Algebra through practice
Book 1: Sets, relations and mappings
Algebra through practice
A collection of problems in algebra with solutions
Bookl
Sets, relations and
mappings
T.S.BLYTH o E.F.ROBERTSON
University of St Andrews
The right of the
University of Cambridge
to print and sell
all manner of books
was granted by
Henry VIII in 1534.
The University has printed
and published continuously
since 1584.
CAMBRIDGE UNIVERSITY PRESS
Cambridge
London New York New Rochelle
Melbourne Sydney
CAMBRIDGE UNIVERSITY PRESS
Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo
Cambridge University Press
The Edinburgh Building, Cambridge CB2 8RU, UK
Published in the United States of America by Cambridge University Press, New York
www.cambridge.org
Information on this title: www.cambridge.org/9780521272858
© Cambridge University Press 1984
This publication is in copyright. Subject to statutory exception
and to the provisions of relevant collective licensing agreements,
no reproduction of any part may take place without the written
permission of Cambridge University Press.
First published 1984
A catalogue record for this publication is available from the British Library
Library of Congress Catalogue Card Number: 83-24013
ISBN 978-0-521-27285-8 paperback
Transferred to digital printing 2007
Contents
Preface vii
Background reference material
l:Sets 1
2: Relations 6
3: Mappings 13
Solutions to Chapter 1 24
Solutions to Chapter 2 37
Solutions to Chapter 3 56
Test paper 1 90
Test paper 2 92
Test paper 3 94
Test paper 4 96
IX
Preface
The aim of this series of problem-solvers is to provide a selection of worked
examples in algebra designed to supplement undergraduate algebra courses.
We have attempted, mainly with the average student in mind, to produce
a varied selection of exercises while incorporating a few of a more challenging
nature. Although complete solutions are included, it is intended that these
should be consulted by readers only after they have attempted the questions.
In this way, it is hoped that the student will gain confidence in his or her
approach to the art of problem-solving which, after all, is what mathematics
is all about.
The problems, although arranged in chapters, have not been 'graded'
within each chapter so that, if readers cannot do problem n this should not
discourage them from attempting problem w + 1. A great many of the ideas
involved in these problems have been used in examination papers of one sort
or another. Some test papers (without solutions) are included at the end of
each book; these contain questions based on the topics covered.
TSB,EFR
St Andrews
Background reference material
Courses on abstract algebra can be very different in style and content.
Likewise, textbooks recommended for those courses vary enormously, not
only in notation and exposition but also in their level of sophistication. Here
is a list of some major texts that are widely used and to which the reader
may refer for background material. The subject matter of these texts covers
all six books in the Algebra through practice series, and in some cases a great
deal more. For the convenience of the reader there is given below an indication
of which parts of which of these texts is most relevant to the appropriate
chapters of this book.
[1] I. T. Adamson,Introduction to Field Theory, Cambridge University Press,
1982.
[2] F. Ayres,to,Modern Algebra, Schaum's Outline Series, McGraw-Hill, 1965.
[3] D. Burton, A First Course in Rings and Ideals, Addison-Wesley, 1970.
[4] P. M. Cohn, Algebra Vol. I, Wiley, 1982.
[5 ] D. T. Finkbeiner II, Introduction to Matrices and Linear Transformations,
Freeman, 1978.
[6] R. Godzment, Algebra, Kershaw, 1983.
[7] J. A. Green, Sets and Groups, Routledge and Kegan Paul, 1965.
[8] I. N. Herstein, Topics in Algebra, Wiley, 1977.
[9] K. Hoffman and R. Kunze, Linear Algebra, Prentice Hall, 1971.
[10] S. Lang, Introduction to Linear Algebra, Addison-Wesley, 1970.
[11] S. Lipschutz, Linear Algebra, Schaum's Outline Series, McGraw-Hill, 1974.
[12] I. D. Macdonald, The Theory of Groups, Oxford University Press, 1968.
[13] S. MacLane and G. Birkhoff, Algebra, Macmillan, 1968.
[14] N. H. McCoy,Introduction to Modern Algebra, Allyn and Bacon, 1975.
[15] J. J. Rotman, The Theory of Groups: An Introduction, Allyn and Bacon,
1973.
[16] I. Stewart, Galois Theory, Chapman and Hall, 1973.
Book 1
Background reference material
[17] I. Stewart and D. Tall, The Foundations of Mathematics, Oxford University
Press, 1977.
References useful to Book 1
1: Sets [2, Chapter 1], [6, Chapters 1, 3], [7, Chapter 1], [17, Chapter 3].
2: Relations [2, Chapter 2], [6, Chapter 4], [7, Chapter 2], [17, Chapter 4].
3: Mappings [6, Chapter 2], [7, Chapter 3], [17, Chapter 5].
In [2] the author writes mappings on the right, and uses 'one-to-one' for
injective and 'onto' for surjective. In [6] the author uses C for set inclusion
(where we use c). In [7] mappings are written on the right. In [17] the
definition of (partial) order differs from ours in that the axiom of reflexivity
is missing.
1: Sets
We assume that the reader has a basic knowledge of elementary set theory
and we shall use standard (i.e. the most commonly accepted) notation. Thus,
for example, we shall denote the complement of a subset A of a set E simply
by A' except when confusion can occur in which case we shall write CE(A).
If A and B are subsets of E then the difference set AC\B' will be denoted
by A \ B (some authors use A—B), and the symmetric difference set
(A C\B') U (A* C\B) will be denoted by.4 AB.
Some questions in this section are best dealt with using the algebra of
set theory, with which we assume that the reader is familiar. For example,
this includes the distributive laws
A n(Buc) = (A nB)U(A nc)
and
AU(BC\C) = (AUB)n(AU C),
and the de Morgan laws
(AnB)'=A'UB'
and (AU B)' =Af HB'.
Other questions, particularly those dealing with set-theoretic identities, are
best dealt with using Venn diagrams.
Other standard notation that we shall employ includes P(iT) for the power
set of E (i.e. the set of all subsets of E); \A\ for the number of elements in
the set A; A x B for the cartesian product of A and B (i.e. the set of ordered
pairs (a, b) with a £A and b€B); and the following for particular subsets
of the number system:
IN = {0, 1, 2,...} for the set of natural numbers;
Z = {..., — 2, - 1 , 0, 1, 2,...} for the set of integers;
<D = {a/b \a,beX,b=£0}
for the set of rationals;
1
Book 1
Sets, relations and mappings
IR for the set of real numbers;
(D for the set of complex numbers;
]a,b] = {xe\R
\a<x<b};
]a,b[={xG\R
\a<x<b},etc.
Finally the usual logical abbreviations 3 (there exists), V (for all),
=> (implies), o (if and only if) will be used throughout, both in the problems
and in their solutions.
1.1
Let ,4 = {0, {0}, (0,{0}}}. Which of the following are true?
0Ci4, QeA, {Q}GA, {0}CA, {{0}}CA, {{0
{{0},0}€EA
1.2
List the elements of P(P(0)) and of P(P(P(0))).
1.3
For the setE = {1,{1}, 2, {1,2}} determine ?(E) and E n
1.4
Find four examples of a set A with the property that every element of A is
a subset of A.
1.5
Can you find sets A, B, C such that
ACBGC
and AGBCC1
1.6
Which of the
(a)
(b)
(c)
(d)
(e)
(/)
(g)
(h)
1.7
Show that {x,y}D {y,z} = {y} maybe false.
1.8
Let A, B, C be subsets of a set X. Simplify the expressions
(a) 04 U 04 U £)')';
(b) ((A U 0)n (BUA') D(AUB'UX))';
(c)
(Au(Bnc)u(B'ncf)uc)'.
1.9
Let A,Bbe subsets of a set X. Prove, using a Venn diagram, that
(AnB')U(A'
following hold for all sets A, B and C?
IfA£BmdB<£CthenA(£C
UA^BmdB^CthenA^C
IfAGBmiBgLCihenA^C
HACBmdBCCthenC&A.
Ifyl C fl and JBGC then i 4 £ C
If A r\CCB then (AnB)U(BnC)=
B.
If A
nceBthenAGBUC
1: Sets
1.10
Let A, B, C be subsets of a set X. Prove, using Venn diagrams, that
(a) AA(BAC) = (AAB)AC;
(b)
AUB=AABA(AnB);
(c) An(BAC) = (AnB)A(ADC);
(d) AA(BDC) = (A &B) n(AAC) if and only if A C\B = A n C
1.11
If A, B are subsets of a set E let A \B = CE(A OB). Show that CE(A)=A \A
and that A HB = (A\B)\(A\B).
Express U in terms of | alone.
1.12
Let A,B, C,D be sets with {A, B} = {C, D}. Prove that ,4 nB = CCiD and
1.13
If A, B, C are sets prove that A n(BU C)C(A HB)U C, with equality if
and only if CCA.
1.14
For sets E, F, G prove, using a Venn diagram, that
(E n F)U (F n G)u (G n E) = (E u F) n(Fu
G) r\(G u E).
1.15
Give an example of sets A,B,C,D
with
1.16
Given two objects x, y one may define the ordered pair (JC, y) by
(*,JO = {{*},{*,;>}}.
Use this definition to prove that (x,y) = (**, >>*) if and only if x = x* and
j ; = j ; * Prove also that, for every object x,
{*}x {*} = {{{*}}}..
1.17
Prove that if A, B, C are sets with^4 andi? not empty then
1.18
Let & be a collection of sets such that
X, Ye&=>X\ Ye&
Prove that if X, 7 G ^"then X Pi 7 G J^
1.19
Let «^be a collection of sets. Define
Show that / ° C (^"°)°. Give an example to show that it is possible to have
1.20
Let A, B be sets. Are the following true?
(a) ?(A) n ?(B) = ?(A n B);
(b) ?(A) U ?(B) = ?(A U B).
Book 1
1.21
Sets, relations and mappings
If A, B, C are sets prove, using Venn diagrams, that A \ (BUC) = (A\ B)U
(A \ C) if and only if A A (B U C) = (A AB) U (A A C). Find sets A, B, C
with
A A(BU C) = (A AB)U (A AC);
,4 A (B U C) =£ (A A 5) U 04 A C).
1.22
Let^x,... ,Am be subsets of a set #. Define^, ...,Bm
Bt=Au
(\/n>2)Bn=An
recursively as follows:
\ "Q Ak.
Show that Bt n £ y = 0 for i =£/ and that U^x Bt = U-^i y4f.
1.23
Let A, B be sets with A CB. Prove that there is a unique subset XofB such
1.24
An examination in three subjects algebra (A), biology (B), chemistry (C)
was taken by 41 students. The following table shows how many students
failed the various combinations of subjects.
Subjects
A
B C A,B A,C
No. of failed students 12 5 8
2
6
How many students passed all three subjects?
B,C
3
A,B,C
1
1.25
At least 70% of a class of students study algebra, at least 75% study calculus,
at least 80% study geometry, and at least 85% study trigonometry. What
percentage (at least) must study all four subjects?
1.26
Let E be a set consisting of n elements. If X, Y are subsets of E such that the
number of elements in the sets i n r j ' n y j n Y\ X' n Y' are p, q, r, s
respectively, prove that p + q + r + s = n.
In a sixth form of n girls and boys each pupil is either an arts student or
a science student. If the proportion of arts students among the girls is greater
than the proportion of arts students among the boys, show that the proportion
of girls among the arts students is greater than the proportion of girls among
the science students.
1.27
UA = { J C G Z | Qy G Z)x = 2y) and£ = {fl£Z I (3b, cG Z)a = 6b + 10c}
prove that A = B.
1.28
For whichS G {IN, Z, CD, IR, C} are the following statements true?
1: Sets
(c){xGS\x2 = -!} = $.
1.29
Given n G IN defineriZ.= {nx \ x G Z}. Is it true that given nu n2 G IN there
exists m G IN with
Is it true that, for some m G IN,
1.30
Let .4 =
(a) How many subsets does A have?
(Z>) How many subsets of A contain at least one even integer?
(c) How many subsets of A contain exactly one even integer?
(Hint: consider separately the cases n even, n odd.)
1.31
Let i4 = { ^ E I N | K x < « } . What is the maximum possible k for which
A( CA (i = 1 , . . . , k) 3ndAtCAf
1.32
i f / < / ? Find 2f =1 \At\.
Express as a union of intervals
1
xGIR\{-l,4}
^
i
1.33
4
Express as a union of intervals the set of real numbers k for which
{x-lf
(x + l)(x + 3)
= * =0.
2: Relations
A relation R between a set E and a set F is a subset of E x F, and we shall
use the notation (x, y)€R or xRy with the same meaning. When we think
of a coordinate pictorial representation of E x F we refer to {(JC, y) \ xRy} as
the graph of R, to {x GE | (3j> £ F ) ( x , .y) £/?} as the domain of R, and to
{^ £ F | QxeE)(x,
y) GR} as the image of R. A relation between E and
F is called a (binary) relation on E.
If the relation R on F is reflexive (xRx for all x G ^ ) , symmetric (if xRy
thenyRx), and transitive (if xRy andyRz then xRz), then/? is an equivalence
relation on E. When R is an equivalence relation on E we sometimes write
x=y(R) instead of xRy. For xGE the /?-class of x, i.e. {.yGJ?! j>/ta}, is
denoted by [X]R or simply [x] when no confusion can arise. The following
are equivalent:
x=y(R),ye[x]R,
[x]R=[y]R, [x]Rn[y]R±Q.
It follows that two i?-classes either are disjoint (i.e. have empty intersection)
or are identical. This leads to the notion of a partition of E as a collection of
non-empty subsets of E which are pairwise disjoint and whose union is the
whole of E. If R is an equivalence relation on E then the /^-classes form
a partition of E. Conversely, every partition of E defines an equivalence
relation = on E by
JC =y o x, y belong to the same subset in the partition.
An example of an equivalence relation is the relation mod n defined on Z
by a = b(mod n) if and only if n divides a — b. The corresponding partition
consists of the equivalence classes
2: Relations
[0]
={...,-2n,-/i,0,?i,
[1]
= { . . . , - n + l, l,w +
A relation R on a set E that is reflexive, anti-symmetric (if xRy and jyito:
then x = y), and transitive is called an order (or a partial order) and is often
written <. When a < b and a^b we write a Kb. Examples of order relations
are
(a) Con?(E);
(b) | on Z, where a | b ^> a divides b.
An order relation can often be represented pictorially by a Hasse diagram.
In this, a < b is exhibited by joining the point representing a to the point
representing b by an increasing line segment. For example, the Hasse diagrams
for the set {1, 2, 3, 4} ordered first in the usual way and then by divisibility
are shown in Figs 2.1 (a) and (b) respectively.
Fig.2.1
(a)
O4
(b) 4O
O3
O2
O 1
2.1
Let S be the relation defined on IR by
xSy<*x2 =x\y + 1|.
Sketch the graph of S.
2.2
Sketch the graphs of the following relations o n IR:
(a) {(je,jOII*+;p|<lh
(ft) {(JC,^)|2X 2 + 3 ^ - 2 / < 0 } ;
(c)
{(x,y)\(x-y)(x-2y)(x-3y)>0};
(d) {(x,y) | x + y - 4 < 0, 2x -y - 4 < 0, 2x - 5y - 10 < 0,
3x-y + 3>0}.
2.3
Let A ={1, 2, 3, 4}. Determine the graphs of the relations JR, S defined
Book 1
Sets, relations and mappings
on A by
2.4
RelationsR t andi? 2 are defined on IR by
x2 +y2 <4,x>y.
Sketch the graphs of these relations.
2.5
Let the relation p on a set A have the properties
(a) apa for every a E A;
(b) ifapb and bpc then cpa.
Prove that p is an equivalence relation on A. Does every equivalence relation
on A satisfy (a) and (b)l
2.6
Consider the relation R = {(a, b), (a, c), (a, a), (b, d), (c, c)} defined on the
set X = {a, b, c, d}. Find the minimum number of elements of X x X which
must be adjoined to R in order to make R
(a) reflexive;
(b) symmetric;
(c) an equivalence relation.
Answer the same questions for S = {(a, b), (a, c), (a, a), (c, c)}.
2.7
How many different equivalence relations can be defined on the set {a, b, c}l
2.8
Given relations R, S on a set A, define the product relation RS by
(x,y)eRSo(3zeA)((x,z)eS*nd(z,y)eR).
Give an example of relations R, S with RS = SR and an example of relations
Prove that if R and S are equivalence relations then RS is an equivalence
relation if and only if RS = SR. Deduce that RS is an equivalence relation if
and only if SR is an equivalence relation.
2.9
2.10
Let R1, R2 and S be relations on a set X. Prove that
(a)
ifRlCR2thenSRlCSR2andR1SCR2S;
(b) S(RlUR2) = SRlUSR2.
If Ri and/? 2 are equivalence relations on a set X prove that
(a) Ri CiR2 is an equivalence relation;
(b) Ri U R2 need not be an equivalence relation.
2: Relations
Give an example of equivalence relations/?! and R2 with/?! ¥=R2 and/? r UR2
an equivalence relation.
2.11
Let a be the relation on IN defined by
Show that a is an equivalence relation. Into how many equivalence classes
does a partition IN?
2.12
Let S = IR \ {0}. Define a relation p on S x S by
(a,b)p(c,d)*>c2b=a2d.
Prove that p is an equivalence relation. Describe geometrically the p-classes.
If o is defined on S x S by
(ayb)o(c,d)oc*b2=aAd2,
show that a is an equivalence relation. Describe geometrically the a-classes.
Explain how the equivalence classes of p and of o are related.
If the relation r is defined on IR x IR by
is r an equivalence relation?
2.13
Consider the relation ~ defined on € \ {0} by
zi ~ 22 o \Zi\(\z2 \2 4-1) = \z2 \(\Zl\2 + 1).
Prove that ~ is an equivalence relation. If aG IR is such that 0 <a<\,
sketch on the Argand diagram the ~-class of a.
2.14
Consider the relation ~ defined on C \ {0} by
zx~z2<* zxzx(z2 + z2) = z 2 z 2 (zi + zO.
Prove that ~ is an equivalence relation. If a is a non-zero number on the real
axis, give a geometrical description of the ~-class of a.
2.15
Let S = {(x,y) G IR x IR | JC =£ 0,y ¥= 0} and define a relation - on S by
(xuyO-ix^y^ox^iixl
- y\) = x2y2{x\ -y\).
(a) Show that ~ is an equivalence relation.
(b) If (a, b) is a fixed element of S show that
y
a
- = --.
x a
x
b
{c) Sketch the ~-class containing (2,1).
(x,y)~(a,b)o-=-
2.16
y
b
or
If a is a given real number consider the relation/on IR given by
Book 1
Sets, relations and mappings
(x,y)ef*>y = x2 +ax+a2.
(a) By considering the graph o f / prove that if A ={xG IR | (x,
then
(i) A = 0 if and only if \a \ > 2/\fa;
(ii) \A | = 1 if and only if \a | = 2/V3;
(iii) | A | = 2 if and only if |<z | < 2/V3.
(Z?) Prove that the relation S defined on IR by
x = y(S)<>x3 — y3 =x~y
is an equivalence relation. Deduce from the above that the £-class of x E IR
consists of
(i) a single element if and only if | x | > 2/\/3;
(ii) two elements if and only if \x | = 2/\/3 or \x \ = l/\/3;
(iii) three elements otherwise.
2.17
(a) Prove that trie relation/? defined on IR by
xRy<>x2 —y2 = 2(y—x)
is an equivalence relation. Determine the R-class of 0 and the /^-class of 1.
(b) The following argument leads to a false conclusion. Explain where it
is incorrect.
Since x2 —y2 = (x + y)(x —y) it follows that if x2 —y2 = 2(y—x) then
(x + y)(x —y) = —2(x —y) and so x + y = —2. Hence the relation S defined
by xSy^x 4- y — — 2 is also an equivalence relation, and from lSl we have
2=-2.
2.18
Let M be a set of mn + 1 positive integers. Let = be the relation on M
defined by
a = b<*a | b,
and let S be the relation on M defined by
aSb <>a^b
and b ^ a .
Show that M contains either a subset {ax, a2, . . . , tfm+i) with #,- = 0/+i for
Ki<m
or a subsetj^, b2,. ..,bn+1} with bjSbk for ji=k. (Note: this
question is quite hard.)
2.19
Let Au A2,.. .,An be subsets of a set X. For each ^4,-, let A® denoteylf and
let A} denote the complement of Af in X. A constituent of X with respect
1 ? . . .,An is defined to be a non-empty subset of the form
10
2: Relations
where each e,- is either 0 or 1. Prove that the constituents of X are disjoint
and that they partition X.
Suppose now that A, B, C are subsets of X. Write the subset
A\(B\C)
as a union of constituents of X with respect to A, B, C. If = is the equivalence
relation defined on X by the partition of constituents, is it possible to have
x = y when x EA \ (B \ C) andj> G {A Pi B) \ C?
2.20
Define a relation p on IR + = {x G IR | JC > 0} by
(a,b)Gp^a-y/(a
+ 1)<b -\ <a + y/(a 4-1).
Is p reflexive? Is p symmetric? Is p transitive?
2.21
A set contains 1000 elements and is partitioned into m 4- 1 subsets. The
smallest subset in the partition contains n elements, the largest contains
« + m elements, and no two members of the partition contain the same
number of elements. Find all possible positive values of m and n.
2.22
Let£ = {O,.y)GIR x IR | x¥- 0,y =£0}. Define a relation ~ on E by
(x,y)~(a9b)*(xb)2=(ya)2.
Verify that ~ is an equivalence relation. Prove that (x,y) ~ (a, b) if and only
if there is a non-zero real number k such that x = ka, y = ±kb. Sketch the
^-class of the element (2, 1).
2.23
Let£ = {(*,>>)G IR x IR \x^O,y =£0}. Define a relation^onEby
(x,y) = (z,t)o xy(z2 + zt + r2) = zf (x2 + xy + j>2).
Verify that = is an equivalence relation on E. If m ^ 0 find the equivalence
class of (1, m). Hence describe geometrically all the equivalence classes.
2.24
Let a, Z? G Z and suppose that a, b are coprime with a>b>0.
recursively fi,f2,... byf1 = a,f2 =a — b and, forfc>2,
•a
if / * < & ;
-b
if
Define
/fc > Z?.
Prove that {fl9 / 2 , . . .,fa + b) forms a 'complete set of representatives' mod
a + b, in the sense that each (mod a + Z>)-class contains one and only one ft.
What happens if a and b fail to be coprime?
2.25
Draw the Hasse diagrams for the set E = {1, 2 , . . . , 10} when ordered by
(a) divisibility;
(b) the relation < defined by x y if and only if either x = y or
• where, for all positive integers p and q,
11
Book 1
Sets, relations and mappings
2p< 2qo
(c) the relation < defined by p < q if and only if p = q or p<q
where
[ there is a prime t G E such
2.26
Draw the Hasse diagrams for ?(E) ordered by C when
(a) E = (D;
(6) £" = {0};
(c) E = {{0}};
(e) /f = {0,{0},{0,
2.27
Draw the Hasse diagram for the set of positive divisors of 210 ordered by
divisibility.
2.28
Let E = {xtj | 1 < / < m , Kj<n)
be a set of mn distinct positive real
numbers. For every / € [1, m] define j>z- = max {xtj \ 1 < / < « } , and for
every/ G [1, n] define z;- = min {xy 11< i < m}. Prove that
min {j;,- | 1 < i < m} > max {ZJ | 1 < / < «}.
A regiment of soldiers, each of a different height, stands at attention in
a rectangular array. Of the soldiers who are the tallest in their row, the
smallest is Sergeant Mintall; and of the soldiers who are the smallest in their
column, the tallest is Corporal Max Small. How does the height of Sergeant
Mintall compare with that of Corporal Max Small?
2.29
Let E be a set on which there is defined an order relation <. Let = be an
equivalence relation on E. Suppose that the conditions x = z and x < y < z
together imply that x=y = z for x, y, zGE. Let E={[x]\xGE}
and
define a relation R on E by
Prove that R is an order relation on E.
12
3: Mappings
A mapping (or function) / from a set X to a set Y is a relation between X and
Y (i.e. a subset of X x Y) with the properties
(0) given x€iX there is some y EY with (x, j>) E /,
(b)
if(x,y1),(x,y2)<=fthenyx=y2.
We denote such a mapping / by writing f:X-+Y, and if (x, y)Efwe use
the notation j ; = / ( * ) . The set X is called the domain of/; and the subset
{y E y I (3x EX).y = /(x)} of y is called the image of/ and is denoted by
Im / When it is clear what the sets X and Y are, we often write the mapping
/ in the form x -»/(x). If A C X then f(A) is defined to be {yE Y \
(3xG^4)>> = /(x)}. In particular, /(X) = I m / Two mappings f:A-*B
and
£ : C -> Z> are said to be equal if A = C, B = D and (Vx G A)f(x) = g(x).
Given / : X-> y and g : y-> Z, the composite ^ o / is the mappings o / :
X-+Z defined by (g o f)(x) = g[f(x)] for every x E X For mappings
/, g and ft we recall that the associative law (fog)oh=fo(goh)
holds
whenever these multiple composites are defined. Also, when/: X->X we use
the usual notation/ 2 f o r / o /, and in general/71 f o r / 0 / 0 ... o f(n terms).
A mapping / : X-> Y is said to be bijective if it is injective (if f(pCi) = f(x2)
then Xi = x2) and surjective ( I m / = Y). In this case there is a unique mapping
f~l : y ^ - X such that f~l 0 / = id^ and / o / " 1 = idy where the notation
id x denotes the identity mapping on X, namely the mapping id^: X-*X
given by id^(x) = x for all x E X The composite f o g of two bijections / and
g is also a bijection; we have (fog)~1=g~1
of'1. Some authors use the
terms one-one to mean injective, and onto to mean surjective.
A relation on IR is a mapping if and only if every line parallel to the j>-axis
meets the graph of the relation precisely once. Other geometrical criteria in
terms of graphs can also be useful. For example, a mapping / : IR -> IR is
13
Book 1
Sets, relations and mappings
injective if and only if every line parallel to the x-axis meets the graph of
/ in at most one point; and / is surjective if and only if every line parallel
to the x-axis meets the graph of / in at least one point. Hence/is bijective if
and only if every line parallel to thex-axis meets the graph of/precisely once.
By way of example, consider the mapping/: \R -> CR given by x ->f(x) =
Ixj where [JC] is defined to be the greatest integer that is less than or equal
to JC. The graph of / i s as shown in Fig. 3.1 (in which each arrow Indicates
a constant value throughout an interval of the form [n,n + 1 [).
Fig.3.1
/(*)•
This mapping is not injective (for example, the linej> = 1 meets the graph
in infinitely many points), nor is it surjective (for example, the line y = \ does
not meet the graph at all). Note, however, that I m / = Z and that /induces
a surjective mapping from [R to Z also given by the prescriptions -> \x\.
3.1
Let S, P, C be the functions from IR to IR given by Six) = x2, P(x) = 2X and
C(x) = cos x. Express each of the following functions/ : IR —> IR in terms
of S, P, C using • and o (where / • g is defined by ( / • g)ix) = /(x)g(x)):
(a)
(f)f(x) =
(b)
(g) /(x) = cos(cosx) 2 ;
2
(c) /(x) = (cosx) ; (h) fix) = (cos 2COS*)2;
(J) f(pc) = cos x2;
(i) /(x) =
(e) f(x) = cos cos x;
3.2
Let X = [R \ {0, 1}. Define^ : X^X
14
for 1< i < 6 by
3: Mappings
A(x) = x,
x-l
f2(x)=l-x,
f3(x) =
x
1—x
x—1
Show that/- o ff G {fk | 1 < k < 6} for all ij.
3.3
Let / : [R -• [R be defined by
Express / as a composite of four maps none of which is the identity.
3.4
Let / : X^X and define a relation R on X by xRy if and only if y —fix).
Prove that
(a) R is reflexive if and only if/ = idy;
(b) R is symmetric if and only if/ 2 = id^;
(c) R is transitive if and only if/ 2 = /
3.5
If / : X -> X and if g is defined by
g={(y,x)GXxX\y=f(x)}
prove that g o f is an equivalence relation, where g o f denotes the product
relation defined in 2.8.
3.6
If/: A-> B prove that the relation Rf defined on A by
xRfyof(x)=f(y)
is an equivalence relation.
L e t / : [R x [R -> CR x CR be defined by
x
y
if
(x,y)*(0,0);
«>)'
1(0,0)
if (x,y) = (0,0).
Find theify-class of (1, 0). Describe geometrically the jR^-class of (a, b).
3.7
For each of the following relations on [R
(i) sketch its graph;
(ii) find its domain and image;
(iii) say whether or not it is a mapping.
(a) x2<y; (b) sinx=^; (c)x = siny;
(d) x-2<y<x + l\ (e)y = \x\\ (/) x+y = l;
(g) I x | + ^ = 1;
(/) I^KI*I<1;
15
(h) x + | ^ | = l;
(k) y =
(i) 1*1 + |;H = 1;
\x\-lxl
Book 1
3.8
Sets, relations and mappings
Find the domain and image of the following relations on IR. Are any of
them mappings?
(a) {ix,y)\x2 + 4y2 = l};
(b){(x,y)\x2=y2};
(c)
{(x,y)\y>O,y<x,x+y<lh
id) {(x9y)\x2
+y2<\,x>Q}\
00 {(x,y)\y = 2x-l).
3.9
Let/be the subset of <Q x Z given by
/ = {(x,y) I y is t n e least integer withj> >x}.
Determine the domain and image of / Is / a mapping? Answer the same
question for the subset of Z x Q given by
/ = {(x,y) I x is the least integer withx >y}.
3.10
Sketch the function/: [R -> [R given by
= 11*-II-II.
3.11
If / : m -+ [R is defined by
f(x) = 2x2 + 6x + l
determine the set {x G [R | fix) < x + 5}.
3.12
Let / : [R -> [R. Show that there exist mappings g, h : [R -> IR such that
f = g + h with £(x) = g(-^:) and /z(x) = - A ( - J C ) for allx G IR.
3.13
Let/,g : [R -> IR be given by
fix) = ix + \)2, gix) = 2x-L
Determine the mappings fog,gof
and the set
Given the subsets of \R x [R described by
A={(x,y)\x>0,y>0}',
B = {(x,y)\y<f(x)h
C={(x,y)\y>g(x)};
D = {ix,y)\y<l-x}9
describe the set A n B n C n Z).
3.14
Sketch the subset 5 of IR x IR given by
16
3: Mappings
If the mapping d : S -> CR is given by
determine Im d.
3.15
Sketch the subset S of IP x IP given by
S = {(x9y)\\x\ + \y\>\
If the mapping / : S -» IR is given by
x2+y2<\).
and
determine I m /
3.16
Let S= {(x,y)G
given by
IP x IR \x2+y2-
10x4-16 = 0) and let / : S-> IR be
y
x
By considering this mapping geometrically, determine I m /
3.17
Let S= {(x, y)e\R x IP | x 2 4 / -6xbe given by
8y 4 21 = 0) and let / : £ - * IR
By considering this mapping geometrically, determine I m /
3.18
Each of the following describes a mapping f:A-+B
by exhibiting the set
A, the set B, and the image f(x) for every xEA. Determine which of the
mappings are injective or surjective.
(c) A = \R, B=\R,
f(x)=x3;
(d)A = (L, £ = I P ,
f(x)=\x\\
(e) A = \R, B= IP, f(x) = xsinx;
(g)A = m, B=\R, f(x) = x2+x + i.
3.19
If/: X-> YmdA,B are subsets of Xprove that
(a) f(AnB)Cf(A)nf(B)(b)f(AUB)=f(A)Uf(B).
Prove also that
(c) f is injective if and only iff (A CiB)= f(A) n /(£) for all A, B\
(d) /is surjective if and only if/(X\ A) D Y\ f(A) for all A\
(e) /is bijective if and only if f(X\ A) = Y \ f(A) for all A
17
Book 1
Sets, relations and mappings
3.20
If X and Y are sets denote by Yx the set of all mappings from X to Y. Show
that, for all sets A, B and C, it is possible to find a bijection between
(a)
(AxB)cmdAcxBc;
(b) (A ) and A x ;
(c) ABUC mdABxAc
if B nC=ty.
If there is a bijection between ABUC and AB x Ac is it necessary for B n C
to be empty?
3.21
Describe explicitly a bijection
0 ) from [0,1] to [1,2];
(b) from [0, l ] t o [0,2];
(c) f r o m ] - l , l [ t o IR;
(d) from [0, l ] t o [0,1 [;
(e) from [ - 1 , l ] t o IR.
(Hint: for (d) consider a map that acts as the identity map except on some
set of rationals.)
3.22
The functions/: IR -> IR andg- : IR -> IR are defined by
' Ax 4- 1
if
x '..
x
if
x
3x
if
x>0;
if
x < 0.
1x4-3
Show that g of is a bijection and give a formula for (g of)'1.
t h a t / o #• is neither injective nor surjective.
3.23
Show also
The functions/: IR -> IR and# : IR -> IR are defined by
(f 1l - x;
if x > 0 ;
I Xx2
x
if
x < 0,
if
x>0;
\x - 1 if x < 0.
Find a formula for / o g and draw its graph. Show t h a t / o g is a bijection
and find its inverse.
Find also a formula forg-o /and draw its graph. Show thatg o / i s neither
injective nor surjective.
3.24
L e t / : IR -> IR be given by
f(x) = x\x\.
18
3: Mappings
Show that / is a bijection and determine f~l. Is the mapping g : IR -> IR
given by
g(x) = x2\x\
a bijection? If so, determine g"1.
3.25
Using calculus, or otherwise, prove that the mapping/: IR -» IR described by
/ ( * ) = x3 + ax2 +bx + c
ia, b,cE IR)
2
is a bijection if and only if a < 3b.
3.26
Let IR + = {xG IR \x>0} and l e t / : IR + -> IR+ be given by
1
/(*) =
Prove t h a t / is injective and find t w o distinct m a p p i n g s g , h : I R + ^ IR + such
th2itgof=h o / = id|R+.
3.27
L e t / : IR -> IR be defined by
/(JC)
= cos [TT(JC — I J C ] ) ] — 2|[JC]|.
Prove that / is injective and find two distinct mappings g, h : IR + -> IR+ such
a bijection?
3.28
Prove that if IN* = IN \ {0} then the m a p p i n g / : IN* -> Z given by
is a bijection and obtain a formula f o r / 1.
3.29
Letg : IR -^ IR be given by
g(x) = 3 + 4JC.
Prove by induction that, for all positive integers n,
g"(x) = (4" -1)4- 4nx.
If for every positive integer k we interpret g~k as the inverse of the function
gk, prove that the above formula holds also for all negative integers n.
3.30
Use induction to prove that if Au ..., An are subsets of a set E then
n
n
1=1
19
!</
Book 1
Sets, relations and mappings
Let A and B be finite sets with \A\= m and \B\ = n. How many mappings
are there from A to Bl How many injections are there from A to Bl Show
that there are
surjections from A to B.
3.31
Give an example of non-empty sets A, B and C with the property that there
are injections
none of which are bijections.
3.32
Consider the mapping/: IR -> IR given by
Sketch the graph of/. Find an interval^ = [—k9 k] on the x-axis such that
(a) \f(x) \xE.A) = Im/;
(b) g : A -* Im/given by g(a) =f(a) for every a GA is a bijection.
Obtain a formula forg-"1 : Im/-» A
3.33
Sketch the graph of the function/: IR -> IR given by
Show that / is not injective. Determine Im / and find a subset A of IR such
that the restriction of/ to A induces a bijection g : A -> Im / Obtain a formula
for the inverse of this bijection.
3.34
Let p be a fixed positive integer. Prove that the mapping/: Z -> Z given by
\n+p
if H is divisible by p,
H
if
« is not divisible by p,
is a bijection, and determine/" 1 .
3.35
Prove that the mapping/: IR -MR given by
A.)-1
I ( 2 )
if
x<0,
is a bijection, and find its inverse.
20
3: Mappings
3.36
Define / : IN -> IN by /(O) = 0, / ( I ) = 1, and f(n) =f(n - 1) +/(w - 2)
for « > 2. Prove that
(fl) / ( i ) </0" + 1) for aU i > 2;
(b) there exist precisely four i G IN with (/ o / ) ( / ) = /(/)•
Writing ^ for/0*), prove also that
(c) fSn is divisible by 5;
(£0 fn+2 = 1 + S ? = 1 / / ;
(e) / w - i / w + i - / S = (-1)" for /i > 1;
(/) 2% = [(1 +x/5) w -(l -V5) w ]/V5.
3.37
L e t X = {1, 2, 3, 4} and define/: X
f/(jc) = jt + l
if
x<3,
1/(4) = 1.
Show that there is only one mapping g : X -> X with the property that
g{\) = 3 and f o g = g o f. Find g. Is it true that there is only one mapping
h :X->Xwith/*(l) = l a n d / o A = ft o / ?
3.38
(a) If a : IN -^ IN is given by a(«) = « + 1 show that there is no mapping
g: IN —> IN such that a o g = idj^ but that there are infinitely many mappings
k : IN -> IN such that k o a = id jN .
(6) If 0 : IN -* IN is given by
n/2
if n is even;
l(l)/2
if ?i is odd,
show that there is no mapping / : IN -> IN such that / o |3 = id|N but that
there are infinitely many mappings k : IN ^ IN such that 0 o k = id|N.
3.39
Let ^ be a set and let / : A -> P(,4). Define the subset X of A by X = {a G ^ |
a^fiq)). Can there exist aGy4 with/(x) = X? Can/be a surjection? Can
/ b e an injection?
3.40
Let IN* = {1, 2, 3,...} and for every k G IN* define
4 = {X G IN* | £*(* - 1) < x < £*(* + 1)}.
(flf) Show thatIk has/: elements and that {Ik \ kG IN*} is a partition of IN*.
(b) Define/: IN* x IN* -> IN* by
/(m, n) = \{m +n- 2){m +n-l)
Show that/(m, n) G/ m + w _ x and deduce that
21
+ m.
Book 1
Sets, relations and mappings
Hence show that/is injective.
(c) For 1 < r < k show that f(r, k + 1 — r) G Ik and deduce that / is also
surjective.
3.41
Let S = I R \ { 1 , - 1 } . Find a mapping f'.S^S
such that fof=
(Hint: try mapping ]—1, 1 [ to its complement in S.)
3.42
Given mappings / \ A -+B, g : B -> C, h : C -> Z>, suppose that g o f and hog
are bijections. Prove that f,g, h are all bijections.
3.43
Let (Q+ = {x G (Q | JC > 0}. If^/Z?, c/dG(D.+ prove that
C
a+b
hcf(a,b)
d
Deduce that the prescription
a+b
—id^.
c+d
hcf(c, d)
'€)•
hcf(fl, b)
defines a mapping/: (Q+-* Q+. I s / a bijection?
3.44
For mappings / , g : IR -HR and every X G IR define the mappings/ + g, f • g
and X/from IR to IR in the usual way, namely by setting
for every x G IR.
(V) Show that there are bijections /, g w i t h / + g not a bijection. Show
also that there are bijections / g with f-g not a bijection. Do there exist
bijections/,# such that neither / + g norf-g is a bijection?
(b) Prove that if X =£ 0 then X/is a bijection if and only if/is a bijection.
(c) Define [/#] : IR -> IR by [/,
rf=/og-go/
Do there exist
bijections f,g with [/^] a bijection?
(d) If 0 denotes the mapping from IR to IR described by x ->0, prove
that, for all mappings/,g, /z : IR -HR,
[[fg]h]+[[gh]f]+[[hf]g] = 0.
3.45
(a) Define a relation /? on C* = C \ {0} by
(a + ifc)/*(c + \d) *>a y/(c2 + d 2 ) = c \/(a2 + ^ 2 ).
Show thati? is an equivalence relation and describe geometrically the i?-classes.
(Z?) Let U= {[x]R U G C * } be the set of ^-classes. Show that if xRy
22
3: Mappings
thenx 2 /?^ 2 and deduce that the relation / o n U given by
/={([*]*, [*2k)lxec*}
is a mapping. Is/injective? Is/surjective?
(c) For each of the following relations f on U determine whether / is a
mapping U-*U. For those relations which are mappings, determine which
are injective and which are surjective.
( 0 / = { ( [ * ] * , [2x]*)UGC*} ;
(ii)/={([*]*,[* +2]*) |xGC*};
(iii)/={([*]* j x - ^ U e e * } .
3.46
(a) Define a mapping/: IN -> IN by
f(n) = the sum of the digits of n.
Is/injective? Is/surjective?
(b) For every i > 1 let Rt be the equivalence relation on IN defined by
xRiyofi(x)=fi(y).
Describe the class [ l ] # r Prove that if l < / < / then [l]i^.£ [!]#•• Is it
true that [n]R.C [n]Rj for every n G IN? Is [ 1 ] ^ = [ 1 ] ^ ?
(c) Prove that n is divisible by 9 if and only if n G [9]#. for some / > 0.
Find a similar criterion for n to be divisible by 3.
(d) Define a relation^ on IN by
xRy o xRfy for some / > 0.
Prove that R is an equivalence relation. Show that there are infinitely many
R/-classes for each / > 0 but that there are only finitely many R-classes.
How many i?-classes are there?
23
Solutions to Chapter 1
1.1
Each of the statements is true. Since 0 is a subset of every set we have 0 C A.
From the definition of A we see that 0GA and {0}GA, and these give,
respectively, {$}CA and {{0}}CA Since 0 and {0} are elements of A it
follows that {{0}, 0 } C A Finally, {{0}, 0}={0, {0}} is, by definition, an
element of A.
1.2
The only subset of 0 is 0 and so P(0) = {0}. It follows that P(P(0)) = {0, {0}}.
Finally,
P(P(P(0))) = {0,{0},{{0}},{0,{0}}}.
13
The subsets of E = {1, {1}, 2, {1, 2}} are
0, {1}, {{1}}, {2}, {{1,2}}, {1,{1}}, {1,2}, {1, {1,2}},
«1},2}, {{1}, {1,2}}, {2, {1,2}}, {1,{1},2}, {{1},2,{1,2}},
{1,{1},{1,2}}, {1,2, {1,2}}, E
From this list we see that E n ?(E) = {{1}, {1, 2}}.
1.4
Four examples are
0,
7.5
1.6
{0},
I0,{0}},
{0,{0U0,{0}}).
Yes. For example, take A = 0, B = {0} and C = {0, {0}}.
(a)
(b)
(c)
(d)
(e)
(/)
(g)
24
False; consider, for example,^ = {1},B = {2}, and C = {{I}}.
False;consider^ = {l},£ = {2},C = A
False; considers = {1},£= {{1}, 1},C= {{1}, 2}.
False;consider^ =B = C={1}.
False; consider A = {1},B,= {1, 2}, C= {{1, 2}, {1}}.
False; consider^ = {1, 2},B = {2, 3}, C= {2, 4}.
False; consider A = {!},£ = {0}, C = {2}.
Solutions to Chapter 1
(h) True; using the distributive law we have
(A UB) n c = (A n c) u (B n c) = 0 u 0 = 0.
1.7
1.8
Take* = z^.y.
(a) Using the de Morgan law we have
(AU(AUB)'y=A'n(AU B)"
= A'D(AUB)
= (A'r)A)U(A'nB)
= $U(A' HB)
= A'nB.
(b) The expression clearly reduces to
= A'U(BUA')'
= A'U(B'nA)
= (A'UB')n(A'UA)
= (A'UB')nX
(c) Since ( 5 H C ) U C = C , the expression reduces to
(A U (Br Pi C') U C)1 = {{A UB'UC)n(AUC'U
C))'
= ((AUB'uc)nx)'
= A'
1.9
f
The expression (A nB')U(A
HB) is none other than A AB. The Venn
diagram for A A B is as shown in Fig. Sl.l.
Fig.Sl.l
25
Book 1
Sets, relations and mappings
It is clear from this that A A B = A U B if and only if A n B = 0.
1.10
(a) Note that another expression for A AB is (AU B)\ (A O5). Using
this, we see that the Venn diagrams for A A(B AC) and (A AB) AC are the
same: see Fig. SI.2.
Fig.S1.2
(b) Using the fact that A A B = (A U B) \ (A n B) we have
AABA(AHB) = ((A AB)U(AD B))\((A AB)n(A OB))
= (AUB)\9
= AUB.
(c) This follows from the fact that the Venn diagram for A n (B A C) is
the same as that for(AnB)A(AHC): see Fig. SI.3.
Fig.S1.3
(d) The Venn diagram for A A (B O C) is shown in Fig. SI.4 and that for
(AAB)D(AA
C) is shown in Fig. SI .5.
26
Solutions to Chapter 1
Fig.S1.4
Fig.S1.5
These are the same if and only if
i.e. if and only if A n B C C and A PiCCB, which is the case if and only if
A n B = A n c.
7J7
We have
Also,
I B) | 041 5) = CE(A n 5) | CE{A n 5)
= AC)B,
27
Book 1
Sets, relations and mappings
and
= CECE{A UB) = CE(CE(A) n CE(B))
= CE(A) I CE(B)
=
(A\A)\(B\B).
1.12
{A,B} = {C, D} implies (i) A = C and B = D,ox (ii) A = D and B = C. In
case (i), ^ O 5 = C n i ) ; and in case (ii), APiB = DC)C=C(~)D. Argue
similarly for U.
1.13
We have A n(BU C) = (A DB)U (A DC) the right hand side of which is
contained in (A O B) U C since i O C C C If now C C A then ,4 n C = C
and we have equality in the above. Conversely, if the equality^ fi(5UC) =
(A C\B)U C holds then since the left hand side is contained in A and the
right hand side contains C we have that CCA.
1.14
As is readily verified, each side of the equality is represented by the Venn
diagram shown in Fig. SI .6.
Fig.S1.6
1.15
Take, for example,^ = C= {VhB = D = {2}. We have
(AUB)x(CUD)
1.16
= {(1, 1), (1, 2), (2, 1), (2, 2)},
Suppose first thatj> ¥= x andj* ¥=x*. Then we have
} = {**} and{*,;;}={
= x* and>> =y*,
so the result holds in this case.
28
Solutions to Chapter 1
Suppose now that y = x. Then we have
(x,y) = (x,x) = {{x},{x,x}} ={{x},{x}} =
and so in this case
(x9y) = (x*,y*)o{{x}}={{x*},{x*9y*}}
=7* =x(=y).
This establishes the result in this case. The only other case to consider is
that in whichy* = x*9 and this is similar to the case in whichy — x.
Finally, we have
{x} x {x} = {(x9 x)} = {{{x}, {x, x}}} = {{{x}}}.
1.17
If A =B = C then clearly (A xB) U {B x A) = Cx C Conversely, suppose
that this equality holds and let aGA and bGB. Since (a, b) belongs to
the left hand side, it belongs to the right hand side, s o c G C and b EC.
Thus A C C and B C C. However, if xGC then (x, x)ECxC,
so either
(x, x)GA x B or (x, x)GBx A. In either case, xGA a n d x G B from which
we conclude that C = A and C = B.
1.18
Observe that
XC\Y = X\(X\Y).
IfX, YE & then (X\Y)\(Y \X)G(J^°)°. But clearly
1.19
Thus ^
Consider now &= {X, 7, Z} where X = {1,2}, Y = {2, 3, 4}, andZ = {4}.
We have
^o = { { 1 } ) { i 5 2},{4},{2,3},{3,4},0}.
Since (e^"0)0 contains {2} and^" 0 does not, the desired inequality follows.
1.20
(a) True;for
(b) False. For example, take A = {\}, B = {2). Then we have
{l,2}GP(,4U£)but
29
Book 1
1.21
Sets, relations and mappings
The Venn diagrams for A\(BUC)
Figs. SI.7 and SI.8, respectively.
and (A\B)U(A\C)
are shown in
Fig,S1.7
Fig.S1.8
Clearly, we have equality if and only if A n (B A C) = 0.
As for AA(BUC)
and (A A B) U (A A C), the respective Venn diagrams
are shown in Figs. SI.9 and SI. 10.
Fig.S1.9
30
Solutions to Chapter 1
Fig.Sl.10
Again we have equality if and only if A n (B A C) = 0; hence the result.
Consider now^4 = {1},B = C= {2}. We have
AA(BUC)={l}A{2}={\,2}
and
(A AB) U (A A C) = {1, 2} U {1, 2} = {1, 2}.
However, taking^ =B= {1}, C = {2}we have
AA(BUC)={l}A{\,2}={2}
whereas
7.22
Assume, without loss of generality, that / < / . From the definition we have
Bi CAt and Bj=Aj \ UfetiAk. Now BfCAi C UfclV** so jfy ££,- and it
follows that 5 / 0 ^ = 0.
Since Bt CA( for every i, we have U ^ ^ f C UfLiAf. Now let x G UgL^,-.
Let ^ be the least integer such that x G i f and x^At for z <t. Then xGBt
and so x G UJ^i^/, showing that UgL^/ C U ^ ! ^ .
7.25
Consider the subset of 5 given by X = B\A. Clearly, X n ^ 4 = 0 and
I U i = 5 , so a subset with the given properties exists. To show that it is
unique, let Y be a subset of B with these properties. Since YC\A = 0 we
have that YCB\A. But 7D ( 7 U ^ ) \ ^ = B\A whence we have that
Y = X.
1.24
Draw a Venn diagram to illustrate the given information, putting numbers
in the appropriate places to indicate the failures (see Fig. Sl.l 1).
31
Book 1
Sets, relations and mappings
Fig.Sl.ll
From the diagram we see that the number of students who failed in any
of the subjects is 15. Consequently 41 — 15 = 26 passed in all three subjects.
1.25
Let A, C, G, T signify algebra, calculus, geometry and trigonometry respectively, and let \X\ denote the number of students who study subject X.
Assume, without loss of generality, that there are 100 students in the class.
Then from the fact that
\A orB\ = \A\ + \B\-\A
and£|
we have, from the information given,
\A andC| = | ^ | + | C | - | i 4 o r C |
> 70 + 75 - 1 0 0 = 45,
\GdndT\ = \G\ + \T\-\GoiT\
> 80 4- 85 - 1 0 0 = 65.
Consequently we have that
\A and C and G and T\
= \A andCI + \G and T\ - \(A andC) or(G and T)\
> 45 + 65 - 1 0 0
= 10.
A Venn diagram illustrating this minimum percentage of 10 is shown in
Fig. SI.12.
1.26
For each subset A of E let \A\ denote the number of elements in A. We
make use of the formula
32
Solutions to Chapter 1
Fig.Sl.12
T
—i
20
15
10
30
25
I
Note that, from this formula,
Now let A=XnY,
B = X1 D y, C = XC\YI and D = X'nY'.
Since
A n B = 0 and A U B = Y v/e have, from the above, \Y\=p + q. Similarly,
since C n D = 0 and C U Z) = 7 ' we have that | Y' \ = r 4- s. Again, using the
fact that Y n y ' = 0 and Y U y ' = JS1, we deduce that « = p + # + r + s.
In order to relate this to the problem, consider Fig. SI. 13 in which X
denotes the set of arts students and Y denotes the set of boys.
Fig.S1.13
Boys
Girls
Arts
xn Y
XHY'
Science
x'n Y
rnY'
33
Book 1
Sets, relations and mappings
Let \X\ = x and \Y\=y.
What is given is that
\XHY'\ \XnY\
\Y'\
so that we have
>
in '
x-\XC\Y\ \XHY\
n-y
y
What we have to prove is that
urnr'i \x'nY'\
\x\
\x'\ '
i.e., we have to show that
x - \X n Y\ n - x -y + \X D Y\
x
n—x
This is clearly equivalent to
\xnY\
1
1.27
y-\xnY\
^> 1
x
n —x
which is the same as
\XC\Y\ y-\XDY\
x
n —x
which reduces to
n\XHY\<xy.
Returning now to what we are given, we see that
xy-y\XDY\>n\XnY\-y\XnY\,
which also reduces to xy > n \X n Y\. Hence the conclusion holds.
Let x GA. Then we have
x = 2y for some y E Z
= \2y-l0y
= 6b + 10c where b = 2y G Z, c = —y G Z
and soxGB.
Conversely, if x £ B then
x = 6Z> 4- 10c
= 2y where y = 3b + 5c E Z
and
34
Solutions to Chapter 1
1.28
7.29
(a) S=\R andS=(D;
(b) 5 = IN and£ = Z;
(c) None; for 0 ? IN,0£Z, etc.
Yes; in fact ^ Z H ^ Z ^ m Z where m = lcm(«i,« 2 ). To see this, let
^ E « 1 Z O « 2 Z . Then f is a common multiple of n^ and n2, whence it is
a multiple of m and hence belongs to mZ. Thus ^ Z H ^ Z C m Z . The
reverse inclusion is immediate from the fact that every multiple of m is
a multiple of both nx and n2.
In contrast, we have, for example, 2Z U 3Z ¥= mZ for any zrz. To see that
this is so, suppose that we had 2 Z U 3 Z = mZ. Then from 2 G m Z and
3GmZ we would have 2=pm, 3=qm whence l = 3 - 2 = feIt follows from this that we must have m = 1; but clearly 1 $ 2Z U 3Z.
(0) Given any xGA and any B CA we have either x E2? orx ^i?. Since
there are these two possibilities for each of the n elements of A it follows
that there are 2n subsets of A.
(b) Suppose that n is even. There are n\2 odd integers in A and hence
2n/2 — 1 subsets with only odd integers. (Note that 0 is a subset of this set
of odd integers and must be excluded.) Hence {2n — 1) — (2n/2 — 1) =
2n — 2n/2 subsets contain at least one even integer. Suppose now that n is
odd. There are (n + l)/2 odd integers in A and hence 2n - 2 ( w + 1 ) / 2 subsets
contain at least one even integer.
(c) Suppose that n is even. We have observed above that there are
n/2
2 — 1 subsets with only odd integers. Now there are n/2 even integers in
A. Thus there are (n/2)(2n/2 — 1) subsets consisting of odd integers and
a single even integer. Add to this the n\2 singletons of even integers and we
have a total of (n/2)2n/2 subsets containing exactly one even integer. If now
n is odd then a similar argument applies : there are 2^" +1 ^ 2 — 1 subsets with
only odd integers, and (n —1)/2 even integers in A. Hence there are
^n _ j ^ 2 subsets with exactly one even integer.
1.31
One is looking for the longest chain of subsets
in PG4). Take Ai =ty.A2 must consist of a single element, A3 must consist
of two elements, etc. Thus we see that k = n + 1.
= Z 0*-O=
35
Book 1
1.32
Sets, relations and mappings
One must treat separately the cases where (x 4- l)(x — 4) > 0 ar
(x + l)(x — 4) < 0. The given set can be expressed as A U B where
A = {x G IR | (x + 1)(JC - 4) > 0}
O {x G IR I (JC + l)(x - 4) > -4},
B={xe\R |(X + 1 ) ( J C - 4 ) < 0 }
n {* G IR | (x + l)(x - 4) < -4}.
Now
A={xG\R |(JC + 1 ) ( J C - 4 ) > 0 }
= {*GIR |JC<-1}U{JCGIR
|JC>4}
= ]-oo,-l[U]4,oo[,
and similarly
B={xG\R | (JC + 1)(JC — 4) < —4>
= {JCG
IR U 2 - 3 J C < 0 }
|JC(X-3)<0}
Thus the required union of intervals is
]-oo,-l[u]0,3[U]4,°o[.
1.33
We have (JC - I)2 = k(x + 1)(JC 4- 3) if and only if
(k - l)x2 + 2(2k + 1)JC + 3k -1 = 0.
There is no real solution to this quadratic in x if and only if
4(2k + I)2 - 4(fc - l)(3Jfc - 1) < 0,
i.e., if and only ifk2 + 8k < 0, which is the case if and only if k E ]—8, 0[.
36
Solutions to Chapter 2
2.1
We have that
x2 = x\y + 1\ *x(x
- \y + l\) = 0
<>x = Oorx = |^ + 1|
whence we deduce that the graph of S is as shown in Fig. S2.1.
Fig.S2.1
2.2
(a) Since
o—l — x<y<
1 —x
we see that the graph is as shown in Fig. S2.2.
37
Book 1
Sets, relations and mappings
Fig.S2.2
(b) 2x2 + 3xy -2y2<0
if and only if (2x ~y)(x + 2y) < 0, which is
the case if and only if either 2x — y > 0 and x + 2y < 0, or 2x — y < 0 and
JC 4- 2y > 0. The required graph is therefore as shown in Fig. S2.3.
Fig.S2.3
(c) Consider the expression
38
y - 2x
Solutions to Chapter 2
if and only if jy<
Suppose first that x > 0. Then we have
\x < y < x. Suppose now that JC < 0, say x = — z where z > 0. We have
or
from which we see that E > 0 if and only if y < — z = x or \x = —|z
—\z = | x . Thus the graph of the given relation is as shown in Fig. S2.4.
Fig.S2.4
y=x
(d) The various inequalities are satisfied by the points (x,y) contained
in the region indicated in Fig. S2.5.
Fig.S2.5
39
Book 1
Sets, relations and mappings
Fig.S2.6
2.3
The graphs of R and S are depicted in Fig. S2.6 in which those points (a, b)
that belong to the graph of R are denoted by (•) and those that belong
to the graph of S are denoted by X •
2.4
The graph of Rx is shown in Fig. S2.7 and that of R2 is shown in Fig. S2.8.
Fig.S2.7
-10
2.5
Property (a) is clearly the reflexive property. To prove that p is symmetric,
suppose that apb. Combining this with the fact that bpb (from (a)), we
obtain from (b) that bpa. As for transitivity, suppose that apb and bpc. By
(b) we have cpa, whence ape since we have shown that p is symmetric. Thus
apb and bpc together imply that ape, so p is transitive, and hence is an
equivalence relation on A
40
Solutions to Chapter 2
Fig.S2.8
yy =x
Every equivalence R relation on A clearly satisfies (a). The condition
(b) is also satisfied; for if aRb and bRc then by transitivity aRc, and by
symmetry cRa follows.
2.6
(a) To make R reflexive on X we must adjoint the elements (b, b) and
(d,d).
(b) To make R symmetric we must adjoin (b, a), (c, a), (d, b).
(c) Note from the definition of R that we have aRb, aRc mdaRd. Thus
if R is an equivalence relation on X it must comprise all the elements of
X x X. So in order to make R an equivalence relation on X we must adjoin
the remaining 11 elements of X x X to R.
Consider now the relation S = {(a, b), (a, c), (a, a), (c, c)}.
(a) To make S reflexive on X we must adjoin the elements (b, b) and
(d9d).
(b) To make S symmetric we must adjoin the elements (b, a) and (c, a).
(c) Note from the definition of S that we have aSb m&aSc. The minimum
number of elements that must be added to S in order to make S an equivalence
relation is readily seen to be 6, namely the elements (b, b)9 (b, a), (c, a),
(b, c), (c, b) and (d, d), the ^-classes then being{0, b, c} and {d}.
2.7
The answer is five. To see this, consider the number of equivalence classes
that are possible. There is clearly only one equivalence relation with a single
class; there are three distinct equivalence relations with a partition consisting
41
Book 1
Sets, relations and mappings
of two classes; and only one with a partition consisting of three classes.
These are depicted in Fig. S2.9.
(•b) [cm)
2.8
If R is the identity relation on A, i.e. R = {(x, x) \ xEA}, then clearly we
have RS = SR for every relation S on A. For an example of relations R, S
with RS ^SR consider A = {1, 2,3} and/* = {(l, 2)}, S= {(2,3)}. We have
that SR = {(1,3)} and RS = Q.
Suppose now that R, S and RS are equivalence relations on A. If
(x, y)GRS then (y, x)GRS and there exists zGA with (j>, z)G5, (z, x)G/?.
Now (z, j ) G 5 and (x, z) G 7?, and so (x,y)E. SR. This shows that RS C £R.
A similar argument gives S!R C/?£ and equality follows.
Conversely, suppose that R, S are equivalence relations with RS = SR.
Since (x,x)GR and (x, x) G S for every x G ^ w e see that (x, x) GRS, so /?£
is reflexive. To show that it is symmetric, let (x, y) GRS. Then (x, y) GSR
(since by hypothesis RS = SR) and so there exists zG^4 with (x, z)ER,
(z,y)GS. Now (z, x)Gi?, (y, z)GS since JR,^ are symmetric, so (y,x)GRS
as required. As for transitivity, let (x,y)ESR, (y, z)ESR = RS. Then for
some t, u GA we have
(x,i)eR9(t9y)es,(y,u)es,(u9z)eR.
Since S is transitive we deduce that (t, u) G S and hence
(x,w)^^5,(w,z)JR.
The former gives (x, w)GS, (w, u)GR for some w G i . The transitivity
of R now yields
from which it follows that (x, z)GRS = SR as required.
Finally, it is clear from the above that RS is an equivalence relation if
and only if RS = SR if and only if SR is an equivalence relation.
2.9
(a) Suppose that RlCR2 and let (x,y)GSR1. For some tGXwe have
(x, t) G Rx and (t, y) G S. Since then (x, i) G i? 2 we deduce that (x, >') G ^ 2
and hence that SRlCSR2.
Similarly, if (x, y)GR1S then (x, t)GS,
42
Solutions to Chapter 2
0 , y) G R ! C R2 whence (x, y) G R2S and consequently i^S C R2S.
we deduce from (a) that ^ C ^
(b) Since RlCRlUR2
Similarly SR2CS(R1UR2)
and hence SRrU SR2CS(R1UR2).
To
obtain the reverse inclusion, let (x, y) G SRX U £/?2- Then either (x,y) G £7?!
or (x, y) G S#2; i.e. either there exists f G I such that (x,
t)GR1,(t,y)GS
or there exists fGXsuch that (x, t)GR2, (t,y)eS.
Since RUR2
CRlUR2
it follows that in either event we have (x, y) G £(7?! U R2).
2.70
(a) ForallxGXwehave(x,x)G J R 1 and(x,x)G J R 2 and so(x,x)GR1 DR2
whence Ri CiR2 is reflexive on X. Suppose now that (x,y)GR1 C\R2. From
(JC, y) G7?i we have (y, x)G7? 1 ; and from (x, y) GR2 we have (y,x) £R2.
Thus (y, x)GRiC\R2
and so RiHR2 is symmetric. As for transitivity,
suppose that (x, y), (y, z)GR1HR2.
From (x, y), (y, z)GR1 we deduce
that (JC, z)GRl9 and from (x, y), (y, z)GR2 we deduce that (JC, Z)GR2.
Thus (JC, Z) G RX O i? 2 and so /?! n /^2 is transitive.
(b) Take, for example, X = {1, 2, 3} and let
^ = {(1,1), (2, 2), (3, 3), (1,2), (2,1)},
R2 ={(1,2), (2, 2), (3, 3), (1,3), (3,1)}.
Then clearly JRX and R2 are equivalence relations on X. However, RXUR2 is
not an equivalence relation since we have, for example, (3, 1), (1,2)GR l UR2
( , ) ^
l
2
For the last part, consider X = {1,2} and let
tf! = { ( l , l ) , (2, 2), (1,2), (2,1)},
* 2 = {(1,1),(2,2)}.
Clearly, Rx and R2 are equivalence relations on X with Rl¥=R2. Here we
have R1U R2 = Ri which is an equivalence relation.
2.11
That a. is an equivalence relation on IN is immediate from the fact that the
relation mod 7 is an equivalence relation. Now we have
aab o {a - b)(a + 6) = 0 (mod 7)
<>a = b (mod 7) or 0 = — b (mod 7).
Since, modulo 7, we have 6=—1, 5 =—2, 4 ——3 it follows that there are
four a-classes, namely [0], [1], [2], [3].
2.12
It is immediate from the definition that p is reflexive and symmetric on
S x S. To see that it is also transitive, we note that if (a, b)p(c, d) and
(c, d)p(e, / ) then c2b = a2d and e2d = c2f. Since by definition a, c, e ^ 0
43
Book 1
Sets, relations and mappings
we deduce that
b _d _ /
2
2
7 ~7~7 '
whence e2b = a2fand hence (a, b)p(e,f).
Now, since
Thus p is an equivalence relation.
we see that the p-classes can be described geometrically as parabolae passing
through (0, 0) with (0, 0) deleted (since 0 £ S by definition).
The proof that o is also an equivalence relation on S x S is entirely
similar to the above. In this case, we have
(x,y)o(c,d)<>c4y2=x4d2
>y=-x*
4*aW
C
- - - x
~
c2
I
2
•
Thus we see that each a-class consists of a pair of parabolae, one of these
being above the x-axis and the other consisting of its reflection in the x-axis
(with, as before, the point ( 0 , 0) deleted). Hence each a-class consists of
two p-classes.
Finally, r is not an equivalence relation on IR x IR since, for example,
we have ( 1 , l ) r ( 0 , 0) and (0, 0 ) r ( 0 , 5), but ( 1 , 1), (0, 5) are not r-related.
2.13
That ~ is reflexive and symmetric on € \ {0} is immediate from the definition.
To show that ~ is transitive, let z x ~ z 2 and z 2 ~ z 3 . Then we have
|Z 1 | 2 + 1 = | Z 2 1 2 + 1 _ | Z 3 | 2 + 1
1*1 I
1*2 I
1*3 I
from which we see that zx~ z3.
If now a E IR is such that 0 < a < 1 we have, writing z = x 4- iy, that
o {x2 + y2)(a2 4-1)2 = a2(x2 + y2 + I) 2
44
Solutions to Chapter 2
o (x2 + y2-a2)(a2x2
+ aV - 1 ) = 0
ox2 +y2 =a2 or x2 + y2 =— .
a
Thus we see that the ~-class of flE]0, 1[ consists of two circles in the
Argand diagram (Fig. S2.10). These are concentric at the origin, one passes
through a and the other through \\a.
Fig.S2.10
2.14
Observe first that for zG(C\{0} we have zz — |z| 2 =£0. Consequently we
can write
zx + zi
z2 + z 2
z2z2
from which it is easy to see that ~ is an equivalence relation.
If now a is a non-zero number on the real axis we have, writing
z — x + iy,
2x
2a 2
x'+y"
a'
_
+y22 =ax
a
Thus the ~-class of a is a circle with centre at the point ($a, 0) and of radius
^ ( F i g . S2.ll).
45
Book 1
Sets, relations and mappings
Fig.S2.11
2.15
(a) Since ;c =£ 0, j> =£ 0 => ;cy =£ 0 we can write
from which it is immediate that ~ is an equivalence relation on S.
(b)
x2-y2
a2-b2
(x,y)~(a,b)o
- = — —
xy
ab
o abx2 — aby2 — xya2 — xyb2
o (ay — bx)(ax + by) = 0
o ay — bx = 0 or ax + by = 0
y b
y
a
* > _ = - or - = --•
x a
x
b
(Note: Another way of obtaining this result is to write (x, y) ~ (a, b) as
x
y
a
b
y x b a
Under the definitions
y
a b
t = — and OL =
—,
x
b a
this can be transformed into the quadratic equation t2 + ta — 1 = 0 whose
roots are b/a and —a/b.)
46
Solutions to Chapter 2
(c) We have
y
i
y
- = x 2
x
so the ~-class of (2, 1) can be described geometrically as the perpendicular
line-pair y = \x, y = — 2x with the origin deleted (since (0, 0)&S). See
Fig.S2.12.
- = -
or
Fig.S2.12
y =
2.16
-2x\
(a) Use calculus to sketch the graph of / (Fig. S2.13). The minimum
value of x2 4- ax + a2 is attained when 0 = 2x + a, i.e. when x = —\a. The
minimum value is then \a2.
Fig.S2.13
V-
-ia
Al
X
Book 1
Sets, relations and mappings
(i) The statement A = 0 is equivalent to saying that the line y = \ does
not meet the graph of/ This is the case if and only if \a2 > 1, i.e. | # | > 2 / \ / 3 .
(ii) 1^41 = 1 is equivalent to saying that the line y = 1 meets the graph
of/exactly once. This is the case if and only if \a2 = 1, i.e. \a\ = 2j\/3.
(iii) \A\ = 2 is equivalent to saying that the line y = 1 meets the graph
of/in two points. This is the case if and only if la2 < 1, i.e. \a\ < 2\\]i.
(b) That S is an equivalence relation on IR is immediately seen from the
fact that
x=y(S)^x3
—x =y3 —y.
As for the S-class of x E IR, we observe that
y =x(S) o (x ~y)(x2 4- xy + y2 - 1 ) = 0
oy = x or x2 +xy + y2 — 1 = 0.
Consequently we have the following.
(i) The S-class of x consists of a single element if and only if the equation
x + xy + y2 = 1 has no solution. From part (a), this is the case if and only
if |JC| > 2 / V 3 .
(ii) The S-class of x consists of two elements if and only if either the
equation x2 +xy +y2 = \ has only one solution (which is the case precisely
when \x\ = 2/\/3), or if the equations y —x and x2 + xy + y2 — 1 have
a solution in common (which is the case precisely when |JC| = l/\/3).
(iii) The S-class of x consists of three elements otherwise.
2
2.7 7
(a) We have that
xRy o x2 + 2x = y2 + 2y,
from which it is immediate that R is an equivalence relation on IR. Since
it is clear that thei?-class of 0 is {0, —2}. Likewise,
so thei?-class of 1 is {1,-3}.
(b) The given argument breaks down at the point where cancellation by
x — y takes place. This step is valid only when x — y ¥= 0. In fact the relation
S is far from being an equivalence relation since, for example, we have xSx
if and only if x = —1.
2.18
Suppose that M contains no subset {bx, b2,..., bn+1} ofn + 1 elements with
48
Solutions to Chapter 2
the property that bjSbk for / =£ k. Let B be the subset of M consisting of the
n-\-\ smallest integers in M. There must be a smallest bnEB such that
bn divides some element of B. Replace bn inB by the next smallest element
of M and place bn into a set ^4X. Again there is a least element b inB which
divides some other element of B. lfbn = b then denote b by bn and put it
into Ax\ if bn^b then denote b by Z?2i and put it into a set^ 2 - Now add to
B the next smallest element of M and continue the process.
Note that at most n sets Ai,A2,.. -,An can be built up in this way since
each time we remove an element b from B we leave an element b G B with
b = b and at this stage B contains n elements.
Since M contains mn + 1 elements, one of the subsets At must contain
at least m 4-1 elements when the process is complete. If this subset is
{au a2,..., am + 1} then by its construction we have at = ai+i for 1 < i < m.
2.19
Let U, V be distinct constituents. Then there is an At with^4f appearing in
U and Af appearing in V. Then UCAt and VQA\ and so
E/Ti VCAt r\A't = 0,
whence the constituents are pairwise disjoint. Consider now any J C G I
Either x E^4X or x G^4[; and either XELA2 OV X GA[', and so on, so for each
/ there is an e/E{0, 1} with xEAp. Thus x belongs to some constituent
and the constituents partition X.
The Venn diagram for A \(B\C) is as shown in Fig. S2.14.
Fig.S2.14
C
Expressed as a union of constituents with respect toA,B, C,
A\(B\C) = (AnBnc)u(AnB' nc)u(AnB' nc').
Also, (A PiB)\C is the constituent ADBC\Cf. Hence, since ,4 \ ( B \ C) is
the union of constituents different from A C\B C\C', x and y must lie in
49
Book 1
Sets, relations and mappings
different members of the partition. Then x, y lie in different ^-classes and
so cannot be equivalent.
2.20
By definition,
p = {(a, b)\a-y/(a + l)<b-i<a+y/(a
+ 1)}.
We thus have that
(a, b) G p o - y/(a 4- 1) < b - a - \ < y/(a 4-1)
It is clear from this that p is reflexive (since a, b > 0) and symmetric. It is
not transitive, however, since for example (0, 1), (1, 2) G p but (0, 2) $ p.
2.27
The number of integers in the interval [n, n 4- m] is m 4- 1 and so, from the
information given, for every integer t G [«, « 4- m] there is a member of the
partition that has precisely t elements. We thus have
n 4- (n 4- 1) 4- (n 4- 2) 4- . . . 4- (n 4- m) = 1000.
Now the expression on the left is
m
(m 4- \)n 4- £ r = (m 4- \)n 4- ^m(m 4- 1)
r=l
= £(/w + l)(2w + /w)
and so we have to find all integer solutions of the equation
(m + 1)(2H 4- m) = 2000 = 2 4 5 3 .
Suppose that m is odd. Then m 4-1 must be a power of 2. The only possibility
is m 4-1 = 2 4 since, as is readily verified, the other potential values give the
contradiction that 2n is odd. Thus the only solution in this case is m = 15,
n = 55. Suppose now that m is even. Then m + 1 is a power of 5. Listing the
various possibilities, the reader will have no trouble in seeing that the other
possible solutions to the problem are m = 4, n = 198 and m = 24, n = 8.
2.22
Since ~ can be expressed in the form
x2 a 2 x 2 y 2
(x,y)~(a,b)*—=-<>-=
—
y 2 b 2 a2 b 2
it is clear that ~ is an equivalence relation. Moreover, we have that
(x, y) ~ (a, b) if and only if
50
Solutions to Chapter 2
\a b )\a
b,
which is the case if and only if
x y
x
y
—= a b
or
—=
a
.
b
Writing x/a = k, these equations can be written in the form x — ka,y — ±kb,
which is of course the line-pair j> = ±(b/a)x.
The ~-class of (2, 1) is represented in Fig. S2.15 as a line-pair with the
origin deleted.
Fig.S2.15
2,1)
2.23
From the definition ofE we see that
x2+xy+y2
xy
zt
whence = is clearly an equivalence relation on E.
If now m ^ O w e have
x2 4- xy + y2
1 + m 4- m2
xy
m
2
o mx + my = xy + xym2
o (mx — y)(x — my) = 0
1
2
<> y = mx
51
or
y = — x.
m
Book 1
Sets, relations and mappings
Thus the = -class of (1, m) consists of the line-pair y = mx, y = x/m with the
origin deleted (Fig. S2.16). The equivalence class containing a general point
(a, b) is the same as that containing the point (1, m) where m = b/a.
Fig.S2.16
2.24
Modulo a 4- b, subtracting b is the same as adding a. Hence we have that
fk = ka
mod a + b.
Since a, b are coprime, so also are a,a + b. Hence there exists c such that
ac = 1 (mod a + b). Now
Since we can assume that 1 < / : < # + b and 1 < k' < a + b, this gives k = kf
as required.
If a, b fail to be coprime then by definition each fk is a multiple of
d = hcf(#, b) and the set of fk does not form a complete set of representatives, since the class of 1 contains no integer divisible by d and hence no
fk. In this case there exists c such that ac = d (mod a + b) so that, modulo
a + b, d =fc. It follows that pd =)^c ( m °d 0 + £) and hence every multiple
of d is some fk (mod 0 + b).
2.25
(a) See Fig. S2.17.
52
Solutions to Chapter 2
Fig.S2.17
• 10
(b) See Fig. S2.18.
Fig.S2.18
2«
Am
6J
8«
\0m
m 1
9m
lm
si
A
(c) See Fig. S2.19.
Fig.S2.19
• 6
53
• 10
Book 1
2.26
Sets, relations and mappings
(a) \E\ = 0 so \?(E)\ = 1. Hasse diagram : •
(b) \E\ = 1 so \?(E)\ = 2 1 = 2. Hasse diagram :
(c) 1^1 = 1 so the Hasse diagram is the same as (b).
(d) \E\ = 2 so \?(E)\ = 2 2 = 4. Hasse diagram :
(e) \E\ = 3 so \?(E)| = 2 3 = 8. Hasse diagram :
2.27
/
•
|\/
•
<i
The set of prime factors of 210 is {2, 3, 5, 7}, and there are 16 positive
divisors, as shown in the Hasse diagram (Fig. S2.20).
Fig.S2.20
• 210
2.28
•
Consider the elements oiE arranged in a rectangular array
54
Solutions to Chapter 2
Xn
X\2
•••
%in
By definition, yt is the greatest element in the zth row; and Zj is the least
element in the /th column. Thus we have
It follows that, for all/, min {j>z- | 1 < i < m > zj and hence that
min {yt | 1 < / < m\ > max {ZJ \ 1 < / < « } .
Applying the above argument to the problem of the soldiers, it is clear
that Sergeant Mintall is taller than Corporal Max Small.
2.29
It is immediate from x <x that [x\R [x\, so thatR is reflexive. Suppose now
that [x]i?[j>] and [.y]i?[x]. Then for every aE. [x] there exists b E [y]
with a<b; and for every bf E [j>] there exists 0' E [x] such that b' <a .
Taking b'= b and using the given property, we obtain a = b and hence
[x] — [y]. Thusi^ is anti-symmetric. Finally, if [Jt]/?[.y] and [.y]Z?[z] then
for every 0 E [x] there exists b E [y] with a<Z?, and for every b E [j;]
there exists c E [z] with b < c. Thus, for every 0 E [JC] there exists c E [z]
such that a<c, whence [x]7^[z] and R is transitive. Thus R is an order
relation on E.
55
Solutions to Chapter 3
5.7
3.2
(a) PoC;(b)Co />; (c) SoC;(d)Co
(?)CoSoC;(/i)5
(POS).
We have, for example,
(f6°fs)(x)=f6
/
S; (e) C o C; (/) C o [i> • (P o C)];
1 \
1—x
=—
\ 1—xI
1
-1
1-JC
l-(l-jc)
and hence / 6 o / 5 = ^ . Proceed similarly with all other composite pairs to
show that, for all i and /, ft o fj belongs to the set of six given mappings.
Check your answers with the following 'composition table' of which the
interpretation is that fjOfj appears at the intersection of the row headed
by fi and the column headed by fj\
h h u fs n
/l
h k h u fs fe
k h /i U h h fs
h h U fs h fx U
h U fs h h h h
fs fs n fx h h k
f6 h h h fs u /l
o
56
/i
Solutions to Chapter 3
3.3
Let g i , £ 2 , £ 3 , # 4 • I R ^ I R b e defined b y
Then we have/ = g4 o g3 o g2 o gx.
3.4
(a) R is reflexive on X if and only if xRx for all x E X, i.e. if and only if
x = f{x) for a l l x G I , i.e. if and only i f / = id x .
(Z?) The criterion for symmetry is
xRy^yRx
which is equivalent to
y=f(x)=>x=f(y).
Suppose that R is symmetric and let x G I If /(JC) = J ; then from the above
we have x =f(y). But then / 2 (x) =f(y) =x and consequently f2 = id x .
Conversely, if/ 2 = id x then from >> = f(x) we deduce that/(j>) = / 2 (x) = x
whence we see that/? is symmetric.
(c) The criterion for transitivity is
(xRy and yRz) =• xi^z
which is equivalent to
(y =f(x) andz =f(y)) ^z =f(x).
Suppose that R is transitive and let xGX. Let f(x) —y and let f(y) = z.
By the above we have z=f(x). But z -f{y) = / 2 ( x ) . Hence we see that
f = f2. Conversely, if f = f2 then the conditions^ =f(x) and z =f{y) give
z = / 2 ( x ) =f(x) and hence 7? is transitive.
5.5
We are given that
f={(x,y)eXxX\y=f(x)},
g=
{(y,x)GXxX\y=f(x)}.
Now (a, b)Ggof if and only if there exists tGX with (a, t)Gfdnd (t9 b)Eg.
For this to be so, it is necessary and sufficient that/(a) =f(b)(=t).
we have
Thus
(a9b)egofof(a)=f(b).
Now since clearly f(x) — f(x) we have (x,x)Gg o f and sog o f is reflexive.
Also, if (*, y)<Egof then f(x) =f(y) whence f(y) =f(x) and (7, x)eg o / ,
so that # o / is symmetric. Finally, if (JC, y), (y, z)Gg o /then/(jc) =f(y),
f(y) = f(z) so/(x) = /(z) and (x, z) E#• o /, whence g o / i s transitive.
5.(5
The given relation is clearly reflexive since /(x) = / ( x ) holds trivially for all
x E A It is also symmetric since /(x) =f(y) gives f(y) =f(x). That it is
57
Book 1
Sets, relations and mappings
transitive is equally immediate :/(x) =f(y)
Now (x,y)Rf(l,
x
2
and /(j/)=/(z) imply that
0) if and only if fix,y) = / ( l , 0) = (1, 0), i.e.
y
„ = 1
and
. „
yj(x +y2)
y/(x2+y2)
It follows that the Rf -class of (1, 0) is
= o.
For a general (a, b) ^ ( 0 , 0) we have that (x, >>) belongs to the ify-class of
(a, b) if and only if
x
2
_
a
(a)
2
2
y
2
b
y/(x +y )~y/(a +b )
™ V(* + / ) ~V(^ + ^ 2 ) '
These conditions give j / x = &/# where x, a have the same sign andj, b have
the same sign. This last equation may be written y = (b/a)x. We conclude
that the Rf -class of (a, b) is the half-line from (but excluding) the origin
which passes through (a, b). For (a, b) = (0, 0) the ify-class is the singleton
3.7
2
2
x2<y
(i) See Fig. S3.1.
(ii) The domain is IR.
The image is {y G IR | y > 0}.
(iii) It is not a mapping.
(b) sin x = j>
(i) See Fig. S3.2.
(ii) The domain is IR.
The image is [—1, 1].
(iii) It is a mapping.
Fig.S3.2
/ ^
58
Solutions to Chapter 3
(c) x = sin y
Fig.S3.3
(i) See Fig. S3.3.
(ii) The domain is [—1, 1].
The image is IR.
(iii) It is not a mapping.
(d) x-2<y<x + l
(i) See Fig. S3.4.
(ii) The domain is IR.
The image is IR.
(iii) It is not a mapping.
(e) y = \x\
(i) See Fig. S3.5.
(ii) The domain is IR.
The image is {>>E IR
(iii) It is a mapping.
59
Fig.S3.5
y>0}.
Book 1
Sets, relations and mappings
Fig.S3.6
(i) See Fig. S3.6.
(ii) The domain is IR.
The image is IR.
(iii) It is a mapping.
(g)
\x\+y
=l
Fig.S3.7
y'
(i) See Fig. S3.7.
(ii) The domain is IR.
The image is {y G IR | y < 1}.
(iii) It is a mapping.
(h) x + | j ; | = l
(i) See Fig. S3.8.
(ii) The domain is {x E IR | x < 1}.
The image is IR.
(iii) It is not a mapping.
60
Fig.S3.8
Solutions to Chapter 3
Fig.S3.9
(i) See Fig. S3.9.
(ii) The domain is [—1, 1].
The image is [—1, 1].
(iii) It is not a mapping.
Fig.S3.10
(i) See Fig. S3.10.
(ii) The domain is [—1, 1].
The image is [—1, 1].
(iii) It is not a mapping.
-l
(k)
j ^ = IJCI — I J C J
(i) See Fig. S3.ll.
(ii) The domain is IR.
The image is the infinite union
[0,l[U]l,2]U]3,4]
U]5,6]U..
(iii) It is a mapping.
l
Fig.S3.11
\
\
\ :
61
x
Book 1
3.8
Sets, relations and mappings
(a)
(b)
2
+ 4y2 = 1 (Fig. S3.12)
The domain is [—1, 1].
The image is [—\, \
It is not a mapping.
x
Fig.S3.12
y
x2=y2(Fig.S3A3)
The domain is IR.
The image is IR.
It is not a mapping.
Fig.S3.14
(c)
(Fig. S3.14)
The domain is [0, 1].
The image is [0, \\.
It is not a mapping.
62
Solutions to Chapter 3
Fig.S3.15
(d)
(Fig. S3.15)
The domain is [0, 1],
The image is [—1, 1].
It is not a mapping.
(e) j> = 2 x - l ( F i g . S 3 . 1 6 )
The domain is IR.
The image is IR.
It is a mapping.
Fig.S3.16
f= {(Xiy) e (Q x Z | y is the least
integer withy >x) (Fig. S3.17)
The domain is (Q.
The image is Z.
It is a mapping.
Fig.S3.17
63
Book 1
Sets, relations and mappings
f= {(x,y) £ Z x (Q | x is the least
integer with x > y} (Fig. S3.18)
The domain is Z.
The image is (D.
It is not a mapping.
Fig.S3.18
The graph of the relation^ = |JC | is
Fig.S3.19
as shown in Fig. S3.19, whence that
of y = \x I — 1 is as shown in Fig.
S3.20, whence that ofy = \x - 1| - 1
is as shown in Fig. S3.21, whence that
of the function f(x) = 11x - 11 -11 is
as shown in Fig. S3.22.
Fig.S3.20
64
Solutions to Chapter 3
Fig.S3.21
y'
-l
Fig.S3.22
3.11
We have
2x2 4- 6x + 7 < x + 5 o 2x2 + 5x + 2 < 0
and so the required set is [—2, —^
5.72
For every * G IR define
Then clearly (g + h)(x) = g(x) + h(x) = f(x)
65
for every x G I R
and so
Sets, relations and mappings
Book 1
g + h=f. Moreover,
g(-x) = \[f(-x)+f(x)]=g(x)9
3.13
We have
f[g(x)]=f(2x-l)
=
2
g[f(x)] =g[(x + I) ] = 2(x + I)2 - 1 = 2x2 + Ax + 1,
and so
/ [g(x)] = g [f(x)] * 2x2 - Ax - 1 = 0
The set A H B n C n D is the subset shown in Fig. S3.23.
Fig,S3.23
3.14
S is the square shown in Fig. S3.24.
d(x,y) is the distance from (0, 0)
to (x,y). The maximum distance
from (0, 0) to a point (x,y) on S is
clearly 1, and the minimum distance
is l/\/2. Since all values between
these extremes are attained we have
thatlm<7= [1/V2, 1].
Fig.S3.24
-l
66
Solutions to Chapter 3
3.15
S is the shaded region shown in
Fig. S3.25.
f(x,y) is the distance from (0, 0)
to (x,y). The maximum distance
from (0, 0) to a point (x, y) on S is
clearly 1, and the minimum distance
is \\\Ji. Since all values between
these extremes are attained we have
t h a t l m d = [1/V2, 1].
Fig.S3.25
-1
3.16
3.17
S is the circle shown in Fig. S3.26.
f(x,y) is the gradient of the line
joining (0, 0) to (x,y). The maximum value occurs when tan & = f
and the minimum value when
tan d- = — I. Hence we see that
Fig.S3.26
y'
Fig.S3.27
S is the circle shown in Fig. S3.27.
yl
f(x,y) is the distance from (0, 0)
to (x, y). Now the radius of the
circle is 2, and the centre of the
circle is at the point (3, 4) which is
at distance 5 from the origin. Thus
4the maximum distance from the
origin to a point on S is 7, and the
minimum distance is 3. Consequently,
Im/=[3,7].
67
Book 1
3.18
Sets, relations and mappings
(a) Injective, since 7x\ + 1 = 2x2 + 1 implies xx = x2. Not surjective: for
example, ^ G 5 = (D but there is no x E Z with/(x) = \.
(b) Injective since (xx — 1, 1) = (x2 — 1, 1) impliesxl = x2. Not surjective:
for example, n o x G Z maps to (0, 0).
(c) Injective, in fact
x\ —x\ =(x1 — x2)(xl + xlx2
+x\)
and so if x\ = x\ we must have either Xi = x2 or x\ + X\X2 + x\ = 0, and
the only solution to the latter equation in IR is xx = x2 = 0. Also surjective:
given anyj G 5 = IR we have/Xjy173) = y.
(d) Not injective: for example,/(I) = / ( i ) = 1. Not surjective: there is
no c E C such that/(c) = —1 since |c| > 0 for all c E (D.
(e) Not injective: for example, /(0) =/(TT). Surjective; probably the best
way to prove this is to use continuity : the function * ->x sin x is continuous
and, given any k E IR we can find xx E IR with f(xi) > A: (take for example
= 2rii7T + \n > /:), and x 2 E IR with
(take for example
*2 = 2«27T + 2n < /:), whence by continuity/(x) takes the value k.
(/) Not injective: for example, / ( l ) = / ( 0 ) = 0. Surjective; in fact the
graph of/is as shown in Fig. S3.28, from which we see that every line parallel
to the x-axis cuts the graph at least once.
Fig.S3.28
/(*)'
(g) Not injective: for example, /(0) =/(—1) = 1. Not surjective: for
example, there is no x E IR such that JC 2 +JC + 1 = 0.
3.19
(a) H yGf(A HB) then J ; = / ( J C ) for some xGACiB. From x<EA we
have y E/(y4); and from x <EB we havej> E / ( £ ) . Hence y €f(A) C\f(B).
68
Solutions to Chapter 3
(b) Since ACAUB
it is immediate that f(A)Cf(AUB).
Similarly,
f(B) Cf(A U B) and hence f(A)Uf(B) Cf(A U B). To obtain the reverse
inclusion, let y£f(A U B). Then y = f(x) for xGAUB. If x EA then
y ef(A), and if x EB theny G/(5). In either case, >> G/C4) U/(£).
(c) Suppose that / is injective. In view of (a), we need only prove that
f(A)nf(B)Cf(AnB).
Suppose that yEf(A)nf(B). Then y=f(x1)
for XiEA and y = / ( x 2 ) for x2GB. Since / is injective by hypothesis,
we havexx = x 2 G^l Pi5 and so>>G/(y4 O5).
Conversely, suppose that f(A)Df(B)Cf(A
C\B) for all subsets A, B of X.
lff(xl)=f(x2)=y
then from j> =f(xl)Gf({xl})
and j ; = /(* 2 )G/({x 2 })
we deduce that ^^/({xx) n{x 2 }), whence it follows that {xx} n{x 2 } ^ 0
and hence JCX = x2.
(J) Suppose that / is surjective. If .y G F\/(^4) then there exists x with
fix) =y, and clearly x GX\A. Thus .y =f(x) Gf(X\A)
and consequently
Conversely, suppose that Y\ f(A) Cf(X\A)
for all subsets A of X.
Then in particular we have F \ / ( 0 ) C / ( X \ 0), i.e. YCf(X). It is immediate
from this that every y G Y is of the form / ( / ) for some t G X, so that / is
surjective.
0 ) Suppose that/is a bijection. Then by (d) we have/(AT \ 4 ) D F X / U )
for all subsets A. It therefore suffices to show the reverse inclusion. Suppose
then thatj> Gf(X\A). We havej> =f(x) for some J C G X \ ^ 4 . Now we cannot
have y =f(a) for any a GA; for / is injective and f(x) =f(a) would imply
x — a where x G X \ A and a G A, a contradiction. Hence we see that j> ? f(A)
and consequently j> G Y\ f(A).
Conversely, suppose that f(X\A) = Y\ f(A) for all subsets A of X.
Then / is surjective by (d). To show that / is also injective, suppose that
xlizx2. Then
f(x1)ef(X\{x2})=Y\f({x2})
and SO/(JCI) £ /({x 2 }) which shows that/(Xi)
3.20
(a) LetfG(A x £ ) c and define/x : C->AJ2 :B->A by
fi(x) = first component of f(x);
f2{x) = second component of f(x).
Define v?: (A x B)c -+AC x Bc by <p(f) = (/x ,/ 2 ). Then y is clearly a bijection.
(b) If/G (yl B ) c t h e n / : C~>AB and, for each c G C,/(c) : 5 ^ A Thus
(Vcec)(vbeB)[f(c)](b)eA.
69
Book 1
Sets, relations and mappings
We can therefore define a mapping y: (AB)C -+ABxC by / -»</>(/) where
(Vc e C)(V6 6 5 ) [<p(/)](Z>, c) - [/(c)](&).
Clearly, y is a bijection.
(c) Let f<EABUC, so that f:BUC->A.
Let fx\B-*A
be given by
(VbeB)fl(b)=f(b)
and let / 2 : C->,4 be given by (VcGC)/ 2 (c)=/(c).
Now define ^ : y 4 s u C - > ^ 5 x ^4C by <p(/) = (/i, h). Then ^ is clearly
a bijection when B 00=0.
Although B H C= 0 is a sufficient condition for the existence of a bijection
from ABUC to v4B x ^4C, it is not a necessary condition. For example, take
A=B = C={1}; we have ABUC={f}
where / is described by 1 -> 1,
B
c
A = {f} = A , and there is a bijection from {/} to {/} x {/}.
3.21
(a)
(b) x -> 2x.
(c) Let A be the point (1, —1) and let B be the point (—1, 1). For
XiG [0, 1[ let f(xi) be the point of intersection on the j>-axis of the line
through A and Xi\ and for x2 €=] —1, 0] let/(x 2 ) be the point of intersection
on the j-axis of the line through B and x2. This describes a bijection / :
]—1, 1 [-> IR; as is readily verified, for^xE [0, 1 [ we have/(xi) = # i / ( l — Xi)
and forx 2 G ] - l , 0] we have/(x 2 ) = x2l(l + x2). See Fig. S3.29.
Fig.S3.29
14 \
B
\
\
\
-1
-1
»
\A
(d) Define/: [0,1]-* [0, 1 [ by
if
/(*) =
70
JC=
1/2";
otherwise.
Solutions to Chapter 3
See Fig. S3.30.
Fig.S3.30
0
i
(e) Note that in (d) the formula for / can be extended to negative values
of x thereby providing a bijection from [—1, 1] to ]—1, 1 [. Composing this
with the bijection in (c), we obtain a bijection from [—1, 1] to IR.
3.22
Using the formulae given, we have
(4* + l)
*[/(*)] =
g(x)
f 12x4-3
+ 3
x>0;
x<0JC>0;
JC<0.
The graph ofg o / i s as shown in Fig. S3.31.
Fig.S3.31
71
Book 1
Sets, relations and mappings
Since each line parallel to the x-axis meets the graph exactly once it
follows that g o / is a bijection. The graph of (g o f)'1 is the reflection in
the line y = x of the graph ofgof It is immediately seen from this that
_t
( x-3
x<3;
l^(x-3) x>3.
As for/o^, we have
, /(3x)
f[g(x)] = l
13
x>0;
-3<x<0;
The graph o f / o ^ is as shown in Fig. S3.32.
Fig.S3.32
It is clear from this that fog is neither injective nor surjective (since
there are lines parallel to the jc-axis that meet the graph more than once,
and others that do not meet the graph at all).
3.23
Applying the formulae we have
x>0;
fix-I)
JC<0,
l-x
x>0;
x<0.
(x-l)
2
The graph o f / o g is therefore as shown in Fig. S3.33.
72
Solutions to Chapter 3
Fig.S3.33
Since each line parallel to the x-axis meets this graph precisely once, it
follows that / o g is a bijection. The formula for ( / o g)'1 is
l-x
x<l;
l—y/x
x>\.
As for g o / we have
1gO- -x)
2
g(x )
—x
= . l-x
x2
The graph of g o
0<JC<1;
x<0.
n 5 as shownin Fig. S3.34.
Fig.S3.34
73
x>0;
x<0,
x>l;
Book 1
Sets, relations and mappings
It is clear from this that g o / i s neither injective nor surjective.
3.24
Since f(x) = x \x I we have
x2
/(*) =
-x2
if
if
x>0;
x < 0.
The graph of/is therefore as shown in Fig. S3.35.
Fig.S3.35
/(
Since every line parallel to the jc-axis meets the graph exactly once, / is
a bijection.
For x > 0 we have y = x2 and so x = \Jy\ and for x < 0 we have j> = — x2
and so x = —y/{—y). Hence/" 1 is given by
Vx
if
The graph of g is as shown in Fig. S3.36.
Fig.S3.36
74
Solutions to Chapter 3
g is not surjective : for example, the line j> = —1 does not meet the graph.
g is not injective : for example, the line>> = 1 meets the graph twice.
Hence g is not a bijection.
3.25
In general, the cubic x -> x3 + ax2 + bx 4- c has a graph of the form shown
in Fig. S3.3 7.
Fig.S3.37
fix)
For this function to be a bijection, it is necessary and sufficient that every
line parallel to the x-axis meets the graph precisely once. This is so if and only
if the gradient of the function is greater than or equal to 0. Now the derivative
is 3x2 + 2ax 4- Z?, which is a quadratic. The condition that / be a bijection is
therefore (WxG\R)3x2 + 2ax + b>0. Now the quadratic x->3x2 + 2ax + b
attains its minimum value when 6x 4- 2a = 0, i.e. when x = —\a, this minimum
value then being — \a2 + b. This is greater than or equal to 0 if and only if
a2 < 3b.
3.26
=f(x2) then from
1
1
l+xj
l+x22
we deduce x\ = x\ whence X\ = ±x2. Since xu x2 £ IR+ we have xx = x2 and
so / i s injective.
Since / is injective, the assignment x -*f(x) defines a bijection & : IR + -*
I m / a n d the inverse of this bijection is given by
To find two distinct mappings #, h :IR + -HR + such thatg o / = h o / = i d J R + , it
suffices to find two mappings which agree with # - 1 on Imf. Thus, for example,
75
Book 1
Sets, relations and mappings
we can consider
0
x
1
1)
1
if
x=0
or
if
0«< x < l .
if
x=0
if
0 < J C < 1.
> 1;
or
h(x) =
~-l|
3.27
The graph of / is as shown in Fig. S3.38. Clearly, every line parallel to the
x-axis meets the graph precisely once, so / is a bijection. The derivative
Fig.S3.38
fix)
f of/is given by
Since, for example,/'(0) = / ' ( l ) = 0 we see t h a t / ' is not a bijection.
3.28
The mapping can be described by
1 2 3 4 5 6
7
...
0
1 - 1 2 - 2 3 - 3 . . .
which suggests that/is a bijection whose inverse is given by
76
Solutions to Chapter 3
[
if
if
2m
=
m>0;
m<0.
Now with this definition of/"1 we have
2m
I (-l))2mlm}
J U
V/'vJ
\,
1>v2|m
I+11
I
|W2|
\y—1)
m
-\m\
if
T
m>0;
^* r\
II
YYl ^s U,
if
m\
if
m:
= m.
Consequently we see that / o / - i = i d z and so / i s certainly surjective. Since
/ is clearly injective by definition, it follows that / is indeed a bijection with
Z"1 as described above.
3.29
Since g(x) = 3 + 4x the formula ^ ( x ) = (4n - 1 ) + 4wx certainly holds for
n = 1. For the inductive step, suppose that gn(x) = (4n — 1)4- 4nx. Then
gn + l(x)=g[gn(x)] = 3 4- 4[(4" - 1 ) + 4nx]
A _L yl^ "Mv
- 1 ) + 4 W+1 JC.
For the last part, we have
1
4"
= 4-"x + ( 4 " " - l ) ,
whence we see that the formula holds also for negative integers n.
3.30
For n = 1 the formula is trivial, and for n = 2 it reads
For the inductive step, suppose that the result holds for n subsets. Then
using the result for two subsets we have
[H + l
I
\ I
n
= (lU/
* ii+ii
/=i
77
n +l
Book 1
Sets, relations and mappings
Using the formula again, this last component (which is a union of n sets)
can be written
l
p<q
which, when taken with the other terms, gives the required formula for
n 4- 1. This completes the inductive step.
\i \A\ — m and \B\ = n then in defining a mapping from A t o B we have,
for every element xofA,n choices of image in B. Thus there aren m mappings
from A to B.
Suppose now that A = {ax, a2,..., am) and that / : A ->B is an injection.
There are « possible images of 0l9 then « — 1 possible images of #2> and so
on. Hence there are
n\
\)(n - 2) • • • (n - m + 1) = —
ml
injections from A to B.
To determine the number of surjections, let A = {ai,a2, • • -,tfm} and
5 = {Z?l5 b2, •. •, * w } . For each ^ G 5 let
n(n-
Then the number of surjections from A to B is the total number of mappings
from A to B less |L)p =1 ^^|, i.e. it is nm - |U? = ii4/|. It is at this stage that
we require the general formula established above. To apply this, we observe
that \At\ is the number of mappings from A to B\{bi}9 {AfHAj-l is the
number of mappings from A to B \{bh bj}, etc. Thus we have that
\At\ = (n- l ) m ,
\AtC\Aj\ ={n- 2 ) m , . . .
Since we can remove i elements of B in precisely (") ways, each sum
2 | (^iterms^k I contains (") terms equal to (n — i)m and the result follows.
3.31
Let A = B = C = [0, 1 ] and let each mapping to be given by f(x) = \x. Then
/is an injection which is not a bijection and which has the required properties.
For another example, consider A = [0, 1] x [0, 1], B= [0, 1] x [0, §],
C= [0,|] x [0, l]zndf:A^B,g:
B-+C,h :C-+A given by
f(x,y) = Or, \y), g(x,y) = (y,x), h(x,y) = (x,y).
3.32 Use calculus to determine the graph o f / (Fig. S3.39). From this graph we
see that I m / = [—2, 2] and that the required interval^ is [—1, 1]. To find
g~l we set y — 4x/(x2 + 1), so that yx2 — 4x + y = 0, and solve this quadratic
78
Solutions to Chapter 3
Fig.S3.39
for x in terms of y. We get
4±V(16-4y2)
if
x—
y
0 if y = 0.
This suggests either that
y
o
if
}
if
y = 0,
or that
if
0
if y = 0.
The first possibility is excluded since, fory E [—1, 1], we have
- ? [-2, 2].
3.33
The graph of/is shown in Fig. S3.40.
79
Book 1
Sets, relations and mappings
Fig.S3.40
fix)
Since, for example, /(3) =/(—1) = 0 we see that / is not injective.
Im / = {x G IR \x < 4}. Consider ,4 = {x G IR | x > 1}. Clearly, the restriction
of / to A is a bijection from A to Im /. To determine the inverse of this
bijection we must find the solution of y — 3 + 2x — x2 which lies in A. The
solutions of this quadratic equation are x = 1 ±yJ(A—y) of which only
1+V(4— y) lies in A. Hence the required inverse is described by
3.34
Note from the definition of / that /(ft) is divisible by p if and only if n is
divisible by p. Thus if/(ft^ = f(n2) then either /(fti) is divisible by p, in
which case n1 =f(n1) — p z=f(n2) — p = n2, or /(fti) is not divisible by
p, in which case nx = /(fti) = f(n2) = n2. Hence/is an injection.
To show that / i s also surjective, suppose that fcG Z. If k is not divisible
by p then f(k) = k\ and if A: is divisible by p then so is k—p whence f(k—p) =
(k — p) + p = k. Hence / i s a surjection.
As for/" 1 : Z ->• Z, this is given by
divisible by p\
{ n—pft if ft ifis ftnotis divisible
by p.
3.35
It suffices to find g : IR -> IR such that / o # = # o / = id )R . Then / i s
a bijection with/" 1 = g. Now if jy = x 4 we have JC = j>1/4 so we can define
g(x) = x1/4
Also, if y = x(2—x)
80
if
x>0.
then x 2 — 2x +y = 0 and we have x = 1 ±V(1 ~~y)-
Solutions to Chapter 3
Since 14-\J{\ — y)> 0 we must choosex = l — y/(l—y)
g(x)=l-y/(l-x)
and define
if x<0.
Then it is easy to check that / [g(x)] = x and g[f(x)] = x for every x G IR,
so that g is the inverse off.
3.36
(a) Since we have 1 = / ( 2 ) < / ( 3 ) = 2 the result holds for i = 2. By
way of applying the second principle of induction, suppose that the result
holds for 2 < i < n. Then we have
/(H + l) =/(/!) + / ( * - 1 )
whence we see that it also holds for n + 1.
0 ) I f / [ / ( 0 ] = / ( 0 then, by (a), if / > 2 we must have / ( / ) = i and the
only possibility is / = 5; and if / < 2 then it is readily seen that the possibilities
are/ = O, 1,2.
(c) For n = 1 we have / 5 = 5. Assume by way of induction that the
result holds for n. Then
whence it also holds for n 4- 1.
(d) Since / 3 = 2 = 1 + / i , the result holds for n — l. Assume by way of
induction that it holds for n, so that/,2 + 2 = 1 + 2f =1 ^. Then we have
f
f
+ fn + 1
= 1+ X //+/„
1
1=1
whence it also holds for n + 1.
(e) The result is clearly true for n = 1. Assume by way of induction that
the result holds for n, so that fn^xfn+1 —f\= (—1)". Then we have
fnfn + 2 ~fn + l ~ fnifn ~*~fn + l) ~~fn + l
=
fn +fnfn + l ~fn + l(fn + fn—l)
81
Book 1
Sets, relations and mappings
— f2
f
i
~ J+
f
l J l
whence it also holds for n 4- 1.
(/) The result clearly holds for n = 1. By way of applying the second
principle of induction, suppose that it holds for all n < k. Then we have
= 2(1+V5) f c -2(1-V5) f c
+ 4(1 + V5) fc-1 - 4(1
whence we see that the result holds also for n = k + 1.
337
Suppose thatg(l) = 3 and t h a t / o ^ = ^ o / Then we have/[^(l)]=/(3) = 4
andhenceg[/(l)]=4, i.e.#(2) = 4. Now/[#(2)] =f(4)= 1 sog[/(2)] = 1,
i.e. £(3) = 1. Finally, /[g(3)] = / ( l ) = 2 so g[/(3)] = 2, i.e. g(4) = 2. Thus
we see that g is determined uniquely and is given by
g(x) = x + 2 (mod 4).
A similar argument shows that if h(l) = 1 then h is uniquely determined,
and/* = idj^.
3.38
(a) Clearly such a mapping g must satisfy a [g{0)] = 0. But there is no
m G IN for which a(m) = 0. Hence such a mappings cannot exist. For every
p G IN let kp : IN -> IN be given by
(n-1
if
n>\;
kJn)
=
p
\ p
if n = 0.
Then for every m G IN we have
kp[a(m)] = kp(m + l) = m,
so that each mapping kp is such that kpo a = id )N . Clearly there are
infinitely many such kp.
(b) Suppose that there exists / : IN->• IN such that / o |3 = id|N. Then
we have
3=/[0(3)]=/(l)
82
Solutions to Chapter 3
and from this contradiction we conclude that no such / can exist. For every
p G IN let kp : IN -> IN be given by
kJn)
=
p
f 2i2 + l
\ 2p
For m ^ p w e have
if
n±p\
if
n=p.
and for m = p we have
Thus ]S o /:p = id|N • Clearly there are infinitely many such kp.
3.39
3.40
Suppose that there exists x€A such that f(x) = X where, by definition,
X = {aE:A \a £/(#)}. Then if JCE/(JC) we have xGX whence the contradiction x&f(x). Also, if x&f(x) then we have x&X whence the contradiction xGf(x). We conclude that there can be no x GA with f(x) = X. It
is immediate from this observation that / cannot be surjective. There does
exist, however, an injection/: A -+ ?(A): for example, x -+f(x) = {x}.
(a) It is helpful to compute a few of the Ik. For example,
h={\l
/2 = {2,3}, / 3 = {4,5,6}, 74 = {7,8,9,10}.
Now by the definition of Ik we have \k(k + 1) GIk and \k(k + 1) - k =
\k(k — 1) ? Ik. It is clear from this that Ik consists of the k elements
Consequently we see that the Ik form a partition of IN*.
(b) We observe that
\(m + n- 2){m + n - 1) + m = \{m + n)(m + n - 1) + 1 - n
and that this belongs to Im + n^.l since the largest integer in Im + n_1 is
\{m + ri)(m + n — 1) and 7m + w _! contains m + ^ — 1 integers. Thus
Since / ( p , #) G / p + Q _ i and the 4 fa™ a partition, we have
f(P>Q)
^Im+n— l ^ ^ p + Q—1 = Im + n—1
^p + g - 1 = m +n— 1
To show that / is injective, suppose that / ( p , #) = / ( m , «). Then p + q =
m + n and so
83
Book 1
Sets, relations and mappings
= f(rn9n)
n- 2)(w + n -1) + m
Consequently we have that m=p, and hence also q = n. Thus (p, q) = (m, n)
andso/isinjective.
(c) We note that if 1 < r < fc then
/(r,fc+ 1 - r) eir+(k+l_r)_1
= 4.
But / is injective, Ik contains k elements, and there are k values of r; therefore we must have that every element of Ik is the image of some element
(r,k + l—r) under /. Since this is true for every Ik in the partition, it follows
that/is also surjective.
3.41
Consider the mapping f:S-*S
/(*)=•
defined by
l/x
if
x G ] - l , l[and*=£0;
0
if
x = 0;
—1/JC if x g ] - l , l [ .
It is readily seen that for every x E S we have / [f(x)] = —x. The graph of
/ i s shown in Fig. S3.41.
Fig.S3.41
/(*)•
-1
84
Solutions to Chapter 3
3.42
Since go f is a bijection we have that g is surjective; and since h o g is
a bijection we have that # is injective. Thus g is a bijection. Since g o f=k
where k is a bijection, we thus have that f = g~l o A: is also a bijection.
Likewise, since h o g = m where m is a bijection, we have that ft - m o g~lis
also a bijection.
3.43
Let a = hcf (a, b). Then we have a = da, b = Z?'a and a/Z? = d/b\ the latter
quotient being 'in its lowest terms' in the sense that hcf (a', b') = l. Similarly,
if /3 = hcf (c, d) then c — c'j3, d — d'$ and c/d — c'/df the latter being in its
lowest terms. Thus, if a/b = c/d we have a jb' — c jd' so that either a = c\
b' = d' ox a' — —c\bf = —df. It follows that
a +b
a +b \ = \c +d \ =
c+d
The first part of the question is precisely the condition that is necessary
to ensure that the given prescription defines a mapping from (Q+ to itself.
This mapping is not a bijection. For example, it fails to be injective: we
have/(2/3) = 5 / l = / ( 3 / 2 ) .
3.44
(a) Take, for example, / = idjp and g = —idjp. Then (f + g)(x) = 0 for
every x G IR and so / + g is not a bijection.
Consider n o w / = idjp and defineg : IR -> IR by
1
-
g(x)= x
if
JC^O;
0 if x = 0.
Theng is a bijection; its graph is shown in Fig. S3.42.
Fig.S3.42
85
g(x)
Sets, relations and mappings
Book 1
Now
1
x • - = 1 if xi=0;
(f'g)(x)=f(x)g(x) = ( x
0 - 0 = 0 if x = 0,
sof-g is not a bijection. For this/andg we also have
1
+
(/+*)(*)= * * ^ * ^ ° '
0
if * = 0.
The equation x 4- 1/JC = 1 has no solution in IR, so there is no x G IR such
that ( / + g)(x) — 1. Thus/ + g is not a bijection.
(b) Suppose that / i s a bijection. Then for X ^ 0 we have
and so X/ is injective. Also, since/is surjective, given any y G IR, there exists
t G IR with f{t) = y/X. Then (X/)(0 = y and hence X/is also surjective. This
shows that if / is a bijection then so is X/for every X =£ 0. Suppose now that
X/is a bijection with X ^ 0 . Applying the above result to X/and/i = l/X,
it follows that ju(X/) = (1/X)(X/) = / i s also a bijection.
(c) Consider, for example, the mappings f,g: IR IR given by f(x) = x2,
g(x) = x + 1. We have
[fg](x)=f[g(x)]-g[f(x)]
so
[fg] is a bijection.
(J) We have
llfg]h]=[(fog-gof)h]
=
(fog-gof)oh-ho(fog-gof)
=fogoh-gofoh-hofog+hogof
and similarly
= hofog-fohog-gohof + g
Adding these together, we obtain the required result.
86
Solutions to Chapter 3
3.45
(a) We have that
(a + ib)R(c + id)
yj(a2+b2)
yj(c2+d2)
from which it is immediate that R is an equivalence relation on (D*. Now
R can also be expressed in the form
Re z Re w
\z\
\w\
from which we see that zRw=>zRw. We now observe that (z/\z\)Rz for
every zGC*, where w = z/\z\ lies on the circle \w\ = 1 and so can be written
w = e1^. Since Re w/\w\ = cos #, it is readily seen that in the Argand diagram
the R-class of z E (D* consists of a pair of half-lines emanating from the origin
(with, of course, the origin deleted), one of these lines passing through z and
the other through z (Fig. S3.43).
Fig.S3.43
(b) Suppose that x = a 4- Ib and y = c + id in C* are such that xRy.
Then we have
a2+b2
c2+d2
from which it follows that
2c2
a2-b2_
2a2
- 1 =a2+b2~a2+b2
87
•
-
!
=
•
c2-d2
' c2+d2
Book 1
Sets, relations and mappings
Now x2 = a2 — b2 4- 2\ab and \x21 = \x | 2 = a2 + b2 and hence
x2Ry2
Rex 2
2
\x2\
a2-b2
\y2\
c2~d2
a2 + b2
c2+d2
Consequently we see that xRy => x2Ry2.
It is clear from this that{([x]#, [x 2 ]^) | xG (D*} describes a mapping
/ : U-+U. As observed above, we can represent every R-class as [el^]R
where 0 < # < 7 r . Using this representation, we see that this mapping / is
surjective: for clearly [e 1 ^]^ =/([e l d ^ 2 ]^). It is also injective since if
[^]R = 1^]R then, by the definition of R,
cos 2# = Re e 2i * = Re e2i<^ = cos 2y
whence # = </? since each belongs to [O,TT], and consequently [e ld ]# = [e1</3]#.
(c) (i)f={([x]R,
[2x]^)UG(D*} is also a mapping. Indeed, if x=a + \b
then
Re x
a
2a
Re 2x
+ 4b2)
\2x\ '
so that [;*:]# = [2x]^. In other words,/is the identity mapping.
( i i ) / = {([*]#, [X + 2]JR») |xE(D*} is not a mapping. For example, we
have [1 + i ] * = [2 + 2i]# but [3 + i ] ^ =£ [4 4- 2i]^ since
Re(3 + i)
3
Re(4 + 2i)
2
|3 + i|
(iii)/= {([x]R, [x" 1 ]^) | x G €*} is a mapping. Indeed, if JC = a + iZ? E (D*
we have JC"1 = (a — \b)/ \x\2 and so
Re x~l
a
1
a
Re x
\x~x\ ~\xT2 \x~ri~\x~\~~\x~\'
hence/is the identity mapping.
3.46
(a) The mapping / is not injective: for example, / ( I ) =/(10) = 1. It is
surjective, however, since /(0) = 0 and for every « > 1 we have n =f(m)
where m = 11 . . . 1 (n terms).
(b) n G [1]R if and only if f(n) = / ( l ) = 1, which is the case if and only
if n = 10* for some/E IN.
If x E [1]*. then A x ) = 1 so if/ > i we have
88
Solutions to Chapter 3
and hencex G [1]#..
If xG [n]R.thenfi(x)=fi(n) so
W
whence x G [ft]#.. Thus [«]#. C [ft]#. whenever i </.
[l] jRi ¥= [ I ] * 7 since, for' example, /(55) = 10 so / 2 (55) =/(10) = 1
whence 55 G [l]# 2 ;but clearly 55
^
(c) Since 10w = 1 (mod 9) for every n, it follows that if
1
m = 10"xw + l O " " " ^ - ! + - • • + IOJCJ + x 0 ,
where each xt G { 0 , 1 , . . . , 9}, then m is divisible by 9 if and only if f(m) =
x n 4- x n _ ! 4- • • • 4- x0 is divisible by 9. Writing /(m) in terms of the base 10
and applying this observation again, we see that m is divisible by 9 if and only
i f / 2 ( m ) is divisible by 9. Applying this argument repeatedly we eventually
reach some fl(m) = 9 =/ I '(9). Thus m is divisible by 9 if and only if
raG [9]^.for some/.
A similar argument holds for divisibility by 3. In this case, however, at
the final stage of the argument we reach some fl(m) G {l, 2 , . . . , 9} which is
divisible by 3. The only possibilities for f\m) are 3, 6, 9 and so we conclude
that m is divisible by 3 if and only if m G [3]R. U [6]R.U [9]^. for some /.
(d) The reflexivity and symmetry of R are obvious. For transitivity, note
that if xRty then xRjy for every/ > I Suppose then that xRy and>>i?z. Then
xR{y and yRjZ for some f, /'. If t = max {/, /} it follows that xRty andyR t z
whence xRtz and consequently xRz.
Since / i s surjective, so also is/* for every I It follows that Rt has infinitely
many classes. However, for every n we have nRfk for some k with 1 < k < 9
and some large enough /. Hence nRk for some k with 1 < f c < 9 , and there
are nine R-classes.
89
Test paper 1
Time allowed: 3 hours
(Allocate 5 marks for eachvl question; 20 marks for each B question.)
Section A
Al
VLA,B,C are subsets of a set E prove that
A nBCC^ACCUB'.
A2
If A, B are subsets of a set E simplify the expression
AL [Bn(ALB')].
A3
Express as a union of intervals
3*
I*GIR\{I}
A4
1 -x
< 8 + 3x .
Draw the graph of the relation p defined on IR by
xpy o (y2 < \x\< 1 and* 2 < | ^ | < 1).
A5
Determine which of the following relations, defined on (D, are equivalence
relations:
(a)
xRy*>x-y=y-x;
(b) xSy<>x—y=x—y.
A6
A relation ~ is defined on (D \ {0} by
a~bo-e\R.
b
Show that ~ is an equivalence relation and describe the ~-classes.
A7
L e t / : IR-> IR be given by
90
Test paper 1
-
if
/(*) = '
1
if x = - 2 .
Prove that/is a bijection and find its inverse.
A8
Given mappings f.A^B
and g : B^C prove that
(a) if g o / i s injective then/is injective;
(&) if g o / i s surjective theng is surjective.
Section B
Bl
In a group of 75 students, each of whom studied at least one of the subjects
mathematics, physics, chemistry, it is known that 40 studied mathematics,
60 studied physics and 25 studied chemistry. Only 5 studied all three. Show
that
(a) at least 25 studied mathematics and physics;
(b) at least 10 studied physics and chemistry;
(c) at most 20 studied mathematics and chemistry.
B2
Let IN* = {1, 2, 3 , . . . } . Define a relation ~ on IN* by
Show that ~ is an equivalence relation and that in each ~ -class there is
precisely one odd integer. If a, b £ IN* are in the same ~-class, prove that
either a divides b or b divides a.
Let ao,ai, ...,an be n + 1 distinct positive integers each less than or
equal to 2n. Prove that, for some /,/ with / ¥=j, at divides a^.
B3
Let a, b, c, d G IR with c =£ 0. For x =£ —d/c establish the identity
ax + b a ad —be
1
ex 4- d
c
c2
x 4- d/c
Deduce that if X = IR \ {-d/c} and if/: X^
IR is given by
ex + d
then/is either a constant mapping or is injective. What is Im/in each case?
91
Test paper 2
Time allowed: 3 hours
(Allocate 5 marks for each^l question; 20 marks for each B question.)
Section A
Al
Let ,4 ={(x,y)G IR x IR | x2 + y2 - 6x + Ay + 14 = 0} and let £ = {(
IR x IR \x2 4- xy + y2 4-1 = 0}. Prove that A = B.
kl
Forn G IN d e f i n e n Z = { n x \ x G Z}. Prove that
4 Z A 6 Z = [ 4 ] 1 2 U [6] 1 2 U[8] 1 2 .
A3
Express as a union of intervals
jxSIR\{-l,2}
\
A4
>j
2
—x — 2
4
Let R be the relation defined on IR by
Sketch the graph of R.
A5
Draw the Hasse diagram of the order relation < defined on the set of positive
divisors of 300 by
a<b^a
A6
divides b.
Prove that the relation = defined on IR by
x=yox—yGX
is an equivalence relation. Using the decimal representation of real numbers,
show that for every xG IR the = -class [x] contains one and only one real
number in the interval [0,1 [.
92
Test paper 2
A7
Let f:X->Y
be a mapping. For CCY define the subset Cf of X by
xeCfo f(x) G C. Prove that if A, B C F then
04 C\B)f = Afr\Bf.
A8
The map / : IR -> IR is given by f(x) = ax + b for constant real numbers
a, b. For what values of a and Z> i s / a bijection w i t h / o / = idjp ?
Section B
Bl
Show that the relation p defined on IR x IR by
(xl9yl)pipc29y2)ox1{xl +y\ + 3) = x2(x? + .y2 +3)
is an equivalence relation.
Show that distinct points (x, 0) and (z, 0) belong to the same p-class if
and only if xz = 3. Find the p-class of (0, b). Describe geometrically the
p-class of (#, b) when a
B2
id) Show that
x2 + xy + y2 + 6x + 6y + 14 =
(JC
+ \y + a) 2 4-1(> + Z>)2 + c
for some a, b, c € I R . Deduce that the mapping / : IR ^ IR defined by
fix)
= x3 4- 6x2 + 14JC + 3 is injective.
ib) Show that the mappings :IR\{—1,1}->IR given by
x2 + 3x + 1
x —1
is surjective. Find{x | gix) = 1 } .
B3
Let s : IR ->• IR and t : IR -> IR be the mappings given by six) =x2 and
tix) = xix — 4). Compute s o t and r o s . Sketch the graphs of these composites and show that neither is an injection or a surjection.
Let f=t o s. Find the smallest kG IR such that # : [fc, °°[-» IR given
by #(x) =fix) for JCE [A:, °°[ is an injection. Find a mapping <p : Im #->
[/:, °° [ such that # o cp = id Im ^.
93
Test paper 3
Time allowed: 3 hours
(Allocate 5 marks for each A question; 20 marks for each B question.)
Section A
Al
If A, B, C are subsets of a set E prove that
(AAB)\C = (A\C)A(B\C).
A2
If A, B, Care subsets of a set E such that A UB=AUC
prove that B = C.
and A
DB=AnC,
A3
Express as a union of intervals
3
A4
Give two examples of relations that are both reflexive and symmetric but
not transitive.
A5
Let R be the relation defined on IR by
Sketch the graph of R.
A6
For every r G (Q with 0 < r < 1 let Ar = {x G <Q | x - [x] = r}. Prove that
{,4,. I 0 < r < 1} is a partition of (D.
A7
If/: IR\{0, l}-> IR\{0, 1} is given by/(jt) = 1 - 1/JC w h a t i s / o / o / ?
A8
Prove t h a t / : IN x IN -> IN given by
94
Test paper 3
0
m
\2 -\2n-l)
if
otherwise,
is a bijection.
Section B
Bl
Given / : A -+B, let Rf be the relation defined on A by
xRfyof(x)=f(y).
Prove that Rf is an equivalence relation.
Let IR* = I R \ { 0 } and d e f i n e / : IR* x I R * ^ I R * x IR* b y
What is the R-class of (— 1, 1)? Describe geometrically the R-class of (a, b).
B2
B3
Determine equivalence relations on IR x IR whose equivalence classes are
(a) all lines parallel to 3x + Ay = 5;
(b) all circles with centre (1,2);
(c) all squares with vertices on the coordinate axes.
/
s
Given mappings X —> Y—>Z, prove that the following statements are
equivalent:
(a) t h e r e is a m a p p i n g h : Z - + X s u c h t h a t f o h O g = i d Y ;
(b) / i s surjective andg is injective.
Deduce that if a : X^Z is any mapping then there is a mapping ]3 :Z->X
such that a o 0 o a = a. (//inf: Take F = Im a.)
95
Test paper 4
Time allowed: 3 hours
(Allocate 5 marks for each A question; 20 marks for each B question.)
Section A
Al
If A, B, C are subsets of a set E prove that
(A\B)\CCA\(B\C).
Find a necessary and sufficient condition for equality to hold.
A2
If A ={xE\R
A=B.
A3
Express as a union of intervals the set of real numbers k for which
xGIR
|-2<x<3}
and B = {yG IR \y2 <y + 6} prove that
=k 1 = 0.
A4
Let JR be the relation defined on IR by
xRy^x + |JC| = 7 + 1^1.
Sketch the graph of R.
A5
Let X be the set of mappings / : [0, 1 ] -> IR. Define a relation R on X by
Show that R is an equivalence relation. Give two examples of elements in
the R -class of the mapping described by t -> t2.
A6
For every real number r let Ar = {(x, y) E IR x IR I x — y = r}. Prove that
{Ar\r£. IR} is a partition of IR x IR.
A7
Show that if x > 1 + yjl then (x +
96
1)/(JC
- 1 ) < 1 + yjl. Show also that if
Test paper 4
A = {x G IR | x > 1 + y/2} and B ={x G IR | 2 + 2\/2 <*} then the mapping
/ : ,4 ->B given b y / 0 ) = O 2 + l)/O ~ 1) is injective.
A8
Let g : IR -HR be defined by
1 — JC
2
-x)
if
if
x>0;
x<0.
Show that g is a bijection. What is g~ll
Section B
Bl
If E is a non-empty set then a non-empty collection ^"of subsets of E is
called 2L filter if
(a) F#=0 for every F G J ^ ;
(ft) i ^
(c)
ifGDj
Verify that the collection of all subsets of E that contain a given element
of Eis a filter.
Show that if & is a filter and X is a subset of E such that X n F ^ 0 for
every F G ^ , then
is also a filter. Show also that ^ C J f ^ , and that the inclusion is strict
whenever X ^ J ^ .
B2
Let E be the set of months in the year. Show that the relation R defined
on E by xRy if and only if x, y start on the same day of the week is an
equivalence relation on E. Determine the equivalence classes for an ordinary
year and for a leap year. How many Friday the thirteenths can there be
in a year?
B3
Let X={1, 2 , . . . , « } . Show that a mapping / : X -• X is such that / o / = /
if and only if the restriction o f / t o Im/is the identity map.
Suppose now that, for 1 < r < n,
Prove that
\Er\ = r\r"-r.
97