an elementaG.(ii) Let H be a subgroup ofG. Is it true that, for every elementhH, the order

ofh in H is the same as the order ofh in G ?Justify your answer by either giving a proof or by providing acounterexample.

(iii) LetH and Kbe subgroups ofG such that|H|= 20 and|K|= 11. Prove thatH K={e}.

See Chapter 13 of the subject guide for relevant reading.

(i) We say that aG has finite order if there is a natural number m such that am =e.Otherwise, we say that a has infinite order. Ifa has a finite order, then the order ofa is

min{mN|am =e}.(ii) Since His closed under the group operation, for every hHand for every nN, wehave thathn H. Hence, hn =e inH if and only ifhn =e in G. From this it follows that,for every hH, the order ofh is the same in both H and G.(iii) One must realise that H K is a subgroup of both H and K. Then, by LagrangesTheorem (see Chapter 14 of the subject guide),|H K| divides|H| and|K|, butgcd(20, 11) = 1, so|H K|= 1. We know that the identity e is in H K, henceH K={e}.We have seen that H Kis non-empty.Ifa, bH K, then a bH (becauseH is a subgroup anda, bH) and a bK(becauseKis a subgroup and a, b

K). So, a

b

H

K.

Similarly, ifaH K, then a1 H (because His a subgroup and aH) and a1 K(becauseKis a subgroup and aK). So, a1 H K.

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Question 8

(a) Let (F, ) be a group with identity e. Let a,b,c be elements ofF such thata c2 =b and c3 =e. Find c in terms ofa and b.Sincec3 =c2 c= c c2 =e, we have that c2 =c1. Hence, b = a c2 =a c1, from whichwe get that a = b cor c = b

1

a.(b)

(i) Explain what is meant for a group to be abelian.

(ii) What is meant by an isomorphism from a group (G, ) to a group (H, )?(iii) Let (G, ) and (H, ) be two isomorphic groups. Prove that (G, ) is abelian if,

and only if, (H, ) is abelian.Isomorphisms are discussed in Chapter 14 of the subject guide.

(i) A group (G, ) is abelian if, for all a, bG, a b= b a.(ii) An isomorphism from a group (G, ) to (H, ) is a bijection f :GH such that alla, bG, f(a b) = f(a) f(b).Letf :GHbe a bijection such that all a, bG, f(a b) =f(a) f(b).Suppose thatHis abelian. Take any a, bG, then

f(a b) = f(a) f(b) = f(b) f(a) = f(b a).

Sincefis a bijection, we must have a b= b a.Suppose thatGis abelian. Take any a, b H. Sincefis a bijection, there exists a, bGsuch that a =f(a) and b =f(b). Then,

a b =f(a) f(b) = f(a b) = f(b a) = f(b) f(a) = b a.

(c) Let M=GL(n,R) be the group of invertiblen n real matrices, with thegroup operation being matrix multiplication. (You may assume this is a group.)Define : M M by

(A) = (AT)1,

where AT denotes the transpose of the matrix A. Prove that is anisomorphism.

is a bijection, because (A) =(B) implies that (AT)1 = (BT)1, implying AT =BT,and henceA = B .

is surjective: for, ifA is invertible, then A1 and (A1)T are also invertible and((A1)T) = (((A1)T)T)1 = (A1)1 =A.

We also have

(AB) = ((AB)T)1 = (BTAT)1 = (AT)1(BT)1 =(A)(B).

So is a bijective homomorphism; that is, it is an isomorphism.

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Important note




Comments on specific questions Zone B

Candidates should answerSIXof the following EIGHTquestions: THREE from Section A andTHREE from Section B. All questions carry equal marks.

Section A

Question 1

(a) Let Sbe the following statement about natural numbersm and n:

Ifm is even or n is even, then mn is even.

(i) Prove that S is true.

(ii) Write down the converse ofS.

(iii) Write down the contrapositive of the converse ofS.

(iv) Is the converse ofS true or false? Justify your answer.

(v) Write down the contrapositive ofS.This question builds on the material of Chapter 1 of the subject guide.

Suppose (without loss of generality) that m is even. Then m = 2k, for some integer k .Then,mn = (2k)n= 2(kn), which is even.

The converse is: ifmn is even, then m is even or n is even.

The contrapositive of the converse is: ifm and n are odd, then mn is odd.

We prove the contrapositive of the converse: ifm= 2k+ 1 and n = 2l+ 1, then

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

which is odd. It follows that the converse is true, since it is logically equivalent to its

contrapositive.The contrapositive ofS is: ifmn is odd, then m is odd and n is odd.

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(b) Let A and B be subsets of a set E. The symmetric difference ofA and B,denoted by AB, is the set of elements ofE that belong toA or B, but not bothA and B. That is, using set notation,

AB= (A B) \ (A B)

or, equivalently,AB= (A \ B) (B\ A).

Prove that, ifX=AB, then AX=B.

(Hint: show thatAXB and BAX.)This is quite a challenging question. SupposeX=PQ. We follow the hint (which isalways a good idea): we show first that PXQ and then that QPX. LetzPX.Then either (i) zP\ Xor (ii)zX\ P. In case (i), given that X=PQ, we must havezQ (otherwise we would have zX). In case (ii), given that X=PQ, we must havezPQ and hence zQ (sincezP). This shows PXQ.Now suppose zQ. Since X=PQ, ifzX, then zP and so zPX. IfzX,

thenzP(otherwise we would have zPQ= X), so zPX.

Question 2

(a) Prove by induction that, for all natural numbersn,

nr=1

r(n2 r2) =14

n2(n2 1).

You may use the fact thatk

r=1

r= k(k+ 1)/2.

Proof by induction is discussed in Chapter 3 of the subject guide.

The statement is easily seen to be true when n = 1. Both sides are equal to 0. This is thebase case.

Suppose the statement is true for n= k, and consider n = k+ 1. We have

k+1r=1

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

r=1

r(k2 + 2k+ 1 r2)

= (2k+ 1)

kr=1

r+

kr=1

r(k2 r2)

= (2k+ 1) 12

k(k+ 1) +14

k2(k2 1)

= 1

4

2k(2k+ 1)(k+ 1) + k2(k2 1)

= 1

4

k4 + 4k3 + 5k2 + 2k

.

For the case n = k+ 1 to be true, this should be the same as

1

4(k+ 1)2((k+ 1)2 1).

We have

(k

2

+ 2k+ 1)(k

2

+ 2k) =k

4

+ 4k

3

+ 5k+ 2,as required. So the result is true, by induction, for all values ofn.

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(b) Suppose f is the function f : N Z given by

f(n) =

n/2 ifn is even,(n 1)/2 ifn is odd.

Prove that f is a bijection and determine a formula for the inverse functionf1.

Functions and their properties are dealt with in Chapter 4 of the subject guide.We need to show that f is injective and f is surjective.

Injective: (We need to prove iff(x) =f(y), thenx = y.) Supposef(x) = f(x). Either: (i)f(x) =f(x)> 0, or (ii)f(x) = f(x)0. (This split into two cases is a crucial step.) Incase (i)x and x are both even and x

2 =f(x) = f(x) = x

2, so x = x. In case (ii) x, x are

both odd andx12

=f(x) =f(x) =x12

, so x= x. So fis an injection.

Surjective: (We need to prove that for any y Z, there is some xNsuch that y = f(x).)Ify1, let x = 2y. Then x is even and so f(x) = x/2 =y. Ify0, let x = 1 2y. Then xis odd and so f(x) =(x 1)/2 =y.The formula for the inverse function follows immediately from the proof that the function issurjective: we have

f1(n) =

2n ifn11 2n ifn0.

(c) The relation R is defined on N by: a R b ab= n2 for somenN. Provethat R is an equivalence relation.

Equivalence relations are discussed in Chapter 5 of the subject guide. Recall what it meansto say that a relation R on a set Ais an equivalence relation. It means three things:

R is reflexive: for all xA, x R x. R is symmetric: for all x, yA, x R y implies y R x.

aa= a2 is a square, so it is reflexive. ab= ba so it is symmetric. SupposeaRb andbRc, soab= n2, bc = m2 for some integers m, n. Then

ac=(ab)(bc)

b2 =

mnb

2.

Now, because ac is an integer, mn/bmust be an integer too. (If it were not, there would besome prime p and some integer k such that pk dividesb but not mn. Then the denominatorof (mn/b)2 could not be reduced to 1, as it would have to in order to give an integer.) So acis a square and the relation is transitive.

Question 3

(a)(i) Use the Euclidean algorithm to prove that25 and 572 are coprime, and

express 1 in the form 1 = 25x + 572y for some integers x, y.

(ii) Determine the inverse of25 in Z572.

(iii) Use your answer to (ii) to solve the congruence

25x2 (mod 572).

This is a standard, easy, question. We use the Euclidean algorithm (of Chapter 6 of thesubject guide).

By the Euclidean algorithm, we have

572 = 22.25 + 2225 = 1.22 + 3

22 = 7.3 + 1,

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so 572 and 25 have gcd equal to 1 and are therefore coprime.

Then,

1 = 22 7.3= 22 7(25 22) = 8.22 7.25= 8(572 22.25) 7.25 = 8.572 183.25.

It follows that 25 is invertible in Z572and 251 =183 = 389.

The solution to the congruence equation is given by

x251 2 = 389 2 = 778 = 206.

(b) Let a, b and c be natural numbers such that a | c and b | c. Prove that ifa and bare coprime, then ab | c. (You may use, without proof, the fact that ifa and b arecoprime, then there are integers x and y such that xa + yb = 1.)

The relevant background material can be found in Chapter 6 of the subject guide.

For this part, we could invoke the Fundamental Theorem of Arithmetic. Or, more directly,as hinted, we use the fact that there are x, y such that 1 = xa + yb. Now, the facts aboutcimply that there are integers k, l such that c = ka, c = lb. Then,

c= c.1 =c(xa + yb) = (lb)xa + (ka)yb = ab(lx + ky),

soabdivides c.

(c) Prove that

5 is irrational. (You may use, without proof, the fact that ifn isan integer and 5 | n2, then 5 | n.)Reading related to this question can be found in Chapter 8 of the subject guide. This proof

closely resembles the standard proof that 2 is irrational.Suppose there are a, bsuch that a/b =

5 and assume, without loss, that gcd(a, b) = 1.

Then we have a2 = 5b2, so 5 divides a2. By the Fundamental Theorem of Arithmetic, 5 | a,soa= 5c for somec. Then a2 = 25c2 = 5b2, sob2 = 5c2. So 5 | b2 and hence 5 | b. Thiscontradicts the fact that a, b are coprime.

Question 4

(a) Express the recurring decimal0.3122 as a rational number p/q, where p and qare integers.

See Chapter 8 of the subject guide for a discussion of how to solve questions of this kind.Suppose the number is x. Then 1000x= 312.2122 and

1000x x= 999x= 312.2 0.3 = 311.9 = 3119/10,

so

x=3119

9990.

(b) DeMoivres formula tells us that

(cos + i sin )4 = cos(4) + i sin(4).

Use this to prove that, for any angle,

sin(4) = 4 cos3 sin 4cos sin3 .

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Show, further, by using the facts sin2 + cos2 = 1 and sin(2) = 2sin cos , that

sin(4) = 2 sin(2) 4 sin(2)sin2 .

For this, and the next, part of the question, see the discussion of complex numbers inChapter 8 of the subject guide.

Letc = cos ands= sin . Then we have

(c + is)4 =c4 + 4c3(is) + 6c2(is)2 + 4c(is)3 + (is)4 =c4 + 4ic3s 6c2s2 4ics3 + s4.

Equating imaginary parts in (c + is)4 = cos(4) + i sin(4), we have

sin(4) = 4c3s 4cs3,

as required. Furthermore, sinces2 = 1 c2 and since 2cs= sin(2),

cos(4) = 2c2(2cs) 2s2(2cs) = 2 sin(2)(c2 s2) = 2 sin(2)(1 2s2),

which gives the required identity.

(c) Let X= C \ {0}, and let f :XXbe given by f(z) = z|z|2, forzX. Provethat f is a bijection.

First, we prove the function is injective. Supposef(z) =f(w). Then z |z|2 =w|w|2. Takingthe modulus of each side,|z|3 =|w|3, so|z|=|w|, Thus (since|z|2 =|w|2 = 0) we maycancel in z |z|2 =w|w|2 to obtain z = w.Surjectivity is a little harder. Supposez is given. We need to find w so thatw |w|2 =z . Thissuggests takingw = z for some R. Ifw = z then

f(w) = z|z|2 =||3|z|2z.

So this will work if||3

|z|2

= 1. So we may take =|z|2/3

. So, z = f(w) wherew=|z|2/3z.

Section B

Question 5

(a) Define what it means for a sequence (an)nN to converge to LR.

Use this definition to prove that if (an)nN

is decreasing and the subsequence(a2n)nN converges to L, then (an)nN also converges toL.

See Chapter 10 of the subject guide for the background material to this part of the question.

A sequence (an)n1 converges to L Rif, for every > 0, there is an Nsuch that for alln > N we have|an L|< .Sincea2nL, for > 0, there is an N such that, for n > N, we have|a2n L|< . SetN= 2N + 2 and suppose n > N. Then we have ana2n > L . But we also haveana2n < L + , wheren =n/2> N.

(b) Define what it means for a function f: R R to be continuous at c.Let f: R

R and g : R

R be functions that are continuous at0, with f(0)> g(0).

Prove that there is an >0 such that f(x)> g(x) for all x(, ).See Chapter 11 of the subject guide for background material on limits of functions andcontinuity.

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(c) Show that the polynomialp(x) =x3 + x2 6x + 3 has two real roots in theinterval (0, 2).

Sincep(x) is a polynomial function, we know that it is continuous on [0 , 1] and [1, 2]. Wenote thatp(0) = 3> 0, p(1) =1< 0 and p(2) = 3> 0, so, by the IVT, there are c1(0, 1)and c2(1, 2) such that p(c1) = p(c2) = 0.

Question 7

(a) Let F = R \ {0} and for each a, bF define a b=ab.(i) Show that (F, ) is a group.

(ii) Give an example of a finite subgroup of (F, ) of order larger than 1 and anexample of an infinite subgroup of (F, ) not equal to (F, ).

See Chapters 12 and 13 of the subject guide for relevant reading. Recall, from Chapter 12 ofthe subject guide, that (S, ) is a group if and only if the following properties hold: x, yS, x yS[G0, closure property] x,y,zS, (x y) z = x (y z)[G1, associativity property] eSsuch thatxS, e x= x e= x [G2, identity property] xS,x1 Ssuch that x x1 =x1 x= e [G3, inverse property]

We verify the group axioms, as follows.

(G0): for every a, bG, we have a, b= 0, hence a b=ab= 0 and a bG.(G1): for every a, b, cG, we have

(a b) c=(a b)c=(ab)c= abc =a(bc) =a(b c) = a (b c).

(G2):1G is the identity because a (1) =a(1) = a and (1) a=(1)a= a.(G3): for every aG, 1/aG (becausea= 0 and 1/a= 0 has no solution) anda (1/a) =a(1/a) =1 and (1/a) a=(1/a)a=1.H={1, 1} is a finite subgroup ofG with group table

* -1 1

-1 -1 11 1 -1

H = Q \ {0} is an infinite subgroup that is not equal to G

(b) Let (G, ) be a group with identity e.

(i) State the definition of the order of an elementaG.(ii) Let H and Kbe subgroups ofG such that|H|= p and|K|= q, where p and q

are primes such that p=q. It is known that H K is a subgroup ofG. Provethat H K={e}.

(iii) In addition, assume that G is finite. Prove that, for everya, bG, the orderofa b is the same as the order ofb a.Hint: notice that (b a)n =b (a b)n1 a.

See Chapter 13 of the subject guide for relevant reading.

(i) We say that aG has finite order if there is a natural number m such that am =e.Otherwise, we say that a has infinite order. Ifa has a finite order, then the order ofa ismin

{m

N

|am =e

}.

(ii) SinceH Kis a subgroup of both H and K, by Lagranges Theorem (see Chapter 14 ofthe subject guide),|H K| divides|H|= p and|K|= q, but gcd(p, q) = 1, so|H K|= 1.We know that the identity e is in H K, hence H K={e}.

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(iii) Suppose that, for some nN and a, bG, we have that (a b)n =e. Hence,(a b)n1 = (a b)1 =b1 a1 and(b a)n = (b a) (b a) (b a) =b (a b)n1 a= b (b1 a1) a= e.From this it follows that: Ifa b has order n, then (b a)n =e, and ifb a has order m,then (a b)m =e. So, we must have mn and nm.

Question 8

(a) Let (G, ) be a group. Let a,b,c,x be elements ofG such that x2 a= b x c1and a c x= x a c. Find x in terms ofa, b and c.From the first equation, we see x (x a c) = b x. Since x a c= a c x, we have thatx a c x= b x. We multiply both sides by x1, x a c= b, from which we get thatx a= b c1 or x = b c1 a1.

(b)

(i) Explain what is meant for a group to be abelian.(ii) What is meant by an isomorphism from a group (G, ) to a group (H, )?

(iii) Let (G, ) and (H, ) be two isomorphic groups. Prove that (G, ) is abelian if,and only if, (H, ) is abelian.

Isomorphisms are discussed in Chapter 14 of the subject guide.

(i) A group (G, ) is abelian if, for all a, bG, a b= b a.(ii) An isomorphism from a group (G, ) to (H, ) is a bijection f :GH such that alla, bG, f(a b) = f(a) f(b).Letf :GHbe a bijection such that all a, bG, f(a b) =f(a) f(b).Suppose thatHis abelian. Take any a, b

G, then

f(a b) = f(a) f(b) = f(b) f(a) = f(b a).Sincefis a bijection, we must have a b= b a.Suppose thatGis abelian. Take any a, b H. Sincefis a bijection, there exists a, bGsuch that a =f(a) and b =f(b). Then,

a b =f(a) f(b) = f(a b) = f(b a) = f(b) f(a) = b a.

(c) Let M=GL(n,R) be the group of invertiblen n real matrices, with thegroup operation being matrix multiplication. (You may assume this is a group.)Define : M

M by

(A) = (AT)1,

where AT denotes the transpose of the matrix A. Prove that is anisomorphism.

is a bijection, because (A) =(B) implies that (AT)1 = (BT)1, implying AT =BT,and henceA = B .

is surjective: for, ifA is invertible, then A1 and (A1)T are also invertible and((A1)T) = (((A1)T)T)1 = (A1)1 =A.

We also have

(AB) = ((AB)T)1 = (BTAT)1 = (AT)1(BT)1 =(A)(B).

So is a bijective homomorphism; that is, it is an isomorphism.

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