INSTRUCTOR'S
SOLUTIONS MANUAL
DISCRETE AND
COMBINAtORIAL MATHEMATICS
FIFTH EDITION
Ralph P. Gritnaldi
Rose-Hulman Institute o/Technology
Boston San Fra.'1cisco New York
London Toronto Sydney Tokyo
Mexico City Munich Paris
Town
Madrid
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2004 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted,
in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior
Printed in the United States of America.
written permission of the
ISBN
0-201-72660-2
1 2 3 4 5 6 CRS 06 05 04 03
PEARSON
Addison
Wesley
Dedicated to
the memory of
Nellie and Glen /Fuzzy/ Shidler
CONTENTS
PART 1
FUNDAMENTALS OF DISCRETE
1
Chapter 1
Fundamental Principles
Counting
Chapter 2
:Fundamentals of Logic
26
Chapter 3
Set Theory
59
Chapter 4
Properties of the Integel's: Mathematical Induction
95
Chapter 5
Relations and Functions
134
Chapter 6
Languages: Finite State Machines
167
Chapter 7
Relations: The Second Time Around
179
3
PART 2
FURTHER TOPICS IN ENUMERATION
207
Chapter 8
The Principle of Inclusion and Exclusion
209
Chapter 9
Generating Functions
229
Chapter 10
Recurrence Relations
243
GRAPH 1.'HEORY AND APPLICATIONS
281
4:
Chapter 14
Riugs and Modular Arithmetic
369
Chapter 15
Boolean Algebra and Switching Functions
396
Chapter 16
Groups, Coding Theory, and Polya's Method.
Enumeration
413
Chapter
Fini te Fields and Comhinatorial Designs
440
THE APPENDICES
459
Appendix 1
Exponential and Logarithmic Functions
461
Appendix 2
Properties of Matrices
464
Appendix 3
Countable and Uncountable Sets
468
PART 1
FUNDAMENTALS
OF
DISCRETE MATHEMATICS
CHAPTER 1
Sections
and
1.
( a) By the rule of sum, there are 8 5 = 13
for the eventual winner.
(b) Sim~e there are eight Republica,ns and
Democrats,
the rule of product we have
8 X 5 ::;:;; 40 possible pairs of opposing candidates.
(c) The rule of sum in part (a); the rule of product in part (b).
2.
By the rule of product there are 5 x 5 x 5 x 5 x 5 x 5 = 56 license plates where the first
two symbols are vowels and the last four are even digits.
3.
By the rule of product there are (a) 4 X 12 X 3 x 2 = 288 distinct Buicks that can be
manufactured. Of these, (b) 4 x 1 x 3 x 2 = 24 are blue.
4.
( a) From the rule of product there are 10 X 9 X 8 X 7 = P( 10, 4) = 5040 possible slates.
(b) (i) There are 3 x 9 x 8 x 7 = 1512 slates where a physician is nominated for president.
(ii) The number of slates with exactly one physician appearing is 4 x [3 x 7 x 6 x 5] = 2520.
(iii) There are 7 x 6 x 5 x 4 840 slates where no physician is nominated for any of the
four offices. Consequently, 5040 - 840 = 4200 slates include at least one physician.
=
5.
Based on the evidence supplied by Jennifer and Tiffany, from the rule of product we find
that there are 2 x 2 x 1 x 10 x 10 x 2 = 800 different license plates.
6.
(a) Here we are dealing with the permutations of 30 objects (the runners) taken 8 (the first
eight finishing positions) at a
trophies can be awarded in P(SO,S) = 30!/22!
ways.
(b) .n.!.Hl>en;(:!, and
runners in 6 ways.
each
these 6 ways, there are P(28,6) ways for the other 6 finishers (in the top 8) to finish the
race.
the
are 6·
to
with
two runners a.U:lon,~ the top
are (a.) 12!
re31~nCl"lOl,LI); (b) (4!)( 8!) ways so that the four ..... ,." ....."'. .
and (c) (4!)(51)(3!)
where the
top .,..""",.,.... tare PT(:JCe:8SE:<1
3
9.
(a) (14)(12) = 168
(b) (14)(12)(6)(18) = 18,144
(c) (8)(18)(6)(3)(14)(12)(14)(12) = 73,156,608
10.
Consider one such arrangement - say we have three books on one shelf and
on the
other. This can be accomplished
15! ways.
fact for any subdivision (resulting in
two nonempty shelves) of the 15 books we get 15! ways to arrange the books on the two
shelves. Since there are 14 ways to subdivide the books so that each shelf has at least one
book, the total number of ways in which Pamela can arrange her books in this manner is
(14)(15!).
11.
( a) There are four roads from town A to town B and three roads from town B to town
C, so by the rule of product there are 4 x 3 = 12 roads from A to C that pass through B.
Since there are two roads from A to C directly, there are 12 + 2 = 14 ways in which Linda
can make the trip from A to C.
(b) Using the result from part (a), together with the rule of product, we find that there
are 14 X 14 = 196 different round trips (from A to C and back to A).
(c) Here there are 14 x 13 = 182 round trips.
12.
(1) a,c,t
13.
(a) 8! = P(8, 8)
14.
(a) P(7,2) = 7!/(7 - 2)! = 7!/5! = (7)(6) = 42
(b) P(8,4) = 8!/(8 - 4)! = 8!/4! = (8)(7)(6)(5) = 1680
(c) P(lO, 7) = 10!/(1O - 7)! = 101/3! = (10)(9)(8)(7)(6)(5)(4) = 604,800
(d) P(12, 3) = 12!/(12 - 3)! == 12!/9! = (12)(11)(10) = 1320
15.
Here we must place a,b,c,d in the positions denoted by x:
of product there are 4! ways to do this.
16.
(a) With repetitions allowed there are 40 25 distinct messages.
(b) By the rule of product there are 40 x 30 x 30 x ... x 30 x 30 x 40 = (40 2 )(30 23 )
messages.
(2) a,t,c
(3) c,a,t
(4) c,t,a
(5) t,a,c
6!
(b) 71
e A e A e A e A e. By the rule
Class
(21 - 2)(224 - 2) = 2,1
928,964
14 16
Class B: 2 (2 - 2) = 1,073,709,056
C:
= 532,
608
system.
(a) 7! == 5040
(3!)(5)( 4!) =
20.
Since
(b) 4 x 3 x 3 x 2 x 2 x 1 x 1 =
are three A's, there are 8l/3! =
4
(6) t,c,a
arrangements.
(b) Here we arrange the six symbols D,T,G,R,M, AAA in 6! = 720
21.
(a) 121/(3!2!2!2!)
(b) [11!/(3!2!212!)] (for AG) + [11!/(31212!2!)] (for GA)
(c) Consider one ca.se where
the
are adjacent: S,C,L,G,C,L, OIOOIA. These
seven symbols can be arranged in (7!)/(2!21) ways. Since O,O,O,I,I,A can be arranged
in (6!)/(3!2!) ways, the number of arrangements with all
vowels adjacent is
[7!/(2!2!)}[61/(3!2!)].
22.
(Case 1:
leading digit is 5) (6!)/(21)
(Case 2: The leading digit is 6) (6!)/(2!)2
(Case 3:
leading digit is 7) (61)/(21)2
In total there are (6!)/(2!)][1 + (1/2) + (1/2)J = 6! = 720 such positive integers n.
23.
Here the solution is the number of ways we can arrange 12 objects - 4
the first type,
3 of the second, 2 of the third, and 3 of the fourth. There are 12!/( 4!3t2!31) = 277,200
ways.
24.
Pen + 1,r) = {n + l)!/(n 1- 1')1 = [en + l)/Cn
fen + 1)/Cn 1- r)]P(n,r).
25.
( a) n = 10
(b) n = 5
(c) 2nl/(n - 2)1 + 50 = (2n)!/(2n - 2)! = } 2n(n -1) + 50 = (2n)(2n -1) = } n 2 = 25 = }
n=5.
26.
Any such path fro:m, (0,0) to (7,7) or from (2,7) to (9,14) is an arrangement of 7 R's and
7 U's. There are (141)/(7!7!) such arrangements.
In general I for m, n nonnegative integers, and any real numbers a, b, the number of such
paths from (a, b) to (a + m, b n) is (m n)!/{m!n!).
21.
(a)
1 - r)] . [nt/en - 1')IJ =
path consists
2
1 V, and 7 A's. There are 10!/(2!1!7!) ways to arrange
these 10 letters and this is the number of paths.
numbers and m~ n, and p are nonnegative integers,
to (a m,b+ c
+n
b
j
times, while those for j and k are py';·rut.PI'I 10-5+ 1 = 6
o
)
we
the instrudio:o.s in.
k ............'1-'''', resPei:trv'elY,
5
29.
(3) & (b) By
rrue
product
print statement is executed
x 6 x 8 = 576 times.
five '.<>1'.1':,0,...,
are 26x26x
x1x1=
3
letters.
X 1 x 1 x 1 = 26
(b) When letters may not appear more than two times, there are 26 x 25 x 24 = 15,600
palindromes for either five or six letters.
31.
By
rule
product
are (a) 9 X 9 X 8 x 7 x 6 x 5 = 136,080 six-digit integers
with no leading zeros and no repeated digit. (b) When digits may he repeated there are
9 X 105 such six-digit integers.
(i) (a) (9 x 8 x 7 x 6 x 5 x 1) (for the integers euding in 0) (8 x 8 x 7 x 6 x 5 x 4) (for
the integers ending in 2,4,6, or 8) = 68,800. (b) When the digits may be repeated there
are 9 x 10 x 10 x 10 x 10 x 5 = 450,000 six-digit even integers.
(ii) (a) (9 X 8 x 7 x 6 x 5 x 1) (for the integers ending in 0) + (8 x 8 x 7 X 6 x 5 x 1)
(for the integers ending in 5) = 28,560. (b) 9 x 10 x 10 x 10 x 10 x 2 = 180,000.
(iii) We use the fact that an integer is divisible by 4 if a.nd only if the integer formed by
the last. two digits is divisible by 4. (a) (8 x 7 x 6 x 5 x 6) (last two digits are 04, 08, 20,
40, 60, or 80) + (7 x 7 x 6 x 5 x 16) (last two digits are 12, 16, 24, 28, 32, 36, 48, 52, 56,
64, 68, 72, 76, 84, 92, or 96) = 33,600. (b) 9 x 10 x 10 x 10 x 25 = 225, O~~.
32.
(a) For positive integers n, k, where n = 3k, n!/(3!)k is the number of ways to arrange
the n objects Xl! Xt, Xb Xz , X2, X2,' .• ,Xk, Xk, XI;. This must be an integer.
(b) If n,k are positive integers with n = mk, then n!/(m!)" is an integer.
33.
(a)
(b)
34.
With 2 choices per question there are 210 = 1024 wa.ys to answer the examination.
Now there are 3 choices per question and 310 ways.
(41/2!) (No 7'8) (4!) (One 7 and one 3) + (2)(4!/21) (One 7 all.d two 3's) (4!/2!) (Two
7'13 and no 3's) (2)(41/20 (Two 7'8 and one 3) + (41/(2121» (Two 7'8 and two 3'15). The
total gives us 1.02 such four-digit Intc~l!elrS
(a) 6!
36.
Aon
6
38.
The nine women can be situated around the table in 8!
Each such arrangement
provides nine spaces (between women) where °a man can
placed. VVe can
of these places
situate a man
each of them
(:)6! = {). 8 . 7· 6 . 5· 4 ways.
Consequently,
2,438, 553, 600.
seating arrangements
the
18
39.
procedu.re SumOfFad( i, sum: positive integers; j,k: nonnegative integers;
factorial: array [0 .. 9] of
positive integers)
begin
factorial [0] := 1
for i := 1 to 9 do
factorial [~1 := i * factorial [i - 1]
for i := 1 to 9 do
for j := 0 to 9 do
for k := 0 to 9 do
begin
sum := factorial [iJ + factorial [y]
factorial [kl
if (100 :+: i + 10 * j + k) = sum then
print (100 * i + 10 * j + k)
end
end
The unique answer is 145 since (II) + (41) + (5!) = 1
24
120 = 145.
Sedion 1.3
a
a
a
a
a
2.
b
b
c
b
b
b
d
e
f
c
Order is not relevant
(b)
-
--
--
1.
=
-.
c
d
e
f
d
c
c
d
d
e
e
f
e
f
f
selection in
can
= (10)(9)(8)(7)/(4)(3)(2)(1) = 210
12!/(7!5!) == (12)(11)(
7
5
-
ways.
(:) 6!
=
(c) C(14,
= 14!j(1212!) = (14)(13)!{2)(1) c::; 91
(d)
= 151/(10!51) = (15)(14)(13)(12)(11)/(5)(4)(3)(2)(1) = 3003
G!)
+ (:)
(:) = 31
4.
(a)
5.
(a)
are peS,3) = S!j(5 = 51/2! = (5)(4)(3) = 60 permutations size 3 for the
five letters
l, a, f, and t.
(b) There are C(5, 3) = S!j[3!{5 - 3)IJ = 5!/(312!) =
combinations of size 3 for the five
letters m, l', a, f, and t. They are
-1=
a,f,r
f,m,r
a,f,m
a,r ,t
a,f,t
f,m,t
a,m,r
f,r,t
a,m,t
m,r,t
6.
(;)
(n ~
1) (~)(n)(n 1) (~)(n l)(n - 2) (~)(n
_
=
-
=
-l)[n + (n - 2)J =
n)(n - 1)(2n - 2) = (n - 1)2.
7.
(a) (~)
(b)eaO) (~)
(c) C20) G~)(2 women) (~O) (~O) (4 women) ... + G~) e!O) (10 women) = Ef=l (;~) (121~2i)
(d)
C50) (7 women) + (~O) (::) (8 women) + (~) (9 women) +
G~) (;0) (10 women) =
C?) (l;~i)'
e:)
e:)
(e) E!!s eiO) (li~i)
(b)
(!) (~8)
(c)
e13) (!) (~)
(d) (:) (;)
(f) (~3) (!) C12) (~) = 3744
(Division by 2 is needed since no distinction is made for the order
54,
result
are
=
1
(!) (~) -3744 =
+
,
r,
1
S
11.
(:)
(four
six) = (15)(4)
ways.
set
8
G) ways. vOllSe<luelllUJI the 12
14.
.l
= (1
= 97
(b)
+
-1) =
_m 1)
=
-- 2--1
10
(c)
}J1 +(-
=2+0+2+0+2+0+2+0+2+0+2=
i",{l
2'1
(._6
+
+
= -1 + 2 - 3 + 4 - 5 + 6 = 3
~)(
i;:::l
7
(b) L>~
i;;;;;l
n
i +1
i=O n
1)'
+i
+
+
1, one
9
+
0+1
(;~)
an even
locations for
(~~) ways
for 0 < i < 5. Then for the
positions selected there are tvvo choices; for the 10 - .
'C.u,~C:WJ,,,,.uJ'5 positions there are also two choices 1,3.
20.
(a) We can select 3 vertices from A, B, C, D, E, F, G, H in (:) ways, so there are (:) = 56
distinct inscribed triangles.
(b) (:) = 70 quadrilaterals.
(c)
total number of polygons is (~)
256 - [1 8 + 28] = 219.
21.
(:)
(:)
(:)
(~) + (:) = 28 - [(~)
(~) + (~) J =
There are (;) triangles if sides of the n-gon may be used. Of
(;) triangles,
when n
4 there al'e n triangles that use two sides of the n-gon and n(n - 4)
trian~les that use only one side. So if the sides of the n-gon cannot be used, then there
are
n - n( n - 4), n 2 4, triangles.
G) -
22.
(a) From the rule of product it follows that there are 4 x 4 x 6 = 96 terms in the complete
expansion of (a + b + c + d)( e f 9 + h)( 'It + V w x y + z).
(b) The terms bvx and egu do not occur as summands in this expansion.
23.
(a)
en
(b)
e~2)(23)
(c) Let a = 2x and b = -3y. By the binomial theorem the coefficient of a9 b3 in the
expansion of (a + b)12 is (~). But
a9 b3 = (~2)(2x)9(_3y)3= (~)(29)(_3)3x9y3,so
e;n
the e.oefficient of x 9 y3 is
(a)
(c)
(1'~'2) = 12
(1,i,2)(!~)( -1)( -1)2 =
e:)(29 )( _3)3,
4. ) ( 0,1,1,2
-
12
(d) (1,:,2)<-2)(3)2 = -216
(3,2~1,Z)
(d) 4/'
TO
28.
29.
1
E
i=O
a)
m+n) _
-
n( m
1)
(m
= f1
=-
= f
n. i=O
n!
(m+n)! _
n m!n!
E (n). =2n/nl
=-
-
(m+n)!
(m+1)(m!)(n-l)!
n
n·i=o ;
1
(m+n)! _
m!(n-l)! (
1)
= m
E(--lt (n)
: = 1
n
(a) 1=[(1
(b) 1=[(2
x)-
=(1+x)n_-(7)
1
x)-(x+lW'·
(a) I:(ai - ai-l) = (al - ao)
(a2 - al)
x)n-1
(;)x2(1 x)",-2_ ...
(c) 2'l = [(2 + x) - x]n
+ (aa - (2) = as - G()
i:;;;:l
11,
(b) E(ai - ai-d = (al - 0.0) + (a2 - al) + (as - (2)
i=l
an -0.0
100
1
1
1
1
1
1
1
1
(c) ~(i + 2 - i + 1) = (3 - "2) + (4 - 3) + (:5 - 4) + ...
1
1
1 - 51
-50
-25
102 -"2 = 102 = 102 ::::: 5"134.
procedure Seled2(i,j: positive integers)
begin
for i:= 1 to 5 do
for i:= i + 1
6
end
begin
for *:= 1 to 4, do
j:= i + 1
-
( m+n)
= (m
m+l
+ 2)n = 3"',
3
33.
~
(m+",)!
(m+l)!(n-l)!
The sum is the binomial expansion of
31.
.
5
Ii::= j
end
11
(_l)n(:)x n .
Section 1.4
1.
Let Xi, 1 :s; i :5 5, denote the arnounts given to the five "'u.J'~'UJ.""·"'4
(a)
integer
Xl
X2 + X3 + X4 + Xl) =
0 :5 Xi, 1 :5 i :5 5, is
(5+i~-l) = G~). Here n = 5, r = 10.
(b) Giving each child one dime results in the equation Xl + X2 X3 X4 X5 = 5, 0 <
1
Xi, 1 :5 i
5. There are
distribute the remaining five dimes.
) = (:) ways
(c) Let Xs denote the
for the oldest child.
number of solutions to Xl +
X2
Xa X4 X5 = 10, 0 :5 Xi, 1 :s; i :5 4, 2 :5 X5 is the number of solutions to
Yl Yz + + Y4 + Ys = 8, O:S; Yi, 1 i :5 5, which is
= (~2).
e+:-
C'+:-1)
2.
Let Xi, 1
i:5 5, denote the number of candy bars for the five children with Xl
the number for the youngest. (Xl = 1):
X2 + Xs + X4
Xs = 14. Here there are
(4+~:-1) = (i:) distributions. (Xl = 2): X2 + X3 + X4 + X5 = 13. Here the number of
distributions is (4+!~-1) =
3.
_. (23)
( ~H20-1)
20
20
4.
(a)
(~;)
G:). The answer is G~) G:) by the rule of sum.
(b)
e
1 12 1
+1 2 - )
=
G;)
(c) There are 31 ways to have 12 cones with the same flavor. So there are (:;) - 31 ways
to order the 12 cones and have at least two flavors.
5.
(&) 25
(b) For each of the n distinct objects there are two choices. If an object is not selected,
then one of the n identical objects is used in the selection. This results in 21'1. possible
selections of size n.
1.
(a)
e+:!--l) = (;~)
e+:-1)= C;)
(e)
X4
=
(b) e+;:~-l) = (;!)
(d) 1
1
OVJ'U",~V£.I":> to
Ya + 'lI4
= 40, Yi
8.
12
10.
Here we want the number of integer
for Xl
X2
X3 + X4
X.r;
X6 = 100,
Xi ?: 3, 1 :s; i :s; 6. (For 1 ::; i :s; 6, Xi counts the number of times the face with i
dots is rolled.) This is equal to the numher of nonnegative integer solutions there are to
'113
+'115+116 =
> 1:S i
Consequently the answer is e+:;-l) = (:;).
11.
(a)
en
('0+55-1) = e}4)
(h)
('+:-1) SC'+:-l) + 3(7+;-1) + e+~~l) =
-t 3(~O) 3(:) (~), where the first summand accounts the case where none of
1,3,7 ap~ars, the second summand for when exactly one of 1,3,7 appears once, the
summand for th~ case
exactly two of
digits appearing once each, and the last
summand for when all three appear.
12.
(a) The number of solutions for Xl + X2 +.. . X5 < 40, Xi ?: 0, 1 < i 5, is the same as
the number for Xl + X2 ••• + X5 < 39, Xi ?: 0, 1 :S i < 5, and this equals the number of
solutions for Xl + X2 ••• + X5 + X6 = 39, Xi 2: 0, 1 :S i ::; 6. There are
= (::)
such solutions.
(b) Let 'IIi = Xi 3, 1 :S i :5 5, and consider the inequality '111 + '112 + ... + '115 :s; 54, 'IIi > O.
There are
in part (a)]
= (!:) solutions.
C,+:-l)
e+::-1)
13.
(a)
e+:- 1) = G).
e+!-l) (container 4 has three marbles)
e+~-l) (container 4 has one marble)
e+~-l) (container 4 has five marbles) + (3+~-1)
(b)
(container 4 has seven marbles)
= Lr:o (~=~~).
14.
(a) (2t4.~,O,1)(3r~(2)4
vaw bxCyd ze
where a, h, c, d, e are
1
e+:- ) == e:) terms.
(b)
The terms in the expansion have the forID
nonnegative integers that sum to
There are
are six books on tP..ach of
four
Consider one sum dist:dbutioll - the one where
shelves. Here there are 24! ways for this to happen. And we see that there a:re also 24!
books
any other
lS
+ Tl?, + + '114 = 20.
13
For
WI
(1) we
W:;t
+ Ws +...
the
nonnegative 1U~"'K'C;"
W19 = n -19, where Wi
0 for alll
(:"::-1~)' The number of positive integer solutions
ative integer solutions for
equation
= n - 64,
Z64
and t
hi .
S IS
(64+(n-S4)--1) _
(n-M)
-
number of nonneg-
IS
(n-1)
n-64 •
So (:.:-~~) = (::;4) = (n~l) and n - 19 = 63. Hence n = 82.
(b)
18.
e+:-
1) = (:) solutions for Xl +X2 X3 = 6 and (4+i~-l) = (~) solutions
(a) There are
for X4 + XI) x~ + X7 = 31, where Xi 2: 0, 1 ::; i ::; 7. By the rule of product the pair of
equations has \:) (~) solutions.
(b)
19.
(~) (~i)
Here there are r = 4 nested for loops, so 1 < m ::; k ::; j ::; i ::; 20. We are making
selections, with repetition, of size r = 4 from a collection of size n = 20. Hence the
print statement is executed
= (~3) times.
eO+,t1)
20.
Here there are r = 3 nested for loops and 1::; i ::; j < k ::; 15. So we are making
selections, with repetition, of size r
3 from a collection of size n = 15. Therefore the
statement
counter := COftnier +1
=
is executed
690.
21.
24.
C5+aS-l) =
en
times, and the final value of the variable counteris 10+
The begin..end segment executed e°",;a-l) =
this segment the value of the variable Bum is
(a)
:3
= 220 times.
i = (220)(221 )/2
=
e:)
=
execution of
24,310.
for i := 0 to 10 do
for j := 0 to 10 - i do
. .\
10 - t - J)
end
For all 1:5 i ::; 4
= Xi + 2 2:: O.
the number of integer solutions to
Xl + X2 + X3
X4 = 4, where -2::; Xi for 1::; i :5 4, is the number of integer solutions
to !II + !l2 + '!J3
where !Ii 2:: 0
1 ::; i < 4.
use this observation
the
following.
(b)
procedure Selectio'n$2(i,j,k: nonnegative integers)
begin
for i := 0 to 12 do
for j := 0 to 12 - i do
for k := 0 to 12 - i - j do
print (i,j,k, 12 - i - j - k)
end
25.
If the smmnands must all be even, then consider one such composition - say,
20 = 10 + 4 + 2 + 4 = 2(5 + 2 + 1 + 2).
Here we notice that 5 + 2 + 1 + 2 provides a composition 10. Further, each composition
of 10, when multiplied through by 2, provides a composition 20, where each summand is
even.
we see
the number
20,
e8,(".h summand
10 1
even, equals the nUlnher of compositions of 10 ~ namely, 2 - = 29.
m. Consequently,
a
is a
26.
Consequently,
c)
6 = 10
an~~enlellts that
runs.
the
run
we
number of
solutions for Xl
Xz + X3
X4 = 12, whe,re Xl + xa = 5, XI, X3 > 0 and X2
X", =
XZ, X4 >
This
number is e+~-l)
= (!) (:) = 4·6 = \Vhen the first run consists of tails we get
e+:-1)
(:) (:) = 6 . 4 = 24 arrangements.
all there are 2(24) = 48 arrangements with four runs.
d) If the first run starts with an H, thel1 we need the number of integer solutions for
Xl + Xz + Xs
X4
Xs =
Xl
Xs
X5 =
Xl) Xa, X5 >
and X2
X4 =
Xz, X4 >
This is e+~-l)
= (:) (:) = 36. For the case where the first run starts
O.
+ +
+ +
C+:-l)
with a T, the number of arrangements is
hI total tha'e are 36
0
e+:-1)e+;-l) =
+
7,
(!) = 60.
60 = 96 ways for these 12 t()sses to determine five runs.
1) e+~-l) = (:) = 90 - the number of arrangel'uents which result in six runs, if
e)
the first run starts with an H. But this is also the number when the first run starts with
a T. Consequently, six runs come about in 2·90 = 180 ways.
e+:-
f)
G)
2C+:-l) e~:-1)+2e+~-1) e+:- 1) +2(3+;-1) e+:-1)+2(4+!-1) (4+~-1)+2(5+6-1) e+~-l) =
2 L,t:::o (i~i) (6~i) = 2fl . 1
28.
4 . ()
() . 15 + 4 . 20 + 1 . 15] = 420.
(a) For n
4, cIOn sider the strings made up of n bits - that is, a total of nO's and 1's.
In particular, c()nsider those strings where there are (exactly) two ()ccurrences ()f 01. For
example, if n = 6 we want to include strings such a..~ 010010 and 100101, but not 101111
or 010101. How many such strings are there?
(b) For n ~ 6, how many strings ()f nO's and l's contain (exactly) three occurrences of 01?
(c) Provide a combinatorial proof for the foll()wing:
For
(a)
n~ 1, 2ft = (n ~ 1) + (n 3 1)
x" O's followed by
type
of :1::1
followed
X2
followed
X3
1'8 foll()wed by
Xs 1 '8 followed by X6 O's, where,
Xs
of ~J""""""""lD
..."n",,,,.,,, the
o for 1
Y5 + Ya = n-
Us = n - 6,
> O•
o
1
t
~
6.
_ (n+l)
'i
•
-
(C) There are
strings
n 1 strings where there are k
followed by n ,- k
for k = 0,
.. '/: n.
n 1 strings contain no occur:rence~ of 01, so there are
21'1, - (n 1) =
- \ntl) strings that contain at least one occurrence of
are
(n;l) strings that contain (exactly) one occurrence of 01, (nil) strings with (exactly) two
occurrences, (n~l) strings with (exactly) three occurrences, ... ) and for
n odd, we can have at most 1'1.;1 occurrences of 01.
number of strings with
occurrences of 01 is the number of integer solutions for
This is the same as the number of integer solutions for
1116+1 = n - (n - 1) = 1, where !II, 112, ••• ,111'1.+1
0.
1) = (n+l) _ (1'1,+1) _ ( n+l )
This number is (n+1)+11
1
n
2( ~)+1 .
(ii) n even, we can have at most ~ occurrences of
occurrences of 01 is the number of integer solutions for
The number of strings
This is the same as the number of integer solutions for
YI + Y2 +,., + YnH = n - n = 0, where Y' ~ 0 for 1 :::; i :::; n + 2.
This number is
Consequently,
(1'1.+2)+0-1)
= (1'1,+1)
+1) -_ ( 1'1,+1 ) •
()
() -_ ("'n+1
2(~)+1
.
{
follows.
~;~;~: : odd
n
2"
blO = 16796
bs)j 14(=
(b) For n ;:: 0 there are bn ( =
n 2: 0 the
(e)
(~) such
from (0,0) to (n,n).
move is U and the la.st
(b) bs =
4.
(a) lis = 132
5.
Using the results
07 = 429
third column of Table 1.10 we have:
111000
110010
101010
123
125
346
135
246
456
6.
(a) (i)
(b) (i)
(iii)
2568
(ii) 1 2 5 7
3468
1235
4678
10111000
(ii)
(iii)
11011000
1347
7.
There are bs( = 42) ways.
8.
(a) (i) 1110001010
(b)
9.
11100010
(ii) 1010101010
(iii) 1111001000
(i) «(ab(c(de H> «(ab)(c(de»)f)
Oi) «ab((cd(e H> «ab)«cd)(ef»)
(iii) (a«(bc(de H> (a«(bc)(de»f»
(i) When n = 4 there are 14(= b4 ) such diagrams.
(li) For any n
0, there are bn different drawings of n semicircles on and above a horizontal
line, with no two semicircles intersecting. Consider, for instance, the diagram
part (f)
of the figure. Going from left to right, write 1 the first time you encounter a semicircle
write 0
IS en(:oUltltel~ed.
corresponds with the dn\'wing
such rlr""'i:vn~
""u,::;<"". as
10.
R R U RRR
18
Here
condition is violated, for the first time, after the third U. Transform the
R
U
+-'?RU
U
UUU.
Here the entries up to and including the first violation remain unchanged, while those
following
first violation are changed:
become U's and U's become R'a. This
cOITesponcience sbows us that the number of paths that violate the given condition the
same as the
of paths
up of eight U's and two R's - and there are
=
such paths.
e:) e:)
Consequently, the answer is
_ (10)
_ lQi _ JQl
_ 10!(S) _ 1O!(3} = (2) 1O! _ (1±1-3) (10)
(10)
'1
8 - 7131
8!2! 8!3!
8!3!
8 7!3! 1+1
'1'
(b) (m+n)
n
-
(m+'ll) _ (m+n)!
n+1 n!m! -
(m+n)!
(n+1}!(m-l)!
= (mtn)!(n+1)-(m+».)!m = (n±l-m)«m+n)!) = (n±l-m)(m+n).
(n+l)!m!
'11.+1
n!m!
'1+1
n
[Note that when m = n, this becomes (n!l)(~)' the formula for the nth Catalan number.]
11.
Consider one oUlle C~!l) e~6) =
- say,
(*)
n) e:) ways in which the $5 and $10 bills can be arranged
$5, $5, $10, $5, $5, $10, $10, $10, $5, $5, $10, $10.
Here we consider the six $5 bills as indistinguishable -likewise, for the six $10 bills. However, we consider the patrons as distinct. Hence, there are 6! ways for tbe six patrons, each
with a $5 bill, to occupy positions 1, 2~ 4, 5, 9, and 10, in the arrangement (*). Likewise,
there are 6f ways to locate the othe:r six patrons (each with a $10 hill). Consequently, here
the number of arrangements is
seen in
all
Consequently, the largest UW,UVPJl.
(~2) 495.
=
4.
e:t
(a)
(b) 3 e15):I (~) (four hymns from one book, one from each the other two)
hymn from one
two hymns
a <:>""AJJ.J.U
and
3
:I
( two
from each of the three books).
e15)e:) e:)
6
third book)
5.
(a) 1025
(b) There are 10 choices for the first flag. For the second flag there are 11 choices: The
poles with no flag,
above or below the first flag on the pole where it is situated.
There are 12 choices for the third
choices for
fourth, ... , and 34 choices for the
last (25th). Hence there are (34!)/(91) possible arrangements.
(c) There are 251 ways to arrange the flags. For each arrange:tnent consider the 24 spaces,
one between each pair of flags. Selecting 9 of these spaces provides a distribution among
the 10 flagpoles where every flagpole has at least one flag and order is relevant. Hence
there are (25!)(~) such arrangements.
6.
Consider the 45 heads and the 46 positions they determine: (1) One position to the left
of the first head; (2) One position between the i-th head and the (i l)-st head, where
1 SiS 44; and, (3) One position to the right of the 45-th (last) head. To answer the
question posed we need to select 15 of the 46 positions. This we can do in (::) ways.
In an alternate way, let Xi denote the number of heads to the left of the i-th tail, for
1 siS. 15. Let X16 denote the number
heads to the right of the 15th taiL Then we
want the number of integer solutions for
Xl
+ XzXa + ... + Xli; + XIS = 45,
where Xl > 0, Xi6 2 0, and Xi > 0 for 2 sis 15. This is th.e number of integer solutions
for
111 + Ya '1/3 ••• +1115 Y16 = 31,
with Yi > 0 for 1
z
16. Consequently
P(12, 8)
,N
8.
not adjacent to a
9.
(a) There are two blocks,
only
only
wooden
(i)
S12~e:
there are 1 x 2 = 2
(li) Material, color:
pair yields 1 X 4 = 4 such blocks.
(iii) Material, shape: For this pair we obtain 1 x 5 := 5 such blocks.
we get 2 X 4 = 8 of the
(iv) Size,
(v) Size, shape: This pair gives us 2 x 5 = 10 such blocks.
(vi) Color, shape:
this pair we find 4 x 5 = 20
the blocks we need
ULV....!\.O
:=
+ 20 49 of
5+8
large blue plfUltic hexagonal block in exactly two ways.
In total there are 2 + 4
10.
en
Sinc..e 'R' is the 18th letter of
= (17)(16)/2 = 136 ways.
count.
blocks that differ from the
alphabet, the first and middle initials can be chosen in
Alternately, since 'R' is the 18th letter of the alphabet, consider what happens when the
middle initial is any letter between 'B' and 'Q'. For middle initial 'Q' there are 16 possible
first initials. For middle initial 'P' there are 15 possible choices. Continuing back to 'B'
where there is only one choice (namely' A') for the first initial, we
that the total
number of choices is 1 + 2 + 3
15 16 = (16)(17)/2 = 136.
11.
The number of linear arrangements of the 11 horses is 11!/(513!31). Each circular arrangement represents 11Unear arrangements, so there are (1/11)[111/(5!313!)J ways to arrange
the horses on the carousel.
12.
(a)
13.
14.
P(16,12)
(b)
(;2) P(15, 10)
(ii) e+:-1
) + e+;-l) C'+;-1) e+:-1):= (!) +
(!) e) (!) + (!)
(~) (D + G)
(iii) (:) + (~) (;) + (~) - 9
1
(ii) and (iii) (~) e+:- ) e+;-l) (~) = (D (:) + G) (:).
(b) (i) (n (!) (!) (~)
(a)
(i)
(a)
there are no restrictions Mr. Kelly can nl.ake the a8sigmnenis
12! = 479,001,600
can
in 4 x 3 = 12 ways,
assigned in 10! ways. Consequently, in
UGAU."O
348,364,800.
13.
(a)
the
Mr.
can
be arranged as a decreasing :l:O'!.U'-(l1f!lt
UM,"-'.",,"'oI.
'To complete the solution we ZIlust account for the decreasing four-digit integers where the
units digit is
are (:) = 84 of these.
.
Consequently there are 2 (!)
(:) = 343 such four-digit integers.
(b) For each ~ondecrea8ing four-digit integer we
allowed. These four digits can be selected
. four nonzero digits, with repetitions
=
ways. And these same four
e+!-l) en
digits account for a nonincreasing four-digit integer, So at this point we have 2cn - 9 of
the four-digit integers we waut to count. (The reason we subtract 9 because we have
counted the nine integers
2222,3333, .. " 9999 twice
2(~2).)
We have not accounted for those nonincreasing four-digit integers where the units digit is
O. There are
1 =
1 of these four-digit integers. (Here we subtracted 1
since we do not want to include 0000.)
Cil+3S-1) -
(;2) -
Therefore there are [2C42) - 9} +
integers.
16.
{(;2) -- 1] = [2e,n
(~)] - 10 = 1200 such four-digit
(a) (~,~,2)(1/2)2( _3)2 = 135/2
(b) Each term is of the form :e nt yn2 zrl.S where each ni, 1 < i :s; 3, is a nonnegative
integer and nl + n2 + n3 = 5. Consequently, there are
=
terms.
ec) Replace x, y, and z by 1. Then the sum of all the coefficients in the expansion is
«1/2) + 1 - 3)5 = (-3/2)5.
e+:-1) G)
17.
(a) First place person A at the table. There are five distinguishable places available for
A (e.g., any of the positions occupied by A,B,C,D,E in Fig. 1.11(a». Then position the
other nine people relative to A. This can be done in 9! ways, so there are (5)(91) seating
arrangements.
(b) There are three distinct ways to position A,B 80 that
are seated on longer sides
table across from each other.
other eight people can then be
8!
different ways, so the total nw::-qber of arrangements is (3)(8!).
18.
(a) For Xl
Xl
X2
3''>;1
X3:= 6
solutions for
X4
+ X3 = 6
are
X2
e+:- = (:)nonnegative ultel:€~'
1
)
X4
-
= 15,
. The number of
scores for each set
<>'-P.... ",.""...''''
of
scores to
can win
So if A wins in four or five
sets
e) ways,
soores can be """I"nr-d"l1'>F1
the scores can be recorded in [(;)
Since B may be the winner, the final answer is 2[(~)74
20.
(:) 75 ] ways.
(~)75].
We can choose r objects from ,n in (:) ways. Once the r objects are selected they can
be' arranged
a circle
(1" - 1)! ways. So there are (~)
of the n objects taken 1" at a
21.
(~)74 ways.
circular arrangements
For every positive integer n, 0 == (1 - l)n = (~)(1)o - (~)(1)1 +
(-l)n(:)(l)\and
(~)+(~)
(~)+ ... =(7)+(;)
(~)
(b) 5!
7!/3!
(;)(1Y' """' (;)(1)3 + ...
(~)(4!)
22.
(a)
23.
(a) There are P(20, 12) == 2~! = (20)(19)(18)··· (11)(10)(9) ways in which Francesca can
fill her bookshelf.
(b)' There are
ways in which Francesca can select nine othe~ books. Then she can
arrange those nine books and the three books on tennis on her bookshelf in 12! ways.
Consequently, among the arrangements in part (a), there are e:)(12!) arrangements that
include Francesca's three books on tennis.
(c)
en
24.
Following the execution of this prognuu segment the value of counter is
10+(12-1+1)(1"-1 1)(2) [3+4 ... +(s-3+1)](4)+(12-3+1)(6)+(t-7 1)(8) ==
10 + (12)(1')(2)
22
25.
[(1/2)(8 - 3 + 1)(8 - 3 + 2) - 2 - 1}(4) + (10)(6) + (t - 6)(8) =
24,., + 8t + 2(8 - 2)(8 -1) - 12 = 14 + 24r + 8t 2S(8 - 3).
(a) For .17 there must be an odd munber, between 1 and 17 inclusive, of l's.
For 2k + 1 1'5, where 0:::; k 8, there are 2k + 2 locations to select, with repetitions
allowed. The selection size is the number of 2'£1, which is (1/2)[17 - (2k + 1)) = 8 - k. The
selection can be made in (2k+2+(S-k)-1)
= (9+k).
wavg and so the answer is ",8_
(9+k) =
a-Ie
8-k" ,
'-'k_O 8-/;
2584.
(b)
18
even:2k, for 0:::; k :::; 9. there are 2k
1
(1/2)(
= 9- k
n as an
(b)
1 (six
. (four 3's)
(:)
5- 1 = 4 horizontal moves
in 111/(4!71) ways.
move
diagonal moves is hetween 0
4
9 -2 = 7
xnov\,:,
Cl'lS(:"ii
are M
7
:3 R's, {}
2
5
1 R,
4,
o R'g, 3
sum of the results:
29.
8!j(3!1!4J)
71/( 410!3!)
2.:1=0[(11- i)!j(i!(4--
1
[11
- [4!/(2!2!)][4!/(3!l!)]
[11!/(7!4!)J + [10!/(6!3!1!)] + [9!/(5!212!)] +
1!/('7I4!)) + [101/(613!1!)] [9!j(5!2!2!)] + [8!/(4!1!3!)] +
[([4!/(2!2!)] + [3!j(1!1!1!)] + [2!j2!]} x ([4!/(3!H)] + [3!/(2!1!)J}]
Hexe we want certain paths from (1,1) to (14,4) where the moves are of the
(m, ~ (m + 1, n + 1), ifthe (n + l)-st ballot is for Katalin.
(m, -+
+ 1,11. --1)) iHhc (n + l)~st ballot is for Donna.
paths are
ollesthat never touch or cross the horizontal
...",...!I""".,,,,,, pair (Tn, n) here indicates that m ballots have been counted
hy n votes.
number of ways to count the ballots according to
18
corners
O$a<c~8
b,d
O$b'<fl~
answer IS
33.
FrY['
are (:) =
wa.ys to choose the
of these choices ofiour (lUarh'!I's, there are 12·11, 10·9 ways to """"",,'p.,u
in total, there axe (~) ·12 ·11 ' 10·9 = 178,200 ways for
to
Consider the
as one unit. Then we are trying to arra:n.ge
family and the eight other people -- around the table. This can be dOllEl in
the family unit can be arranged in four WdYS,
total number
'U'~"'f'"""'U"T' conditions is 4(8!).
2
FUNDAMENTALS OF LOGIC
Section 2.1
1.
The sentences in parts (a), (c), (d), and (f) are statements.
2.
The statements in parts (a), (c), and (f) are primitive statements.
3.
Since p -+ q is false the truth value for p is 1 and that of q is O. Consequently, the truth
values for the given compound statements are
(a) 0
(b)
0
(c)
(d)
1
(b) q -+ p
0
4.
(a) r -+ q
5.
(a)
(b)
(c)
(d)
(e)
6.
(a) True (1)
7.
(a) If Dard practices her serve daily then she will have a good chance of winning the
tennis tournament.
(c) If Mary is to be allowed on Larry's motorcycle, then she must wear her hehnet.
(c)
(sAr)-+q
If triangle ABC is equilateral, then it is isosceles.
If triangle ABC is not isosceles, then it is not equilateral.
Triangle ABC is equilateral if and only if it is equiangular.
Triangle ABC is isosceles but it is not equilateral.
If triangle ABC is equiangular, then it is isosceles.
8.
(a)
P--HI
1
q-+p
1
1
1
0
o
1
1
o
P-Hl
(c) True (1)
(b) False (0)
-+(q-+p)
pA
1
0
1
0
1
1
1
1
1
0
p
q
0
0
0
0
1
1
0
'0
1
1
1
1
9.
10.
0
0
1
1
q -} -r , (b) p -} (q-} '1 p-}q (c) (p-}q)-}r (h)
i 1
1
1
0
1
0
!
1
1.
1
1.
1
1
1
1.
1
0
0
0
1
1
1
1.
1
1
1
1
1
1
0
0
I
1
1
1.
0
1
1.
I
1
0
0
1
0
0
1
1
1.
1
1
1.
r
I
I
Propositions (a), (e), (f), and (h) are tautologies.
t
'"
.----.
p q r p-}(q-}r) (p -} q) -} (p-+ r)" s -} t
II
r
0
0
0
0
1
0
1
0
1
0
0
1
1
0
1
1
1
1 0
1 0
1 1
1 1
1
1.
0
1
1
1
1
1
1
1
1
1.
1.
0
1
1
0
1
1
1
1
1
1
11.
(a) 25 = 32
12.
(a) [(p 1\ q) 1\ rJ -} (8 V t) is false (0) when (p 1\ q) 1\ r is true (1) and s V t is false (0).
Hence p, q, and r must be true (1) while s and t must be false (0).
13.
p: 0;
14.
(a) n = 9
15.
(a)
r ; 0;
8: 0
(b) n = 19
m = 3, n = 6
m=4, 11>=9
(b)
m
=
~n=
(c) n=1.9
n =9
n=9
(c)
m = 18, n = 9
16.
=90
-10=
11.
Consider the
possibilities:
(i) Suppose
the
or the second statement is
true one.
and (4) are false - so
are true. And we
eat the piece of pie (4) we conclude that Tyler
Now
(3) is the
and (4) no longer contradict
other. But now statement (2)
27
statements
guilty
statement (2» and Tyler guilty (from statement (3».
(iii) Finally,
the last possibility - that is, statement (4) is
true one. Once
again statements (3) and (4) do not contradict each other, and here we learn from statement
(2)
Dawn is the vile
Section 2.2
(a)
(i)
p
0
0
0
0
1
1
1
1
q
r
0 0
0 1
1 0
1 1
0 0
0 1
1 0
1 1
qAr
0
0
0
p -+ (q !\ 1»
p-+q
1
1
1
1
,
I
0
1
0
0
1
1
1
I
1
I 1
0
0
0
1
0
0
p -+ r I (p -+ q) A (p -+ r)
1
1
1
1
1
1
0
1
1
I
0
0
0
1
1
0
1
1
1
(ii)
p
0
0
0
0
q r pVq (p V q) -+ r
1
0
0 0
p-+r
1
q-+r
(p-+ r)A(q -+ r)
1
1
0
1
1
1
1
0
1 1
1 0 0
1 0 1
1 1 0
1 1 1
1
1
1
1
1
1
0
1
1
1
0
1
0
0
1
1
1
0
1
0
0
1
0
1
1
0
1
0
1
0
1
,
1
1
(iii)
Ip q r l q V r
1010 0
0 o! 1
0 1
0 1
1 0 0
1 0 1
1 1 0
1 1 1
~I
0
1
1
1
0
1
1
1
p -+ (q V 'r)
p"~
1
1
1
1
1
1
0
1
1
0
q
-'1' ->
(p -+ q) I
1
1
1
1
0
0
1
1
1
1
1
I
I
b)
[p -l> (q V 1")]
{::=::}
{::=:;}
[.1' -l> (p -l> q)]
[-.1' -l> C.p V q)]
¢=::} [( -"p A -.q) -l> 1'}
{::=::}
[(p A -.q) -l> r]
From part (iii) of part (a)
By the 2nd Substitution Rule,
and (p -l>
{::=:;} (.p V
By the 1st Substitution Rule,
and (8 -l> t) {::=::} (--.t -l> -'s), for
primitive statements s, t
By DeMorgan's Law, Double Negation
and the 2nd Substitution Rule
By Double Negation and the
2nd Substitution Rule
2.
q pAq
0
0 0
0
0 1
0
1 0
1
1 1
p
3.
P V (p A q)
0
0
1
1
a) For a primitive statement 3, 8 V -'8 {=:::} To. Replace each occurrence of s
p V (q A r) and the result follows by the 1st Substitution Rule.
by
b) For primitive statements s, t we have (s -l> t) ¢=} (-,t -l> -'8). Replace each
occurrence of .5 by p V q, and each occurrence of t by r, and the result is a consequence
of the 1st Substitution Rule.
4.
(1) [(p A '1) A rJ V [(p A q) A -,r] {=} (p A q) A (r V -,r) ¢=} (p A q) A To ¢=} P A q.
(2) [(p A q) V """Iq] {=:::} (p V .q) A (q V .q) {=:::} (p V -,q) A To ¢=} p V -'q.
Therefore,
given statement simplifies to (p V .q) -l> S or (q -l> p) -l> S
5.
a) Kelsey placed her studies before her interest
a good
cheerleading, but she (still) did not get
or ",n.'''''u
practicing
lesson.
6.
(a)
A(qV
A{-.pV.qVr)] ¢=}
V
A.r)V(pAqA
(p A q A .1")] ¢=::}
A -'1") V
A ( ""p V (q A
¢=}
A
V !(-,q V
-'[(p A
pA
V
A -'1') ¢=} -'p V -'1".
-l> 1"] {=}
A V
¢=}
V
(.q A -'1' ) V [..,p V
V
A -,r)] '¢=:}
7.
v q) A (p A (p /\ q»
p /\ q
0
o
o
o
o
o
1
1
b) (.,p /\ q) V (p V (p V q»
8.
( a)
{::=:}
p Vq
q ---;. p {:=} -'q V p, so (q ---;. p)d <==} 'q A p.
(b) p---;.(qAr){:::::?
V(qAr), so(p---;.(qAr)]d{:=}
A(qVr).
(c) p +-7c q {:::::? (p ---;. q)/\(q -7 p) {:::::=? (-.pVq)A(.qVp), so (p +-7c q)d {:::=::> (-,pAq)V( . . . qAp).
(d) P)/Jl ¢=::} (p A -,q) V ("'p /\ q), 80 (p'i.q)d {:::=::> (p V ..... q) A (,p V q).
9.
0 = 0, then 2 + 2 = l.
Let p : 0 + 0 = 0, q : 1 1 .:. . . 1.
(The implication: p --+ q) - If 0 + 0 = 0, then 1 + 1 = 1. - False.
(The Converse of p ---;. q: q -7 p) - If 1 + 1 = 1, then 0 0 = O. - True
(The Inverse of p -7 q:
-7 -.q) - If 0 + 0 =1= 0, then 1 + 1 =1= 1. - True
(The Contrapositive of p -7 q: -.q -7 .....p) - If 1 + 1 =1= 1, then 0 + 0 =1= o. - False
(a) If 0
(b) If -1 < 3 and 3 + 7 = 10, then sine;) = -1. (TRUE)
Converse: If sine;) = -1, then
< 3 and 3 + 7 = 10. (TRUE)
Inverse: If -1 :;::: 3 or 3 + 7 =1= 10, then sine;) =1= -1. (TRUE)
Contrapositive: If sin(~) =1= -1, then -1 :;::: 3 or 3 7 =1= 10.
10.
(a.) True
11.
a.)
(b)
(c)
True
b)
(q ---;. r) V "'p
12.
( ..... q V r) V 'p
I
q
p'i.q
pA -'q
0
1
1 0
1 1
0
0
0
p
0
0
1
1
0
I 1
I
0
i
-'p A q I (p A '41) V ( "'p A q)
0
True
-'(p +-7c 41)
1
0
1
0
0
0
1
1
1
0
0
13.
p q r
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0
0
0
0
0
0
1
0
0
0
14.
q pAq
0
0 0
(a) 0 1
0
0
1 0
1
1
q -+ (pA q)
1
0
P -+ [q -+ (p A q)]
1
1
1
1
1
1
1
(b) Replace each occurrence of p by pV q. Then we have the tautology (pV q) -+ {q-+
[(p V q) A q]] by the first substitution rule. Since (p V q) A q {::::=} q, by the absorption laws,
it follows that (p V q) -+ [q -+ q) {::::=} To.
p
q
0 0
(c) 0 1
1 0
1 1
pVq
0
1
1
1
pAq
0
0
0
1
q -+ (p A q)
(p V q) -+ [q -+ (p A q)]
1
1
0
1
1
0
1
1
So the given statement is not a tautology. If we try to apply the second substitution rule
to the result in part (a) we would replace the first occurrence of p by p V q. But this
does not result in a tautology because it is not a valid application of this substitution rule
- for p is not logically equivalent to p V q.
15.
(a) -'p ¢:=:} (p i p)
(b) p V q {:=:} ..( -op A -,q) ~ ('p """'q) ~
t (q i q)
pA q <f;=:}
A
t q) {::=} (p i q) i (p i
(d) p-+q4';:;"
Vq~
1\
{=}
i
~pt(qiq)
(e) p H- q ~ (p -+ q) /\ (q --+ p) {:=:} t /\ u {::::=} (t '1£) i
i u), where t
p (q
and u for q (p
r
r r
16.
(a)
¢:=>
(p L p)
V
P A q {::=} ,"'p A
(e)
r
r r
pVq~
for
r
P -+ q ~
pH- q {.=::::}
<==>
<==>
V q {::::::::}
1 '1') 1 (8 !
lq)!p]![(q!q)!
4=}
1 l (p 1
1
{.=::::} (p 1 p) ! (q 1
1 q) 1 (..,p 1
1 p) 1 q] 1 r(p 1 1 q}
r
[(p 1 1 1 {(p 1 p) 1
31
and s
17.
q
-'(p! '1)
11'
0 0
0
1
0 1
1
0
I1
I1 1
18.
1
I
(-'1' r -.q)
-'(p i q) ("p 1 -'(1)
I
0
0
0
0
0
0
0
I
1
1
1
1
(a) [(1'Vq)/\(1'V-,q)]Vq
<=> [p V (q A ,q)J V '1
<=> (1' V Fo) V q
<=> pV'1
Reasons
Distributive Law of V over A
q /\
{:> Fo (Inverse Law)
1'V Fo {:> l' (Identity Law)
(b) (1' - t q) 1\ [-,q A (1' V -.q)J
(1' - t q) A.q
{:>
<=>
¢}
{:>
¢}
('1' V (1) A -''1
-'q A (-,pV (1)
(-.q A -'1') V ( -q A q)
( -.q A "'1') V Fo
{:>
-.q A..,1'
-.('1 VI')
(a)
pV [pA (1'V q)]
{:}
pV1'
{:>
l'
(b)
pV q V ("'1' A -''1 A 1')
{:>
{:}
(1' V q) V [-.(1' V q) A r]
[(1' V (1) V -'(1' V q)] A (p V q V r)
To 1\ (1' V q V r)
{:>
1'VqVr
{:>
19.
¢}
Reasons
Absorption Law (and the
Commutative Law of V)
1'-+q{:>
V'1
Commutative Law of A
Distributive Law of A over V
Inverse Law
Identity Law
DeMorgan's Laws
Reasons
Absorption Law
Idempotent Law of V
(c)
-+(1'AqA
#
#
#
V(pAqAr)
<*
1
Reasons
DeMorgan's Laws
Distributive Law
Law
Law
V
I\qt\
I\ql\
pl\q
(1' 1\
1\ (-,t' V q V -,q)J V
V t V -'1') 1\ -,q) ¢::;::::> [p A (-'1' V
V
1\
~ l' V -q
(b)
V (p A
20.
V over A
V (1' /\ q /\
/\ [{p /\ r 1\
V [(To V t) 1\
V t] <¢=> P /\ t by the Absorption
Section. 2.3
1.
(a)
q
0 0
0 0
0 1
0 1
1 0
1 0
1 1
1 1
p
r
0
1
0
1
0
1
0
1
p -). q I (p V q)
1
0
1
0
1
1
1
1
1
0
1
0
I
1
1
1
1
J
(p V q) -). r
1
1
0
1
0
I
1
I
0
1
The validity of the argument follows from the results in the last row. (The first seven rows
may be ignored.)
'b)
-.q p -). -"'Y'
(p A q) -). r
1
0
1
1
0
1
1
1
1
0
1
1
1 1 0
0
1 1 1
1
p q
0 0
0 0
0 1
0 1
1 0
1 0
r
,pV'(j
1
1
1
1
1
1
0
0
1
1
0
0
1
1
1
0
1
0
1
1
1
1
0
0
The validity of the argument follows from the results in rows 1, 2, and 5 of the table. The
results in the other five rows may be ignored.
pV(qVr)
[pV(qVr)]A,q
pVr
0
0
0
1
1
1
1
1
0
0
0
1
1
1
0
Consider the last
value IF V \j
A
1
1
1
1
truth
truth value
Here we
p Vr
that whenever
2,
2.
rows of
table
and 6. Rows 1, 3,
are
for assessing
7,
argument are rows
8 may
a)
p
q
q---+'r
P-H l
1
0 0 0
1
0 0 1
1
0 1 0
1
0 1 1
0
1,0 0
0
1 0 1
1
1 1 0
1
1 1 1
'r
1
1
0
1
1
1
p->rr
1
1
(p ---+ q) A (q ---+ r)] ---+ (p ---+ r)
1
1
1
1
1
1
0
1
1
1
0
0
1
1
1
1
b)
p
0
q
0
1
1
1
(p ---+ q) A -.q
1
p---+q
1
1
0
1
1
1
0
0
0
0
1
0
r(p ---+ q) 1\ ,qJ ---+ 'p
1
1
c)
p
q
0
0
0p
1
1
0 1
1 0
1
0
0
1
pVq
(pVq)l\.p
0
0
1
1
[(p V q) 1\ -.p] ---+ q
1
1
1
1
1
0
0
1
(d)
p
0
0
0
r
q
p---+'r
1
0 0
0 1
1 0
1
o
1 0 0
0
1
1
1
o 11
1
1 10
0
1
11
I
i
1 1
1
q---+r'
1
1
1
1
I
0
1
1
1
s
....----.
(p V q) ---+ r
1
1
0
[(p ->r r) 1\ (q ---+ r)] ---+ s
1
1
1
1
1
0
0
1
0
1
1
1
1
1
1
80
P A q.
0, tlu:m
the truth
stal~en:Jlent
q Vs
value 0 only
truth value
p (and thai of q) is
[(p V q) 1\ -'PJ IS
regardless of
each of
8
truth
Then
(p
q) has truth value 1 when p has truth value
oS) has truth value 1 when r
has
value O. But then (p V r) must have truth value 0,
---j>
---j>
(e) For (-,p V
the truth value is 0 when both p, r have truth value 1. This then
forces '1, s to have truth 'value 1,
for (p ---} q), ('1' ---}
to have
value 1.
However this results in
value 0
( -.q V
j
4.
(a) Janice's daughter Angela will check Janice's spark plugs. (Modus Ponens)
(b) Brady did not solve the first problem correctly. (Modus ToBena)
(c) This is a repeat-until loop. (Modus Ponens)
(d) Tim watched television ill the evening. (Modus Tollens)
5.
( a) Rule of Conjunctive Simplification
(b) Invalid - attempt to argue by the converse
(c) Modus ToUens
(d) Rule of Disjunctive Syllogism
(e) Invalid - attempt to argue by the inverse
6.
(a)
Steps
(1)
qAr
(2) q
(3) /. q V r
Reasons
Premise
Step (1) and the Rule of Conjunctive Simplification
Step (2) and the Rule of Disjunctive Amplification
Consequently, (q 1\ r) ---} (q V r) is a tautology, or q A r =? q V r.
(b) Consider the truth value assignments p : 0, q : 1, and r : O. For these assignments
[p A (q A r)} V -,[p V (q 1\ r)] has truth value 1, while [p A (q V r)] V -,[p V (q V r)J has truth
value O. Therefore, P ---j> PI is not a tautology, or P '::fo Pl'
1.
(1) & (2)
(3)
(4)
(5)
(7)
(8)
Premise
Steps (1), (2) and the Rule of Detachment
Premise
(4)
Steps (6), (7) and the Rule of Disjunctive
(8)
8.
(1)
(4)
of Detachment
Rule of Conjunctive
Rule
(5)
(7)
(8)
(9)
(10)
Premise
Step
and [r -.~ V {=}
V t) -4
Step (8)
DeMorgan's Laws
Steps (6), (9) and the Rule of Detachment
(11)
(12)
(13)
(14)
(15)
9.
. Step
[(-,p V q) -4 <====} [-,." -4
V q)]
Step (12) and DeMorgan's Laws and the Law of
Steps (10), (13) and the Rule of Detachment
Step (14) and the Rule of Conjunctive Simplification
Negation
(a)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
Premise (The Negation of the Conclusion)
Step (1) and -,(-..,q -4 8) {=} -{"""q V 8) {=} ..(q V 8) {:.=.} -'q A...,s
Step (2) and the Rule of Conjunctive Simplification
Premise
Steps (3), (4) and the Rule of Disjunctive Syllogism
Premise
Step (2) and the Rule of Conjunctive Simplification
Steps (6), (7) and Modus Tollens
Premise
Steps (8), (9) and the Rule of Disjunctive Syllogism
Steps (5), (10) and the Rule of Conjunction
Step (11) and the Method of Proof by Contradiction
(b)
(1)
p-"q
,t} "..~
pVr
-4,-4r
-,rV 8
r-4,fJ
(2)
(3)
(4)
(5)
"'p -1> q
q-1>'r
-'P -1> r
(7) ","" p
10.
Step (2) and the Rule of Conjunctive Simplification
Steps (3), (4) and the Law of the Syllogism
Premise
Steps (5), (6) and Modus Tollens.
(3)
(1)
(2)
(3)
(4)
p/\-,q
p
r
pAr-
(5) /; (pAr)Vq
Premise
Step (1) and the Rule of Conjunctive Simplification
Premise
Conjunction
Steps (2), (3) and the Rule
Step (4) and the Rule of Disjunctive Amplification
(b)
(1)
p, p - t q
(2)
q
(3)
"-Iq V r
(4)
q-1>1"
(5) /" l'
Premises
Step (1) and the Rule of Detachment
Premise
Step (3) and -.q V r <¢::=> (q -1> r)
Steps (2), (4) and the Rule of Detachment
(c)
p - t q, -.q
(1)
(2)
"'p
(3)
-'1'
(4)
-'p A. -l)"
(5) /" -.(p V r)
Premises
Step (1) and Modus Tollens
Premise
Rule of Conjunction
Steps (2), (3) and
Step (4) and DeMorgan's Laws
(d)
Rule of
(3)
p-tq
Premise
(2),
(e)
(1)
(2)
(3)
(4)
(5)
(6)
p
p- q
q
pAq
p - (q - r)
(7)
(p 1\ q) - r
(8) :~ :r
Premise
Premise
Step (2)
(p - q) ¢:::::}
Steps (1), (3) and the Rule of Detachment
Steps (1), (4) and the Rule of Conjunction
Premise
Step (6), and [p - (q - r)] {:::::::> [(p 1\ q) - r]
Steps (5), (7) and the Rule of Detachment
(f)
(1)
pAq
(2)
P
(3)
p-(rl\q)
(4)
rl\q
(5)
r
(6)
r - (.5 V t)
(7)
s Vt
(8)
-'8
(9) ,,".. t
Premise
Step (1) and the Rule of Conjunctive Simplification
Premise
Steps (2), (3) and the Rule of Detachment
Step (4) and the Rule of Conjunctive Simplification
Premise
Steps (5), (6) and the Rule of Detachment
Premise
Steps (7), (8) and the Rule of Disjunctive Syllogism
(g)
(1)
-'S,
(2)
(3)
(4)
(5)
P
pVs
p - (q - r)
q- r
t - q
t -
r
(7) ~"q -,r-
Premises
Step (1) and the Rule of Disjunctive Syllogism
Premise
of Uet,aCIlmeJ[U
Steps (2), (3) and
Premise
of
Syllogism
and
Steps (4),
Step (6) and
Vr
(p-
p-r
(3)
(5)
(6)
-''1'
pVq
-q
,,"'$ q
Premise
Steps
{::}
Vr)
(3)
-q)
q:O
(a) p: 1
q:O
(b) p:O
q:l
p:O
(c) p,q,r: 1 8:0
(d) p,q,t': 1 8:0
12.
r :1
r : 0 or 1
r:l
a) p: Rochelle gets the supervisor's position.
q: Rochelle works hard.
T: Rochelle gets a raise.
s: Rochelle buys a new car.
(p A q) --+ r
r--+s
-'8
.. -'P V -'q
.$ ,.
(1)
(2)
(3)
-'1'
(4)
(pAq) -4 l'
-'8
r -4 8
(5)
.(p A q)
(6) ...." .p V.q
Premise
Premise
Steps (1), (2) and Modus TollellS
Premise
Steps (3), (4) and Modus Tollens
Step (5) and -.(p A q) {:::::::> -'p V -'q.
b) p: Dominic goes to the racetrack.
q: Helen gets mad.
r: Ralph plays cards all night.
s: Carmela gets mad.
t: Veronica is notified.
(q
-> t
Modus Tollens
(9)
'f' ~ s
(10) "'1'
(11) / .. ..,p A . . . r
Premise
Steps (8), (9) and Modus Tollens
Steps (7), (10)
Rule
Conjunction
c) p: There is a chance of rain.
q: Lois' red head scarf is missing.
r: Lois does 110t mow her lawn.
s: The temperature is over 80°
The following truth value assignments provide a counterexample to the validity of this
argument:
p : 0; q : 0; 'f' : 1; s : 1
t
(a)
13.
p
0
0
0
0
1
1
1
1
q r pVq
0
0 0
0
0 1
1
1 0
1
1 1
1
0 0
1
0 1
1
1 0
1
1 1
..,pV r
1
1
1
1
0
1
0
1
(p V q) A ( "'p V r)
0
0
1
1
0
qVr
t~(qVr)
0
1
1
1
1
1
1
1
1
1
0
0
1
1
1
1
1
1
1
truth table it follows that [(p V q) 1\ ( ...... p V r)] ~ (q V
18 a
values that can
Alternately we can try to see if there are
assigned
p, q,
so that
V
has
0 while (p V q), ("'p V r) both have
value
r'
From the last column of
tautology.
l~
we can
it
we have p : 1
that q : 0
q : O.
we cannot
(qV
Steps
pVq
qVp
...,( -,q) V P
-..q ~ p
1.
2.
1.
2.
3.
4.
5. -'P V r
6. p~r
7. -:q ~ r
8. """',, q V 'f'
5.
6.
7.
8.
(b)
(i) Steps
1. pV(qVr)
2. (p V q) A (p V r)
3. pVr
4. p~s
o. ...,pVs
,
6. " ., rVs
..
Steps
1. p+-?q
2. (p ~ q) A (q ~ p)
3. p~q
4 . ...,pV q
5. P
6. pVq
7. [(p V q) 1\ (""p V q)]
8. qVq
.,
9. d!!l1> q
Reasons
1. Premise
2. Step (1) and the Distribution Law of V over A
3. Step (2) and the Rule of Conjunctive Simplification
4. Premise
5. Step (4), p ~ 8 {::} ...... p V s
6. Steps (3), (5), the Rule of Conjunction, and Resolution
(ii)
(iii) Steps
1. pVq
2. p-l>r
"'pVr
V
Reasons
Premise
Step (1) and the Commutative Law of V
Step (2) and the Law of Double Negation
Step (3), -:q ~ p {:} -(,q) V p
Premise
Step (5), p ~ r {:} ""p V r
Steps (4), (6), and the Law of the Syllogism
Step (7), . . . q ~ r {::} q V r
1.
2.
3.
4.
5.
7.
7.
8.
9.
Rea..':lons
Premise
(p +-? q) {::} [(p ~ q) A (q ~ p)]
Step (2) and the rule of Conjunctive Simplification
Step (3), p ~ q {:} . . . p V q
Premise
Step (5) and the Rule of Disjunctive Amplification
Steps (6), (4), and the Rule of Conjunction
Step (7) and Resolution
Step (8) and the Idempotent Law of V.
1.
2.
3.
A
V
5. qVr
6. r~8
5.
-.r V s
""i •
8.
V
6.
A
"ifI., q V s
V
8.
Reasons
Premise
Step (2), p -I> r {:} 'F V r
(1),
Rule
Step (4) and Resolution
(iv)
1.
Reasons
.pV q V r'
q V (.p V r)
1.
2.
3.
4.
-''1
3.
4.
5.
[['1 V
6.
(.p V r)
Commutative and
V
V (.p V r)
V r)] 1\ [.'1 V
'r
8.
-'1' V .p
9.
[(1' V .p) A (-,r V ...,p)]
10.
. "p
!# ..
V r)Jl
Pranlse
Step (3) and
Rule of Disjunctive
Amplification
5. Steps (2), (4),
the Rule of
Conjunction
6. Step (5), Resolution,
the
Idempotent Law 1\
Premise
8. Step (7) and the Rille of
Disjunctive Amplification
9. Steps (6), (8), the Commutative Law
of V, and the Rule of Conjunction
10. Step (9), Resolution, and the
Idempotent Law of V
(v)
..... pVs
l.
2. pVqvt
3. pV(qV
3.
l.
[[PV (q V
1\
V .s)]
5. (qvt)Vs
7.
qV(tVs)
,qVr
8.
[[qV(tVs)J
9.
(tVs)Vr
10.
tV(sVr')
.t V (81\ r)
12. ( -It V /3) 1\ ( ...t V
11.
13 .
• tv s
14.
Ht V (8 V r)J A (..,t V s)]
15.
(sVr)Vs
Ql
16. ,,'" r V 8
Associative
(2) and
Law of V
4. Steps (3), (1),
of Conjunction
5. Step (4) and Resolution (and
the
Substitution Rule)
6. Step (5) and
Associative Law of V
7. Premise
8. Steps (6), (7), and
Rule of Conjunction
9. Step (8) and Resolution (and the
First Substitution Rule)
10. Step (9) and the Associative
Law of V
11. Premise
12. Step (11) and the Distributive
Law of V over A
13. Step (12) and the Rule of
Conjunctive Simplification
14. Steps (10), (13), and the Rule
of Conjunction
15. Step (14) and Resolution (and
the First Substitution Rule)
16. Step (15) and the Commutative,
Associative, and Idempotent Laws of V
( c) Consider the following assignments.
p: Jonathan has his driver's license.
q: Jonathan's new car is
of gas.
VU(l." ...... " ••u likes
drive
Vq
P V-,r
Premise
1.
3.
...,pVq
p V-,r
(p V -'1") A (,p V q)
Premise
Premise
3. Steps (2), (1), and the Rule of Conjunction
Resolution
4. Step (3)
Step (4) and
Commutative Law of V
6. Premise
7. Steps (5), (6), and the Rule of Conjunction
8. Step (7) and Resolution
9. Step (8) and Idempotent La\\' of V
,1" V q
5.
6.
7.
8.
9.
q V-'r
.q V-,r
V
A
-'1' V-'r
V -w)
.,"'" -'1"
Section 2.4
l.
(a)
(d)
2.
False
True
(a) (i) True
(b)
(e)
False
False
(c)
(f)
False
False
(ii) True
(iii) True
(iv) True
(b) The only substitution for x that makes the open statement [pC x) A q( x)] 1\ r( x) into a
true statement is x = 2.
3.
Statements (a), (c), and (e) are true, while statements (b), (d), and (f) are false.
4.
(a) Every polygon is a quadrilateral or a triangle (but not both). (True - for this
universe.)
(b) Every isosceles triangle is equilateral. (False)
(c) There exists a triangle with an interior angle that exceeds 1800 • (False)
(d) A triangle has all of its interior angles equal if and only if it is an equilateral triangle.
(True)
(e) There exists a quadrilateral that is not a rectangle. (True)
(f)
exists a rectangle that is not a square. (True)
(g) If all
sides of a polygon are equal, then the polygon is an equilateral triangle.
its interior angles are
if
44
if
of
are
3x [m(x) A c(x) I\j(x)]
(b)
[sex) A c(x) A -.m(x)J
(c)
'Ix {c(x) -7
V p(x»]
(d)
\Ix [(g(x) A c(x» -7 -,p(x)J,
or "Ix {(p(x) A c(x» -7 -.g(x)J,
or 'Ix [(g(x) A p(x» -7 -,c(x)]
(e)
'Ix [(c(x) A sex»~ -7 (p(x) Ve(x»),
(a)
6.
(a)
(d)
7.
(b)
(e)
True
True
True
False
(c)
(f)
True
True
False
True
True
False
False
(a)
(i)
q(x)
(ii) 3x [P(x) A q(x)]
(iii) 'Ix [q(x) -7 -,t(x)]
(iv) 'Ix [q(x) -7 -.t(x)]
(v) 3x [q(x) A i(x)]
(vi) "Ix [(q(x) A 1'(.1:» -7 sex)]
(b) Statements (i), (iv), (v), and (vi) are true. Statements (ii) and (iii) are false: x = 10
provides a counterexample for either statement.
(c)
(i) If x is a perfect square, then x > O.
(ii) If x is divisible by 4, then x is even.
(iii) If.1: is divisible by 4, then x is not divisible by 5.
(iv) There exists an integer that is divisible by 4 but it is not a perfect square.
(d) (i) Let x O.
(iii) Let x
20.
=
8.
(a)
(c)
(g)
True
True
True
(a)
(i)
=
(b) False: 1<or x = 1, q(x) is true while p{x) is false.
(e) True
(d) True
(f) True
(11) False: For x = -1, (p(x) V q(x» is true but rex) is false.
9.
(c)
10.
(i)
(iii)
(i)
(iii)
True
True
True
True
Ttue
(ii)
(iv)
(ii)
x=
'reue
False -
(")
,11
(a)
70
45
Consider x = 3.
(c) 3m,n A[m,n] >60
(d) "1m
~ n < 19) ---+ (A[m, n] < A[m, n
1])]
(e) "In [(15m < 9) ---+ (A{m, n] < A[m + 1, n])]
V 1 ~ m,i 53 V 15 j 520
n) =1=
---+ (A[m,n]
A[i,jDl
11.
(a)
this case the variable x free while the variables y, z are bound.
(b) Here the variables x, y are bound; the variable z is free.
12.
(a)
(i) False
(iv)
(vii)
(viii)
(ii) True
(iii) True
False, if x = 0 (v) False, if x = 0 (vi) True
False - If y = 0 then x =f. 0; if y =f. 0, let x = 2y.
False - Let x :::::: 2 and y = -2, for example.
(b) Statements (iv), (v), and (viii) are now true - because of the change in universe.
(c) (i) True
(ii) True
(iv) False - For any y consider x = 2y.
(iii) True
13.
(a) p(2,3) Ap(3,3) A p(5, 3)
(b) [P(2, 2) V p(2, 3) V P(2, 5)] V [P(3, 2) V p{3, 3) V p(3, 5)] V [P(5, 2) V p(5, 3) V p(5, 5)]
(c) [P(2, 2) V p(3, 2) V p(5, 2)] A [P(2, 3) V p(3, 3) V p(5, 3)] A [P(2, 5) V p(3, 5) V p(5, 5)]
14.
Statements (a), (b), (e), and (f) are logically equivalent and each may be expressed as
Vn[q(n) ---+ pen)]. Statements (c), (g) are logically equivalent and each may be expressed
as Vn[p( n) ---+ q( n )}. Statement (d) is not logically equivalent to any of the other six
statements.
15.
a) The proposed negation is correct and is a true statenlent.
b) The proposed negation is wrong. A correct version of the negation is: For all rational
numbers x, y, the sum x y is rational. This correct version of the negation is a true
statement.
c) The proposed negation is correct - but false. The (original) statement is true.
d) The proposed negation is wrong. A correct version of the negation is: For all integers
x) 11, if
odd, then xy is even.
The
statement is true.
16.
(a)
or,
17.
dass
student in Professor ....."",'' ' ' 4<:.w..
not majoring in either ,'r.rn n.
history is in PrcXl:es;soI Lenhart's C++
a) There exists an ,nr",,,,.,,,,,.,. n such that n is not divisible by 2
n such
k-m
46
m - n are
n is eyen (that
k-n
odd.
c) For some real number x, X 2 > 16
d) There exists a real number x
s x S 4 (that is,
S x and x
4).
or x :2: 10.
- 31 < 7 and either x
18.
"'Ix [-,p( x) A -,q( x)]
(b)
[-,p(x)v q(x)]
(c) 3x [p{x) A -.q(x)]
(d) Vx [(p(x)Vq(x»I\-,r(x)]
19.
(a) Statement: For all positive integers m,n, ifm > n then m 2 > n 2 • (TRUE)
Converse: For all positive integers m, n, if m 2 > n 2 then m > n. (TRUE)
Inverse: For all positive integers m,11., if m S n then m 2 S n 2 • (TRUE)
Contrapositive: For all positive integers m, n, if m 2 S n 2 then m S n. (TRUE)
(b) Statement.: For all integers a, b, if a > b then a 2 > b2 • (FALSE - let a = 1 and
b = -2.)
Converse: For all integers a, b, if a2 > b2 then a > b. (FALSE - let a = -5 and b = 3.)
Inverse: For all integers a, b, if a S b then a 2 S b2 • (FALSE - let a = -5 and b = 3.)
Contrapositive: For all integers a, 0, if a2 S b2 then a S b. (FALSE - let a = 1 and
b = -2.)
(c) Statement: For all integers m, n, and p, if m divides nand n divides p then m divides
p. (TRUE)
Converse: For all integers m and p, if m divides p, then for each integer n it follows that
m divides nand n divides p. (FALSE -let m = 1, n = 2, and p = 3.)
Inverse: For all integers m, n, and p, if m does not divide n or n does not divide p, then
m does not divide p. (False -let m = 1, n = 2, and p = 3.)
Contrapositive: For all integers m and p, if m does not divide p, then for each integer n
it follows that m does not divide n or n does not divide p. (TRUE)
(d) Statement: \:Ix [(x> 3) ~ (x 2 > 9)] (TRUE)
Converse: Vx [(x 2 > 9) ~ (x > 3)] (FALSE -let x = -5.)
Inverse: Vx [(x S 3) ~ (x 2 ~ 9)] (FALSE -let x = -5.)
Contrapositive: Vx {(x:.'! 9) ~ (x S 3)] (TRUE)
(e) Statement: \:Ix [(x 2 + 4x - 21 > 0) ~
> 3) V (x < -7)]] (TRUE)
Converse: l/x [[(x > 3) V (.1: < -7)J ~ (x 2 4x - 21> O)J (TRUE)
s 0) ~
Vx [(x 2
:::; x <
(x +4x - 21
or
4x -
~
S
(TRUE)
S 3) A (x
2
S 3) 1\:2:
4x - 21 SO»)) or \:Ix
x
0)] (TRUE)
integer is divisible by 21,
by 7,
it is
n.r>~'nn''''' ulteJ~er 18
7. (TRUE)
Inverse:
a positive integer is not
by 21, then it is
divisible by 7. (FALSE
- consider the positive
14.)
Contrapositive: If a positive integer is not divisible by 7, then it is not divisible by 21.
(TRUE)
Implication: If a snake is a cobra, then it is dangerous.
it is a cobra.
Converse: If a snake is
Inverse:
a snake is not a cobra, then it is not dangerous.
Contrapositive: If a snake is not dru.lgerous, then it is not a cobra.
( c) Implication: For each complex number z, if z2 is real then z is real. (FALSE - let
z = i.)
Converse: For each complex number z, if z is real then Z2 is real. (TRUE)
Inverse: For each complex number z, if z2 is not real then z is not real. (TRUE)
Contrapositive: For eaf'.1. complex mnnber z, if z is not real then Z2 is not real. (FALSE
-let z = i.)
21.
(a)
True
(b)
False
(c)
False
(d)
True
(e)
False
22.
(a) True
(b)
False:
(c) True
(d)
True
(e)
True
23.
(a) Va 3b fa b = b + 0,= 0]
(b) 3u Va [au = ua = a]
(c) Va =1= 0 3b lab = ba =:: 1]
Cd) The statement in pru.·t (b) remains true but the statement in part (c) is no longer
true for this new universe.
24.
(a) True
25.
(a) 3x 3y {(x> y) 1\ (x - y :S 0)1
(b) 3x 3y [(x < y) A Vz[x > z V z 2
(c) 3x 3y [(lxl = Iyl) 1\ (y f ±x)]
(b) False
yn
1+1
at most three perfect
",",,"U"""''''.u we may write 28 = 25
sunl
(c) False
1
=
4
4+
7
7 as
the integers 15
48
(d)
True
3.
Here we find that
30 = 25
32 = 16
34 =
36
4 +1
16
9
50 = 25 + 25
52 = 36 16
54=25254
56 = 36+ 16+4
58 = 49 + 9
40 = 36 4
= 25+
1
44 = 36+4+4
46 = 36 + 9+ 1
48 = 16 16 16
= 36
38=36+1+1
4.
16 = 13 + 3
18 = 13 + 5
20 = 17 +3
22 = 17 5
24 = 17 7
26 = 19+7
4=2+2
6=3 3
8=3+5
10=5+5
12 = 7 +5
14 = 7 + 7
28 = 23 5
30 = 17
32 = 19 + 13
34=17+17
36 = 19 17
38 = 19 + 19
5.
(a) The real number 1f is not an integer.
(b) Margaret is a librarian.
(c) All administrative directors know how to delegate authority.
(d) Quadrilateral M N PQ is not equiangular.
6.
(a) Valid - This argument follows from the Rule of Universal Specification and Modus
POllens.
(b) Invalid - Attempt to argue by the converse.
(c) Invalid - Attempt to argue by the inverse.
7.
( a) When the statement 3x [PC x) V q( x)] is true, there is at least one element c in the
prescribed universe where p( c) V q( c) is true. Hence at least one of the statements
p( c), q( c) has the truth value 1, so at least one of the statements 3x p( x) and 3x q( x) is
true. Therefore, it follows that 3x p(x) V 3x q(x) is true, and 3x [P(x) V q(x)J = }
3x pC x) V 3x q( x). Conversely, if
p( x) V 3x q( x) is true,
at least one of
pea), q(b) has truth value 1, for some a, b in the prescribed universe. Assume without
that it pC a).
V
truth value 1 so 3x [PC x) V q( x)] is a
true statement, and
p(x) V
q(x) = }
[P(x) V q(x)].
OvVJ.j,,,,,,,,,,',,,,, when the statement \ifx [P(:1:) A
occurs
a
prescribed
is true (as is q( a))
a in the unh'el'se, so
A
statements Vx pC x),
and
A
p(x) A Vx
the
,,,,,,,,,,,,<>,p that the statement Vx p( x) V "'Ix
p(x) is
true, as
p(c) V q(c). Hence 'Ix [P(x) V q(x)] is true and Vx p(x) V 'Ix q(x) :::=:;} 'Ix [P(x) V q(x)].
(0) Let p( x): x > 0
q( x): x < 0 for the universe of all nonzero integers. Then
Vxp(x),Vxq(x) are false, so
p(x) vVx q(x) is false, while
fp(x)Vq(x)] is true.
9.
(1)
(2) Premise
(3)
the Rule of Universal Specification
(4) Step (2) and the Rule of Universal Specification
(5) Step (4) and the Rule of Conjunctive Simplification
(6) Steps (5), (3)1 and Modus Ponens
(7) Step (6) and the Rule of Conjunctive Simplification
(8) Step (4) and the Rule of Conjunctive Simplification
(9) Steps (7), (8), and the Rule of Conjunction
(10) Step (9) and the Rule of Universal Generalization
10.
11.
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
Step (1) and the Rule of Universal Specification
Steps (3), (4), and the Rule of Disjunctive Syllogism
Premise
Step (6) and the Rule of Universal Specification
Step (7) and -.q(a) V f'(a) <=> q(a) -~ rea)
Steps (5), (8), and Modus Ponens (or the Rule of Detachment)
Premise
Step (10) and the Rule of Universal Specification
Step (11) and sea) --I> -,r(a) <=> -,-,r(a) ~ -,s(a) <=> rea) ~ -,s(a)
Steps (9), (12), and Modus Ponens (or the Rule of Detachment)
Consider the open statement.s
w(x): x works for the credit union
£( x ): x writes loan applications
c(x): x knows COBOL
q(x): x knows Excel
and let r
i represent Imogene.
SYlnoi)!1c form
"Ix
-4 c(x)]
\Ix [('W(x) A e(x)} ~
A
as
Vx [w(x) ~ c(x)]
q( i) A ,c( i)
-.c(i)
w(i)-I>c(i)
(1)
(2)
(3)
(4)
(5) -,w( i)
(6) Vx [(w(x) Al(x»
(7) w(r) A .q(r)
(8) -.q(r)
Premise
~ q(x)]
(9)
(w(r)t\l(r»~q(r)
(10)
-,(w(r) Al(r»
w(r)
(11)
-.w(r) V .l(r)
,l(r)
(12)
(13)
(14) .. .,,, """Il(r) A .w(i)
Step (2) and the Rule of Conjunctive Simplification
Step (1) and the Rule
Universal Specification
Steps (3), (4), and Modus Tollens
Premise
Premise
Step (7) and the Rule of Conjunctive
Simplification
Step (6) and the Rule of Universal Specification
Steps (8), (9), and Modus Tonens
Step (7) and the Rule of Conjunctive
Simplification
Step (10) and DeMorgan's Law
Steps (11), (12), and the Rule of Disjunctive
Syllogism
Steps (13), (5), and the Rule of Conjunction
12.
(a) Proof: Since k,£ are both even we may write k = 2c and l = 2d, where c,d are
integers. This follows from Definitioll 2.8. Then the sum k + l = 2c + 2d = 2( c + d) by the
distributive law of multiplication over addition for integers. Consequently, by Definition
2.8, it follows from k + e= 2(c + d), with c d an integer, that k + l is even.
(b) Proof: As in part (a) we write k = 2c and l = 2d for illtegers c, d. Thenby the commutative and associative laws of multiplication for integers - the product
kl = (2c)(2d) = 2(2cd), where 2cd is an integer. With (2c)(2d) = 2(2cd), and 2cd an
integer, it now follows from Definition 2.8 that kf is even.
13.
(a) Contrapositive: For all integers k and l, if k, l are not both odd then kl is not odd.
- OR, For aU integers k and i, if at least one of k, i is even then ki is even.
Proof: Let us assume (without loss of generality) that k is even. Then k =
for some
integer c - because of Definition 2.8.
ki = (2c)l = 2(d), by the associative
of
multiplication for integers - and d is an integer. Consequently, kl is even - once again,
Definition
[Note that
result
not require anything about the integer
Contrapositive; For all integers k
k + " is odd. -
k
l,
k and i, if one
l
51
other even
14.
Proof: Since 11, is odd we may write n = 20, + 1, where a is an integer - by Definition
2.8. Then n 2 =
1)2 = 4a 2
+ 1 = 2(2a 2 2a) + 1, where
2a is an integer.
2
So again by Definition 2.8 it follows that n is odd.
15.
Proof: Assume that for some integer n,
is odd while n is not odd.
n is even
and we may write n = 20" for some integer a - by Definition 2.8. Consequently, n 2 =
(20, y~ = (20,)(20,) = (2·2)(0,' a), by the commutative and associative laws of multiplication
for integers. Hence, we may write n 2 =2(20,2), with 20,2 an integer - and this means that
n 2 is even. Thus we have arrived at a contradidion since we now have n 2 both odd (at the
start) and even. This contradiction came about from the false assumption that n is not
odd. Therefore, for every integer n, it follows that n 2 odd ::::} n odd.
16.
Here we mu.st prove two results - namely, (i) if n 2 is even, then n even; and (ii) if n is
even, then n 2 is even.
Proof (i): Using the method of contraposition, suppose that n is not even - that is, n is
odd. Then n = 20,+1, for some integer a, and n 2 = (20,+1)2 = 4o, 2+4a+1 = 2(20,2+20,)+1,
where 20,2 + 20, is an integer. Hence n 2 is odd (or, not even).
Proof (ii): If n is even then n = 2c for some integer c. So n 2 = (2c)2 = (2c)(2c) =
2(c(2c» = 2«c.2)c) = 2«2c)c) = 2(2c2 ), by the associative and commutative laws of
multiplication for integers. Since 2c2 is an integer, it follows that n 2 is even.
17.
Proof:
(1) Since n is odd we have n = 2a + 1 for some integer o,. Then n + 11 = (20, 1) + 11 =
2a + 12 = 2(a 6), where 0,+6 is an integer. So by Definition 2.8 it follows that n + 11 is
even.
(2) If n + 11 is not even, then it is odd and we have n 11 = 2b + 1, for some integer b.
So n = (2b + 1) - 11 ::::: 2b - 10 = 2(b - 5), where b - 5 is an integer, and it follows from
Definition 2.8 that n is even - that is, not odd.
(3) In this case we stay with the hypothesis - that n is odd - and also assume that
n 11 is not even - hence, odd. So we may write n
= 2b 1, some integer b.
This then implies that 1'1, = 2(0 - 5), for the integer b - 5. So by Definition 2.8 it follows
even. But
n both even
shown) and odd (as in the hypothesis) we have
OOlltr'Ml1Ct1~[)n. So our
was
it now
that n 11
18 even
odd integer n.
18.
19.
integers
=
For example, m = 4 = 22 alld n = 1
are two
m +n
22
=5
a np1ri'P,:"t
=
m
n = 25 = 52, 80 the
18
21.
given
by establishing the truth of its (logically equivalent)
Proof: We shall
contraposi Hve.
us consider the negation
conclusion - that is, x <
and y < 50. Then
x < 50 and 11 < 50 it follows that x 11 < 50 + 50 = 100, and we have the negation
method of proof (by the
of the hypothesis. The given result now follows by
contrapositive).
22.
Proof: Since 4n + 7 = 4n + 6
is odd.
23.
Proof: If n is odd, then n = 2k + 1 for some (particular) integer k. Then 7n + 8 =
7(2k + 1) 8 = 14k + 7 + 8 = 14k 15 = 14k 14 + 1 = 2(7k + 7) + 1. It then follows
from Definition 2.8 that 7n + 8 is odd.
1 = 2(2n + 3) + 1, it follows from Definition 2.8 that 4n + 7
To establish theconveise, suppose that n is not odd. Then n is even, so we can write
n = 2t, for some (particular) integer t. But then 7n 8 = 7(2t) + 8 = 14t + 8 = 2(7t + 4),
80 it follows from Definition 2.8 that 7n+8 is even - that is, 7n+8 is not odd. Consequently,
the converse follows by contraposi tion.
24.
Proof: If n is even, then n = 2k for some (particular) integer k. Then 31n + 12 =
31(2k) + 12 = 62k + 12 = 2(31k + 6), so it follows from Definition 2.8 that 31n + 12 is
even.
Conversely, suppose that n is not even. Then n is odd, so n = 2t + 1 for some (particular)
integer t. Therefore, 31n+12 = 31(2t+l)+12 = 62t+31 12 = 62t+43 = 2(31t+21)+1,
so from Definition 2.8 we have 3In 12 odd - hence, not even. Consequently, the converse
follows by contraposition.
1-
s
0 0
r
0
0
0
o 11
1
0
1
0
1
0
0 1
0 1
0 1
1 0
1 0
1 10
1 10
1 1
1 1
1 1
1
2.
1
1!0 i
1
1
0
0
1
1
0
0 1
1 0
1 1
0 0
0 1
1 0
1 1
(jl\r
0
0
0
0
0
0
1
1
0
0
0
0
0
0
1
1
1
0
0
1
1
1
1
1
1
1
0
0
0
1
0
0
0
0
0
0
1
1
1
1
1
1
1
1
0
0
1
1
1
1
0
0
0
1
0
0
0
1
0
0
0
0
0
a)
p q
0 0
0 0
0 1
0 1
1 0
1 0
1 1
1 1
p-tq
1
0
1
1
1
0
1
1
0
0
1
0
1
0
1
1
r
p,
.....p - t r
(p - t q) 1\ ( ..... p - t r)
0
1
0
1
0
1
!
0
1
0
0
1
1
1
1
1
1
r.
54
3.
~a4-~~____~________~~__~~~__~__~
p q r q+-+r
+-+r) (p+-+q)
,000
1
010 1
0
0100
01 '1 1 1
10
01
1
1 10!1
1
0
,l'O
0
1 1 1
1
0
I
Wlll
1
1
1
1
1
0
1
0
0
0
1
0
1
0
0
1
0
1
1
O~
Jl
,~----~--------~----~~-----
+-+ r)] ¢} [(p H- q) H- r].
the results
follows
The
[p
-jo
p : 0; q : OJ r : 0 result
truth value 1 for
(q --l- r)] and 0 for [(p --l- q) --l- r]. Consequently, these statements are not logically
equivalent.
4.
P f-+ q {::::;:::> (p -jo q) A (q --l- p) ~ (..,p V '1) A (-,q V p), so -{p f-+ q) { = }
-(,p V q) V -,(-,q V p) ¢::::} (p A ""'Iq) V (q A -,p)
5.
Since p V -'q ¢} -'-'p V -'q ¢} 'P --l- -'q, we can express the given statement as:
(1) If Kaylyn does not practice her piano lessons, then she cannot go to the movies.
But p V 'q {::} -'q V P {::} q --l- p, so we can also express the given statement as:
(2) If Kaylyn is to go to the movies, then she will have to practice her piano lessons.
6.
a) p--l-('1Ar)
Converse: ('1 A r) --l- p
Inverse: {"p --l- -,( q A f')] {:} [-op --l- (-oq V -",) J
Contrapositive: (-'('1 A r) -jo -,p] {::} [( -'q V
--+ -,p]
b) (pVq)-."r
Converse: r --l- (p V q)
Inverse: [.,(p V q) --l- -,rJ ¢} ("p A -oq) --l-
V
V
--l-
V q)] ¢:}
A
V
A
--l-
A
Ap
Ap
A
~
A
~
~
A
~
{=}
V
~
pA "'1
A over V
pA
Fo is
{=}
Fo
"
for
(a)
A -.q) V
P A ('1 V -..r V
A
(q A -'r A
[(p A Fo) v
A
¢:}
-'p V
it
A -.r A
1\ [1' V S V
9.
(a)
(d)
10.
contrapositive
(b)
(e)
(c)
contrapositive
converse
Proof by Contradiction
(1)
(2)
-'(p pA -.s
p
(3)
(4)
p-q
q
(5)
(6)
r
qA1'
(7)
(qAr)-+s
(8)
s
(9)
(10)
-'s
sA
-.s
(11)
Fo)
(12) ..4",p _ S
Method 2
(1)
(2)
Ar) -+ s
r -4 (q -+
r
'1 -.j. s
p-+q
Premise (Negation of Conclusion)
Step (1), (p -+ s) ¢=:::} -'p V 5, DeMorgan's Laws, and the Law
of Double Negation
Step (2) and the Rule of Conjunctive Simplification
Premise
Steps (3), (4), and the Rule of Detachment
Premise
Steps (5), (6), and the Rule of Conjunction
Premise
Steps (7), (8), and the Rule of Detachment
Step (2) and the Rule of Conjunctive Simplific.ation
Steps (9), (10), and the Rule of Conjunction
Contradiction
Steps (1), (11), and the Method of Proof
Method 3
)
(2)
(3)
$
V
A r)
(5)
r
(6)
8 V -;q
(7)
q -+ .9
(8)
p -+ q
(9) /" p -+ s
Step (1)
for primitive
u, v
u -+ v {:;> """IV -+ """I'U - and the 1st Substitution Rule.
Step (2) and for primitive statements u, tJ, U -} V {::::::=> -''U V V -and
Substitution Rule. Also, ,-'8 {::::::=> S.
Step (3), DeMorgan's Law, and the
Law
V
Premise
Steps (4), (5), and
Law of Disjunctive Syllogism
Step (6) and s V
{::::::::} -'q V .5 {;::=:} q -+ s
Premise
Steps (7), (8), and the Law of the Syllogism
Method 4 (Here we assume p as an additional premise and obtain s as our conclusion.)
(1)
P
(2)
p -+ q
(3)
q
(4)
r
(5)
qA r
(6)
(q A r) -+ s
(7) /" s
Premise (assumed)
Premise
Steps (1), (2), and Modus Ponens
Premise
Steps (3), (4), and the Rule of Conjunction
Premise
Steps (5), (6), and Modus Ponens
11.
1
0
1 01
1
1
ill 1
(p Y. q) Y '"
q V r
0
0
1
1
1
1
(}
(}
p
(q y. r)
0
1
1
0
(}
1
0 1
1 0
1
(}
()
0
.':L
p:
q:
r :
on
Craig wears his suede jacket.
(of the
IS
are LUC;.uuc;u.
p --t (r --t q)
The argument is invalid. The
counterexample.
13.
(a) True
(e) False
a)
(b)
(f)
This statement is true.
7( -2x) + 5(3x).
False
False
assignments p: 1;
q: 1;
(c)
(g)
(d)
True
False
(h)
r': 0
True
True
Note that 1 = 7(-2) + 5(3), so for each integer x, x =
b) Since 2 divides both 4 and 6, it fonows that 2 divides 4y 6z. Consequently, the result
is false for each odd integer x. [Since 2 = 4( -1) + 6(1), the result is true for each even
. ./,
J
In~eger x.
15.
Suppose that the 62 squares in this 8 x 8 chessboard (with two opposite missing corners)
can be covered with 31 dominos. We agree to place each domino on the board so that
the blue part is on top of a blue square (and the white part is then necessarily above a
white square). The given chessboard contains 30 blue squares and 32 white ones. Ea.ch
domino covers one blue a.nd one white square - for a total of 31 blue squares and 31 white
ones. This contradiction tells us that we cannot cover this 62 square chessboard with the
31 donlln06.
16.
Suppose that the 60 squares in the 8 x 8 chessboard (with two squares - one blue and
one white - removed from each of two opposite corners) can be covered with 15 of these
T -shaped figures. When covering the chessboard we agree to place each T -shaped figure
on the board so that the color of each square in the T -shaped figure matches the color of
chessboard square
it covers. Let n
of T -shaped figures
blue squares (and one white one) used in the covering. The chesshoard contains 30 blue
so it
3n
1·(15n-n)=30.
us
we
CHAPTER 3
THEORY
Section 3.1
1.
They are all the same set.
2.
All of the statements are tnle except for part (f).
3.
All of the statements are true except for parts (b) and (d).
4.
All of the statements are true except for parts (a) and (b).
5.
(a) {O,2}
(b) {2,2~,3~,5~,7~}
(c) {O, 2, 12, 36, SO}
6.
(b)
(e)
(a) True
Cd) False
True
True
(c)
(f)
True
False
7.
(a) "Ix [x E A -+ x E BJ A
[x E B A x ~ A]
(b) 3x {x E A A x If/. B] V Vx [x ~ B V x E A}
8.
(a) 27 = 128
(b) 128 - 1 = 127 (We substract 1 for 0).
(c) 128 ---1 = 127 (We subtract 1 for A)
(d) 126
(e)
= 35
(f) For each of the other
elements of A there are two choices: Include it with
or exclude it, frotn a
that conta,ins 1,2.
the rule of
there are
<>Ut.,,,,,,,,"O C!:JIHan.!Jug 1,2.
(~)
= 64
(i)
= 64
G)
G)
=6
=n
1.
10.
the
11.
one
(b) 30
(c)
28
of one ""....-,n,<r one
(a)
(~2) = 924
(b)
(:) (~) =
25- 1 = 63
(c)
(a) (~O)
(b) Since
smalle,st element in
5 we must select the other four elements in A from
{6, 7,8, ... ,29, 30}. This canbe done in (~5) ways.
(c) Let x denote
element
A. Th.en
are
cases to consider.
(x = 1)
we can choose the other four elements in
ways.
e:)
S
(x = 2) Here there are e4 ) selections.
(x = 3) There are
subsets possible here.
en
(x = 4) In this
case we have
e46)
e:) + (~) + (z:) + e:) subsets A where IAI = 5 and the smallest element
In total there are
in A is less than 5.
14.
(a) There are 211 subsets for {1,2,3, ... ,1l},and 26 subsets for {1,3,5,7,9,1l}. The
26 subsets of {I, 3, 5, 7,9, II} contain none of the even integers 2,4,6,8,10. Hence, there
are 211 - 26 = 1984 subsets of {I, 2, 3, ... ,11} that contain at least one even integer.
(b) 212 - 26 = 4032
(c) For n=2k+l,where k~O,thenumberofsubsetsof {l,2,3, ... ,n} containing
at least one even integer is 2'11. - 2k+ 1 .
For n = 2k, with k ~ 1, the number of such subsets is 2n - 2"',
15.
Let W = {I}, X = {{1},2}, Y = {X,3}.
16.
(n = 6)
(n = 7)
(n= 8)
17.
(a)
and
1
1
21
7
8
15
6
1
28
20
Since A ~ B, x E B. Then
by
35
35
56
15
70
1
6
21
56
1
7
28
8
1
B S; C, xE
So xEA==>xEC
C. With A c
E B =::::} x E
an
19.
k E Z+ with n 2:: k + 1, consider the hexagon centered at (:). This has the
(a)
(n+l)
1.:+1
where the two alterllating triples - namely,
(k~l) (nil) = (n;;l) (~!D
satisfy
(b) For n,k E Z+ with n > k
-
20.
k
1
k
I
n.
1,
(n-l)( +n )(n+l) [
k- 1
(n+1)
(n)
(n-l)
k
k+l' k-l -
(;=D, (k~l)' (ntl)
-
(n-I)l
(k - l)!{n - k)!
1[(k + l)!(nn!- k - I)!1[k!{n(n+l)!
1
+ 1 - k)!
+ I)! 1[
71,!
1
[k!(nen-1- 1)1- k)1 1[(k +(n1)!(71,
- k)! (k - l)!(n - k 1)1 =
(a) Each of these strictly increasing sequences of integers correspollds with a subset of
{2,3,4,5,6}. Therefore there are 25 such strictly increasing sequences.
(b) 25
( c ) 235 and 235
(d) Let m, n be positive integers with m < n. The number of strictly increasing sequences
of integers that start with m and
with n is 2[(n-m)+1]-2 = 2 n - m - 1 •
. (1/4)(;) =: (n~l) :::::} (1/4)[(n!)/(5!(n-5)!)] = (n-l)!/(41(n-5)!):::::}
= 20(n-1)!:::::}
n =20.
=22n
a) 2n
23.
For a
nE
kE N so
(k:1) =2
n2::2
(:), (k:l)'
k;::
to
(k n 2) = 3(~).
so 21:
2=n-
or n =
2
-
- 4 =
or 0 = k 2 - 4)(1e
k ?:
it
Ie = 4
So the 5th, 6th, and 7th entries ill the row for n = 14 provide the unique
(2
24.
0000
1000
1100
0100
0110
1110
1010
0010
25.
0
0011
1011
{w}
{w,
{x}
y}
{w,x,y}
{w,y}
{V}
1111
0111
0101
1101
1001
0001
{V,z}
{w,y,z}
{w,x, z}
{x,y,z}
{x,
{w,x,z}
{w, z}
{z}
A = {x,v,w,z,y}.
As an ordered
26.
27.
(a) If S E S, then since S = {AlA rt. A} we have S rt. S.
(b) If S ¢ S, then by the definition of S it follows that S E S.
28.
(b)
10
20
30
40
50
70
80
90
Random
Dim B(12), S(6)
B(l) = 2: B(2) = 3: B(3) = 5: 8(4) = 7
B(5) = 11: B(6) 13: B(7) = 17: B(8) = 19
B(9) =:: 23: B(lO) = 29: B(ll) = 31: B(12) = 37
I:::::;: 1
6
SCI) : : :;: Int(Rnd*40) 1
J=1
I-I
If S( J) = SCI) Then GOTO 70
=
J
110
I
For 1= 1
130
140
6
J= 1
If 8(1) =
GOTO 170
J
160
GOTO
170 Next I
180
62
For I = 1 To 5
200
Print S(I); '" "i
Next I
220 Print S(6)j "
is a subset of
230
290
240 Print "The subset 8 contains the c;n:;,LU<:;U~O
For 1= 1
1)
260
Print SCI); ", " ;
270 Next I
280
8(6); " but it is not a subset of B"
290 End
29.
procedure Su,bJeis4( i,j,k, 1: positive integers)
begin
for i := 1 to 4 do
for j := i+1 to 1) do
for k := j+l to 6 do
for l:= k+l to 7 do
print ({ i,j,k,l})
end
30.
Program List_subsets4 (Input, Output);
Canst
= 10;
Type
63
If i
S
Begin
S := S - [i];
S <> [},
Write (i:3, "')
Else Write (i: 3);
End;
Writeln
End;
Procedure Subsets (L,R : Set-type; i: Member _type);
Begin
If i <== n then
Begin
Subsets (L [i], R, i+l)i
Subsets (L, R [il, i+ 1);
End
Else
Begin
Wrlte..set (L);
Write..set (R);
End;
Endj
Begin
Write ('What is the value of nT);
Readln (n);
Subsets ([1],[ ],2);
3.2
(by
[2,3]
0) U
U
== {A -
U (A n
we
A=
64
3,4,
9,11}.
we
B=
(b)
= {I, 2, 4, 5, 9},
(a)
(i)
(iv)
True
True
(i)
(iv)
E
-
7,8,9}.
4.
(b)
(ii)
False
(iii)
(v)
(ii)
(v)
(vi)
B
z-
set of all
E
(positive and negative) odd integers
False
False
(iii)
(vi)
D
5.
(a)
(f)
True
True
(b)
(g)
True
True
(c)
(11)
True
False
(d)
(i)
False
False
(e)
True
6.
(a) x E An C => (x E A and x E C) = } (x E B and x E D), since A ~ Band
CD=} x E B n D, so An C ~ B n D.
x E AU C = } x E A or x E C. If x E A, then x E B, since A ~ B. Likewise,
x E C => xED. In either case, x E Au C ==> x E BUD, so AU C ~ BUD.
(b) Let A ~ B. We always have 0 ~ An B, so let x E An B. Then x E A and x E B.
x E A ==> x E B, since A
B. x E B, x E B ==> x E B n B = 0, so An B = 0.
Conversely, for An B = 0, let x E A. If x ¢ B, then x E B, so x E An B = 0. Hence
x E B and A ~ B.
( c) Follows from part (b) by the principle of duality.
1.
(a) False. Let U = {1,2,3}, A = {I}, B = {2}, C = {3}. Thell An C = B n G but
A i= B.
(b) False. Let U = {1,2}, A = {I}, B = {2}, C = {1,2}. Then AUB= AuG but
Ai: B.
(c) x E A ==> x E A U C ==> x E B U C.
x E B or x E C. If x E B, then we
case, x E B so
are finished. If x E C, then x E A n C = B n C and x E B. In
11 E B ==> 11 E B U C = A U C, so 11 E A or y E C. If y E C, then
y E B n G = An G. In either case, 'II E A B A . Hence A = B.
(d) Let x E
Consider
cases: (1) x E C ==> x ¢ AL\C ::=:} x ¢ BL\C ==> x E B.
(ii) x ¢ ==> x E AL\C = } x ¢
=} x E
case
a ",..,.w,u"",we
so
=B.
8.
From the Venn diagrams it follows that ALl(B n C) f:. (A~B) n (ALle), so the result is
false.
(b)
True
(c) True
9.
(A n B) U C = {d, x, z} which has 23 - 1 = 7 proper subsets; An CB U C) = {d} which
has 1 proper subset.
10.
(a) 0
11.
(a)
(b)
(c)
(d)
12.
The dual of the statement An B = A is the statement Au B = A. But AU B = A '¢:::::}
B ~ A, so the dual of the statement
~ B is the statement B ~ A.
13.
(a) False. Let U = { 2,3}, = {I}, B = {2}. PCA) = {0,A},P(B) = {0,B},
peA U B) = {0, {l},
{I, 2}},
{1,2} ¢ peA) U PCB).
X E PCA) n PCB) '¢:::::} X E peA)
E PCB) ¢=} X
A and X
B '¢=}
An ~ X E p(AnB), so peA) np(B) = P(AnB).
14.
(b)
0 and 1
0=(AuB)n(AUB)n(AUB)n(AUB)
A = A U (A n B)
AnB=(AUB)n(AUB)n(AUB)
A=(AnB)U(AnU)
'd)
B
0
0
A
0
0
'0
0
1
AnB AnC (AnB)U(AnC)
1
0
0
0
0
I 1
0
0
1
0
1
0
0
1
0
I
1
0
0
C
0
1
0
I
1 !1
o 0
0 1
1
1
1
1
1
0
1
1
I
1
1
0
0
0
0
AnB AnC (AnB)U(AnC)
1
1
0
0
0
0
1
1
0
0
15.
0
1
0
0
0
0
0
0
1
0
1
1
0
0
(b) 2n
(c) In the columns for A, B, whenever a 1 occurs ill the column for A, a 1 likewise occurs
in the srune position in the
for B.
(a)
~ = 64
67
Ed
0
0
0
1
1
0
1
1
0
0
1
1
1
1
1
0
1
0
1
0
1
1
0
0
1
1
1
1
0
0
0
0
1
1
1
1
1
0
1
0
1
1
1
0
16.
--
11.
Steps
(A n B) U [B n « c n D) U ( C n D» 1
(A n B) U [B n (C n (D U D))]
(A n B) U IB n (C n U)]
(AnB)U(BnC)
(B n A) U (B n C)
Bn(AUC)
DuD=U
Identity Law {C n U = C]
Commutative Law of n
Distributive Law of n over U
U 4.. = A'/' = {I, 2, 3,4,
n = ={
7'
6,7},
1'1,=1
Al
111=1
m
m
-
UAn=
nAn=
... ,rn -
-
}.
n=l
n:::::l
(a) [--6,9}
20.
Distributive Law of n over U
(a) An(B-A)=An(BnA)=Bn(AnA)=Bn0=0
(b) f(A n B) U (A n B n n D)J U (A n B) = (A n B) U (A n B) by the Absorption Law
= (AUA)n B =UnB = B
(e) (A - B) U (A n B) = (A n B) U (A n B) = A n (B U B) = A n U = A
(d) AUBU(AnBnC) = (AnB)U[(AnB)nC] = [(AnB)U(AnB)]n[·-,-(A-n-B-,-)UC] =
[(AnB)UC] =AUBUC
7'
18.
Reasons
(b)
(f)
f-8,
(e) 0
3]
(g) R
at
x E
onei E I~xE
(d)
-6)U
1--2 ,
onei E I ~ x E
12}
1.
universe U comprises
600 freshmen.
we
A,
U be the subsets
A: the freshmen
attended
first showing
the freshmen who attended the second showing,
then lUI = 600,
= 80, IBI = 125, and IA n BI = 450.
Since IA n
= 450, it follows
UBI = 600 - 450 =
Consequently,
IA n Bi =
- IA U Bi = 80 125 - 150 = 55 - that is, 55 the 600 freshmen
attended the movie twice.
2.
Here the universe U comprises the 2000 automobile batteries. If we let A, B ~ U be the
subsets
A: the batteries with defective terminals
B: the batteries with defective plates,
then lUI = 200, 1::4 n 111 = 1920, IBI = 60, and IA n BI = 20.
Since An B, = Au B, it follows that 1.4 UBI = 2000 - 1920 = 80. From IA U BI =
IAI + IBI-IA n BI we learn that IAI = IA U BI- IBI + IA n BI = 80 - 60 + 20 = 40 - that
is, 40 of her 2000 batteries have defective terminals.
3.
There are 29 such strings that start with three 1 '13 and 28 that end in four D's. In addition,
25 of these strings start with three 1 's and end in four D's. Consequently, the number that
start with three 1'13 or end in four D's is
29 + 28 -- 25 = 512 + 256 - 32 = 736.
4.
(a) Here AUBUC = C, so IAUBUCI = ICI = 5000.
(b) Here AnBnC = 0 as well, so it follows from the formula for I.4UBUCI = 1.41+IBI+ICI =
50 500 + 5000 = 5550.
(c) IAUBUCI = IAl IBI ICI-lAnBI-jAuCI-jBnCI IAnBncl = 50+500
5000 - 3 -- 3 - 3 + 1 = 5542.
-81
(a)
F
12
(h) 2
'VU'AJ'"'''''IC''' four cases.
IJ1.Q,.LU.\.!)
can
(36)3 ways.
(c)
(b) _ F U __ . Here the blanks can
fined
(26)(36)2 ways.
(c) __ FUN _. Again there are 26(36)2 ways to fill in the blanks.
(d) ___ F
N. There are also 26(36)2
to fill in the blanks
character variable names
is (36)3
Consequently
number of
3(26)(36)2 -1, because the variable FUN FUN is counted in both case (a) and case (d).
are also (36)3 3(26)(36)2 - 1
these variables that
I P and
contain both F U and TIP. Consequently, the number of these six character variable.':!
that contain
FUN or TIP is 2[(36)3 + 3(26)(36)2 - 1]- 2.
circle labeled (i) is for the arrangements with conS's; circle (ii) is for consecutive E'Si and circle
for consecutive 1's. The answer to the problem
the number of arrangements in region 8 which we
obtain as follows. For region 5 there are 10! ways
arrange the 10 symbols M,I,C,A,N,O,U,SS,EE,LL.
For regions 2,4,6 there are (11!/21) - 1O! arrangements containing exactly two pairs of consecutive letia's. Finally each of regions 1,3,7 contains (12!/(2!2!»!/21)-1O!]-101 arrangements, so region 8 contains
[13!/(2!)3J -3[121/(2121)] 3(11!/21) -10! arrangements.
(i1
num.ber of NTNlgt'lmmts
10.
either H before E, or E before T, or T before
M equals the total number of a:rroogements (i.e., 7!) minus the number of arrangements
E is before H,
T
before E, and M is before T. There are 3! ways
to arrange C, I, S. For each arrangement there are four locations (one at the start, two
between pairs ofletters, and one at the end) to select from, with repetition, to place M, T,
E, H in this prescribed order. Hence there are (3!)e+:- 1) = (3!)G) arrangements where
M is before T, T before E, and E before H. Consequently, there a.re 7! - (3!) (!)
arrangements with either H before E, or E before T, or T before M.
Section 3.4
1.
(a)
=2/8 =
UB) = 5/8
=1U
s=
s=
... ,
...
)
70
= 0.04.
each
likely
= 6 outcomes
A.
4.
The probability of each equally likely outcome is
0.;4 = 0.02 = n Therefore, n =
= 50.
(a) (~)/e;) = 15/66 = 5/22 = 0.2272727 ...
(b)
[(Dell)) + ei)(D (!)(;)l/(;O) = 21/66 = 7/22 = 1- [(;0)/(;2)J
6.
S={{x,y}lx,yE{1,2,3, ... ,
100},x=ly}
A = {{x,x + 1}lx E {1,2,3, ... ,99}}
= e~) = 4950; IAI = 99
Pr(A) = 99/4950 = 1150
7.
8 = {{x, y}lx, y E {I, 2, 3, ... ,99, IOO}, x =I y}
A = Hx, y} I{x, y} E $ 1 X Y is even}
= {{x, y}l{ x,y} E $, x, 11 even} U {{x, y}l{x, 11} E $, x, y odd}
1$1 = (l~) == 4950; IAI = e~)
= 2450
Pr(A) = 2450/4950 = 49/99
e:)
8.
=I
$ = {{a,b,c}la,b,c E {l,2,3, ... ,99,100}, a
b, a"# c, b c}
A = {{a, b, c} a, b, c} E S, a b c is even} = {{ a, b, c} I{ a, b, c} E S, a, b, c are even, or
+
Ii
one of a, b, c is even and the other two integers are odd}
e:) (;0) (~) = 19,600 + 61,250 = 80,850
181 = e~) 161,700; IAI =
+
Pr(A) = 80,850/161,700 = 1/2.
=
9.
The sample space S = {(Xl? X21 X3, X4, Xli, x6)lxi = H or T, 1 i :5 6}. Hence 181 = 26 = 64.
(a) Here the event A = {HHHHHH} and Pr(A) = 1/64.
(b) The event B = {HHHHHT,HHHHTH,HHHTHH,HHTHHH,HTHHHH,
THBHHH} and Pr(B) = 6/64 = 3/32.
(c) There are 6!f(4!2!) 15 ways to arrange two heads and four tails, so the prohability
for
event is 15/64.
( d) 0 heads: 1 arru.ngement
2
[6!/(2!4!)J =::;
arra,ngements
4 !le<l~US;
6 nea~as:
so the
all
=
Here
(~) (~)I
(1
-
(b)
Cs4) (;0) /(~) =
.45)/177100':' 0.09249
(c) (:) (~) (~)/(~) = (4·9·45)/177100 -~
s=
(a.)
$; Xi $;
i = 1,2,
lSI =
:::: 216.
Let A = {(Xl, X',h
1 n < 5, !{(n,xz,xs)!n < Xl and n < xa}l = (6 - n)2.
Consequently, lill =
42 +
22 + 12 = 55.
Therefore, Pr(A) = 55/216.
(b) With S as in part (a.), let
:::: {(XbXZl
< X2 < xs}.
l{(l,xa,xs)ll < X2 < X3}! = 10,
!{(2,X:h X3)!2 < Xa < x3}1 = 6,
!{(3,Xlh X S)!3 < Xz < x3}l = 3, and
l{(4,xz,xa)14 < Xz < X3}1:::;: 1,
so !BI = 20 and Pr(B) = 20/216 = 5/54.
12.
(a.) 10
(b)
1
(e)
4/15
13.
(a) 1j!
= .1.
15!
15
(c) (2)(9)(13!)/(15!) = 3/35
14.
(a) 24/300 = 0.08
(b) (i) There are 180 students who can program in Java. Two can be selected in
e~) ways. The sample space consists of the (~) pairs of students. So the proba-
(b) [(14!)
(141)1/(151) = 2(14!)/(15!) = 2/15
bility that two students selected at random can both program
(180)(179)/{300)(299) == 0.36. (li)
/
Java is e~) e~) =
e~2)/(~)':' 0.29.
15.
Pr(A) = 1/3i Pr(B) = 7/15, Pr(A n B) = 2/15; Pr(A U B) = 2/3. Pr(A U B) = 2/3 =
(1/3) (7/15) - (2/15) =
Pr(B) - Pr(A n B).
16.
(a) 2(5!/2!)J/I(7!/(2!2!»)] = 120/1260 .:. 0.0952
(b) [(7!/{2f2!» - 2«6!/2!) - 5!JI(7!/(2!2!))J =
[(7!/(2!2!)) 5!}![(7!/(2!2!»J ==
= 1-
= 1= 0.6
= 1- Pr(B) = 1- 0.3 =
U
-
n
n - -0.2=
Pr(A n B) = PreB) - P1'(A n B) = 0.3 - 0.2 = 0.1
Pr(A U B) = Pr(A n B) = 1n B) = 1- = 0.9
Pr(A U B) = Pr(A n B) = 1 - Pr(A n B) = 1- 0.2 = 0.8
(a) (:)(
= (!)(t)8,: 109375
(:)ct)6(~? (~)(t)7(!) (:)(V 8 = (t)8[(!)
(c) (~)(V2(t)6 .: 0.109375
(~)(V8 (i)G)(t)" (~)
~)6 = ([(~)
(~)
(:)] .: 0.144531
+ (~)] ~ 0.144531
3.
(a) S= {(x,y)lx,y E {1,2,3, ... ,lO},x#y}.
(b) For 1 :5 y :5 9, if y is the label on the second ball drawn, then there are 10 - y possible
values for x so that (x, y) E S and x > y. Consequently, if A denotes the event described
here, then IAI = 9 + 8 7 ., . + 1 = 45 and Pr(A) = IAI/ISI = 45/90 = 1/2.
(c) Let B = {(v,w)lv even,w odd}. Then we want Pr(B U C) where B n C = 0. So
Pr(B U C) = Pr(B) + Pr(e) = ;;; + ~ = :; = ~.
4.
Here Pr(A) = !~, Pr(B) = ~, Pr(C) = !;, Pr(A n B) = :~,p Pr(A n C) = 532'
Pr(B n C) = !~, and Pr(A n B n C) =
So
13 + 26 + 12
0
:.:I
6 + 0 _ 42 _ 21
P r (A U B U C) - 52 52 52 - 52 - 52
- 52
52 - 52 - 26'
5.
Since A, B are disjoint we know that Pre A U B) = Pre A) + Pre B), so Pre B) = 0.7- 0.3 =
0.4.
6.
Pr(A~B)
1.
(a) Let p be the probability for the outcome 1. Then for 1 :5 n :5 0, the probability for the
outcome n is np and p + 2p + 3p + 4p + 5p + op = 1. Consequently p = 1/21.
:2'
= Pr(A) + Pr(B) - 2Pr(A n B)
is;1 ;1
So the probability for a 5 or 6
= ~i.
(b) The probability the outcome is even is ;1 + ~ + 261 = ~i.
I)
1
3+5
1=21=21 21 :n'
8.
9.
10.
PT(A) =
Pr(B) = ~~73
= 0.84
- 00 _ 12 n B) -lli-'25i
Pr(A U B) = Pr(A) Pr(B) - Pr(A n B) = 0.52 + 0.84 - 0.48 = 0.88
Pr(A) = 1 = 0.48
=1Pr(AU
= 1- Pr(AnB) = 0.52
Pr(A n
= 1 - Pre A U B) = 0.12
Pr(At:.B) = P1'(A) + PreB) - 2Pr(A n B) = 0.52 +
- 2(0.48) = 0.4
11.
- is = ~ :.. 0.710526
;: - is = ;~ -=- 0.710526
(b) (i):';: = SS:1 :: 0.224377
(a) (i) !~
(ii) ~
~
") 18
:2
(,11
38' 38
12.
+ as2 . 3s
18
18· 0 049861
= 3619 + 3619 = 361
= .
(i) Pr(AUB) = 1- Pr(A.UB) = 1- k=~.
(ii) Pr(A U B) = Pr(A) + Pr(B) - Pr(A n B). Here Pr(A) = Pr(B), 60 Pr(A U B)
Pr(A n B) = 2Pr(A), or 2Pr(A) = ~ k = 1. Hence Pr(A) = !.
(iii) Pr(A - B) = peA n B). Since A = (A n B) U (A n B), where (A n B) n (A n B) = 0,
we have Pr(A - B) Pr(A) - Pr(A n B)
t = 1~'
(iv) Pr(AtJ.B) = Pr(A) + P1'(B) - 2Pr(A n B) = ~ +
2(k) = ~.
=
13.
t4'
14.
[(~O) (:)
(~O) (;)
=! -
!-
e:) (~) + (~O) (~)lI((~9)
- (:) - (~)]
+
1 - 10]
= {(252)(126) (210)(84) (120)(36) (45)(9)]/[92378 = (31752 + 17640 + 4320 + 405)/92367 = 54117/92367 ..:.. 0.585891.
15.
(a) Ann selects her seven integers in one of (s:) ways.
tions there are
en Ie:) =
en
Among these possible selec-
that are winning selections. So the probability Ann is a winner is
330/3,176, 716, 400 ...:. 0.000000104. [Using a computer algebra system one
gets 0.1038808501 x 10-6 .]
(h) The probability
having two winners is (0.000000104)2 ...:. 0.1079123102 x 10- 13 NOT very likely.
16.
17.
With A ~ B it follows
In general, B = B n S = B n (A U A) = (B n A) U (B n
that
U (B n
since Ann
- n n
n 0 = 0, we
P'1'(B) = Pt·(A)+Pr(BnA). From Axiom (1), Pr(A), Pr(BnA)
so Pr(B) 2:: Pr(A).
Sediou 3.6
1.
Let A,
be
events
A:
card drawn is a
B: the card drawn is an ace or a picture card.
Pr(AIB) = Pr(A n B)/Pr(B) = (5~)/(~~) = l~ = ~ =
Pr(A n B) Pr(AIB) -
Pr(A) Pr(B) - Pr(A U
0.6 + 0.4 - 0.7 = 0.3
Pr(AnB)/Pr(B) = g:! = ~ =0.75
A = A n (B U B) = (A n B) U (A n B), with (A n B) n (A n B) = A n 0 =
Pr(A) = Pr(A n B) Pr(A n B).
0, so
Therefore, Pr(A n B) = Pr(A) - Pr(A n B) = 0.6 - 0.3 = 0.3, and Pr(AIB) =
Pr(A n B)/ Pr'(B) = 0.3/{1- 0.4] =
= = 0.5.
t
3.
Let A, B be the events
A: Coach Mollet works his football team throughout August
B: The team finishes as the division champion.
Here Pr(BIA) 0.75 and Pr(A) = 0.80, so Pr(AnB) = Pr(A)Pr(BIA) = (0.80)(0.75) =
0.60.
=
4.
Let A, B be the events
A: a given student is taking calculus
B: a given student is being introduced to a CAS.
(a) Here we want Pr(BIA).
Pr(A) = (170 + 120)/420 = 29/42
Pre B n A) = 170/420 = 17/42
So Pr(BIA) = Pr(B n A)/Pr{A) = (!D/(~) = ;~.
(b) In this case the answer is Pre AlB).
.
= 1= 1 - [(170
=1-
- .
n
Pr(AU
P1'(B) - Pr(,,4)Pr(B)
[1Pr(B)
6.
Let A, B denote
events
toss a head
B: three heads are obtain.ed
event B n A we consider
(a) P1'(BIA) =
n
_ (!)4/<!) = ~ = ~.
the :number of ways we ean place two
and two Ts
the last four positions. This
(~).]
(b) pf'(BIA) = Pr(B n A)/Pr(A) = (:)( !)4/(~) = ; = !.
1.
A, B denote the events
A: Bruno selects a gold coin
B: Madeleine selects a gold coin
(a) Pr(B) - Pr(B n A) + Pr(B n A)
= Pr(A)Pr(BIA) + Pr'(A)Pr(BIA)
= (165)(i~) + U}5)(i~) 6~t:O = ~~ =
(b) Pr(AIB) = P~~1~f) = Pr(1~~f'A)
=
8.
!;
= [Ct~)G~ )]/( :;) = le;S = ~!.
A = {TH, TT}, Pr(A) = (~)(~) (~y~ = ~ + ~ = ~ = ~
B = {TT,HH}, Pr(B) = (t? (~)2 = ~ + ~ = ~
An B = {TT}, Pr(An B) = (~)2 = ~
Pr(A n B) = i = 8~ ::/: !~ = (~)(~) = Pr(A)Pr(B), so A, B are not independent.
9.
Pr(A U B)
= Pr(A) + Pr(B) - Pr(A n B)
= Pr(A) Pr(B) - Pr'(A)Pr(B),
because A, Bare in.dependent.
0.6 = 0.3 Pr(B) - (0.3)Pr(B)
0.3 = 0.7Pr(B)
So Pr(B) = ~.
A,
events
A: Alice gets four heads (and three tails)
113 a
Pr(AIB) = Pr{A n B)/ Pr(B) = ~~~ = (~)<l)6 =
..:. 0.3125.
1
12.
(0.95)(0.98) = 0.931
13.
Let
B be
events
A: Paul initially selects a can of lemonade
Betty selects two cans of cola.
_ Pr(A)Pr{BjA)
P r(Al'B) .-. ;. . Pr(An-m
Pr(B) -Pr(B)
Pr(A) =
Pr(B) = Pr(A)Pr(BIA) + Pr(A)Pr(BIA)
= (ll)( 153)( + C~ )C~)<t52)
(.§..)(
)
60
6 _ 1
SO Pr(AIB) -- (i\HrsHTIi)
(rs){rr) - 00+240 - 30 - 5'
1;
fi)
14.
Pr(A U B U C) = Pr(A) Pr(B) + Pr(C) - Pr(A n B) - Pr(A n C) - Pr(B n C)+
Pr(A n B n C) = Pr(A) Pr(B) Pr(C) - Pr(A)Pr(B) - 0 - Pr(B)Pr(C) O.
Note: A, C disjoint::::} An C = 0 ::::} An B n C = 0 ::::} Pr(A n B n C) = O.
0.8 - 0.2 + PreB) 004;"'" 0.2Pr(B) - OAPT'(B)
0.2 = O.4Pr( B)
So Pr(B) =
= 0.5
!
15.
Let A, B denote the events
A: the first component fails
B: the second oomponcllt fails.
Here Pr(A) = 0.05 and Pr(BIA) = 0.02. The probability the electronic system fails is
Pr(A n B) = Pr(A)Pr(BIA) = (0.05)(0.02) = 0.001.
16.
Let
B, W denote
of a
blue, and white marble, respectively.
we are interested ill the following cases (with their corresponding probabilities).
8
Pre RRR) = (:9)( 18)( :7)
P1'(RRB) = (199)(188)({1)
P1'(RBR) = Pr(BRR}]
8
= (:9)( 18)( 1~)
Pr{.RBB) = (~)(l~)U)r)
Pr(BRB) = Pr(BBR)]
P:,,(AUBUC) =
11.
n C)
n n C).
Pr(B) + Pr(C) -- P1'(A n B)B, are
77
n
~ =
= Pr(A)+Pr(B)+Pr(C)-Pr(A)Pr(B)-Pr(A)Pr(C)-Pr(B)Pr(C)+
Pr(A)Pr(B)Pr(C) = (l) (:) + Pr(O) - Cl)(:) - (~)Pr(O) -- (!)Pr(C) + (l)(~)Pr(O).
~
~
= [1
~ + ;2]Pr( 0) and Pre C) =
-1-
18.
;1'
-l-
Let A, B, OJ D denote the
the
comes
source
B: the graphics card comes from the second source
0: the graphics card comes from the third source
D: the graphics card is defective.
Then
Pr(A) = 0.2, Pr(B) = 0.35, PreC) =
Pr(DIA) 0.05, Pr(DIB) = 0.03, Pw'(DIO) = 0.02
=
(a) Pr(D) = Pr{D n A)
Pr(D n B) Pr(D no) = Pr(A)Pr(DIA) Pr(B)Pr(DIB)
Pre C)Pr{ DIC} = (0.2)(0.05) + (0.35)(0.03) (0.45)(0.02) = 0.0295.
So 2.95% of the company's graphics card are defective.
(b) Pr(CID) = P~~£fJ = Pr(~~~fICl = [(0.45)(0.02)]/(0.0295) = 18/59":'- 0.305085
19.
!;
Here A = {HB,BT} and Pr(A) =
B = {HT, TT} with Pr(B) = ~; and 0 =
{HT, TH} with Pr(C) =
Also An B = {BTl, 80 Pr(A n B) = = (t)(!) = Pr(A)Pr(B)j An C = {HT}, so
Pr(A n 0) = = (VG) = Pr(A)Pr(O); and B n 0 = {HT} with Pr(B n C) =
=
<!)(i) = Pr(B)Pr(C). Conseque.ntly, any two of the events A, B, 0 are independent.
However, AnBnC = {HT} so Pr(AnBnO) = ~ -I ~ = n)(~)G) = Pr·(A)Pr(B)Pr(C).
Consequently, the events A, B, C are not independent.
t.
t
t
CD
20.
(0.75)(0.85)(0.9)
(0.75)(0.85)(0.1) + (0.75)(0.15)(0.9) + (0.25)(0.85)(0.9) = 0.57375
0.06375 0.10125 + 0.19125 = 0.93.
21.
(a) For 0 < k S 3, the probability of tossing k heads in three tosses is (!)C!)k(~?-k =
(!)(t)3. The prohability Dustin and Jennifer each toss the same number of heads is
== (V6f(:)2 + (;)2 g)2 GrJ = (
9 + 9 1} = ;~ = 1~ :".0.3125.
(b) Let x count the number of heads in Dustin's three to.~ses and y the number
x = 3: y = I, or 0; x == 2: y == 1 or OJ x =
more heads
Jeunifer is (!)(trS[(~)(l)2(~)
)(
22.
We
== (
One eou~s:m
(equal) probabilities
78
Jennifer's.
The
tails; (2) One cousin gets a tail and the other four get heads.
the other
The probability for event (1) is (5)(!Yi =
23.
So the answer is
5 _
5
+. 32 - 16'
Let A, B denote the following events:
A new airport-security employee has had prior training in weapou detection
new airport-security employee fails to detect a weapon during the first :month on the
job.
Here Pr(A) = 0.9, Pr(A) = 0.1, Pr(BIA) = 0.03 and Pr(BIA) = 0.005.
The probability a new airport-security employee, who fails to detect a weapon during the
first month on the
has had prior training in weapon detectiou = Pr(A\B) = P;f~~f) =
Pr{A}pl"(BJA)
Pr(A)Pr{BIA}
- (0 9)(0 00'"')/[0 9)(0 005) + (0 1)(0 03)] Pr(BnA)+pl"(BnA) - Pr(A)Pr(BiA)+Pr(A)Pr(BI'A'} .
. u
..
..0.0045/[0.0045
24.
0.003] = i~ = l = 0.6.
Let A, B 1 C denote the events
A: the biuary string is a palindrome
B: the first and sixth bits of the string are 1
C: the first and sixth bits of the string are the same
(a) Pr(AIB),= Pr(AnB)fPr(B)
Pr(B) = (i)(l)(l)(l)(l)(!) = ~, where ea~h 1 is the probability that a given position
(second, third, fourth, or fifth) is filled with a 0 or 1.
Pr(A n B)
(V(l)(l)(l)n)(l)
1~' where, for example, the first 1 is the probability
that the second position is filled with a 0 or 1, and the third! is the probability that the
bit in the fifth position matches the bit in the second position.
=
=
1
1
1
Pr(AIB) = Pr(A n B)/ Pr(B) = (-)/C -) = 16 4
4
(b) Pr'( AIC) = Pre A n C) / Pre C)
Pree) = (~)(1)(1)(
H+ n)(l)(1)(l)(l)(~),
for the two disjoint events where the binary
start and end with 0, or start and end
25.
26.
0.3=
n
it follows
79
U B) = 1-
Pr(Al:.BIA U B) = ;:...::...u;~~=u A U B) =
[Pr(A U B) - Pr(A n U B ) = (0.7 - 0.1)/(0.7) = 0.6/0.7 = 6/7.
27.
B1, B 2 , B 3 , and denote A denote the events
B i : for
three
randomly
from urn 1 and transferred to urn 2, i
envelopes
contain $I while the other 3 - i envelopes each contain $5,
0 S; i S; 3.
A: Carmen's selection from urn 2 is an envelope that contains U.
Here, Pr(A) = Pr(A n Bo) Pr(A n B 1 ) P'r(A n B 2 ) Pr'(A n B 3 )
= Pr(Bo)Pr{AIBo) Pr(B1 )Pr(AIB1 ) P1'(B2 )Pr(AIB2 ) Pr(B3 )Pr(AIBs )
(;4)]( l~) [(~) (~) / (;4) J( n) {(~) (~) I (~4)]( 151) + I( (:) (~) I e34)J( 161) =
5
(1~)( 1;) + (13)( Ii) (:i)( 11) + (951)Ull) =
.
(;l)CA )[42 168 + 150 + 30J = 390/1001 = 30/77.
= [(~) (:) /
28.
Pr(BIA) < Pr(B) => Pr(B n A)/ Pr(A) < Pr(B) => Pr(B n A) < Pr(A)Pr(B).
Consequently, Pr(AnB) = Pr(BnA) < Pr(A)Pr(B), so Pr(A/B) = Pr(AnB)/Pr(B) <
Pr(A).
29.
0.8 = Pr(AIB) + P7'(BIA) = P~~~f} + P~~~~} = Pr(A n B)[(1/0.3) + (1/0.5)1, so
(0.15)(0.8) = Pr(A n B)[0.5 + 0.3] = (0.8)Pr(A n B). Consequently, Pr(A n B) = 0.15.
30.
Pr(A U B) - Pr(Al:.B) = Pr(A n B) = 0.7 - 0.5 = 0.2. Since 0.5 = Pr(AIB) = Pr(A n
B)/Pr(B), it follows that Pr(B) = Pr(A n B)/0.5 = 0.2/0.5 = 0.4.
0.7 = Pr(A U B) = Pr(A)
Pr(B) - Pr{A n B) = Pr(A) + 0.4 - 0.2, so Pr(A) = 0.5.
Section 3.7
1.
(a) Pr(X = 3) = ~
(b) Pr{X 4) = 2:!=oPr(X = x) = ~ + ! +
(c) Pr(X >0) = E!=:lPr(X = x) = ~ + 1+
(d) Pr(1 ~
=:; 3) =
= x) = t +
= 21X =:; 3) =
~) =
= ~
7
Pr(X
1
4
= 2 and X
=l
>
Pr(X
= l!X ~
80
1
<4
-41
3.
, x = 0,1,2, ... ,5.
(a) Pr(X = x) =
(b)
275
P.,. ( X = 4 ) = ~~ ,393 ..:. 0.000123.
.
_
__ (140)(1116)
1.0 HOO
__ _ +"
P r (X - 5) -- l. C;O)
1;0) -
(c) Pr(X 2: 4) = Pr(X = 4)
1 1;:9
,
•
~~::.L - - 2,268,786 = 0.000121
23,100+252 _
190,578,024 -
23,352
190,578,024 -
,
(d) Pr(X = llX
2) = Pr(X = 1
::; 2) _ Pr(X = 1)
Pr(X ::; 2)
Pr(X 2)
_
10
1
110
",
I 120
1\
- [(~)e~il)/C;o )+( InC!O)/e;o»+<C~)C;O)/C~Il»l
= et) e!O) (~O) e!O)
elO) C!O) + e20) e;O)]
= (10)(5,773,185)/[(1)(122,391,522) + (10)(5, 773, 185) + (45)(215, 820)J
= 57,731,850/[122,391,522 57,731,850 9,711,900]
= 57,731,850/189,835,272= 2675/8796":'0.304116.
(a)
4.
Xl:
X 2-·
Pr(X1 = xd =
P"(X
,
2 - x)
2
--
X
Pr(X = --3)
Pr(X = -1)
Pr(X = 1)
Pr(X = 3)
i~~(!)Xl(!)3-$1 =
i: l(!)3,
1
Xl
= 0,1,2,3.
2 _.
3 (1)$2(1)3-2:
3 (1)3
2
2
-;V2
2 ) X :2 -- 0 , 1, 2, 3 -
X2
Pr(Xt = 0)Pr(X2 = 31X1 = 0) = (o~(~)3(1) = (~?
Pr(Xt = 1)Pr(X2 . 21X1 = 1) = (; (~)3(1) = (3)(t)3
Pr(Xl = 2)P1-(X:2 = l)iXl = 2) = ~)q)3(1) = (3)(t)3
Pr{X1 = 3)Pr(X2 = 0IX1 = 3) = (;)(!)3(l) = (t)3
=
=
=
=
(b) E(X1 ) = E!1=uxtPr(X1 = xd = 0 - (~)(V3 + 1· G)(!)3 + 2· (n(l)3 + 3· (;)(t? =
o i + i + ~ ::= If = ~[= 3(t) = np, since Xl is binomial with n = 3 and p = !].
E(X 2 ) = ~
E(X) = (-3)(l)3 (-1)(3)(V 3 + (1)(3)(~)3 3(!? = 0[= E(X1 ) - E(X2 )}.
= x) = Pr{X = 3) + Pr(X :::::: 4) + Pr(X = 5) + Pr(X =
5.
Pr(X=
l=c
4+9
'2
Pr(X =
so c::::::
6
81
Pr(X ~
-= 3)+Pr(X = +Pr(X = 5) = c(~
.
Pr(X = 4
~ 3)
Pr(X = 4)
(c) Pr(X = 4!X ~ 3) =
Pr(X ?: 3)
= Pr'(X ~ 3) = (43 24
(d)
-x·
-x'
= (4~) E;=:l ~~ = (4~)[1 + ~ + ~1 ~! + g~]
= i~~ 2.457364
(e) E(X2) = (:3) E;:=l ~~ = (:3)[1 1: +
225: + ~~] (:3)(:'22]) = 34~
Var (X) = E(X2) - E(X)2 =
(~;~? = i:!~
1.100895.
- C!)C::)
.:.
W-
'T.
.:.
_
-
19
43'
=
(a)l=E!=:lPr(X=X)=c
(6-x)=c(5 4+3+2+1)=15c,soc=1/15,
(b) Pr(X :::; 2) = Pr(X = 1) + Pr(X = 2) = (1~)(6 -1) (115)(6 - 2) = = ~
x· Pr(X = x) = I:!:::l x . (115)(6 - x)
= <A)[l ·5+ 2·4 + 3·3 4·2 + 5 ·1] = (115)(35) = ~
(d) E(X2) = 2:;=1 x 2 • (t\)(6 - x) =
(115)[1.5 4·4 + 9·3 + 16 . 2 + 25· 1] = (is)(105) = 7
Vax (X) = E(X2) - E(X)2 = 7 - (D2 = 63;49 = ~4
(c) E(X) =
8.
Let the random variable X count the munber of heads in the 100 tosses. Assuming that
the tosses are independent, this random variable is binomial with n = 100 and p = ~. So
Wayne should expect to see E(X) = np = 100(~) = 75 heads among the results of his 100
tosses.
9.
Since X is binomial, E(X) = 70 = np and Val' (X) = 45.5 = npq. Hence, we find
that 45.5 = 70q, so q = 45.5/70 = 0.65. Consequently, it follows that p = 0.35 and
n = 70/p = 70/0.35 = 200.
10.
Let the random variable X denote the player's net winnings and let C denote the cost of
playing one round of this carnival game. The probability distribution for X is as follows:
x
Pr(X = x)
5-C
8-0
It52 It52 -
~52 -
c=
- 7) + P1'(X Pr(X ~ 6!X ~ 4) = -~---------....:-
- (!)
Pr(X 4) = 1!=4 (!)(0.25)X(0.75)'~-x = (:)(0.25)4(0.75)4 (:)(O.25Y'(0.75)3+
(:)(O.25)6(O.75Y' (~)(O.25Y(0.75)1 (:)(0.25)8(0.75)0 ~ 0.113815
So Pr(X 6!X 2 4)~ 0.004227/0.113815~ 0.037139'0
(e) E(X) = np = 8(0.25) = 2..
(f) Vax (X) = np(l - p) = 8(0.25)(0.75) =
12.
Here O'x = J9 = 3.
(a)
'5:c X :5 23) = Pr(l1-17:5
:5 -17) = Pre -6:5
6)
= Pr'(lX - 171 :5 6) = Pr(jX - E(X)I :5 20'x) 2 1 = ,}z = 1 - ~ = ~
(b) Pr(10 :5 X
24) = Pr(IX - 171
= Pf'(!X - E(X)I G)O'x) 2 1 - l/G? =
1-.2.=1Q
49
49
fi = 1 - ! = ~
(c) Pre8 :5 X :5 26) = Pr(IX - 171 < 9) = Pr(IX - E(X)j :5 30'x) 2 1-
13.
In Chebyshev's Inequality Pr(IX - E(X)I :5 kO'x) 2 1 - k\' If 1 1 - 0.96 = 0.04 == 12' and k2 = O.~4' Since k > 0 we have k == 0:2 = 5.
Here Vax (X) = 4 so O'x = 2 and c = kO'x = 5·2 = 10.
14.
Here X is binomial with n = 20 and p = 1/6. So E(X) = np = (20)(~) = ~ = l~ and
Vax (X) = np(l - p) = (20)(1)(~) = ~.
15.
Let D denote a defective chip and G a good one. Then the sample space S =
{D, GD, GGD, GGG} and XeD) = 1, X(GD) = 2, and X(GGD) = X(GGG) = 3.
(a) Pr(X = 1) = 2~ = t
Pr(X=2) = (i~)(l~)=!~
Pr(X = 3) = G~)(!!)Ct~) + G~)(i!)(i:) =
(b) Pr(X :5 2) == Pr(X == 1) + Pr(X = 2) = 1 + = ~: = 19
(c) Pr(X = llX < 2) = Pr(X = 1 alldX <
= Pr(X = =
= 19
Pr{X :5 2)
Pr(X :5 2)
5 19
35
E(X)
=
,\",,3
•
xPr(X
=
x)
=
1(1)
2(16)
+
3(11)
1.
+
;!a
~
=
19+32+180
- ill d)
(
UX=i
5
95
19 - I)
95
19
95
95 2.431579
2)
(e) E(X2) =
Vi
ax'
x 2 Pr'(X = x) =
!:
k12
= 0.96, thell
Ii
1) (!)/(~)
~
_
19 -
1JJ±64t540 _
623
95
95
(X) = E(X2) _ E( V)2 = 623 _ (23~)2 = 5824 = 5824 ...:. 0.:14 3 9
4
95
95
(95}~ 9025
.\1.5 1
-
18.
Ca) Pr(X > 1) = Pr(X = 2)+Pr(X = 3)
P,'(X = 4) = 0.3+0.2+0.1 = 0.6 = 1-0.4 =
1-·" Pr(X S
~ 2) = Pr'(X;;.(~a;d~ ~ 2) = Pr(X =
2) =
Pr(X = 3)/{Pr(X = 2) Pr(X = 3) Pr(X = 4)] = 0.2/0.6 = 1/3
(e) E(X) = E!",1 XPl'(X = x) = 1(0.4) + 2(0.3) + 3(0.2) 4(0.1) = 2
(d) E(X2) = 2:;",,1 x 2 Pr(X = x) = r~(O.4) 22(0.3) + 32(0.2) + 42(0.1) = 5
Var (X) = E(X2) - E(X?, == 5 - 22 = 5 - 4 = 1
(b) Pr(X -:-
(a)
Word
x, the number of letters and apostrophes in the word
I'll
make
him
4
4
3
an
2
offer
he
5
can't
. refu.se
x
2
3
4
5
6
2
5
6
Pr(X = x)
2/8 = 1/4
1/8
2/8 == 1/4
6(1/8)
= 0.0006
(0.05)(0.9)(0.12)
(0.05)(0.1)(0.88) == 0.0114
0.0054
= 2) =
0.0396 = 0.2258
Pr(X = 3) = (0.95)(0.9)(0.88) =
[NO'te thatE!::o PrC X = x) = 0.0006
? 21X
= 0.1026 + v.v"'.......
0.0212
1) = Pr(xp~(~;~ ~ ;?:
0.2258 + 0.7524 =
12 = Pr(X ~ 2)/Pr(X ? 1) = (0.2258 +
0.7524]/[0.0212 0.2258 + 0.7524) = 0.9782/0.9994 = 0.978787272
(c) E(X) = E!""ox' Pr(X = x) = 0(0.0006) + 1(0.02122) 2(0.2258) 3(0.7524) = 2.73.
(d) E(X2) = E!::;::G x 2 • Pr(X :..-. x) == 02 (0.0006) + 12(0.0212) + 22 (0.2258) + 32 (0.7524) =
7.696
Var eX) = E(X2) - E(X)2 = 7.696 - (2.73)2 = 0.2431.
21.
Pr(X = 2) = [G)(Dl/(~) = 1/10
Pr(X = 4) = [(D (DJ/(~) = 3/10
Pr(X = 3) = [(~)(;)J1e) = 2/10
Pr(X = 5) = [(:) G)J/G) =4/10
E(X) = (1/10)(2)+(2/10)(3)+(3/10)(4)+(4/10)(5) = (1/10)[2 6 12 20J = 40/10 = 4
E(X2) (1/10)(4) + (2/10)(9) (3/10)(16) + (4/10)(25) = (1/1O)f4 + 18 + 48 1001 =
170/10 = 17
Var(X) = E(X2) - E(X)2 = 17 - 16 = 1, so (Tx = J1 = 1.
=
Supplementary Exercises
1.
Suppose that (A - B) ~ C and x E A - O. Then x E A but x ¢ O. If x ¢ B,
then [x E A A x ¢ B] = } x E (A - B) ~ O. SO' now we have x ¢ C and x E O. This
contradictiO'n gives us x E B) sO' (A - C) ~ B.
Conversely,
(A - 0) ~ B, let yEA - B. Then yEA but y ¢ B. If y 1. C, then
[y E A 1\ Y ¢ OJ = } Y E (A - C) ~ B. This contradictiO'n, i.e., y ¢ B and y E B, yields
y E C, so (A - B) ~ C.
2.
Let S = {x,lI,al,aZ , '" ,an}, There are (''/.;:2) subsets O'f S containing 'f' elements,
where v,? 2. These subsets {all into three categories. (a) Neither x nor y is
the
are (~) of these. (b) Exactly O'ne of x and 'II is in the subset. The6e
acC{)Ullt for
x
yare in
are
(a)
m n
t.··,x m }, B:::;: {YI,""
(0 Ie
(r-
... ,xm } U
€'Je111el1ts from
(0
(m;n) =
AuB
85
r - k
(b) Replace m
r by n in part (a), and use the
.5.
( a) 126
teams wear
uniforms) i
are
n
10
1t
n
(b) 2 - 2; (1!2)(2 - 2). 2 - 2 - 2n; (1/2)(2 - 2 - 2n).
6.
False: Let A = {O, 2,3, ... },
= {O, -1,
n BI = I{O}I = 1
(b) False: Let A = {1,2} and B = Z+.
(c) True
=Z+.
(d) False: Let A = {1,2} and
7.
(a)
8.
(a) 27
(a)
(d)
(b)
128
(b) (!) (21)
jAI=8
(c)
(~) G)
10 Random
20
S(8)
30 For I = 1 To 8
40
SCI) = Int(Rnd * 15) + 1
50
For J = 1 To I - 1
60
If 8(1)= 8( J) Then GOTO 40
70
Next J
80 Next I
-gO C= 0
100 Rem C counts the odd elements of the subset
110 For I = 1 To 8
120
If (8(1)/2) <> lut (8(1)/2) Then C = C + 1
130 Next I
Print
this program contains the elements"
1=1
7
S(I); "/';
210 End
u
so
x E
distinguishable) .
y E n B) U
then yEA n B or y E C. (i)
yEA n
:::::::::> y E
(A n B) U (A n C) :::::::::> yEA n (B u C). (ii) y E e =;.. yEA, since C ~ A. Also,
y E C => y E B U C. So yEA n (B U e).
either case «(i) or (U» we have
yEA n
u e), so (A n B) U C ~ An (B U C).
(2) Now
Z E An(BUe). Then Z E An(BU
= (AnB)U
From (1) and (2) it follows that (A n B) U e = An (B U e).
10.
n
~ (AnB)Ue.
BI-/A n BI = 5,80 there are 25 subsets e where An B ~ C ~ AU B.
(a) Here
The number containing an even number
elements is
(for
(:)
= 4)
(;) (for
(~) (for lei = 8) = 16.
(b) 2 ; (~) + (~) (!) == 16.
lei = 6)
5
(b) {0}U(6,12]
11.
(a) [0,14/3]
(e) [0,+00)
12.
(a) AAB = (A - B) U (B - A) == (B ~ A) U (A - B) = BAA
(b) AAA = (A - A) U (A - A) = AU A = U
(c) AAU = (A - U) U (U - A) = 0 U A = A
(d) AA0 == (A - 0) U (0 - A) = A U 0 = A
13.
(a)
A
--+
--+
·0
0
1
--+
1
(c)
A
B
0
AnB
,0
0
0
1
1
0
1
0
0
B i
0
0
()
1
0
1
1
1
0
0
1
I1
1
1
r
1
0
1
0
1
Since A B, we only consider rows 1,2,
and 4 of the table. In these rows A and
A n B have the same column of results, so
A B:::::::::>A=AnB.
AnC AnB BnG (AnB)U(BnC)
0
0
0
0
1
0
1
0
0
0
0
0
0
1.
0
1
1
0
0
0
0
0
1
0
1
0
1
1
1
0
0
0
n
14.
n
U
n
(d) 0
u
15.
n
u
==>(AU
n(CUA) =
n(cu
(a) The r O's determine r 1 locations for the m individual
If r I m , we
can
these locations
(r~l)
(b) Using part (a), here we have k Vs (for the elements of A) and n - k 0'8 (for the
elements in U - A). The n - k O's provide n - k + 1 locations for the k 1'13 so that no
two are adjacent.
k locations can he selected in (n-Z+1) ways if n - Ie + 1 ~ k
(n-Z+l) subsets A of U
or 2k < n 1. So there are
contains no consecutive integers.
16.
u
17.
2/7
(ii)
8
(11 1)
(a) 23
8
= k and
that
18.
7
the given figure let circles (i), (ii), and (iii) denote
subset of a.'>signments where no one is working on
experiments 1,2,3, respectively.
each assistant there
are seven possibilities: the seven nonempty subsets of
{1,2,3}. So there are 715 possible assignments. To determine the number of assignments in region 8 we need
to determine the number of assignments in the union
the three subsets. Region 5 has 0 elements, while
regions 2,4;0 each contain 1 element (e.g., for region 2,
if
assistants are assigned only to experiment 3 then
is the one way that everyone is working on an experiment, but no one is working on experiments 1 and
each ofregiol1s 1,3,7
are
-2 elements (e.g., for regions 1,2,4,5 there are 3 cases
to consider where no one is working on experhnent 1 - for each assistant can be working
on
2 or only experixuent 3 or
experiments 2,3). The number of
assignments where at least one person is working on every
is 715 - 3[315 - 2] - 3.
n
= 0,
U
-
+ 10 =
from AU B. Among these selections
B.
0.3483.
22.
probability
en
There are
er2) ways to
seven elements
(~) contain fow: elements from A and three from
4
e:)/e,n = (495)(120)/(170,544)-=-
(a) P(U) = {O,
,{2}, {3}, {I, 2}, {I, 3},
3},U} and
o-(A) = 1 + 2 + 3 +
(1 + 2) (1 + 3) + (2 + 3) + (1 2 3) = 4(1 + 2 + 3) = 22(1 + 2 + 3) = 24.
(b) ~(1+2 3 4)=
(c) 24(1 2+3 4 5)=240
(d) 21'1,-1(1 + 2 3 ... + n)
Proof (1): Let x E U.
x appears
1 subset by itself, ('l~l) subsets of size 2,
('.;1) subsets of size 3, ... (:=:) subsets of size k, ... , and (:=D subsets of size n.
Hence x appears in a total of [(n~l)
(n~l) (n;l) ... + (:=!)J = 2n-1 subsets. SO
1
EAEP(U) o-(A) = 2",-1 E:Z:EU x = 2"'-1(1 + 2 + 3 + ... + n).
_
Proof (2): Let x E U. For each subset A ~ U, if x ¢ A then x E A. Hence for each pair
(A, A) of subsets of U, exactly one of them contains x. How many such pairs are there?
(1/2)(211.) = 2",-1. Consequently, each x E U can be found
exactly 21'1,-1 subsets of U
and the result for EAE"{U) o-(A) follows.
(e) Using the result from part (d) it follows that EAE1'(u)O'(A) = 2n - 1 s.
(a)
c:) e
6
(18 )
15)
~
~
no
moves
one
move
two diagonal
moves
~
three diagonal
moves
(~) e+:-1)
~
six diagonal
moves
=
~ (~i) (2i
(!) (9+:-1) + (:) C+:-l)
~
~
[oW' diagonal
moves
five diagonal
moves
(~) e+~-l)
(~) e+:~l)
~
~
seven diagonal
moves
eight diagonal
moves
1)
+(8, - i) - 1) = t. (2.i) (8 +~)
8-~
t=O
t
8-f
x 2 - 7x = -12:::::} x 2 - 7x
12 = O:::::} (x - 4)(x - 3) = O:::::} x =: 4,x = 3.
x - X = 6 => x - X - 6 = 0 :::::} (x - 3)(x + 2) = 0 => x = 3, x =
Consequently, AnB = {3} and AUB = {-2,3,4}.
2
2
25.
3
-.,..........
(~) e1+33-1)
24.
c:) e+22-1)
x 2 -- 7x -12:::::} x 2 - 7x 12 :5 0 => (x - 3)(x - 4) :5 0 => [(x - 3) 0 and (x - 4) ~ 0]
or l(x - 3) ~ 0 and (x - 4) :5 OJ =>
S; 3 and x ~ 4) or [x > 3 and x :5 4] :::::} 3
x:5 4,
BoA=
S;x
=[3,41.
2) :5 0 :;} [(x - 3) < 0 and
or
~3
x
::;:}
(2)
one
The
The
OJ or
x :5 3,
Consequently, the answer is the sum of these two probabilities - namely,
28.
S
P ( A tB)
r ·"1.1
&,
sample space for an
S.
A,
that
Pr(A)+Pr(B}-l
Pr(B)
Proof: P(AIB) = Pr(A n
- [Pr(A)
- Pr(A U B)]/Pr(B).
UB
S, it follows that Pr(A U B)
= 1. Consequently, -Pr(A U B) 2 -1 and
Pr(AIB) [Pr(A) +
- 1]/Pr(B).
29.
Pr(An(BUC» = Pr«AnB)U(AnC» = Pr(AnB) Pr(AnC)-Pr«AnB)n(AnO».
Since A, B, C are independent and (A n B) n (A n 0) = (A n A) n (B n C) = An B n 0,
0»
Pr(A n (B U
= Pr(A)Pr'(B) + Pr(A)Pr(C) - Pr(A)Pr(B)Pr(C) = Pr(A)[Pr(B)
Pr(C) - Pr(B)Pr(C)J = Pr(A)[P1'(B) Pr(C) - Pr(E n CJ = Pr(A)Pr(B U 0), so A
and B U C are independent.
30.
Suppose we toss a fair coin n times and we let the random variable X count the number
of heads among the n t,osses. Here we want Pr(X 2 2) 2:: 0.95, or Ek:: 2 (~)H)k(ty~-k =
Ek::2 (;)(l),l 2:: 0.95.
Now E;::::2 (z)(!)n 2:: 0.95
Ek=2 (~)c!Y' :5 -0.95 1 - LI::::2 (~)o)n :5 1 - 0.95
2:1=:0 (;)(l)n :5 0.05
(lY' + n(!)n = (n + l)(!)n :5 0.05
=:} -
=:}
=:}
=:}
For n = 7, (n l)(!YI, = B(t? = 1~8 = 0.0625.
For n = 8, (n + l)<!Y' = 9(~)8 = 2~ = 0.035156
Consequently, the minimum number of tos.'3es is 8.
31.
(a) The probability that both tires in any single landing gear blow out is (0.1)(0.1) = O.Ol.
So the probability a landing geal' will survive even a hard landing with at le&'ilt one good
tire is 1 - 0.01 = 0.99.
(b) Assuming the
independently of each other,
probability
the
will he able to land safely even on a hard landing is (0.99)3 = 0.970299.
32.
At
1=
Pr(A n B) 2: Pr(A)
U
= Pr(A)
n
33.
one
+
tCOll1eS are
omiCOInes are
c: the eight tosses result
heads and
The answer to the problem is Pr·(CIAUB). But Pr(CIAUB) = ~~=u. --
Pr«CnA)U(CnB)}
Pr{AUB)
Since.
B are disjoint, it fonows that C n A, C n B are disjoint. Further,
Pr(A U B) = Pr(A) + Pr(B) = (t}~ + GY~ =
Pr«CnA)U(CnB» = Pr(CnA)+Pr(CnB) = (~)[(;)(!)3(t)3J(~)+G)[(:)(!)5G)](!) =
!
G)8(20
6)= 13(~f
Consequently, Rr(CIA U B) = [13(~f]/(i) = (13)(!Y> =
35.
(;)(0.8)3(0.2)2
(:)(0.8)4(0.2)
36.
Pr(19~ 000:::; X :::; 21,000)
!!.
(:)(0.8)1' = 0.2048 + 0.4096
0.32768 = 0.94208
= Pre -1000 :::; X - 20,000:::; 1000) = Pr(IX - E(X)I < 1000)
Since Vax (X) = 40,000 boxes 2 , we have Ux = 200 boxes. So Pr(lX - E(X)I :::; 1000) =
Pr(IX - E(X)I ::; 5ux) ;;;:: 1 - 512 = 1 - is =
= 0.96.
;!
37.
Success: one head and two tails - the probability for this is
=:
(nn)1(i)2 = ~.
n = 4, P
Among the four trials (of tossing three fair coins) we want two successes. The probability
2
for this is
= (6)(9)(25)/212 = 675/211 = 2~~'
38.
(:)(i)2(V
(a) (16) / (22) = ~ . .12 • 11 = ..i :... 0363636
3322212011'
:5 :. .
(b) (~) (~) (;) / (~) = (3!)C~~)( i1)( :0) =
0.249351
(14) + (8)J/(22) - (3l(S)(7)(14)+(8}(7)(6) - U ..:.. 0 290909
(c) [(8)
2
1
:3
3 (22)(21}(20)
- 55 .
39.
(a) 1 ~ E;=oPr(X = x) = c(O 4) + c(l + 4)
8 13 20J = 50c, so c = 5~ = 0.02
c[4 + 5
c(4 + 4) + c(9 + 4) + c(16
4) =
(b) Pr(X > 1) = Pr(X ;::: 2) = Pr(X = 2) + Pr(X = 3) +P7'(X = 4) = (0.02)(8) +
(0.02)(13) (0.02)(20) = 0.02(8 + 13 + 20) = (0.02)(41) = 0.82
Pr(X = 3
X ;;;:: 2)
Pr(X = 3)
(c) P;r(X =
2) =
Pr'(X ;;;:: 2)
= Pr(X 2) = (0.02)(13)/(0.02)(41) =
3,
•
(a) To
with a
flush,
must draw (i)
4 and 5 of diamonds; (ii)
the 5 and 9
diamonds; or (iii) the 9 and
of diamonds. The probability for each of
these three situations is (;) / (~7) 1 so the answer is 3/
Maureen will finish with a flush if
draws any two
the remaining ten dia.monds,
elO) ways. However, for three choices [as descdbed in part (a)], she
which she can do
(;1).
(;1) .
actually finishes
a. straight flush. Consequently, the answer here is [e~) - 3] I
(c) To finish with a straight from 4 to 8, Maureen must select one of the four 45 and one
of the four 58. This
can do
(:) (~) ways. For the straights from 5 to 9 and 6 to
10 there are likewise (~) (:) possibilities. However, these 3
straight flushes, so the answer is [3 (i) (i) - 3]/ (~7).
43.
G) straights include three
The total number of chips in the grab bag is 1 2 3
n = n( n 1) /2, and the
probability a chip with i on it is selected is 2i/[n(n + 1)]. Let A, be the events.
A: the chip with 1 on it is selected
B: a red chip is selected.
Pr(AIB) = Pr(A n B)/ Pr(B) = Pr(A)/ Pr(B).
Pr(A) = l/[n(n + 1)/2] = 2/[n(n + 1)].
Pr(.lJ) =(1+2+3+·· ·+ml/[n(n+l)j2] = (m(m+l)/21/[n(n+l)/2J = [rn(m+l)]j[n(n+l)J.
Consequently, P1'(A!B). = [m(~1~~ii~~!~1)1 = 2/[m( m 1)].
(a)
(b)
x
1
Pr(X::::: x)
2
. ~ (1/6)3 = 15/36 = 5/12
~ (1/6)3::::: 20/36=5/9
= 2:;:::1 x . Pr(X = x) = (1)(1/36) + (2)(15/36)
:::.: (1/36)[1 30 60] == 91/36
= E;:} x 2 • Pr(X = x) = (1)(1/36) +
= (1/36)[1 60 + ISO} = 241/36
3
E(X)
~
~~(1/6)3 = 1/36
93
(3)(20/36)
(a)
HHH
HHT
HTH
THH
45.
THT
TTH
Probability
of Outcome
(3/4)3 = 27/64
(3/4)2(1/4) = 9/64
(3/4)(1/4)(3/4) = 9/64
(1/4)(3/4)2 = 9/64
(3/4)(1/4)2 = 3/64
(1/4)(3/4)(1/4) = 3/64
(1/4)2(3/4) = 3/64
(1/4)3 = 1/64
x, the number
of runs
1
2
3
2
2
3
2
1
The probability distribution for X:
x
1
2
(b)
3
E(X)
(c)
E(X2)
Var (X)
So (Tx
Pr'(X = x)
(27/64) + (1/64) = 28/64 = 7/16'
(9/64) (9/64) + (3/64) + (3/64)'= 24/64 = 3/8
(9/64) + (3/64) = 12/64 = 3/16
= E!=l x . Pr(X = x) = (1)(7/16) + (2)(3/8) + (3)(3/16)
= (1/16)[7 + 12 9] = 28/16 = 7/4
= 2:;=1 x 2 • Pr(X = x) = (1)(7/16) +,(4)(3/8) (9)(3/16)
= (1/16)[7 + 24 + 27J = 58/16 = 29/8
= E(X2) - E(X)2 = (29/8) - (7/4)2 = (29/8) - (49/16)
= (58 - 49)/16 = 9/16' ,
= }9/16 = 3/4.
CHAPTER 4
OF THE INTEGERS: MATHEMATICAL INDUCTION
Section
1.
(a) Sen) : r~ 3 2 + 52 +... (2n - lY'I = (n)(2n - 1)(2n
S(l) : 12 = (1)(1)(3)/3. This is true.
Assume S(k) ; + 32 ••• + (2k - 1)2 = (k)(2k - 1)(2k 1)/3, for some k ? 1.
Consider S(k+l).
+3 2 +.. .+(2k-l?1+(2k+1)2 = [(k)(2k-l)(2k+l)/3}+(2k+1)2 =
[(2k+l)/3J[k(2k-l) 3(2k+l)] = [(2k+l)/3][2P+5k+3] = (k+l)(2k+l)(2k+3)j3,
so S(k) => S(k + 1) a~d the result follows for all n E Z+ by the Principle of Mathematical
Induction.
(c) Sen) : E~l i(i~l) = n~l
S(l): 2:t::::1 i(i~l) =
=
so S(l) is true.
Assume S(k): E~l i(i~l) = k:1' Consider S(k 1).
d2)
Ef,~l i(i~l) = E7::1 i(i~l) + (k+1)U~+2) = (k!l) + (k+l)(l~+2) = [k(k 2) IJ/[(k + l)(k + 2») =
(k + l)/(k + 2), so S(k) = } S(k + 1) and the result follows for all n E Z+ by the Principle
of Mathematical Induction.
The proofs of the remaining parts are similar.
2.
(a) Sen) : Ei=l 2i - 1 = 2n - 1
S(l): E~=l 2i - 1 = 21 - 1 = 21 - 1, so S(l) is true.
Assume S(k) : E~12i-l = 2k -1. Consider S(k 1).
2i - 1 = E~::12i-l 21.: = 2k - 1 + 2" = 2k+l - 1, so S(k) = } S(k 1)
the result
is true for all n E Z+ by the Principle of Mathematical Induction.
(b)
£(2i) = 2 = 2 + (1 - 1)21+1, so the statement S(l) is
Assume
i(2i)
S(k)
Et:::l i(2i) ::::: 2 (k - 1)2k+1, For n = k 1,
(k
2 + (k .- 1 )2k+ 1
1)2k +1 = 2
= 2+k .
, so Sen) is
by
E?1i
=
3.
(a)
3
we
95
i+
3n + 1) - 3[(n)(n 1)/2] - n - 1
_ n3 (3/2)n 2 + (1/2)n
- (1/2)12n 3 + 3n 2 + = (1/2)n(2n 2
1)
- (1/2)n(» + 1)(2n + 1), so
3
Ei~l i'~
£2 _
= (1/6)n(n
(n 3 + 3n2
1)(2)> + 1) (as shO"wn in Example 4.4).
1)4 =
= Ei""o(i4+4i3 +6tia +4i+l) =
1, it follows that (n+ 1)4 = 4
i 3 +6 E:=l i 2
(b) From
6
£2 4
Consequently,
4
i 3 =(n+l)4-6{n(n 1)(2)> 1)/6]-4[n(n 1)/2J- +1)=n4 4n3 6n 2 4n
1- (2 n 3 3»2 n) - (2n2 +2n) - (n + 1) = n 4 + 2n3 + n 2 = n 2 (n 2 + 2n+ 1) = n 2 (n + 1)2,
So 2.:i=1 = (1/4)n 2(n + 1)2 [as shown
part (d)
Exercise 1 for this section],
1)5 = Ei~o(i
1)1' = Ei~o(i5
5i 4
lOis + 10i 2
5i
1) =
""n
'5 + 5 ",'It
'4 + 10 ""n
'3, 40""1'1
'2 + 5 ",'II.
'+
",n
1
h
5
",'It
'4
L,.,i::::l $
L,.,i=l ~
L,.,i=l ~ T 1 L,.,i=l t
L,.,i==l Z
L,.,i=O) we
ave L,.,i:::::l ~ = ( n
1)15 - (1O/4)>>2(n + lY~ - (10/6)11,(n + 1)(2n + 1) _. (5/2)n(n 1) - (n + 1). So
From
'4
5 ",'It
L--i=l l
is
(n
n 5 + 5n4 + lOn3 + lOn 2 + 5n + 1 - (5/2)n 4
-5n 3 - (5/2)n 'l - (10/3)n 3 - 5n 2 - (5/3)>> - (5/2)>>2 - (5/2)n - »- 1
»5 (5/2)n 4 (5/3)>>3 - (1/6)n.
-
_
Consequently, E~l i4 = (1/30)n(n + 1)(6)>3 + 9»2 + 11, - 1).
4.
Let Xh X'l1 ••• ,Xlii denote the numbers (in their order on the wheel), and assume that
Xl + X2 + 3:3 < 39, X'l
X3
X4 < 39, ... ,X24 + XliS
Xl < 39, and X2S
Xl
X2 < 39. Then
Et!13xi < 25(39). But Et!13xi = 3 E~!l i = (3)(25)(26)/2 = (39)(25).
5.
(a) 7626
6.
a) The typical palindrome under study here has the form abba where 1 :::; a ~ 9 and 0 :::;
b:5 9. Consequently there are 9·10 = 90 such palindromes, by
rule of product. Their
Slin is 2::"",1(2::"",0 abba) =
E~=(llOO1a + 110b) = E:::::1[10(1001a) 110 E::=o h} =
1::=1 (1001Oa 1
·10/2) =
(I,
4950 =
9(4950) =
450450 + 44550 = 495000.
(b)
'begin
sum:= 0
for a:= 1
627,874
9 do
b:= 0
9
BUm.
96
-
4'11, + 110
6+8+10
6'11, [0 + 2
so n = 10 8.
n -110 -
number
layers.
i 2 = (n)('11, +
we
and (n)(n
('11,
9.
1 =6
4
= (2n)(2n + 1)/2 =
+
2n
~,
i=l
1)(2'11, + 1)/6 = (2'11,)(2n
1)/6 = 1 ::::} n
...
('11,2 ... ('11, 6'11,+
6n + 2[('11, - 1)('11,)/2]
- 1)('11,) =
(n + ll)(n - 10) = 0,
n2
... [6 + ('11, -1)2]
1)/2
* (n)(n + 1)/6 = (2n)/2 *
* n = 5.
i - E:~l i = [(33)(34)/2] - [(1O)(11)/2J = 561 - 55 = 506
(a) Er!l1 i =
(b) Er!n i 2 = El;l {l - E!~l i 2 = [(33)(34)(67)/6] - [(10)(11)(21)/6] = 12144
10. Et:l ti = E;!i ti - Ef~l ti = (100)(101)(102)/6 - (50)(51 )(52)/6 = 171,700 - 22,100 =
149,600.
11.
a) Ef=lt2i=
(2'H;i+l1=Ei=l(2i2+i)=2Ei~li2+
i=2[(n)('11, 1)(2n 1)/6J+
[n(n + 1)/2] = (n(n 1)(2n + 1)/3J + [n(n 1)/2] = n(n + 1)f 2nt 1 + ~J = n(n + 1)[4n;5J =
n(n + 1)(4n + 5)/6.
b) E~!:. tzo
= 100(101)(405)/6 = 681,750.
c) begin
sum:= 0
for i := 1 to 100 do
S11m := 811m
+ (2 * i) '*'
*i
1)/2
print sum
97
(a)
f'
7 2) 2 X 2
are
22 + 3
2
...
+ 8 = (8)(8
2
(b) For each 1
there are
+ 22
n
36(=
3 X 3
1)(2·8 + 1)/6 = (8)(9)(17)/6 = 204 squares.
nxn
... +n = n(n
2
are
(n - k 1)2 k X I,; squares. In total
1)(2n + 1)/6 squares.
14.
For n = 4 we have 24 =
< 24 = 41, so
statement 5(4)
true. Assume
truth
5(k) is, 21.; < kL For k ~ 4,2 < k+l, and [(2k < k!)1\(2 < k+1)] = }
(21:)(2) < (k!)(k + 1), or
< (k + 1)1 Hence S(n) is
for
n ~ 4 by
Principle
of Mathematical Induction.
15.
For n = 5,25 = 32 > 25 = 52. Assume the result for n = k(;;::: 5) : 2k > 1.;2. For
k > 2, k(k - 2) > 1, or k2 > 2k + 1. But 2k > k 2 ==> 2k + 2k > 1,;'2 k 2 ==> 2k -+-1 >
k 2 k 2 > 1.;2 (2k 1) = (I.; + . Hence the result is true for n ~ 5 by the Principle
Mathematical Induction.
16.
(a) 3
(c) For n
8
n -
'""
L....,;
I -- n •
PA
0:#A~X ...
t
Proof: For n = 1,81 = = 1, so this first case is true and establishes the basis step. Now,
p~
k.
for the inductive step, assume the result true for n = k(~ 1). That is, Ski =
L
For n = k + 1 we find that Sk+l =
""
L....,;
0t-A~Xk+l
..1..
PA
= ""
L....,; ..1..
Pn
0:¢:Bt;,Xk
=
0¢A~X"
,""...L,
L....,;
Pc where the
{kH}~C~Xk+l
first sum is taken over all nonempty subsets B of X k and the second sum over all subsets
C of XkH that contain I.; + 1. Then Sk+! = Sf.; + [( k!l) + C:!l )8,.] = k CI;!l) + (k!l )k =
k + (1t!1 )(1 k) = Ie 1. Consequently, we have deduced the truth for n = k + 1 from that
of n = k. The result now follows for aU 11, ;;::: 1 by the Principle of Mathematical Induction.
11.
(a) Once again we start at n :::: O. Here we find that 1 = 1 (0/2) $ HI = H2o, so this
first case is true. Assuming the truth for n = k( E N) we obtain the induction hypothesis
1
(k/2) $ Has.
= [(2)(1)/2]Hj -
=1=
98
truth of
Assuming
for n = k, we have
given
= [(k
1)(k)/2JHHl - [(k
1)(k)/4].
]=1
J:;br n = k
1 we now find
k+1
k
EiHj = LiHj
j=l
(k
l)Hk-H
j:::::l
= [(k + 1)(k)/2JHk+1 - [(k + 1){k)/4] + (k + l)Hk+1
= (k + 1){1 + (kI2)]Hk+1 -l(k
= (k + 1)[1
1)(k)j4]
(k/2)][Hk +2 - (l/(k
2»] - [(k + 1)(k)/4]
= [(k + 2)(k + 1)/2]Hk+2 - [(k + l)(k + 2)]/[2(k + 2)] - [(k
= [(k
2)(k
1)/2]Hk+2 - [(1/4)[2(k + 1)
= [(k + 2)(k + 1)/2JHk +2 - [(k + 2)(k
k(k
1)(k)/4J
1)]]
1)/4].
Consequently, by the Principle of Mathematical Induction, it follows that the given statement is true for all n E Z+.
18.
Conjecture: For all n EN, (n
2
+ 1)
(n:.! + 2) + (n
2
3)
... + (n + 1)2 =
2n+l
E (n:.! + i) =
i=l
n 3 +(n+l)3.
Proof:
2r!.+1
2n+l
2n+l
~I
.=1
~1
E (n:.! + i) = n 2 E 1 + L i = n,2(2n
2n3 + n'J + (2n + l)(n + 1) = 2n3
n3
+ 1)3,
19.
n2
2n 2 + 3n + 1. = n 3
true
some k ;::: 1.
«k + k) + (1/4) + 2k 2)/2 I(k
S
2
1) + (2n + 1)(2n + 2)/2 =
[n 3 + 3n 2 + 3n
(1/2)F' /2
(1/4)]/2 = [(k + 1)
=
k
1] =
1) =
(1/2)]2/2.
S(k)
{1/2)]2/2 ==:} 0 =
99
21.
x, n E
! let Sen)
If the program.
the
loop, after the two loop
are
n(> 0) times,
integer variable answer'is zen!).
First
S(l), the
for
case where n = 1. Here
program it rea£ll(;S
the top of the while loop) will result in one execution of the while loop: x will be assigned
the value x·1 = x(l!), and the value n will be decreased to O. With the value of n equal
to 0 the loop is not processed again and the value of the variable answer is xC!!). Hence
S(l) is true.
Now assume the truth for n = k: For x, k E Z+, if the program reaches the top
the
while loop, then upon exiting the loop, the value of the variable an"wer is x(1.!). To
establish S(k + 1), if the program reaches the top of the while loop, then the following
occur during the first execution:
The value assigned to the variable x is x(k 1).
The value of n is decreased to (k + 1) - 1 = k.
But then we can apply the induction hypothesis to the integers x( k + 1) and k, and after
we exit the while loop for these values, the value of the variable answer is (x( k + 1»( k!) =
x(k + I)!
Consequently, Sen) is true for all n ~
and we have verified the correctness of this
program segment by using the Principle of Mathematical Induction.
22.
If n = 0, then the statement 'n f:. 0' is false so the while loop is bypassed and the value
assigned to (J.n.'Jwer is x = x + 0 . 11. So the result is true in the first case.
Now assume the result true for n = k - that is, for X,1I E R, if the program :reaches the
top of the while loop with k E Z, k ~ 0, then upon bypassing the loop when k = 0, or
executing the two loop instructions k(> 0) times, then the value assigned to answer is
x ny. To establish the result for n = k + 1, suppose the program reaches the top of the
while loop. Since k ~ 0, n = k 1 > 0, so the loop is not bypassed. During the first pass
through the while loop we find that
The value assigned to x is x + y; and
value n is decreased
(k 1) - 1 = k.
Now we apply the indudion hypothesis to the real numbers x+y and y and the nonnegative
mte,;~~r n - 1
the loop
k = 0, or
two
inst:w:uctions k( > 0)
value assigned to answer is
=
(x +
= x
(k
l)y.
now
23.
written
=5+5
7 7
=5+5
5
5 5
=5
7 7
7
::;::;5 5 5 5 7
28 = 7 + 7 7 7
Hence the result is true for all 24 :::; n :::; 28. Assume the result true for 24 :5 n :5 28 :5 k,
conside:rn=k 1.
k+l2::
we
writek+l=[(k 1)- +5=(k5,
where k - 4 can be
as a sum of 5'8 and 7's. Hence k + 1 can
expressed as
the result follows for
n 2::
by
the Principle
a sum
of Mathematical
24.
(a) as = 3
a4 = 5
as = 8
as = 13
ar = 21
(b) "1 = 1 < (7/4)1, so the result is
for n = 1. Likewise, il2 = 2 <
= (7/4)2 and
the
holds for n =
Assume the result true for
1 < n :5 k, where Ii; 2:: 2. Now for n = Ii; 1 we have ak+l =
ilk + ak-l < (7/4)k
(7/4)1:-1 = (7 /4)k-1[(7 /4) + 11 = (7/4)k-l(1l/4) = (7/4)k-l(44/16) <
(7/4)k-l( 49/16) = (7/4)k-l(7/4)2 = (7/4)k+l. So by
Alternative
of the Principle
of Mathematical Induction it follows that an < (7/4)n for all n 2:: 1.
25.
E(X)
E(X2)
Var(X)
26.
a) al
az
b)
i=O
- ~'-I
-
as -
n· n '
ai a(1-1)-i = 0 aoGo = au
L;;J ! ai a (2-1)-i = ~ aoal + G)alao = 2ag
~-1
i=O
i
e-
1
i ) aia{S-l}-i
e)
= L2i=O i aja2-i
- (;)GQa2 + (;)a1 a + (i)azao
1
-
a4
ao(2a~) + 2(a5)(a5) + (2a~)ao
-
n
-
= 6a~
Form of the Principle Mathematical ,U"""U'-'~'V'''''
result is true for '11, = 0 and this establishes the basis step. [In fact, the calculations
parts (a) and (b) show the result is also true for n = 1,2,3, and 4.] Assuming the result
- that is, that an = (n!)a~+l
n = 0,1,2, ... ,
0)
for n = 0,1, 3, ... ,
- we find that
- E~:::o ~7~aiak_i
- E:::;o 7 (i!)(a~+1)(k - i)!(a~-i+1)
-
"",k
L-ti=fJ
-
L-tt:::O
-
~~
ki '('!)(k
t.
-
k 1• a0k+2
')1.aok+2
t
(k + 1)[k!a~+2] = (k+ l)!aZ H .
So the truth of ,the result for n = 0,1, ... , k(?: 0) implies the truth of the result for
'11, = k + 1. Consequently, for all n ?: 0, an = (n!)a~+1 by the Alternative Form of the
Principle of Mathematical Induction.
27.
Let = {n E Z+ln > no and Sen) is false}. Since S(no), S(no + 1), S(no 2), ... ,S(n1)
are true, we know that no, no + 1, no 2, ... , n1 ¢ T. If T :f. 0, then by the WellOrdering Principle T has a least element '1", because T
Z+. However, since S(no),
S( no + 1), ... , S( r - 1) are true, it follows that S( r) is true. Hence T = 0 and the result
follows.
28.
(a) (i) The number of compositions of 5 that start with 1 is the number of compositions
of which is 24 - 1 = 23 = 8.
(ii) 23 - 1 = 22 = 4
(iii) 22 - 1 = 21 = 2
(iv) 21 - 1 = 2° = 1
(v) 1
(b)
total, there are 2(n+l)-1 = 2n compositions for the fixed positive integer n + 1.
For 1
i:::; n, there are 2{n+1-i)--1 = 211.-' compositions of n + 1 that start with i. In
addition, there is the composition consisting of only one summand - namely, (n + 1). So
n + 1 -- in
we
c.ounted the same collection of objects - that is, the compositions
two ways. This gives us 2 n =
1 = (211,-1 + 2n - 2 21'1.-3 ••. + + 21 2°) + 1 =
(~+22
2,,-3
1=
1.
Ei~l
= 2"-1.
1.
(a)
(c)
Cl
= 10; and
=C n
Cl
= 3,c:;l = 1; and
= en!
2.
n
n
1.
(1)
(2)
of PllP2 as PI V P2;
the disjunction of PbP2"" ,PmPn+1 by PI V P2 V ... V
VP2V",V
V Pf'l,+1 ¢:::::::>
V
The result is true for n = 3. This is the Associative Law of V of Section 2.2.
assume the truth
result for n = k 3 and all 1 :5 r < k, that is,
(Pt V pz V •.. V Pr) V (PrH V ... V PAl) <==> (Pt V Pa V •.. V V Pr+ 1 V ... V PA;).
"'hen we consider the case for n = k 1 we must account for
1 r < k 1.
1) If r = k, then (Pi V P2 V '" V PI.) V PHI {::::::::} P1 V P2 V ... V Ph V Pk+1! from our
recursive definition.
2) For 1 r < k, we have (PI V P2 V ... V p,.) V (Pi-+I V ... V Pk V PHI) ¢:::::::> (PI V pz V
... V Pr) V [(Pr+l V .. , V pI;) V PkH] {::::::::} [(PI V P2 V ... V p,.) V (Pr+1 V ... V Pk») V PlcH {=::}
(PI V P2 V .. , V Pr V Pr+l V .. - V Ph:) V
{=::} PI V
V ... V Pr V Pr+l V ... V Ph V PhHSo the result is true for all n 3 by the Principle of Mathematical Induction.
(b)
3.
For n E Z+, n ~ 2, let T( n) denote the (open) statement: For the statements p, fl.1, fI.'J, ••• ,qnl
p V (fl.} A ... A qn) {::=::::} (p V q1) A (p V q2) A •.. A (p V qn).
The statement T(2) is true by virtue of the Distributive Law of V over A. Assuming
T( k), for k
2, we now examine the situation for the statements p, qI, Q2, ... ,fl.k' Qk+1'
We find that P V (Ql A q2 A ... A qk A Qk+1) {::=::::} P V [( q1 A q2 A ... A qk) A qkH] ¢:::::?
(p V (qi A q2 A .. , A qk)J A (p V qk+d {::=::::} [(p V ql) A (p V qz) A ... A (p V qk)] A (p V qk+l) {=::}
(p V qd A (p V (2) A ... A(p V qk) A (p V qkH)' It then follows by the Principle of Mathematical
Induction that the statement T( n) is true for all n ~ 2.
4.
(a) For n = 2, the result is simply the DeMorgan Law -'(Vl VP2) {::=::::} -'Pl A "'1'2. Assuming
the truth of the result for n = k, we find for n = k + 1 that -'(Pl V 1'2 V ... PI: V 1'10:+1) {=::}
-'[(Pl V P2 V ... V Ph) V PieHl ~ -'(Pl V P2 V ... V piI;) A "'Pk+l {::=::::} (-'Pl A ""P'l A ... A "'Pk) A
-'Pk+l {::::::::} "'P11\ -'1'2 A ..• A "'Pk A "'Pk+h so the result is true for all n 2:: 2, by the Principle
of Mathematical Induction.
(b) This :result can be obtained from part (a) by a similar argument, or by the Principle
of Duality for statements.
5.
(a) (1) The intersection of At, A2 is Al n A z.
(Ii) The intersection of A J , All) ••• ,An, A n+1
(Ain n ... nAn)nAn+l!
the
n A2 n .,. nAn n
nA 2 n ... nA'n
n
n , .. nAk n
n ... n n
n
n, .. nAr}n
= Al n A2 n ... n A n
1•
103
6.
(i) For n = 2, the result follows from the DeMorgal1 Laws. Assuming the result for n = k ~
2, consider the case for k 1 sets All A:Il ..• , Ale) Ale+!. Then Al n A2 n ... n ii k n Ak+! =
(AI n A2 n " . n Ale) n Ak+! = (AI n A2 n ... n Ak) U Ale+! = [AI U A; U ... U A k ] U Ak+! =
A 1 UA2 U .. .UAkUAk+h and the
all n ~ 2, by the
is true
of
Induction.
(ii)
proof for this result is similar to the one in part (i). - Simply replace each
occurrence of n by U, and vice versa. (We can also obtain (li)
(i) by invoking the
Principle of Duality - Theorem 3.5.)
1.
For n = 2, the truth of the result An (Bl U B 2 ) = (A n B 1 ) U (A n B 2 ) follows by virtue of
the Distributive Law of n over U.
Assuming the result for n = k, let us examine the case for
sets A, Bl, B'}.) .•. , Bk) BkH'
We have An (Bl U B'J, U ... U Bk U B k+1) = An [(Bl U B'J. U ... U B k ) U B kH ] = [A n
(Bl u B2 U ... U Bi.J] U (.4 n Bk+d = [(A n B l ) U (A n B 2 ) u ... U (A n B k )] U (A n Bk+d =
(A n B 1 ) U (A n B 2 ) U ... U (A n B k ) U (A n B k+1)'
8.
(a) (i) For n = 2, Xl x'J. denotes the ordinary sum of the real numbers Xl and X2'
(ii) For real numbers Xl, X:;" .•• ,Xn,Xn+h we have Xl + X2 ••• + Xn + XnH = (Xl +
X2 + ... + Xn) + Xn+h the sum of the two real numbers Xl + X2 +...
Xn and XnH'
(b) The truth of this result for n = 3 follows from the Associative Law of Addition - since
Xl + (X2 + X3) = (Xl + Xl) + X3, there is no ambiguity in writing Xl + X2 + X3.
Assuming the result true for aU k ~ 3 and all 1 ::; r < k, let us examine the case for k + 1
real numbers. We find that
1) When r = k we have (Xl + X2
••• + X,.) + XrH = Xl + X2
••• + X,.
X"+1, by
virtue of the recursive definition.
2) For 1 ::; r < k we have (Xl + X2 + ... + x r ) + (XI"H + ... + Xk + Xk+d = (Xl + Xl
... + XI") + [(X,.+! ••• + Xk) XkH] = [(Xl + X2 + ... + X,.) + (x"H + ... + Xk)] + XkH =
(Xl
+ X2 + ... + X,.
So the
Induction.
18
+ Xk)
XkH
= Xl
•••
for
n ~ 3 and aU 1 S; r < n,
the
n=
X2
+...
XrH
product
the
X1 X 2 •• •
nr~:}Ql1ct of
Xz ••• ;Cn)XnH'
XIX2 '"
Xl'>
104
and
+ X r +l + ... + Xk + Xk+l.
Principle
X2'
n E Z+ with n ~ 2.
Xr
Mathematical
definition.
)
= XIX2 ••• X"Xr+l •••
so the result is true for all n 2:: 3 and all 1 =5 r < n, by the Principle of Mathematica.l
Induction.
10.
The result is iI'ue for n = 2 by the material presented at the start of the problem. Assuming
the truth for n = k real numbers, we have,
n = k + 1, IXl + X2 + ... + Xk + XkHI =
I(xi X2
Xk)+XH1!<l x l+ X 2
Xk!
IXkHI=5lxll IX21
IXkl IXkHI,
so the result is true for all n > 2 by the Principle of Mathematical Induction.
11.
Proof:. (By the Alternative Form of the Principle of Mathematical Induction)
For n = 0, 1,2 we have
(n = 0) a0+2 = a2 = 1 2:: (,,'2)0;
(n = 1) al+2 = as = t'lz ao = 2 2:: v2 = ( ";2)1; and
(n = 2) t'lZ+2 = a4 = a3 al = 2 + 1 = 3 ~ 2 = (J2)2.
Therefore the result is true for these first three cases, and this gives us the basis step for
the proof.
Next, for some k 2:: 2 we assmne the result true for all n = 0,1,2, ... , k. When n = k + 1
we find that
t'l(k+1)+2 = ak+3 = t'lk+2 + ak 2:: (J2)k
(J2)h-Z = [( .,;2)Z + IJ( v2)k-2 = 3(";2)k-2
= (3/2)(2)(.,;2)1:-2 = (3/2)( J2)k :;::: {J2)kH l because (3/2) = 1.5 > J2 (...:- 1.414). This
provides the inductive step for the proof.
From the basis and inductive steps it now fonows by the Alternative Form of the Principle
of Mathema,tical Induction that t'ln+2 2:: (J2)n for all n EN.
o
=0=1-1=
so
this first
n=k
that
- L Now we cousider what na'[.)1)<3118 when :n = k
L
") -- 1 =
Fe=
so
-1
nEN--
for
Proof: CBy Mathematical Induction).
Basis Step: When n = 1 we find that
so the result
in
first case.
.
(
) statement .cLOr n = k ,we h ave.t...."
~ -Fi-I
FH2
. the gIven
Inductive Step: Assummg
open
i - = 1 - -1;;-'
i=l
Whenn= k
2
2
1, we find that.
=1-
= 1 {1/2kH )[Fk - 2Fk+2] == 1 + (1/2k+ 1 )[(Fk - Fk+2) - Fk+2]
= 1 + (1/2 kH )[-Fk +1 - FkH ] = 1- (1/2k+ 1 )(FkH + Fk+2) = 1 - (Fk+3/2kH).
From the basis and inductive steps it follows from the Principle of Mathematical Induction
that
n
Vn E Z+ E(Fi_I/2i) = 1 - (Fn+2/2n).
i==l
14.
Proof: (By Mathematical Induction)
For n = 1 we find
L~ = 12 = 1 = (1)(3) - 2 = L1L2 - 2,
so the result holds in this first case.
k
Next we assume the result is true when n = k. This gives us EL; = LkLkH - 2. Then
.
n=k
i=l
lwe
it follows that
E
15.
(n =
=
Form
the Principle
n = ()
n = 1 I.Jtt::<:!'u~je
SF:! 5(1) 5 = 7 - 2 L4 - Lo =
=
5FoH
= 1)
This establisl:te3 the ba.'!iis
we assu.me
LnH -
n
= 0,
=
=
= 11-1 =
1),
= 5Fk +3 = 5( Fk+:! FhH ) = 5( Fk+2
- Lk-l)
= 5FAl+<! + 5E(k-l)+2 = (LkH - L k ) (L(k-l}H - Lk-d =:.:::
- Lk )
= (LkH + Lk+3) - (Lk Lk-d = Lk+5 - LkH = L(kH)+4 - Lk+l - where we have used the
Fibonacci
and
establish
second
of
eighth equalities.
It then follows by the Alternative Form of the Principle of Mathematical Induction that
16.
(a) Let E denote the set of all positive even integers. We define E recursively by
(1) 2EEjand
(2) For each nEE, n + 2 E E.
(b) If G denotes the set of all nonnegative even integers we define G recursively by
(1) 0 E G; and
(2) For each mEG, m+2 E G.
1 '7.
(a)
Steps
(1) p, q, '1', To
(2) (p V q)
(3) (-w)
(4) (To A (-''1'»
(5) «p V q) -+ (To A (-''1'»)
Co) Steps
(1) p,q,r,s,Fo
(2) (-'1')
(3) « "';1') ~ q)
(4) (3 V Fo)
(5) ('I' A (8 V Fo»
~ q).~
(a;) .\:=0:
Ie = 1.:
.\:= 2:
k:=: 0:
Ie = 1;
V Fo»)
Re.asons
Part (1) of the definition
Step (1) and Part (2-i) of the definition
Steps (1), (2), and Part (2-v) of the definition
Step (I) and Part (2-ii)
the definition
Steps (1), (4), and Part (2-iii) of the definition
of
Steps (3),
1
4
1
1
3142,
4132,4213) 4231,4312
1243,
1342t
2134,2314,
1e:2:
k=3 :
A
Reasons
Part (1) of the definition
Step (1) and Part (2-ii) of the definition
Step (1) and Part (2-0 of the definition
Steps (1), (3), and Part (2-iii) of the definition
Steps (2), (4), and Part (2-iv) of the definition
1
1234
( d) ('In - k = m - k ~- 1 ,.."'''''v'-'u~
(e) (i) Five
(1) In
of 1; (2) Between 1,2; (3)
2,4;
3,6; (5) Between 5,8. [The five locations are determined by the four ascents and the one
location. at
(in front of 1)
Oi) Four locations: (1) Between 4,3; (2) Between 6,5; (3) Between 8,7; (4) Following 7.
[The jour locations are determined by the three descents and the one location at the end
(fonowing 7) of p.]
(f) 1rm ,k = (k + l)1rm -l,k (m - k)1rm -l,k-l'
X : Xl, X2) ••• ) Xm denote a permutation of 1,2,3, ... , m with k ascents (and m - k - 1
descents). (1) If m = Xm or if m occurs ximxiH,l i .$ m - 2, with Xi > Xi+2 then the
removal of m results a permutation of 1,2,3, ... 1 m - 1 with k - 1 ascents - for a total
of [1 + (m - k -1)}1Im -l,k-l = (m - k)1I m -l,k-l permutations. (2) If m = Xl or if m occurs
Ximxi+:h 1 .$ i .$ m - 2, with Xi < Xi+2, then the removal of m results in a permutation
of 1,2,3, ... , m - 1 with k ascents - for a total of (k + l}fl"m-l,k permutations.
Since cases (1) and (2) ha:ve nothing in common and aCCoullt for all possibilities the recursive formula for 1fm,k follows. [Note: These are the Eulerian numbers am,k of Example
4:21.J
19.
(a) (;)
C";1) = [k(k -1)/2J
[(k + 1)k/2] = (k 2- k + k2 k)/2 = k2.
(e) (:)+4e~1)+e~2) = [k(k-l)(k-2)/6]+4[(k+l)(k)(k-l)/6J+[(k+2)(k 1)(k)/6] =
(k/6)[(k -l)(k - 2) + 4(k + l)(k - 1) + (k + 2)(k + 1)] = (k/6)[6k 2 ] = k3 •
(d) E;:=1 k 3== Lk=l (:) + 4 Ei:=l e~l) + Lk::l (~2) = t'!l) + 4 (n;2) + (n!3) =
(1/24)[(n + l)(n)(n - 1)(1'6 - 2) + 4(n + 2)(n + l)(n)(n - 1) (n + 3)(n + 2)(n + 1)(n)1 =
[en 1)(n)/24][(n -l)(n -2) 4(n+2)(n"": 1)+(n+3)(n 2)] = [(n+ 1)(n)/24J[6n2 + 6n] =
n 2 (n + 1)2/4.
(e) k4 = (:) +
In general,
[The givt".Jl
He!l) + ne!2) + e!3)
C,;r) ,
ae/s are
numbers of Example 4.2L
formula is known as Worpitzky's identity.]
20.
some k E
... ,S(k) =*
neo.rem 4.2, Sen)
then S(k) =*
1)]
by
true for aU n.
Bome k E Z+
1 ¢ S,
= {x - 11x E S}
k E and by applying the induction hypothesis to T, T has
a least element t ;:::: 1 alld S has a least element t 1 2.
Cd) From part (c), Theorem 4.1 implies the Well-Ordering
In the solution
of Exercise 27 of Section 4.1 the Well-Ordering Principle implies Theorem 4.2.
Theorem 4.1 implies Theorem 4.2.
Sedion 4.3
1.
a = a . 80 1 0 = a . 0, so alO ..
(b) alb ==} b = ac, for some c E Z. bla ~ a = bd, for some dE Z. So b = ac = b{dc)
and d = c = 1 or -1.
a = b or a = - b.
( c) al b ~ b = ax, ble ~ c = by, for some x, y, E Z. So c = by = a( xy) and ale.
Cd) alb:::=} ac = h, for some c E Z :::::::::} acx = bx ==} albx.
(e) If alx,aly then x = ac,Y = ad for some c,d E Z. So z = x - y = a(e - d), and alz.
The proofs for the other cases are similar.
(g) Follows from part (f) by the Principle of Mathematical Induction.
2.
(a) alb ==} ax = b, for some x E Z+; cld:::::::::} cy = d, for some y E Z+. Then (ac)(xy) = bd,
so aclbd.
(c) aelbe ==} aex = be, for some x E Z+ =:} (ax - b)e = 0 ==} [ax - b = 0, since
e > 0] =:} ax = b ==} alb.
The proof for part (b) is similar.
3.
Since q is prime its only positive divisors are 1 and q. With p a prime, p > 1. Hence
pjq ==} P = q.
4.
No. 61(2·3) but 612 and 6 13.
5.
Proof: (By the Contrapositive)
Suppose that a I b or a I c.
a I h, then ak = b
EZ. But ak = b => (ak)e = a(ke) = bc => a I be.
A similar result is obtained if a I c.
6.
Proof: (By Mathematical Induction)
n=2
2).
109
Proof:
311(54 7b lIe) =* 311(104 + 14b 22c). Also, 311(314 + 31b
1(21a + 1
9c).
- (lOa 14b 22c)].
3le), 80 311[(314
31b
8.
Note that
of ElerulOr's 12 numbers is divisible by 6. Consequently, every sum that
uses any of
must also
divisible by 6 (because of part
of
4.3
- where each Xi = 1, for 1 ::; i ::; n). Unfortunately 500 is not divisible by 6, so Eleanor
has not received a
card.
9.
b14,
2) :::::=:} bl[ax + (a
so b = 1 or
10.
Letn=2k 1,k~0. n2-1=(2k 1)2-1=4k2 +4k=4k(k+l). Sinceoneofk,k+l
must be even, it follows that 81($1,2 - 1).
11.
Let a = 2m + I, b = 2n 1, for some m, $1,
so 21(a2 + b2 ) but 41 (a 2 + b2 ).
12.
(a) 23 = 3 . 7 2, q = 3, r = 2.
(b) -115=(-10)·12+5, q=-lO,r=5.
(c) 0 = 0 . 42 + 0, q = 0, r = O.
(d) 434
= 14·31 + 0,
q
2)y] for all X, Y E Z. Let X = -1, Y = 1.
O. Then a2
b2 = 4(m 2
b> 0 and b12,
m
n2
2,
= 14, r = O.
13.
Proof:
For n = 0 we have 7'n - 4ft = 7° - 4° = 1 - 1 = 0, and 310. So the result is true for this
first case. Assuming the truth for n = k we have 31(7k - 41.;). Turning to the case for
n = k+ 1 we find that 7k +1_4kH = 7(7k ) -4(4.1.1) = (3+4)(7 k )-4(4k) = 3(7 k )+4(7k - 4k).
Since 313 and 31(7 k _4k) (by the induction hypothesis), it follows from part (f) of Theorem
4.3 that 3\[3(71:) + 4(71.: - 41:)], that is, 31(71.1+1 - 4k+l). It now follows by the Principle of
Mathematical Induction that 31(1" - 4n) for all n E N.
14.
(a) 137 :::=; (lOOOlOO1h = (2021)4 = (211)8
= (1201203)4 =
(c) 12,345:::=; (1100000011l001h = (3000321)4 = (30071)8
15.
10
2
(3)
527
10000011
16
(a)
(b)
(c)
(d)
17.
(a)
(b)
(d)
Base
A7
4C2
lCB2
2
10100111
1110010110010
1111110
Base 2
11001110
00110001
11110000
01010111
Base 10
206
49
240
16
CE
31
FO
87
57
18.
The base 7.
19.
Here n is a divisor of 18 - so n E {I, 2, 3, 6, 9,18}.
20.
(a) 00001111
(d)
167
1218
7346
667134
(b) 11110001
Start with the binary representation of 65
(c)
65
1
01000001
1
Interchanges the O's and
l's to obtain the one's compleInent
10111110
1
(e)
Add 1 to the one's complement
01111111
(f) 10000000
10111111
21.
22.
23.
Largest Integer
Smallest Integer
(a)
7=~-1
(b)
(c)
(d)
(e)
127 = 21 - 1
-8 = -(~)
-128 = _(21)
_(2 15 )
_(2 31 )
(a)
215 -1
~1_1
-1
(b)
- y) =
=0
01100100
24.
Program ChangeOffiase (Inpnt,Output);
Var Number, Base, Remainder,
Power,
: Integer;
Begin
Writeln ('Input the base 10 number - positive integer is to be changed.');
Write (,Number = ');
Read (Number);
Writeln ('Input the base - an integer between 2 and 9 inclusive. ')i
Write (' Base = ');
Read (Base);
Keep := Number;
Result := OJ
Power:= 1;
While Number> 0 Do
Begin
Remainder:= Number Mod Base;
Result := Result (Remainder * Power);
Power := Power * 10;
Number := Number Div Base
End;
Writeln (,The number', Keep:O, 'when converted to'
'base', Base:O, 'is written as " Result:O)
End.
j
25.
(i) If a = 0, choose q = r = O.
(ii) Let a > 0, b < O. Then -b > 0 80 there exist q, r E Z with a = q( -Ii)
where
{} r < ( -b). Hence a = (
r with 0 r < Ihl.
(iii) Finally, consider the case where a < 0 and b <
-b > 0 so -a =
+ '1"
{}::5 <
a =
=
b) :::::
'1'
= Ibl.
'1' = -b - '1" <
b
0
~
If 1"1 =f. ra, then 1'1'1 - 1"'21 > {}
<
if But
f'1 ::::::
«/
so
W' are
-
sub2 =
.15;
sub3 = 0.. 8;
sub4 = -1..7;
remainders: sub1;
larger: sub2j
positions: array [0.. 7] of sub Ii
i: sub3;
j: sub4;
m,n: integeri
Begin
Writeln eWhat positive integer do you wish to convert to base 161');
Readln (n);
For i := 0 to 7 do
positiolls [I] := 0;
m:= ni
i:= 0;
While m > Odo
Begin
positions[iJ := ill mod 16;
m:= In div 16;
i := i+l
End;
j := i-I;
Write (,The integer ',n:O, 4 in base 16 is written ');
While j > = 0 do
Begin
If positions[j} < 10 then
""'rite (positionsUJ : 1)
Begin
. 1
J. := J-
Writeln (,.')
End.
27.
Program Divisors (input ,output )i
Var
N, Divisor: Integer;
Begin
Write (,The positive integer N whose divisors are sought is N = ')i
Read (N);
Writelnj
If N = 1 Then
Writeln ('The only divisor of 1 is 1. ')
Else
Begin
Writeln (,The divisors of " N:O, 'are :')j
Writeln (1:8);
If N Mod 2 = 0 Then
Begin
For Divisor := 2.to N Div 2 Do
If N Mod Divisor = 0 Then
Writeln (Divisor:8)
End
Else
For Divisor := :3 to N Div 3 Do
If N Mod Divisor = 0 Then
Writeln (Divisor:8)
End;
Writeln (N:8)
28.
order
Now x
Y E X by the defiu.ition
E Y :::;:}- x =
11 =
n), with m n E Z+ :::;:}- X Y E
(1) or
(2)
De<:aUlre x,y
~
1
(ii)
C
inclusion we
to
every
integer multiple of 3 is in X. This will be accomplished by the Principle of Mathematical
Induction.
Start with the open statement
Sen): 3n is an element in X,
which is defined for the universe Z+.
basis step - that is, S(l) - is
because
3·1 = a is in X by part (1) of the recursive definition of X. For the inductive step of
tllls proof we assume the truth
S(k) for some 0.
and consider what happens at
n = k 1. From the inductive hypothesis S(o.) we know that ak is in X. Then from part
(2) of the
definition of X we find that 3(0.
= ale + 3 E X because 3k, 3 EX.
Hence S( k) ::::;.. S( k 1). So by the Principle of Mathematical Induction it follows that
S( n) is true for all n E Z+ - and, consequently, Y
X.
\Vith X
Y and Y X it
that X = Y.
29.
(a) Since211Ot foralltEZ+,2In iff 211"0. (b) Follows fromthefactthat4110tfort~2.
(c) Follows from the fact that 8110t for t ~ 3.
In general, 2H1 1n iff 2t +1l(rt . lOt + . . . r1' 10 + 1'0)'
Section 4.4
1.
(a)
1820 = 7(231)
203
231 = 1(203) + 28
203 = 7(28) + 7
28 = 7(4), so gcd(1820,23) = 7
7 = 203 - 7(28) = 203 - 7[231 - 203] = (-7)(231) + 8(203) = (-7)(231) + 8[1820 7(231)] = 8(1820) + (-63)(231)
(b) gcd(1369,2597) = 1 = 2597(534) + 1369(-1013)
(c) gcd(2689,4001) = 1 = 4001(-1117) + 2689(1662)
2.
If as + bt = 2, then gcd(a, b) = 1 or 2, for the gcd of a,b divides a, b so it divides
as bt:::;:: 2.
(b) as + bt :.= 3 ==} gcd( Cl, b) = 1 or 3.
( c) as
4 ::::::::}
'6) = 1, 2 or
(d) as + bt = 6 -"-::} gcd(a) b) = 1,2,3 or 6.
(a)
=
=d
::=}
d = ax
for some
11 E
E
(b/d)y =*' gcd(a/d,b/d) = 1.
4.
b) =
+ (nb)t, so
h=
by) ::::::::}
::::::::} n(gh2) = h
nb);:;:;;h=
some 3, t E
11 E Z,
=n
115
5.
Proof: Since c =
(a, b) we have a = ex, b = cy
2
2
c ( xy),
e divides abo
6.
(a) 2 = l(n + 2) + l)n. Since gcd(n, n + 2) is the smallest positive integer that can
be expressed as a linear combination of n and n 2, it foHows that gcd(n, n + 2) :::;
gcd(n, n 2)12. Hence gcd(n, n + 2) = 1 or 2. In fact, gcd(n, n
= 1,
n odd, and gcd( n, n 2) = 2, for n even.
some x,y E
(b) Arguing as in part (a) we have gcd( n, n + 3) = 1 or 3. When n is a multiple of 3, then
gcd( n, n 3) = 3; otherwise, gcd(n, n + 3) = 1.
gcd(n, n + 4) we
to be cautious. The answer is not 1 or 4. Here we have
gcd(n, n + 4) = 1 or 2 or 4. For n a multiple of gcd(n, n 4) = 4. \\Then n = 4t 2,
t E Z+, we find that gcd(n, n + 4) = 2. For n odd, gcd(n, n 4) = 1.
(c) In general, for n, k E Z+, gcd( n, n k) is a divisor of k. Consequently, if k is a prime,
then gcd(n,n + k) = k, for n a multiple of k, and gcd(n,n + k) = 1, for n not a multiple
of k.
7.
Let gcd(a, b) = h, gcd(b, d) = g. gcd(a, b) = h => hla, hlb => h!(a· 1 + be) => hid.
hlb, hid => hlg· gcd(b, d) = 9 => glb,gid => gl(d· 1 b( -c» => gla. glb, gia, h =
gcd(a, b) => gilt. hlg, g!h, with g, hE Z+ => 9 = h.
8.
gcd(a, b) = 1 => ax + by = 1 for some x, y E Z. Then e = acx + bey. ajc => c =
ad, bjc => e = be, so e = abe ex dy) aud abjc. The result is false if gcd( a, b) =I=- 1. For
example, let a = 12, b = 18, c = 36. Then ale, ble but (ab) 1c.
9.
(a) If c E Z+, then c = gcd(a, b) if (and only if)
(1) c I a and c I OJ and
(2) VdEZ[[(dla)A(dlb)]:::?dlc]
(b) If c E Z+) then c =I=- gcd( a, b) if (and only if)
( 1) e J a or c J b; or
(2)
10.
If c = gcd(a - b,a
x = y = 1,
an.d
80
11.
E Z [Cd I a) A (d I b) A
C
1
b) then
In particular1
=2
=2,
x = -l,y = 1,
= 1 or 2.
= 1,
SOl"ne
bE Z.
d2 =
- b,
- b) A dzlbJ :::? [d 2
acx+
(f)
Theorem
<11
- b)
Ib, it
gcd(d1 , da) > 0] ::::} gcd(a, b) = d t =
Ida. vOl:tSC(IUCJIU!), we
= gcd(a - b, b).
A
Proof:
for each n E
, (5n+3)(7)+(7n+4)( -5) = (35n+21)-(35n+20) =
Consequently, it follows that
gcd(5n 3, +4) = I, or 5n 3 and 7n 4 are relatively
pnme.
33x 29y =
gcd(33,29) = and 33 = (1)(29) + 4, 29 = (7)(4) + so 1 = 29 -7[33 - 29] =
8(29) - 7(33). 1 = 33(-7) 29(8) =::::} 2490 = 33C -17430) 29(19920) = 33(-17430
29k) + 29(19920 - 33k), for
k E
x = -17430 + 29k, 11 = 19920 - 33k
x >0 =? 29k ~ 17430 => k;?: 602
y ~ 0 =? 19920 ;?: 33k => 603 ~ k
k = 602 : x = 28,11 = 54; k = 603 : x = 57,11 = 21.
15.
We need to find X , 1l E Z+ where y > x and 20x + 50y = 1020, or 2x + 5y = 102. As
gcd(2,5) = 1 we start with 2(-2) 5(1) = 1 and find that 2(-2) 5(1) = 1 ::::} 102 =
2( -204) 5(102) = 2[-204 + 5kJ 5[102 - 2kJ. Since x = -204 + 5k > 0, it follows
that k > 204/5 = 40.8 and 11 = 102 - 2k > 0 implies that 51 > k. Consequently
k = 41,42,43, ... ,50. Since 11 > x we find the following solutions:
k
41
42
43
16.
x = -204
5k
y
1
6
11
= 102 - 2k
20
18
16
Proof: Suppose that there exist c, d E Z+ with cd = a and gcd( c, d) = b. Since gcd( c, d) =
0, we have c = bet, d = bdl' Consequently, a =
= (be 1 )(bd1 ) = b2 (c l d 1 ), so b2
= a,
some x E
c = ox
d = b.
cd = a
<c<
The solutions for 84x 990y = 12 are x = 118 - 1651:, y = -10 14k, k E Z.
When c =
the solutions are x = 177 - 165k, y = -15 + 14k, k E Z.
18.
a, h, c E
ax by = c: has a solution Xo, Yo E Z, then ax!) byo = c,
since
b)lc.
gcd(a, b)jc. Then c = gcd(a, b)d,
gcd(a, b) divides a and b,
for some d E Z. Since gcd( a, b) = as + bt, for some s, t E Z, we have a( sd) b( td) =
gcd(a, b)d = cor axo byo = c, and ax + by = c
a solution in Z.
gcd( a,
= g, lcru(a, b) = h. gcd(a, b) = 9 =::} as + bt = for some .'I, t E Z.
Iem( a, b) = h =::} h = ma = nb, for some m, n E
. hg = has
hbt = nbas mabt =
ab{ns + mt) .:? abjhg. gcd(a, b) = 9 :::=:} gla,glb, so (alg)b = (blg)a is a common multiple
of a and b. Consequently hl(a/g)b, and hx = (a/g)b, for some x E Z, or ghx = abo Hence
ghlab.
ao
19.
From Theorem 4.10 we know that
= !cm(a, b) . gcd(a, b). Consequently,
b = [km(a, b) . gcd(a,b)]fa = (242,500)(105)/630 = 40,425.
20.
lcm(a, b) = (ab)/gcd(a,b)
(a) lcm(231,1820) = (231) (1820)/7 = 60,060
(b) lcm(1369;2597) = (1369)(2597) = 3,555,293
(e) lcm(2689,4001) = (2689)(4001) = 10,758,689
21.
ged(n, n + 1) = 1, lemen, n + 1) = n(n + 1)
22.
Proof: The result follows from Theorem 14.10 and Exercise 4 for this section. We find
(M}(nb)
,.2 ..b
1 1 ( b) .
tha t 1lcm ( na, n h) = gcd(na,nb)
= ngcd(a,b)
= n [ab
gcd(a,b) = n em a,
Section 4,,{)
1.
2.
(a) 22 . 33 • 53 . 11
(c) 32 .53 .72 .11 . 13
""'u".., .. , 7114800) = 22 .3.52 .11
_
lem(148500, 7882875):= 22 •
ged{71l4800,7882875) = 3 .
= 3300
·lf3 ·7'l·11 2 =
·11 =
= 40425
·13 =
..• Pt<let
The
is true for n =
From
4.2 the
follows
n =
k ~
assume that plala:r"ak::=::::> plail
S01ne 1 :::; i :::; k. Now consider plala2···akak+l'
Then p/(alal" ·ak)ak+1 ::=::::> plaIa'}," 'ak or plak+l (by the case where n = 2)::=::::> pjaj for
some 1 i
k (by the induction hypothesis) or plak+l =:> p!ai for some 1 :5 i ::; k 1.
The general result then follows by the Principle of Mathematical Induction.
S.
Proof:
proof is
to that given in Example 4.41.)
If not, we have vp = alb, where a,b E Z+ and gcd(a,b) = 1. Then v'P = alb::} p =
a 2 /b 2 ::} pb? = 0,2 ::} P I a 2 ::} p I a (by Lemma 4.2). Since p I a we know that a = pk
E Z+, and pb2 = 0,2 = (pk)2 = p2P, 01" b2 = pk 2. Hence pi b2 and so pi b. But if pi 0,
and p ! b then gcd( 0" b) = p > 1 - contradicting our earlier claim that gcd( a, b) = 1.
6.
Here 25n IOn +40n = lOOk, so 75n = lOOk, or 3n = 4k. From Lemma 4.2 it follows that
31k. So k = 3· r. Then 3n = 4(3· r) ::} n = 41". So n is any positive multiple of 4.
7.
(a)
8.
a) There are (15)(10)(9)(11)(4)(6)(11) = 3,920,400 positive divisors of n = 214395871°113135371°.
b) (i) (14-3 1)(9-4+1)(8-7+1)(10-0+1)(3-2 1)(5-0+1)(10-2+1)=
(12)(6)(2)(11)(2)(6)(9) = 171,072
(ii) Since 1,166,400,000 = 29 3655 , the number of divisors here is (14- 9+ 1)(9 -6+ 1)(85 1)(10 - 0 + 1)(3 - 0 + 1)(5 - 0 + 1)(10 - 0 + 1)= (6)(4)(4}(1l)(4)(6)(1l) = 278,784.
(iii) (8)(5)(5)(6)(2)(3)(6) = 43,200
(iv) (7)(3)(4)(6)(1)(3)(6) = 9072
(v) (5)(4)(3}(4)(2)(2)(4) 3840
(vi) (1)(1)(2)(2}(1)(1)(3) = 12
(vii) (3)(2)(2)(2)(1)(1)(2) = 48
3 X 4 x 4 X 2 = 96
(b)
(c)
270
144
=
9.
From Theorem 4.10 we know that mn = lcm(m,n)'gcd(m, n), sogcd(m,n) = mn/lcm(m,n) =
223151 111 = 660.
10.
gcd = 3 . 52 • 11 = 285
km = 24 .
. 72 • 1
11.
. 13 =
, 80
""...""1'",,,,.1"
162,
000
are
1)(6
square we ha.ve a =
some y E
3
= 2 ••
1)(3
= 252 poSSlm:lliil e8 for n.
l
, for some x E Z+.
3a a perfect
(noose:ll = 2
by Lenuna. 4.2 it follows
215b we
or
13.
a.
",...,t, " ' • . , ,
(i)
1
oft
1 divides n.)
Proof: We
(So n is a
uru:e-Iree UJ.I,It:lKt~r -
that is, no
that abcabc = (abc)(lOOl) = (abc)(7)(1l)(13).
15.
Since = 24 . 32 • 5 . 7, the smallest perfect squru:e that is divisible by 7! is
(35) x (7!) = 176,400.
16.
If n E Z+ and n is a perfect squru:e,
n = p~l p~2 ... p7:, where
is prime and ei is a
positive even integer for alII 5 i 5 k. Hence (el + 1)(e2 1)·· ·(e n + 1) is a product of
odd integers. Therefore the number of positive divisors of n is odd.
Conversely, if n E Z+ and n is not a perfect square, then n = p~l p~2 . . .
where each Pi
is prime and ei is odd for some 1 5 i 5 k. Therefore (ei + 1) even for some 1 < i 5 k,
so (el + 1)(ea 1) ... ( ek 1) is even and n has an even number of positive divisors.
11.
For 1260 x n to be a perfect cube, the exponent on each prirn.e divisor must be a multiple
of 3. Since 1260 = 22 . 32 ·5 ' 7, we want 1260 x n = 23 • 33 , 53 , 73 , so n = 2·3 ,52 • 72 = 7350.
18.
(a) Since 200 = 23 .5 2 , the number of times the 200th coin will be turned over is (4)(3) = 12,
the number of divisors of 200.
(b) The following coins will also be turned over 12 times:
(i) 23 .32 =72
(ii) 2 5 .3=96
(iii) 22 .33 =108
(iv) 25 .5=160
6
(c) The 192nd coin is turned over 14 times because 192 = 2 .3.
19.
(a) 4 = 22; 8 = 23 ; 16 = 24; 32::;: 25.
. 3 2 • 52 • 72 =
Considering the powers of 2, there are 5 different sums of two distinct exponents: 5 = 2+3;
6 = 2 + 4; 7 = 2 + 5 = 3 4; 8 = 3 + 5; 9 = 4 5. Henee there are 5 different products
that we can foml.
(b) Here we ha:ve 2 for n = 2,3,4,5 and 6. Now there ru:e 7 different sums of two distinct
exponents: 5 = 2 + 3; 6 = 2 + 4; 7 = 2 + 5 = 3 + 4; 8 = 2 + 6 = 3 + 5; 9 = 3 + 6 = 4 + 5;
4 + 6; 11 ;:: 5 6. Consequently, we can form 7 different products in this case.
(c) The
here may also be represented as AUB where A;:: {2nln E Z+,2 < n 6} and
B
E Z+ ,2 -< k <
-the product uses hvo mt~el:e:rs
there are 7 possibilities. If both irltegers
are5x4=
10
=
=
(1)
(3)
of A, C; 5 x 4: =
(6) One element from each of B, C: 5 x 4 = 20 possibilities.
total there are 7 7 + 5 + 25 20 + 20 = 84 possible products.
(e) This case generalizes the result
( d). Once
there are 84
Program Primefadors (input ,output );
Val'
p, j, k, n, originalvalue, count: integer;
Begin
Write (,The value of 11 is
Read (n);
originalvalue := n;
Writeln ('The prime factorization of " n:O, ' is ')j
H n Mod 2 = 0 Then
Begin
count := 0;
While n Mod 2 = 0 Do
Begin
count := count + 1j
n:= n Div 2
Endj
Write ('2(', count:O, ') ')
End; .
Hn Mod 3 = 0 Then
Begin
count := OJ·
While n Mod 3 = 0 Do
Begin
1-,
cOlmt := count
n:= n Div 3
End;
'Vrite
, count :0, ') 1)
n >= 5 Do
j := 1;
+ 1;
k:= p
J
(k = 0)
(j = Trunc(Sqd(p»)j
<> 0)
p = 0)
j := j
Begin
count :== OJ
While n Mod p = 0 Do
Begin
count := count + 1i
n:= n Div p
End;
Write
')', count:O, ') ')
End;
p:= p + 2
End;
End.
21.
The length of AB = 28 = 256; the length of AC = 29 = 512. The perimeter of the triangle
is 1061.
22.
(a)
10
TI( _l)i = -1
i=l
(b)
II (_1)'i = (_1)(21<+1)(21>+2)/2 = (_1)(2n+1}(n+l) =
2n+l
. {
i=l
(c) n(i+l)(i+2) = (~) (~) (~) (~)
i:::4
(i - 1)( i)
3.4
4.5
5.6
6.7
(d)
for n odd
for n even
1,
(9.10)
= 8·9·9·10 =27
7.8
3.4 .4 .5
IT 2n -i, + 1 = (_n
(2n)
=
n 1) (n+..!)
n (n+2)
n- 1 ... (2n-1)
2
1
~
i=n
[(2n)!/(n - l)!]/(n
23.
_'1
I)! = (2n)!I[(n - l)!(n + l)!J = (:~l) = C~2;1)
(a) From the Fundamental Theorem of Arithmetic 88,200 = 23 •
F =
,32 " 7Z}.
. 52 . 72 • Consider the
where gcd( a, b) == l.
(of
1 < b < n, we remove
-1
case
(c)
nh n~l! .•. I nk
1< b<
2:: 1.
and gcd(a,
i)
(c)
Mathematical
1
For n = 2 we find that
1
1
= (1 - 22 ) = (1 - 4 = 3/4 = (2 + 1)/(2 . 2), so the
- i2
is
in this
case and this establishes the basis step
our inductive proof.
Next we assume the result true for some (particular) k E Z+ where k ;:::: 2. This gives us
k
1
the case for n = k 1, using the inductive
= (k 1)/(2k). \Vhen we
II(1- -;-)
i=2
~
g(1-
i~)) = (U(l- :,)) (1-
step, we find that
1] [(k+l?-l]
+
k
[--v;-
(k
1)2
+
k2+2k
= (2k)(k
= [(k + 1)/(2k)][1 - (k
1) = (k + 2)j(2(k
1» = «k
1
1)2] =
1)
1)/(2(k
1».
The result now follows for all positive integers n ;:::: 2 by the Principle of Mathematical
Induction.
26.
(a) When n is a prime then it has exactly two positive divisors - namely, 1 and n.
(b) If n = p2, where p is a prime, then n has exactly three positive divisors - namely, 1,
2
p, an d p.
( c) Let p, q denote two distinct primes. If n = p3 or n = pq, then n has exactly four
positive divisors - 1,p,p2, and p3 for 17, = pS, and 1,p,q and pq for n = pq.
(d) If n = Ii, where p is a prime, then n has exactly five positive divisors - namely,
1,p,p2,p3, and p4,
27.
(a) The positive divisors of 28 are 1, 2, 4, 7, 14, and 28, and 1 2 4 + 7 +
28 =
56 = 2(28), so 28 is a perfect integer.
The positive divisors of 496 axe 1, 2, 4, 8, 16, 31, 62, 124, 248, and 496, and 1 + 2 4 +
8 + 16 31 + 62 124 + 248 + 496 = 992 = 2(496), so 496 is a perfect integer.
(b)
follows from the Fundamental Theorem of Arithmetic that the divisors of
2m-l(21n -1), for 2m--1
are 1,
,.,., 2m~1, and (2m -1), 2(2m -1),
23 (2 m - 1), ... ,
2m - 1 (2 m - 1).
These
sum [1 2 22
(2m - 1) (2m - 1)(2W~ = (2m - 1)[1
is a 'n..'''t''tprt. "',~'-'&
+
2
-
(2m - 1)] ~ 2m(2m - 1) = 2[2
m
1 m -I)}, so
- (2
n ...
Supplementary Exercises
a
(a
(a
(n - 1)d) = na
n = 1, a
=a
0,
[(k we
so
1)dJ/2,
the ...................J'v.. of
sum = 10+17+
... +(10
we
t
.. .
(10
7(t -1».
From the previous exercise we know that for each t E Z+
a
(a
(a + 2d) + ... + (a (t -l)d) =
sum = lOt (7/2)t(t - 1).
d)
t = 52, we have sum = 9802, and
last summand, is 10 7(52 - 1) = 367.
n
3.
[(t - 1)(t)dj/2, so here we
sum = 10176. Therefore,
t = 53, we
n
L:< -1 )i+li 2 = (-1 t+ 2: i, for all n E Z+.
1
Conjecture:
;=1
i'H
Proof: (By the Principle
.
Mathematical Induction)
lfn = 1 the conjecture provides
1
1
i=1
i=1
2:( _1)i+1i2 = (_1)1+1(1? = 1 = (_1)1+1(1) = (_1)1+1 2:i,
which is a true statement. This establishes the basis step of the proof.
In order
confirm' the inductive step, we shall assume the truth of the result
to
k
k
i=l
i=l
2:( _l)i+l{l = (-l)k+1I: i
for some (particular) k ;;::: 1. When n = k + 1 we find that
k+1
k
L:( _1)'+1i = Cl:( _1)i+1i2) + (_l)(k+1)+1(k
2
i=1
>
1)2
i=l
It
= (-1)1:+1 2 + (_1)k+2(k + 1)2 = (-l)k+1(k)(k + 1)/2 (_l)k+2(k + 1)2
i=1
= (_l)kH[(k + 1)2 -" (k)(k + 1)/2)
= (-1)k+2(1/2)[2(k + 1)2 - k(k + 1)]
== (-1)kH (1/2)[2P + 4k + 2 - P - k]
= (-1)k+ 2(1/2)[k2 3k + 2} = (-1)k+2(1/2)(k + l)(k 2)
k+1
::::: (-1 )k+2 ~~), so the truth of the result
n = k implies the
at n = k
1 -- and
i=l
we have t.he inductive step_
It
nE
ASS1.lIDUlg S( k), rom,lC1er
k
1
61[(k 1)3 + 5(k
Mathematical
18
n
n n2
n n 2 + n + 41
n+
43
4
61
1
7
97
47
5
2
71
8
113
6
9
3
2
(b) 1:01' n '39,n + n + 41 = 1601, a prime. But for n = 40,n 2 + n
(a)
5.
S(39)
6.
41 = (41)2, so
S( 40).
= 119/120 = (51 -1)/51, Ss = 719/720 = (6! - 1)/6!
5039/5040 = (7! - 1)/71
(c) Sn = [en + I)! - 1]/(71 1)!
(b)
$4
86 =
(d) Based on the calculation in part (a) the conjecture is true for n = 1. Assuming that
Sk = [(k + 1)1 -lJ/(k
1)1, for k E Z+, consider Sk+!.
= Sic + (k 1)/(k + 2)1 = [(k + 1)! - 1U(k 1)! + (k l)/(k + 2)1 = [(k + 2)! - (k +
2) + (k + l)JI(k + 2)1 = [(k 2)1 - 1J/(k 2)1, so the result follows for aU n E Z+ by the
Principle of Mathematical Induction.
1.
(a) For n = 0, 22n+! + 1 = 2 + 1 = 3, so the result is true in this first case. Assuming that
3 divides 22k+1 1 for n = kEN, consider the case of n = k 1. Since 2 2(k+1)+1 + 1 =
22k+3 + 1 = 4(2Zk+1) + 1 = 4(22k+1 + 1) - 3, and 3 divides both 22k+1 + 1 and 3, it follows
that 3 divides 22(k+1)+1 + 1. Consequently, the result is true for n = k + 1 whenever it is
true for n =
So by the Principle of Mathematical Induction the result follows for all
nEN.
When n = 0,fr3 + (0 + 1)3 + (0 + 2)3 ' 9,80 the statement is true in this case. We
assume the truth of the result when n = k ~ 0 and examine the result for n = k + 1. We
find that (k + 1)3 + (k 2)3 + (k + 3)3 = (k + 1)3 + (k 2)3 + [P + 9k 2 27k + 27} =
[k3 (k+l)3 (k+2)3J [9(k 2 3k+3)], where the first summand is divisible by 9 because
of
induction
Consequently, since the
is true for n = 0, and
the truth
n = k (> 0) implies the truth for n = k + 1, it follows from the Principle of
ult~ee:e:rs n ;;::
(b)
(1)
(2)
n=
lO(lOm2
[Note: For
n
alld
divisible by
we
find
units
9.
we find that 8Ix 9y z =
By x,
so 80x 3y - 35z =
Converting to base
gcd (3,5) = 1, it follows that 51y. Consequently, y = 0 or y = 5.
Since 5j(80x - 35z)
For y = 0 the equation 80x - 35z = 0 leads us to 16x -7z = 0 and 16x = 7z::::} 161z. Since
o :s; z 5 we find
that z = 0 and the solution is x = y = z = O.
15 - 35z = O::::} 16x+3 -7z = O. With 0 :s; x, z < 5, l6x = 7z --3 ::::} z
does not divide
7(1) - 3) or 18(= 7(3) - 3), and
since 16 does divide 32(= 7(5) - 3) we find that z = 5 and x =
Hence x = 2, y = 5,
z = [And we see that (XYZ)9 = 81x+9y+z = 81(2)+9(5) 5 = 212 = 36(5)+6(5)+2 =
36z By x = {zyx )6']
Ify = 5, then 80x
IS odd and z = 1,3, or 5. Since
10.
Fundamental
of Arithmetic we have 3000 = z3 .31 ·5:-\ so 3000 has
(3 + 1)(1 + 1)(3 + 1) = 32 divisors. Since gcd(n, n + 3000) is a divisor of 3000, there are
32 possibilities - depending on the value of n.
11.
For n = 2 we find that 2'2 = 4 < 6 .
< 16 = 4 2, so the statement is true in this first
case.
Assuming the result tm.e for n = k ~ 2 - Le., 2k <
< 4k, we now consider what
happens for n = k + 1. Here we find that
From
G)
en
e~:11») =
e::12) = [(~Z!m!~~~lJ (~k) = 2[(2k + l)/(k 1)]czn > 2[(2k + 1)/(k + 1)J2k>
2kH , since (2k+l)/(k+l) = [(k+1)+kJ/(k+1) > 1. In addition, [(k+l)+k)]/(k
1) < 2,
e::12)
so
= 2{(2k + 1)/(k + l)]e:) < (2)(2)ekk ) <
. Consequently the result is true
for all n ~ 2 by the Prin.ciple of Mathematical Induction.
12.
=
For n 1,73 83 = 855 = (57)(15). Assuming that 571(7 kH + B2 k+l), since 7{k+1)+2+
82(k+l)+l = 7M3 + 82M3 = 7(7M2) + 64(82I1:H) = 64(7.1:+ 2 ) + 64(8 2kH ) - 57(71:+2), we have
571(7k+3 82k i-3 ), so the result follows by the Principle of Mathematical Induction.
First we
the statement is true for all n E Z+ where 64 < n
:from the calculations:
64 =
n
assume the result is
k 1=
- 4)
bE
64
the
3(5)
for all n where 68 :s; n.
64
k - 4 < k we can
k+ 1 = a(17)
of
14.
-
1·7
t - 1·5+
5 2
-
2·2
2·1
1,
0<5<7
0<2<5
0<1<2
68.
1 = 5 - 2 . 2 = 5 - 2[7 = (-2)7 +
12[3 + 7kJ 7[-5 k E Z. Hence
a =3
b=
= (-2)7
-12k,
3(12 - 7) = 12·3
7(-5) =
k E Z.
-----
15.
(99 ... 9) +1'n =
n 9'8
[91'1 +991'2+" .+(99 ... 9)1'n] +(1'0 1'1 +1'2+', .+rn) . .....""'£'-'''' 911' iff 91(1'0+1'1 +1'2+'" 1"n).
(c) 31t for x = 1 or 4 or 7; 9 l t x = 7.
16.
50x + 20y = 620 ==> 5x + 2y = 62
gcd(5,2) = 1 and 1 = 5(1) + 2( -2) 8062 = 5(62) 2(-124) = 5(62 - 2k) + 2( -124 + 5k),
k E Z. x = 62 - 2k ~ 0 ==} 31 2:: k; 11 = -124 5k 2:: 0 ==> k 2:: 24.8
Solutions: (1) k = 25: x = 12,y = 1;(2)k = 26: x = 10,y = 6; (3)k = 27: x = 8,y =
11; (4)k = 28: x = 6,y = 16: (S)k == 29: x = 4,11 == 21; (6)k = 30: x = 2,y = 26; (7)k =
31 : x = 0, y = 31.
17.
(a)
Let n = 2e1 • 3e2 • seg ·7e4 • n e5 where C1 + e2 + C3
1S i
5. The number of solutions to this equation is
(h)
18.
19.
20.
C4
+ es = 9, with Ci 2:: 0 for all
e+:- = (~)
1
)
(:)
(a) 24(1+2+ 3)54(1+2+3)
(b) 2 5 (1+2+3) 5 4(1+2+3+4)
(c) 22(4)(1+2+3)52{4)(:1.+2+3) 74(4)(1)
(d) 23(4)(H2+3) 34(4)(1+ 2}53(4)(H2+3 )
(e) peqf , where e = (n + 1)(1 2 + ... + m) = (n + l)(m)(m 1)/2 and
f = (m + 1)(1 + 2 +... n) =(m l)(n)(n + 1)/2
(f) peqi r 9 , when e = (n + l)(k 1)(1 + 2 + ... m) = (n + l)(k l)(m)(m + 1)/2,
f = (m + l)(k 1)(1 + 2 + ... + n) == (m + l)(k+ l)(n)(n + 1)/2, and
9 = (m+ l)(n 1)(1 + 2 ... + k) =: (m + l)(n l)(k)(k + 1)/2.
(a)
(b)
1,4,9,16, ... , k where k is the largest square less than or equal to n.
1
i
5, it
4, So now we shall '-'V,j.'l"'~UIVI.
a
A will also sum to a ffirutiple of
~=~=:~=~=~S
and the result
le,ast two
~
~
~=
we now "narrow our attention to
occur.
least three 4'8). Here
4 4 3-
have one or two
If
u;
least one or at least one 2 and one 3,
Otherwise we get one of the
possibilities: (i) 4
2
2 2
or (ii) 4 + 3 + 3.
and at least one 3.) Then we
(Now
are no
2j (ii) 3
have 0) 3
+ 1 + 1; or (iii) 3 + 3 + 3 1.
(We now have o:nly 1'8
2'8 as sum:m.ands).
possibilities are
2
1+1+
and (ii) 2 + 2 + 1.
21.
(a) For all n E Z+, n ~ 3, 1 + 2 + 3 +... + n = n(n + 1)/2. If {l,2, 3, ... ,n} = AU B with
SA = Slh then 2s A = n(n + 1)/2, or 4s A = n(n
1). Since 4ln(n
and gcd(n, n 1) = 1
then either 4!n or 41(n 1).
(b) Here we are verifying the converse of our result in part (a).
(i) If 4171 we write 71 = 4k. Here we have {I, 2, 3,. -. ; ) k, k+ 1, ... ,3k, 3k+ 1, ... ,4k} = AUB
where A = {I, 2,3, ... ,k,3k + 1,3k + 2, ... ,4k -1,4k} and B = {k + 1, k +2, ... ,2k, 2k
1,3k-l,3k},withsA =(1+2+3+ ... k)+[(3k 1)+(3k 2) ... +(3k+k)]=
[k(k+ 1)/2] + k(3k) + [k(k+ 1)/2] = k(k 1)+3k2 = 4P+k, and BB = [(k+l)+(k+2)+
.. . +(k+k)] [(2k 1)+(2k+2) .. . +(2k k)] = k(k)+[k(k+l)/2J+k(2k)+[k(k 1)/2J =
3k2 +k(k+l}=4k2 k.
(ii) Now we consider the case where n + 1 = 4k. Then n = 4k - 1 and we have
{I, 2, 3, ... , k - 1, k,. .. , 3k - 1, 3k, ... ,4k - 2,4k - I} = AU B, with A = {I, 2,3, ... 1 k 1,3k,3k 1, ... ,4k -I} and B = {k, k+ 1, ... ,2k-l, 2k, 2k+l, ... ,3k-I}. Here we find
SA = 11+2 3 ... (k-l)]
[3k+(3k+l)+ ... (3k+(k-l»] = [(k-l)(k)/21+k(3k)
[(k-l)(k)/2) 3P+k2_k = 4P-k, andsB = [k+(k 1)+ ... (k (k-l))]+[2k+(2k
1)+ ... +(2k+(k-l»] = k2+[(k-l)(k)/2]+k(2k)+[(k-l)(k)/2] = 3k2+(k-l)k = 4k 2 -k.
=
Let n be one such integer. Then on - 4 = 68 and 7n
22.
1 = 4t, for some 8, t E Z. Since
214 and 216, it follows that 210n because on - 4 = 68. From Lemma 4.2 we have 21n.
as
+ 1 = 4t, we
that 211. This
us that no such
int,eger 71 exists.
23.
(a) The
a = 1, so "AJ'AJ''''A'''~'C;.t
of
nn",.."," we can write a =
:$ i
(a)
a3 +b3 _
as + 1:/' aT'
+ liP -
b)(a 2 - ab + b2 )
b)(a 4 -- a3 b + a2 1iJ - ab3 +
(a
(a
( a + b)( a P- I -
b + ... +
)
p
+ b)EaP-iC -hi-I,
i=l
for p an odd prime.
Since k is not a power of 2 we write k = .,.. p, where pis an odd prime and r 2 1. Then
p
air. Ii = (al'y' (0'")1' = (a r + ll).L:ar(p-i)( _by(i-l), so ak f} is composite.
i=l
(b) Here n is not a power of 2. If, in addition, n is not prime, then n = r·p where p is an odd
prime. Then 2n + 1 = 2n
so 2n + 1 is composite 26.
p
p
i=l
i=l
.
= 21'"1' + r"p = (2'" + lr)E2 r(p-i)( _ly(i-l) = (2'" + 1)L:2r(p-i) ,
not prime.
Proof: Here the open statement S( n) represents: H2n::; 1 + n, and for the basis step we
consider what happens at n = O. We find that H 2" = H'),o = HI = 1 ::; 1 0 = 1 + n, so
S( n) is true for this first case (where n = 0).
Assuming the truth of S(k) for somek in N (not just Z+), we obtain the induction
hypothesis
S(k) : Hak::;:; 1 k.
Continuing with the inductive step we now examine S( n) for n = k + 1. We find that
H 2 1<+1
-
Since
fl + ! + ~ + ... + ;,.] [(2k~1) + (2k~2)
Hzk +
(2k~2) ... + (Alk!2!;}}'
< ;'" for all 1
j::; 2k, it follows that
1
now
= Hah + 1.
l.'"','.U-''''''' we
n E
n = 1 we
21.
... ,Ii: -
we now
_
:::; (5/3)1. + (5/3)/;--1 = (5/3)k-1 [(5/3) + 1) =
= (5/3)1:-1(24/9) :::; (5/3)k-1(25/9) = (5/3)1<:-1(5/3)2 = (5/3)k+ 1 •
It
follows
the
of
F'n (5/3)n for all n E N.
When n = 0 we find
1)
Lo =
L
= 2 = 3 - 1 = L2 - 1 =
- 1,
i::::O
so
claim
in this
case.
For some kEN, where Ie 2:: 0, now we assume true that
k
Lk = L Li = Lk+2 - l.
i""O
Then for n = k
HI
(*)
2:
i=O
k
1(2:: 1) we have
L i = 2: Li+LkH = (Lk+2
)+Lk+l = (LI:+2+ L kH)-l = LI<:+3- 1 = L(,I;+1)+2- 1,
i=O
and so we see how the truth at n = k implies that at k + 1. Consequently, the summation
formula is ""alid for all n E N by the Principle of Mathematical Induction.
[Note that for the equations at (*), the first equality follows from the generalized associative
law of addition -- and the fourth equality rests upon the given recursive definition of the
Lucas uumbers siu(''€ k + 3 2:: 3(> 2).}
29.
a) There are 9·10·10 = 900 such palindromes and their sum is 2:::::1 E:=o E~;.;o abcba =
E:::1 E~=o E~o{lOOOla + 1010b+ IOOe) = E:=1 E:::;o{lO(lOOOla 1010b) +100(9 .10/2)] =
E:=1 El::::o(100010a + 10100b+ 4500) = E:",,1[10(100010a) + 10100(9 ·10/2) + 10(4500)] =
1000100 E:.:<1 a+9( 454500)+9( 45000) 1000100(9·10/2)+4090500+405000 :;;; 49,500,000.
=
b) begin
sum:= 0
for a:= 1 to 9
for b:= 0
9
for c := 0
9 do
sum:= Sillll
print sum
end
Pz'oof:
c = gcd( (1, b») d = gcd(
with
E Z+. Also,
Since a, b are odd, it follows
a - b is even and a - b = 2(
=>cj(a-b)::.:}cj(
c is odd since a, b are odd. Now c = gcd(a, b) => c I a
cI
because gcd(2, c) = 1. Consequently, c I band c I (
=> c I
As d = gcd(
d I c.
, b), it follows that d I 2( ";b) + b, that is, d I a. Since d I a and d I b, we
Since c I d and die
c, d > 0, it follows t·hat c = d.
:n.
Proof: Suppose that 71n. We see that 71n ::.:} 71(n-21u) => 71[(n-u)-20'u] => 71[1O(n~u)_
20u] => 71[10C\'~U - 2u)] => 71(n~u - 2u), by Lemma 4.2 since gcd(7,1O) = 1. [Note:
!!.fii! E Z+
the
digit of n - u is 0.] Conversely, if
- 2u), then since
nw'ti - 2u = n~~lu we find that 71(n-;.~lU)::.:} 7·10· x = n - 21u, for some x E Z+. Since 717
and 7121, it then follows that 7in - by part (e) of Theorem 4.3.
32.
a) If 19m 90 + 8n = 1998, then m = (1/19)(1908 - 8n). Since 1908 = 19(100) + 8, the
remainder for 8n/19 must be 8. This occurs for n = 1, and then m = (1/19)(1908 - 8) =
(1/19)(1900) = 100.
b) In a similar way we have n = (1/8)(1908 - 19m). Here 1908 = 8(238) + 4, so the
remainder for 19m/8 must
4. This occurs for m = 4 (and not for m = 1,2, or 3), and
then n = (1/8)(1908 - 76) = 229.
33.
H Catrina's selection includes any of 0,2,4,6,8, then at least two of the resulting three-digit
integers will have an even unit's digit, and be even - hence, not prime. Should her selection
include 5, thell two of the resulting three-digit integers will have 5 as their unit's digit; these
three-digit integers are then divisihle by 5 and so, they are not prime. Consequently, to
complete the proof we need to consider the four selections of size 3 that Catrina can make
from {1,3, 7, 9}. The following provides the selections - each with a three-digit integer
that is not prime.
(1) {1,3, 7} : 713 = 23·31
(2) {1,3, 9} : 913 = 11· 83
(3){ 7,9}:
=7·131
{3,
9}:
b,c,
we use.
s = (a
s.
e,
1
s = (77 - i)
29 so 77 - i
29 ==
or
--- i = 12k.
As we
gIven mtle~ers
106 - 2 = 104 = 12(8) 8
106 - 7 = 99 = 12(8) + 3
106 - 12 = 94 = 12(7) 10
106 - 3 = 103 = 12(8) 7
106 = 12(8)
106 - 13 = 93 = 12(7) + 9
106 - 4 = 102 = 12(8) + 6
106 - 11 = 95 = 12(7) + 11
106 - 15 = 91 = 12(7) + 7
Therefore we do not placA'l 10
table. So
= {2,3,
11,12,
and the 12
pnt.'."PI;1 in the table total 67 + 29 = 96. It then follows that a + b
14 + c = d 5 + e + 9 =
1 f 9 h =
a d + 1 = b 5 + f = 14 + e + 9 = c 9 + h = 24.
From column 3 we have
+ e g == 24, so e + 9 =
The entries
T imply that
{e,g} = {3,7}; e = 3 ::::} d = 15 (from the equation d + 5 + e 9 = 32). With d = 15,
from a d + 1 = 24 we have a = 8,
8 tt. T. Consequently, e = 7 and g = 3, and
d = 32 - 21 = 11. Column 1 indicates that a d 1 = 24 so a = 12. J:;rom column 2 it
follows that b f =
so {b, f} = {4,15}. As b = 15 ::::} a b + 14 + c > 32, it follows
that b 4 and f =
Row 1 then indicates that c = 32 - a - b - 14 = 2 and from row 3
(or column 4) we deduce that h = 13. The completed table is shown in the figure.
=
12
11
1
35.
4
5
15
14 t 2
7
9
3 13
Let x denote the integer Barbara erased. The sum of the integers 1,2,3, ... , x-I, x
1,x + 2, ... ,n is [n(n + 1)/2j - x, so Un(n + 1)/2] - x]/(n - 1) = 351~' COllsequently,
[n(n 1)/2] - x = (35 177 )(n -1) = (602/17)(n -1). Since [n(n + 1)/2J - x E Z+, it follows
that (602/17)(n - 1) E Z+. Therefore, from Lemma 4.2, we find that 17j(n - 1) because
17 does not divide 602. For n = 1,18,35,52 we have:
x = [n(n + 1)/2} - (602/17)(n - 1)
1
n
1
18
35
52
-428
When n =
x=7
~-
17k, k 2:: 1, we Imve
x
=
= 7+
+
= [7+
.u.<J,..........:;
the ar.l.swer is U""'''H4',.
u'-·~;y
I!........
x = 7.
C: Leslie's selection is divisible by 5: {5, 10, 15, ... ) 95, IOO}
U
-
n
-
Here An B = {6, 12, 18, ... 1 96}, the set of integers between 1 and 100
divisible by 6 by both 2
3.]
(b) PrC A U U C) = Pre A) + Pre B) + P7'( C) - Pre A n B) - Pre A n C) - Pre B n C) +
50
33
20
16
10
6 + 3 74 - 0 "'4
P r (A n B n C) -- 100
100
100
- 100
- 100
-- 100
100 - 100 •i .
31.
A common divisor for m, n has the form p~lp~:1p;ll, where 0
'1'i
min{€il fil, for all
1 <i
3.
mi = min{ei'
1
i:::; 3.
the number of common divisors is
(m! 1)(m2 + l)(ms + 1).
CHAPTER 5
RELATIONS AND FUNCTIONS
Section.
A x B = {(I, 2), (2, 2), (3,2),
(1, (2,5), (3, 5), (4, 5)}
B x A = {(2, 1), (2,2), (2, 3), (2, 4), (5,1), (5, 2), (5,3), (5, 4)}
AU (B x C) = {l, 2, 3, 4, (2,
(2,4), (2, 7), (5, 3), (5,4), (5,
(AUB) x C = (A x C)U(B x C) = {(I, 3), (2,3), (3, 3), (4,3), (5, 3), (1, 4), (2,4), (3, 4), (4,4),
(5,4),(1,7), (2, 7),(3,7),(4, 7),(5,7)}
{(1,2)}; {(1,2),(1,4),(1,5),(2,2),(2,4)};A x B
(b) {(1,1),(2,2),(3,3)};{(1,1),(1,2),(1,3),(2,2),(3,3)}j{(1,2),(2,1),(3,3)}.
2.
(a)
3.
(a) IA x BI = jAIiBI = 9
(b) Since a relation from A to B is a subset of Ax B, there are 29 relations from
A to B.
(c) Since IA x AI = 9, there are 29 relations on A.
(d) For the other seven ordered pairs in A x B there are two choices: include it in the
relation or leave it out. Hence there are 27 relations from A to B that contain (1,2)
and (1,5).
(e) (!)
(f) (~) + (:) + (:)
4.
If either A or B is 0 and when A = B.
5.
(a) Assume that A x B ~ C X D and
G E A and b E B. Then
b) E A x B, and since
A xB
C x D we have (G, b) E C x D.
( tl, b) E C x D :::} a E C
bED. Hence,
a E A:::} a E C, 80 A ~
and bE B:::::> bED, 80 B
y) E x
~
~ C)
E
x
[(A ~ C) f~
x
134
xD.
6.
1.
(a) Since IAI = 5 and jBj = 4 we have IA x BI = IAIIBI = 5,4 = 20. Consequently, A X B
has 220 subsets, so IP(A x B)l = 220,
(b) If !AI = m and IBl = n, for tn, n EN, then IA x BI = mn. Consequently, jP(A x B)I =
27111< •
8.
A x (B U e) =
E A y E (B U C)} =
E
E
y E e)} =
y)l(x E
and 11 E
or (x E A and 11 E C)} = {(x,y)lx E
y E B} U {(x, y)Ix E A and y E C} = (A x B) U (A x e).
( c) &
proofs
are
that given part (b).
or
and
1 + 2 + 2(3) + 2(3)(5) = 39; 38
11.
(x, y) E A x (B - C) ~ x E A and y E B - C ~ x E A and (y E Band
y if C) {:==} (x E A and y E B) and (x E A
'II ¢ C) ~ (x,
E
x
and
(x, y) ¢ A x C ~ (x, y) E (A x B) - (A x C) ..
12.
2(3!BI) = 4096 = }
13.
(a) (1) (0,2) E ni and
(2) If (a, b) E 'R, then (a + 1, b 5) E'R.
(b) From part (1) of the definition we have (0,2) E 'R. By part (2) of the definition we
then find that
(i) (0,2) E 'R ==} (0 + 2 5) = (1,7) E 'R;
eli) (1,7) E 'R :::} (1
7 5) = (2,12) E 'Rj
(iii) (2,12) E 'R :::} (2 1,12 + 5) = (3,17) E 'R; and
(iv) (3,17) E 'R:::} (3 + 1,17 + 5) = (4,22) E 'R.
14.
(a) (1) (1,1),(2,1)E'Rjand
(2) If (a, b) E 'R, then (a
31Bl =
=::}
IBI = 4.
1, b + 1) and (a + 1, b) are 'R.
(b) Start with (2,1) in.
from part (1) of the definition. Then hy part (2) we get
(i) (2,1) E 'R:::} (2 + 1,1 + 1) = (3,2) E 'Rj
(ii) (3,2) E 'R:::} (3 + 1,2) = (4,2) E 'Rj and
(iii) (4,2) E 'R:::} (4+ 1,2) = (5,2) E 'R.
n-
Start with (1,1) in 'R - from part (1) of the definition. Then we find from part (2) that
(i) (1,1) E 'R,:::} (1 1,1 1) = (2,2) E 'Rj
(ii) (2,2) E 'R:::} (2+ 1,2 1);: (3,3) E 'Rj and
(iii) (3,3)e'R:::}
+1,3+1)=(4, ER"
1
~)
~
(c)
(e)
>
{(I, (2, (3,x),(4,x)},
(2, (3, (4,
{(1,x),(2,y),(3,x),(4,y)},{(1,x),(2,y),(3,z),(4,x)}
(3,z),(4,z)}
(a)
(b) 34
= 2187 =?
5.
(a)
(d) 43
0
(c)
(f)
(e) 24
=7
nB={(x,
--
landy=3x}
2x + 1 = 3x => x = 1
So An = {(1,3)}.
(b) B n C = {(x, y)l!! = 3x and y = x - 7}
3x = x - 7 => 2x = -7, so x =
COl18equently, B n C = {( -7/2, 3{ -7/2)} = {( -7/2, -21/2)}.
(c) AUC=AnG=AnG={(x,y)ly=2x landy=x-7}
Now 2x + 1 = x - 7 => x = -8, and so An G = {( -8,-I5)}.
(d) We know that B U G = B n G, and since BnG = {(-7/2, -21/2)} we have BU C =
R2 - {(-7/2,-21/2)} = {(x,y)lx f. -7/2 or II f. -21/2}.
6.
(a)
(b)
1.
8.
9.
(i)
(iii)
An B = {(1,3)}
U == {( -8, -I5)}
(ii)
(iv)
BU
(i)
(iii)
An B = {(1,3)}
(ii)
(iv)
BnG={}==0
Bu C = Z+ X z+
AUG=@
(d)
False: Let a = 1.5. Then 1.1.5J = 1 f. 2 = f1.51
Let a = 1.5. Then -1 = r-al.
U [0,1/7) U [1,8/7) U
(a)
8/7)
15/7) U . "
Z
2JU ... =
-1]
(h) ... U« -2n-
(m-
= Z2 = Z X Z
(b) l2.3J - l1.6J = 2 - 1 = 1
(d) l3.4H6.21 = 3·7 = 21
(f) 2f1rl = 8
(a) l2.3 -1.6J = lO.7j = 0
(c) r3.41l6.2J = 4:·6 = 24
(e) L21rJ = 6
(a) True
(c) True
BnG={}=0
-n-
0)
-l)!n,l] U«2n-
2] •..
(Case 1: kIn) Here n = qk
qE
(n - l)/k = (qk - I)/k = qwith q -1 q - (11k) < q. Therefore fn/kl = rql = q = (q 1 = L(n - l)/kj + 1.
(Case 2: k In) Now we have n = qk+r, where q,1' E Z+ with l' < k, and n/k = q+(r/k)
vvith 0 < (r/k) < 1. So n - 1 = qk (1' - 1) and (n - l)jk = q + [(1' - l)jk] with
0:5
- l)/k] < 1.
rn/kl =
(rlk)l = q 1 = l(n -l)/kJ + 1.
rq
Proof (i): Ifo,EZ+,
=0,
o,l/o,J =
=1. Ifa¢Z+,writea=n c,
where n E Z+ and 0 < c < 1. Then fal/o, = (n + l)/(n
= 1 + (1 - c)/(n c), where
0< (1 - c)/(n + c) < 1. Hence l /aJ =
(1 - c)/(n c)J = 1.
(E):
a E Z+,
= a and flaJ/al = fll = 1.
a ¢ Z+, let a = n+c, where
n E Z+ and 0 < c < 1. Then laJla = n/(n+c) = 1- [c/(n+c)], where 0 < c/Cn+c) < 1.
Consequently rlaJ/al = fl - (c/(n enl = l.
b) Consider a = 0.1. Then
(i) lfo,l/aJ = LI/0.IJ = llOJ = 10
(ii) rlaJ/al = fO/O.ll = 0 1.
1; and
In fact (ii) is false for all 0 < a < 1, since rlaJ/al = 0 for all such values of a. In the case
of (i), when 0 < a :5 0.5, it follows that ral/a > 2 and Lfal/aJ
2 i= 1. However, for
0.5 < a < 1, ral/a = I/o, where 1 < I/o, < 2, and so lfo,l/aJ = 1 for 0.5 < 0,< 1.
14.
(a)
= 2al2/2J = 20,1 = 2
0,3 = 2al3/2J = 20,1 = 2
0,4 = 2al4/2J = 20,2 = 4
as = 2al5/2J = 20,2 = 4
as = 2o,l6/2j = 20,3 = 4
0,7 = 2at'I'/2J = 20,3 = 4
as = 2o,is/2J = 20,4 = 8
a'};
Alternative Form
the Principle of Mathematical
For n = 1 we have 0,1 = 1 :5 1, so the result is
this
case.
for the proof.)
1 we
, where the ____ ",. ~."._.
<k+L
that
E
an
n.
(b)
(c)
the set of
One-to-one. The
One-to-one. Range = Q
Since J(l) = J(O),J is not one-to-one. The
±60, ... } =
J = {O,
3
{n - nln E Z}.
(d)
Range =
(e) One-to-one. Range =
+(0) = R+
1,1]
of J =
f IS
(a)
(d)
{4,9}
[0,9)
(b) {4,9}
[0,491
(c)
(f)
1].
[0,9)
[9,16) U [25,36]
The extension must
J(I) and J(4). Since
=4
are four choices for
each of 1 and 4, so there a,l'e 42 = 16 ways to extend the given function g.
18.
Let A = {1,2},B = {3,4} and f = {(1,3),(2,3)}. For Al = {1},A2 = {2}, f(A1nA z ) =
J(0) = 0 while f(Al) n f(A~) = {3} n {3} = {S}.
19.
(a) f(Al U A 2 ) = {y E BIY = f(x), x E Al U A 2 } = {y E Ely = f(x), x E Al or
x E A 2 } = {y E Bly = f(x), x E Ad U {y E ElY = I(x), x E A 2 } = I(Al) U J(A z ).
(c) 11 E f(Al} n I(A').) .> 11 = I(X1) = I(X2), Xl E At, X2 E Aa ==> y = I(xt) with
Xl = Xa, since f is one-to-one =:::} y E f(Al n Aa).
20.
The number of injective (or, one-to-one) functions from A to B is (lBI!)!(lB! - 5)1 =
6720, and IBI = 8.
21.
No. Let A = {I, 2}, X = {I}, Y = {2}, B = is}. For 1= {(I, 3), (2, 3)} we have fix, Ily
one-to-one, but f is not one-to-one.
22.
(a) A monotone increasing function I : X 7 ~ Xs determines a selection, with repetitions allowed, of size 7 from {l,2,3,4,5}, and vice versa. For example, the selection
1,1,2,2,3,5,5
to the mouotone increasing function 9 : X 1 -+
where 9 =
{(l, 1), (2, 1), (3, 2), (4, 2), 3), (6,5), (7, 5)}. (Note the second components.) Consequently,
the
functions .f :
-+ XI)
(H~-l) =
= 330.
en
= 6 =
nnc;tlOllS f :
llUl:U.Dler of monot-one
1S k
m
1
In.
f :
increasing and f(k} = i, then f( {I, 2, ... , k - I})
{.e, l 1, ... , n}, So there are
{I, 2, ... ,f} and f( {k + 1, ... , m}) ~
H(k-l)-l) (n-l+!}+(m-(k+1Hl)-l) (1t-1)
(m-(k+l}+1)
-
(n+m""J,_-~-k) such functions.
23.
(a) f(ai;) = 12(i - 1) + j
) = lO(i - 1)
24.
g(aij)=m(j-
25.
(a) (i) f(aij) == n(i - 1)
(ii) y(aii} = m(j - 1) +
(b) k + (mn - 1) :;; r
26.
(a) There is
one function in Sh namely f:
~ B where f(a) = feb) = 1 and
fCc) = 2. Hence lSll = l.
(b) Since fCc) = 3 we have two choices - namely 1,2 - for each of f(a) and feb).
Consequently, IS21 = 2 2 ,
(c) With ftc) = i 1 there are i choices - namely 1,2,3, .. . ,i -I,i - for each of f(a)
and feb), so ISil = i 2 •
( d) Any function f in Tt determined by two elements x, y in
where 1 :s; x < y :;; n + 1
and f(a) = feb) = x, ftc) = y. We can select these two elements from B in C1-~l) ways,
(
'"
j
f(o,ij) = 7(i -1)
J
t
(k - 1) J
-1) +i
so ITtl = (n~l).
(e) For T2 we have f(a) < feb) < ftc), so we need three distinct elements from B, and
these can be chosen in (n;l) ways. The argument for Ta is similar.
(f) S = 8 1 U S2 U Sa U ... U Sf!.' where Si, n Sj = 0 for all 1 :;; i < j < n, and S = Tl U Tz U T3
with Tl n T2 = Tl n Ts = T2 n Ts = 0.
(g) From part (f) we have lSI == ~ISil ==
n
t; i = t;lTil = (n +2 1) 2 (n +3 1) . Hence
n
2
n
3
E i = (n + 1)(n)/2 + 2(n + l)(n)(n - 1)/6 == (n + 1)(n)[(1/2)
2
£,,::1
(n + 1)(11.)[(3
21.
2n - 2)/6] = n(n
1)(2n + 1)/6.
= A(1,2) + 1 = A(O, A(l, 1» + 1 - fA(l,l)
2=
1J 2 =
+3 =
3=
(a)
)3) = A(O, A(l, 2»
A(l,
2=
(7
1 = A(O,
= A{l,
A(l, 2» = A(l, 2)
~= A(O, A(l,
= A(l,
2» + 2:= ACl,
3 = A(O,A(l,l»
7) ==
1 == 9
-
(n - 1)/3] =
I}
1 == 5
3)+ 2 =
1==
(b) Since A(l,O) = A(O,l) = 2 = 0 2, the result holds for the ('.-ase where n = O.
of
(open)
some k
0) we have A(l, k) = k
we find that A(l, k + 1) = A(O, A(l, k» = A(l, k) + 1 = (k 2) 1 = (k + 1) 2,80
at n = k implies
truth at n = k + Consequently, A(l, n) = n + 2 for all
by the Principle of Mathematical Induction.
(c)
we find that A(2,0) =
= 1 + 2 = 3 (by the
in
So
A(2, 0) = 3 2 ·0 and the given (open) statement is true in this first case.
we assume the result
for some k (2: 0) - that is, we assume that A(2, k) = 3+2k.
For k 1 -we then find that A(2,k + 1) = A(1,A(2,k» = A(2,k) + 2 (by part (b» =
(3 2k) 2 (by the induction hypothesis) = 3 + 2( k 1). Consequently, for all n EN,
A(2, n) = 3 2n - by the Principle of Mathematical Induction.
(d) Once again we consider what happens for n = O. Since A(3, 0) = A(2, 1) = 3 2(1)
(by part (c» = 5 = 20 +3 - 3, the result holds in this first case.
So now we assume the given (open) statement is true for some k (2: 0) and this gives
us the induction hypothesis: A(3, k) = 21:+ 3 - 3. For n = k 1 it then follows that
A(3, k + 1) := A(2, A(3, k» = 3 + 2A(3, k) (by part (c» = 3+ 2(2k +3 - 3) (by the induction
hypothesis) = 2(k+1)+3 - 3, so the result holds for n = k + 1 whenever it does for n = k.
Therefore, A(3, n) = 2n +3 - 3, for all n E N - by the Principle of Mathematical Induction.
28.
e)42 + (D41 = (4 + 1)5 - (:)45 - (~)4° = 55 - 45 -1
(b) (m~l)nm-l + (m~2)nm-2 + ... + (7)n1 = (n l)m - nm - 1.
(a)
(:)44
(~)43
Section 5.3
Let A={1,2,3, ,B={v,w, y,z}: (a) f={(1,v),(2,v),(3,w),(4,x)}
(b) f = {(1,v),(2,x),(3,y),(4,z)}
(c)
A = {1,2,3, S},B = {w,x,y,z}'f = ((l,w),
w),
x),(4,y),
(d) Let A = il,2, 4},B = {w,x,y, ,/ = {(l,w), (2,x),(3,y), (4,z)}.
range
(a)
5.
(b) 46 ; (4!)S(6,4); 0
; 0
l)k C'~k)(5
For n = 5,m = 3,
(-
(_1)2 (~)(3)3+
- k)3 = (-1)O(!)(5? + (_1)1 (:)(4)3
(D(1)3
(~)
(i)
1)
=125-320+
-80
5=0
1)
6.
(a) l:(!)(i!)8(7, i) =
(;)(21)8(7,2)
(;)(3!)8(7,3)
(:)(4!)8(7,4) +
(5)(24)(350)
(1)(120)(14) =
i=l
(;)(5!)8(7,5) = (5)(1)(1)
78,125 = 57.
(10)(2)(63)
(10)(6)(301)
(b) The expression mil counts the number of ways to distribute n distinct objects among
m distinct containers.
For 1 < i ::; m, let i count the number of distinct containers that we actually use - that
is, those that are not empty after the n distinct objects are distributed. TIllS number of
distinct containers can be chosen in
ways. Once we have the i distinct containers we
can distribute the n distinct objects among these i distinct containers, with no container
(7')
m
left empty, in (i!)S(n, i) ways - where Sen, i) = 0 when n < i. Then L:(7)(i!)8(n, i) also
i=l
counts the number of ways to distribute n distinct objects among m distinct containers.
m
Hence m!l = L('~)(i!)8(n, i).
i=l
1.
(a)
(b)
8.
(i)
(iv)
218(7,2)
[3!S(7, 3)]
g)
(ii)
(v)
e) [2!S(7, 2)]
(iii)
(vi)
4!S(7,4)
3!8(7,3)
(!) [4!S(7, 4)]
(~)[k!S(m,k)]
Let A be the set of com.pounds and B the set of assistants. Then the number of
assignments wit,h no idle assistants is
number of onto functions from
to
B.
There are 5IS(9,5) such
that
real ~;,LU~~U:;~,"bO. t;o:nse:qt:l.4ent,i~
10.
+
n
1
2
9
10
1
1
255 3025
9330
4
3
6
5
7
8
9
10
462
5880
36
750
1
45
1
rn
7770
6951
2646
34105 42525 22827
(a)
100,905 = 5 x 11 X 17 X 29 X 31 x 37, we
there are 8(6, = 90
unordered factorizations of 31,100,905 into three factors --- each .. ,~.r.... than 1.
(b) If the order the fadors part (a) is considered relevant
there are (3!)S(6, 3) =
540 such factorizations.
T ....
43
(c) I:S(6,i) = S(6,2) + 8(6, 3)
8(6,5)+8(6,6) = 31
8(6,4)
T
65 +
90
+ 1 = 202
i='2
6
(d) L:(i!)S(6, i) = (2!)8(6, 2) + (31)S(6, 3) + (4!)S(6,4) + (51)8(6,5)
(2)(31) + (6)(90)
13.
(24)(65)
(120)(15)
(61)8(6,6) =
(720)(1) = 4682.
(a.) Since 156,009 = 3 x 7 x 17 x 19 x 23, it follows that there are 8(5,2) = 15 t"\N'O-factor
unordered factorizations of 156,009, where ea.ch factor is greater than 1.
5
(b) E8(5, i) = S(5, 2) + 8(5,3) + S(5, 4)
S(5, 5) = 15 + 25 + 10
n
(c) ES(n, i).
i:;;;:2
14.
10
20
Dim 8(12, 12)
For I = 1 To 12
S(I,I) = 1
Next I
Print "M =
::;;;; 2
12
30
50
60
100
120
Pliut
M
>(
140
15.
n=
n
= 5:
i)
1 = 51.
answel'18
i:::::l
12
16.
a) (i) 1O!
(ii) The given outcome - namely, {C:;l,C3 ,Or}, {Cl,C4 ,CS ,ClO }, {CIS}' {C6 ,CS} - - IS
an example of a distribution
ten distinct
ru.nong four distinct
with
no container left empty. [Or it is an example of an onto function f : A --+ B ",.,1"\",...,.
A = {C1 ,C:;!"",CUI }
= {1,2,3,4}.J There are 4fS(1O,4) such distributions [or
functions] .
10
The allswer to the question is
(iii) (~O)
i!S(lO, i).
'i
L: i!S(7, i).
i=1
"(
b) (~)
L: i!S(7, i)
i=l
c) For 0 :::; k :::; 9, the number of outcomes where C 3 is tied for first place with k other
candidates is (!)
9-k
L: i!S(9 - k, i). [Part (b) above is the special case where k = 3 - 1 = 2.)
i=l
Summing over the possible values of k we have the answer
t (!) ~
k=O
1 T.
i!S(9 - k, i).
1::::1
Let at, a2, ••• , am) X denote the m
1 distinct objects. Then B,,(m + I, n) counts the
number of ways these objects can be distributed among n identical containers 80 that each
container receives at least r of the objects.
v ...,,.v~.'" falls into 'VA'''''-'''''5 one of two cal;e:e.:,orl,es
of
1) The element x is in a conta,iner with Y' or more other objects:
n>mwe
Here we start with
(b) For m ~ 1,
sCm, m) = 1 because
ordering
the m tables is not taken into
account; and, (ii) sCm, 1) = (m - 1)1, 8.'3 in Example 1.16.
are two people at one table and one at each of the
m - 1 tables.
(c )
There are (';) such arrangements.
(d) When m people are seated around m-2 tables there are two cases to consider: (1) One
table with three occupants and m - 3 tables, each with one occupant - there are
such arrangements; and, (2) Two tables, each with two occupants,
m - 4 tables each
with a single occupant are (1/2)(~)
of these arrangements. We
find
C:)(2!)
(m;2)
that (r;)(2!)+(1/2)(~)
(m;2) == (1/3)(m)(m-l)(m-2)+{1/2)[(1/2)(m)(m-l)J[(1/2)(m-
2)(m -- 3)] = (m)(m -l)(m - 2)(1/3) +(1/8)(m - 3)] = (1/24)(m){m -l)(m - 2)(3m -1).
19.
(a) We know that sCm, n) counts the number of wa,ys we can place m people - call
them PI, P2, ... ,Pm - around n circular tables, with at least one occupant at each table.
These arrangements fall into two disjoint sets: (1) The arrangements where PI is alone:
There are sCm -l,n -1) such arrangements; and, (2) The arrangements where PI shares
a table with at least one of the other m - 1 people: There are s( m - 1, n) ways where
112, PS,' .. ,Pm can be seated around the n tables so that every table is occupied. Each
such arrangement determines a total of m - 1 locations (at all the n tables) where PI
can now be seated - this for a total of (m - 1)8(m - 1, n) arrangements. Consequently,
sCm, n) = (m - l)s(m - 1, n) + sCm n - 1), for m ;?: n > 1.
m-1l
(b) For m = 2, we have s( m, 2) = 1 = 1!(1/1) = (m -1)! E -;-. So the result is true in this
,:::::1 Z
case; this establishes the basis step for a proof by mathematical induction. Assuming the
A.-II
result for m = k(;?: 2) we have s(k, 2) = (Ie - l)!E -;-. Using the result from part (a) we
i=l
z
1;-1
now
s(k
1,
= ks(k,
+s(k, 1) = k(k·-l)!
1
-:-+(k-·
i_I I
now follows
1.
c) = c, while
= a, so f is
1;-1 1
= k!
"7+(l/k)k! =
I
all m ;?: 2 by the Principle of Mathematical
while,
1(3.2, f(4.7, 6.4» = f(3.2, r4.7+6.41) == f(3.2, r11.11) = 1(3.2,12) = [3.2+121 = r15.21 = 16.
z.
(c) There is no identity element. If a E R - Z then for any b E R, fa + b1 E
So if
x were
identity
we would have a = I( a, x) = a
xl with a E R - Z and
r
fa xl E Z.
3.
(a) I(x, y) = x y - xy = y x - yx = I(y, x), so the binary operation is commutative.
1(J(w,x),y)=/(w,x) y-f(w,x)y=(w+x-wx)+y-(w x-wx)y=w x ywx-wy-xy wxy.
l(w,/(x,y»=w+/(x, -w'f(x,y)='w+(x y-xy)-w(x y-xy)=w+x ywx - wy - xy + wxy.
Since f(J(w, x), y) = few, f(x, v»~, the (dosed) binary operation associative.
(b), (d) Commutative and associative
(c) Neither commutative nor associative.
4.
(a) The identity is z = o.
(d) The identity is z = 3.
(b), (c) Neither of these (dosed) binary operations has an identity.
5.
(a)
6.
(b) 515
(a) 524
(c) 3.515 , because neither a nor b can be an identity.
(d) 3.59
7.
ea ) Yes
8.
Each element in A is of the form 2i for some 1 $; i $; 5, and gcd(2i,25) = 2i = gcd(25,2i),
so 25 = 32 is the identity element for f.
9.
(a)
(b)
25
(c) 525
525
(b) Yes
=
510
(c) No
f is
f:AxA-+
(d)
n
f
(a)
(c)
'lfA(D) = [0, +00) 1rB(D) = R
1rA(D) =
,1]
1rA(D) = [-1,1]
1rB(D) = [-1,1]
13.
(a) 5
(c) AI,
14.
(a) 5
(b) {(I, A), (1, D), (1, E), (2, A), (2, D), (2, En;
{(IOOOO, 1, 100), (400,1,100), (30,1,100), (4000,1,250),
(e) A 1 xA 2 ; A2XAS; A3 X As
(b) {(25,25,6), (25,2,4), (60,40,20), (25,40,10)}
250),(15,l,250)}
Section 5.5
1.
Here the socks are the pigeons and the colors are the pigeonholes.
2.
The result follows by the Pigeonhole Principle where the eight people are the pigeons and
the pigeonholes are the seven days of
week.
3.
262 + 1 == 677
4.
Subdivide the set S into the 14 subsets:' {3}, {7, l03}, {ll, 99}, {15, 95}, ... , {43, 67},
{47, 63}, {51, 59}, {55}. By the Pigeonhole Principle if we select at least 15 elements of S
then we must have the elements in one of the two-element subsets and these sum to 110.
5.
For each x E {I, 2, 3, ... , 300} wrote x = 211.· m, where n ~ 0 and gcd(2, m) = 1.
There are 150 possibilities for m; namely, 1,3,5, .. , ,299. In selecting 151 numhers from
{I, 2, 3, ... , SOD} there must be two numbers of the form x = 2"· m, Y = 2t • m. If x < Y
then xlll; otherwise y < x and ylx.
(b)
11,
1 integers are selected from
set {I, 2, ... I 211. }, then there
be two
integers x, y in the
xlY or
(a)
6.
n
even or
2
(d)
, a2, .. , I an)!ai E Z+, 1 ::; i
...v .... ""',,......'"
VI::; i
two ""...,,1 ",....:.rl
,X2,"'Xn),
"',Yn)
Xi
1, then S
'IIi 18 even
n.
5 - as
9.
(a) Forauy t E {I, 3, ... ,lOO},1::; 0::; 10. Selecting elementsfrom {1,2,3, ... ,
there must be two, say x
where l JXJ = l y'iiJ, so that 0 <
< 1.
2
(b) Let n E Z+. If n+ 1 elements are selected from {1,2,3, ... ,n }, then there exist
two, say x and 1/,
0 < I.JX - 071 < 1.
. .- -....- -.. C
I
J
triangle ABC, divide
side
three
parts
nine congruent triangles shown in the figure. Let Rl be
tria..l1g1e
with
on segment
DE, excluding D,E. Region R.l. is the interior of triangle DFG
with the points on segments
excluding D,F.
R 3 , ••• ,R9 are defined similarly so that the interior of
ABC is the union
these nine regions and ,~n
= 0, for
j.
if 10 points are chosen in the interior of D. ABC, at
two of these points are in ~ for some 1 < i :5 9, and these
two
are at a distance less than
from eac.h other.
1
E
H
t---_ C
i -_ _
G
12.
the interior of the square into four smaller congruent squares
as shown in the figure. Each smaller square has diagonal length
1/.;2. Let regi.on Rl be the interior of square AEKH together
with the points on segment EK, excluding point E. Region R'}, is
interior of square EBFK t~gether with the points on segment
excluding points F ,K. Regions R31 R4 are defined in a simibl.-t' way. Then if five points are chosen in the interior of square
ABCD, at least two are in ~ for some 1 < i :5 4 and these
points are within 1/12 (units) of each other.
five-element subset E of A we find that 1 + 2 + 3 4 5::::; 15 :5 SE
115:5
21 + 22 + 23 + 24 + 25, so there are 116 possihle values for such a sum BE. Since IAI = 9,
there are (:) = 126 five-element subsets of A.
For
The result now follows
the Pigeonhole Principle where the 126 five·element subsets of
A are the pigeons and the 116 possible sums are the pigeonholes.
A
s
3.
lSI =
axe
A.
>
-1)
... +
So
the
21.
(~)
s
Proof:
k+lintegers:
3; (2)33;
333;
(k+l)
.,.3,
where for all 1 :5 i .:s; k 1, the i-th integer·has i digits - each of which is a 3. Since
are k 1 integers, it follows from
Division Algorithm and the Pigeonhole Principle that
two of these integers,
a and b, have the same remainder when divided by k. Suppose
that a : : : : . <]1 k r, b : : : : . q'J,k + r, and
a > b. Then a - b = (<]1 - (2)k, so k I(a - b) and the
only digits in a - b are O's and 3's. [Note: The integer 3 is not special.
result is also
if we replace 3 by any of the
1, 2, 4,
6, 7, 8,
However, we
obtain
the result
using the
0.)
"'j
(a) 2,4,1,3
(b) 3,6,9,2,5,8,1,4,7
( c) For 1'1,
2, there exists a
of 1'1, 2 distinct
numbers with no
increasing subsequence oflength n+l. For example, consider 2'11,3'11, • .. ,
or
1)'11, '11 2 , ('111), (2n-l), ... , (n 2 -1), (1'1,-2), (2n-2), .. . ,
-2)".,,1, ('11+1), (2n+ 1), ... , ('11-1)'11+1.
(d) The result
Example 5.49 (for n ~ 2) is best possible - in the sense that we cannot
reduce the length of the sequence from '112 + 1 to '112 and still obtain the desired subsequence
of length '11 1.
18.
This follows from the result due to Paul Erdos and George Szekeres: A sequence of SO( =
72 + 1) distinct real numbers contains a decreasing or increasing subsequence of length
8(= 7 + 1).
19.
Proof: If not each pigeonhole contains at most k pigeons - for a total of at most kn
pigeons. But we have kn + 1 pigeons. So we have a contradiction and the result then
follows.
20.
(a) 7
21.
( a)
(b) 13
(c) 6(n-1}+1
1001
(b) 2001
Let k,n E Z+. The smallest value for IS! (where S C Z+) so that there exist '11
Xl, X2, ••• ,X n E S where all '11 of these integers have the same remainder upon
division by k is k( n 1.
(c)
a total of
23.
Section 5.6
(a)
- of these, 6! satisfy J(I) = 1. Hence there
1.
There are 7! bijective functions on
7! - 6i = 6(6!) bijective functions J: A --t A where J(I)
(b) n! - (n - I)! = Cn - l)(n - I)!
ru:e
2.
(a) Here J,g have the same domain A and some codomain R, and for all x E A we find
that
8
g(x) = 2;2 -2
=
x ;2
= 2(x ~2)(~)
2) = 2(x - 2) = 2x - 4 = lex).
Consequently, J = g.
(b) Here there is a problem and f :f. g. In fact for any nonempty subset A of R, if -2 E A
then 9 is not defined for A because g('. -2) = 0/0. [We note that ~~4 = x - 2, for x :f. -2.]
3.
9x 2 -9x+3=g(f(x))=1-(ax b)+(ax+b)2=a 2 x 2 +(2ab-a)x+(b2 -b+l). By
comparing coefficients on like powers of x, a = 3, b = -lor a = -3, b = 2.
4.
go 1
5.
= {(l, 4), (2,6), (3,10), (4, 14)}
g2(A) = geT n (8 U A» = Tn (8 U [T n (8 U A)1) =
Tn [(8 U T) n (8 U (8 U A»] = Tn [(8 U T) n (8 U A)] =
[T n (5 U T)} n (8 U A) = Tn (S U A) = g(A).
6.
(f 0 g)(x) = J(cx + d) = a(cx + d) b
(g 0 J)( x) = g( ax + b) = e( ax b) d
(f I) 9 )( x) = (9 0 1)(x) ¢:.:::::} acx + ad b = acx
(g 0 f)(x) = 3(x - 1);
(a.) (f 0 g)(x) = 3x-
x even;
(90 h)(x) = {
(;;
8.
={
0
-
=x-
(a) If c E C, there
E
0 g)(x)
x
(} h»(x)::::
(I (}
so 51
d -¢=} ad + b = be + d
be
I)
h)(x)) = {
I)
x even
(;;
x
={
x
1, x
x even;
x
, :x even
={
x
=x-
-
aE
that (51 0
-
-
= c.
-
=c
)=
X,Y E
f} 0
f is
(a)
(b)
_
==} x =
0
.ot~-5+5
-<.-
_
DenX' _
-
....
"..
-"')
5 - 5] = x.
(a)
= {(x, y)12y 3x = 7}
+ lIx = c, b #: a O}
1
3
(c)
= {(x, y)ly = X / } = {(x, y)lx =y3}
(d) Here f(O) =
= 0, so f is not one-to-one, and consequently f is not invertible.
11.
9 invertihle ==} each of
9 is both
and onto = : } f} 0 f is
onto ==} go f invertible. Since (g 0 f) 0 (I-I 0 g-l) =
0
) 0 (g 0
1
lA, f- 0 g-1 is an inverse of go I. By uniqueness of inverses 1-109-1 = (g 0
12.
(a) I- I ( {2}) = {a E AlfCo,) E {2}} = {a E Alf(o,) = 2} = {l}
(b) f-l({6}} = {a E Alf(a) E {6}} = {a E Alf(a) = 6} = {2,3,5}
(c) f-l({5,8}) = {a E AlfCo,) E {6,8}} = {a E Alf(a) = 6 or f(a) = 8} = {2,3,4,5,5},
because f(2) = f(3) = f(5) = 5 and f(4) = f(6) = 8.
(d) f-I( {6, 8, 1O}) = {2, 3, 4,5, 6} = f-I( {5, 8}) since f-I( {1O}) = 0.
(e)
lO,12}) = {2,3,4,5,6, 7}
(f)
12}) = {7}
13.
x
7=
(b) (i)
E R i x S; 0 and
S; x
7 S; -1} U
E
I0 < x < 3
and -5 S; -2x + 5 S; -I} U {x E R I 3 S; x and
S; x-I S; -I} = {x E R I x S; 0
and -12
x
-8} U {x E RIO < x < 3 and 3 S; x S;
U {x E
I 3 S; x and
x :::; O} =
U 0U 0 =
12~ -8]
([-5,0]) =
12,-7] U [5/2,3)
/-1([-2,4])=
E
Ix::;Oand
:5x 7:; U{xE
!O<x<3
-2x + 5 ::; 4} U {x E R I 3 :; x and -2 :::; x-I < 4} = {x E R I x
0 and
x
-3} U
E RIO < x < 3 and 1/2 ::; x :5 7/2} U {x E R I 3 :5 x and
x :;
-3J U [1/2,3) U [3,5] = [-9,
U [1/2,5]
«5,10» = (-2, U (6, 1
1-1 ([11,17» = {x E R I x ::; 0 and 11 :S x + 7 < 17} U E RIO < x < 3
:;
5 < 17} U
E R ! 3 :5 x and
:; x-I < 17} = {x E R I x
0
and 4 :; x < IO} U {x E RIO < x < 3 and -6 < x
-3} U {x E R I 3 :; x and
:; x < 18} = 0 U 0 U [12,18) = [12,18)
(b) {-I,D,l}
(c) [- 1]
(d) (-1,
(f) (-3, -2) U [-1,0) U (0, I} U (2,3)
14.
(a) {-I,O,l}
(e) [-2,2J
15.
Since 1-1 ({6, 7,8}) = {1,2} there are three choices for each of J(l) and /(2) - namely, 6,
7 or 8. Furthermore 3,4,5 ¢ f-I( {o, 7, 8}) so 3,4,5 E f-I( {9, 10, 11, 12}) and we have four
choices for each of f(3), f( 4), and f(5). Therefore, it follows by the rule of product that
there are 32 .43 = 576 functions f : A --t B where f-I( {6, 7, 8}) = {I, 2}.
16.
(a) [0,2)
(e) [-1,3)
17.
(b) [-1,2)
(f) [-1,0) U [2,4)
(c) [0,1)
(d)
to, 2)
(a) The range of f = {2,3,4, ... } = Z+ - {l}.
(b) Since 1 is not in the range of f the function is not onto.
(c) For all x,y E Z+, f(x) = f(y)::::} x + 1 = y + I::::} x = y, so f is one-to-one.
(d) The range of 9 is
(e) Since g(Z+) =
,
codomain of g, this function is
(f)
g(::::; 1 ::::::: g(2),
1 ::;£ 2, so 9 is not one-to-one.
(g) For aU x E Z+, (gof)(x) = g(/(x)):: g(x 1) = max{l, (x+l)-l} =
1,x} = x,
x E Z+.
go/ = 1z+ .
(f 0
o
(f 0 g)(4) =
o
(f 0 g)(12) =
(fo
(i) No, oe(:aUI'le
fog
18.
(a)
1(0,0) = 0 = 1(0, {I}) and (0,0)
I}, {2}) = {I,2} = g( {I,
(O, {I}), so I is not one-to-one.
) {2}) and ({I}, {2})
({I,
,{2}), so 9
one-to-
one.
h({l}, {2}) = {l,2} = h( {21, {I}) and ({I}, {2}) # ({2}, {I}), so h is not one-to-one.
For each
A of Z+, I(A, A) = g(A, A) = h(A,
= A, so each
the
functions f, g, and h, is an onto function.
(c) From the
in part (a) it follows
none
these functions invertible.
1
1
1
(d) The sets 1- (0), h- (0),
I}), h- ({3}), I-I
7}), and h- 1 {{5,9}), are all
infinite.
(e) Ig-1(0) = {(0,0)}, so Ig- 1(0)1 = 1.
g-l( {2}) = {(0, {2}), ({2}, 0), ({2}, {2} H, so Ig- 1 ( {2})1 = 3
!g-1({8,12})1 = 9.
19.
(a) a E 1- 1 (B 1 nB2 ) ~ I(a) E B I nB] { = } I(a) E
and I(a) E B2 {=} a E j-I(BI )
1
and a E I-I(B'};) { = } a E
(Bd n 1- (B 2 )
(c) a E l-l(B!) ¢=? I(a) E BI {.=> I(a) ¢ Bx {=} a ¢ I-I(Bt ) {.=> a E I-I(B!)
20.
(a)
(b)
21.
(i) I(x) = 2x;
(ii) I(x) = lx/2J
No. The set Z is not finite.
(a) Suppose that Xl, X2 E Z and I(x!) = l(x2)' Then either I(xl), f(x'},} are both even or
they are both odd. If they are both even, then I(xt} = I(X2) =:> -2Xl = -2X2 =:> Xl = X2Otherwise, l(x1), 1(x2) are both odd and l(xl) = l(x2) =:> 2X1 - 1 = 2X2 - 1 =:> 2X1 =
2X2 =:> Xl = 2:2' Consequently, the function f is one-to-one.
In order to prove that f is an onto function let n E N. If n is even, then (-n/2) E Z and
(-n/2) < 0, and 1( -n/2) = -2( -n/2) = n. For the case where n is odd we find that
(n + 1)/2 E Z and (n + 1)/2> 0, and I«n + 1)/2) = 2[{n + 1)/2] -1 = (n + 1) - 1 = n.
Hence f is onto.
(b)
1-1:
--+ Z, where
1),
are
1""",,,,,,,1'1 9 {) f
= h (') f =
kol =
o g)(1) = f(l1/3J) = f(O) = 3,0 = 0
-
(iii)
) = f(O) = 3·0 = 0 1, so f 0 h
) = f(l} = 3·1 = 3 #
k,
f
not
it
1.
(a)
1 E O(n)
(b)
1 E 0(n )
(g) 1 E 0(n 2 )
2
2.
fED )
f E 0(n3 )
m = 1 and k = 1 in Definition 5.23.
'in ~ k l/(n)1 = n < n + (l/n) = Ig(n)l, so
1 E O(g).
Now let m
2 and k = 1. Then Vn ~ k Ig(n)1 = n (l/n)::; n n = 2n = 211(n)I,
9 E O(!).
=
3.
(a) For all n E Z+,O::; logan < n.
let k = 1 and m = 200
I/(n)1 = lOOlog2 n = 100«1/2) 10g2n) < 200«1/2)n) = 200Ig(n)l, so IE O(g).
(b) For n = 6, 211 = 64 < 3096 = 4096 - 1000 = 212 - 1000 = 22n - 1000. Assuming
that 2k < 22k - 1000 for n = k 6, we find that 2 < 22 ==> 2(21:) < 22(22k - 1000) <
2222k - 1000, or 2k+1 < 2 2(k+1) - 1000, 80 fen) < g(n) for all n ~ 6. Therefore, with
k = 6 and m = 1 in Definition 5.23 we find that for n
k !f(n) I ~ mly(n)1 and
IE O(y).
(c) For all n ~ 4, n 2 ~ 211. (A formal proof of this can be given by mathematical
induction.) So let k = 4 and m = 3 in Definition 5.23. Then for n ~ k, 11(n)1 = 3n 2 <
3(2n) < 3(2'1), + 2n) = mlg(n)l and 1 E O(g).
4.
Let m = 11 and k = 1. Then \In 2:: k If(n)1 = n 100::; 11n 2 = mly(n)/, so I E O(g).
However, Vm E R+ Vic E Z+ choose n> max{k, 100 + m}. Then n 2 > (100 + m)n =
lOOn mn> 100m mn = m(lOO + n) = ml/(n)l, so y f: O(f).
5.
To show that f E O(y), let k = 1 and m = 4 in Definition 5.23. Then for all
n 2:: k, If(n)1 = n 2 + n ::; n 2 n 2 = 2n 2 ::; 2n 3 = 4«1/2)(n3 )) = 419(n)l, and f is
dominated by 9.
show that 9 ¢ O(f), we follow
in Exa.mple 5.66 -- namely that
VmE
E Z+ [(n ;::: k) A (l9(n)1 > mif(n)l)J.
\IkE
n > max{4m,
00
>m,=1!1,·l=
moose n>
6.
"'J<UJ'Ua...
9 f/.
vr.:ray, 'rim E
9 ¢ O(f)·
!>m=m·l=
and
111,
= 1
we
nenee 9 E O(f).
'OO<i!'"".... J.J''''M'<'U
f E
can he
by using
we
that for
m E
and k E
is an n E
If(n)1 = 11, > mIo!!;;;! 11, = mlg(n)!. Hence f f/; O(g).
8.
J E 0(9) ==:} 3 m l E R+ 3k1 E
E R+ 3k2 E Z+ so that
~ mllg(n)J S
9.
Since
n
~ k1 If(n)1 ~ mllg(n)j.
so
\/11,
such
g E O(h) = }
2: k2 Ig(n)1 :::; m2Ih(n)l. Therefore, \/11, ~ max{kb k 2 }
f E
m2!h(n)!
f E O(g), there
But then I/{n}1
> m, or
m E R+, k E Z+ so that 1/(71,)1:::; mlg(n)1
[mllclJjcg(n)! for all n 2: k, so f E O(cg).
all n ~ k.
10.
(a)
k = 1 and m = 1 in Definition 5.23.
(b) If h E O(f) and IE O(g), then hE O(g) by Exercise 8. Likewise, if hE O(g)
and 9 E
then h E 0 (/) - again by Exercise 8.
(c ) 1~hi8 follows
parts (a) and (b).
11.
(a) For all n 2: 1, fen) = 511,2 + 311, > 11,2 = g(n). So with M = 1 and k = 1, we have
If(n)1 > Mlg(n)1 for aU n 2: k and it follows that f E neg)·
(b) For all n
1, g(n) = n 2 = (1/1O)(5n2 + 5n 2 ) > (1/1O)(5n 2 3n) = (1/1O)f(n). So
with M = (1/10) and k = 1, we find that Ig(n)1 ~ Mlf(n)1 for all n ~ k and it follows
that 9 E n(f).
(c) For all n
1, fen) = 5n 2 + 3n > n = hen). \Vith M
If(n)1 2: Allh(n)! for all 11, 2: k and so f E n(h).
= 1 and k = 1, we have
(d) Suppose that h E fl(f). If so, there exist M E R+ and k E Z+ with 11, = /h(n)1
Mlf(n)l == 11/(5n 2 +3n) for all n 2: k. Then 0 < M < n/(5n2+3n) = 1/(5n+3). But how
can M be a positive constant while 1/(5n + 3) approaches 0 as n (a variable) gets larger?
From this contradiction it follows that h ¢ fl(/).
12.
Suppose that f E fl(g). Then there exist M E R+ and k E Z+ such that
If(n)1 > Mlg(n)! for aU n ~ k. Consequently, jg(n)j (l/l!l)lf(n)1 for all n 2: k, so
Proof:
9 E 0(/).
Conversely, 9 E 0(1) =:;}
3k E Z+ Vn 2:: k (If(n)1
E
> k (lg(n)1
E
(l/m)lg(n)!) ::::::} 3M E R+
M = 11m.] [Note:
if'
without the
mlf(n)i)::::::}
E R+
E Z+ \/n
k (li(n)1 >
fn/2p =
3
Mln 1
n
+
>
With k = 1
]v! = 1/6, we
that ly(n)1 ~ Mln 3 1 for all n :2:: k - 80 9 E 0(n 3 ).
(c)
it
> rn/21 t +..
> fn/2V+···+rn/21t = fCn+l)/2Hn/21 t >
(n/2)1+1. With k = 1 and Ai = (1/2)H1, we have Ih(n)1 2:: MlnH1 1for all n
k. Hence
h E 0(n1:+1),
=
14.
Proof: I E a{g) :::::> 3fil,m2 E R+ 3k E Z+ \in 2:: k fitlg(n)1 :5 !f(n)1 < mz/g(n)j :::::>
3m} E R+
E Z+ Yin
k mllg(n)j :5 If(n)1 and 3m2 E R+ 3k E Z+ \in ~ k I/(n)1 ~
mzJg(n)1 :::::> f E neg) and I E O(g).
Conversely, f E neg) => 3 m l E R+ 3k1 E Z+ \in :2:: kl mllg(n)1 S I/(n)l. Likewise,
I E O(g) =;>
E R+ 3k2 E Z+ \in
k2 I/(n)1 < fi2Ig(n)l. Let k = max{kl1 kz}. Then
for all n :2:: k, mllg(n)! S; I/(n)1 S; fi2Ig(n)l, so I E 0(g).
15.
Proof: f E 0(g) => f E O(g) and I E O(g) (from Exercise 14 of this section) => 9 E OU)
and 9 E n(f) (from Exercise 12 of this section) => 9 E 0U).
16.
Proof: Part (a) follows from Exercises 14 and 13(a) of this section and part (a) of Example
5.68.
The situation
similar for parts (b) and (c).
Section 5.8
1.
(a) f E O(nZ)
(b) f E O(nS)
(d) f E o(loga n)
(e) f E O(nlogz n)
(c) f E O(n:!)
2.
(a) f E O(n)
(b) f E O(n)
3.
(a) For the following program segment the value of
integer n, and the values of the
array entries All), A[2}, A[3}, . .. , A[n} are supplied beforehand.
the variables i, Max,
are integer variables.
Location that are used
If II, > 1 then
Begin
i := 2 to n
Max:= A[iJj
Location := i
End;
Writeln ('
Write C
first occurrence
the maximum ');
in the array is at position "
j. ' )
(b) If, as in Exercise 2, we define the worst-case complexity function f( n) as the number
of times the comparison Max < A[i]
executed,
f (n) = n - 1 for all n E Z+, and
f E 0(»).
4.
(a.) For the following program segment the value of the integer n, and the values of
the array entries A(11, A[2J, A[3], ... ,A[nJ are supplied earlier
the program. Also the
variables i, Max, and Min that are used here are integer variables.
Begin
Min:= A[l];
Max:= A{1];
For i := 2 to n do
Begin
If A[iJ < Min then
Min := A[i};
If A[i] > Max then
Max := Ali];
End;
'Writeln (' The minimum value in the array is ',Min :0);
Write C and
maximum value is " Max:O, '.')
End;
(b) Here ~ve
U'::;' .•u.'<J
parisOD.s that are
the worst-case time-complexity function
as the number of com~
the For loop.
fen) = 2(n - 1)
all n E Z+
andfE
For n = 1, we find that a1 = 0 = l0j =
case.
Now assume
n=k
2,
... ,
.
15
.
12
(i) n = k 1 = 2m , where m E Z+; Here an = 1 +
= 1 + a2".-1 = 1 Llog22m-1J =
1 (m - 1) = rrl, =
; and
(ii) n = k + 1 = 2m + r, where m E Z+ and 0 < r < 2m : Here 2m < n < 2m +\ so we
(1) 2m - 1 < (n/2) < 2m;
(2) 2m - 1 == l2m - 1 J S In/2J < l2m J =
1 and
m I
(3) m -1 =
2 - :::;; log2ln/2J < logz 2m = m.
Consequently~ l10gz
== m -1 and an = 1 + aln /2J = 1 +
In/2JJ = 1 + (m -1) =
m = llog2 nJ.
Therefore it follows from the Alternative
the Principle Mathematical Induction
thai an = lloga nJ for all n E Z+.
8.
We claim thai an =
nl all n E Z+.
Proof: When n = 1 we have at = 0 = rOl = pog:! ,and this establishes our basis step.
For the inductive step we assume the result true for all n = 1,2,3, ... , k (;;::: 1) and consider
what happens at n = k + 1.
(i) n = k + 1 = 2m , where m E Z+: Here an = 1 + arn/21 = 1 a2m--1 = 1 + flogz 2m - 11 =
1 + (m - 1) == m = nog2 2m 1 = flogz n 1·
(ii) n = k 1 = 2m + r, where m E Z+ and 0 < r < 2m : Here 2m < n < 2m +1 and we
find that
(1) 2m - 1 < n/2 < 2m;
(2) 2m - 1 = r2 m- 11 < n/21 S f2ml = 2mi and
(3) m - 1 lOg22m-1 < logz fn/21 S logz 2m = m.
Therefore, flog2fn/211 = mandan = 1 arnl21 = 1+flog2fn/211 = l+m = floglnl, since
2fT~ < n < 2m+1 :::} log22m = m < logz n < m + 1 = logll 2 m+1 ::::} m < pogz n1
m + 1.
Consequently, it foHows from the Alternative Form of the Principle of Mathematical Induction that an = flog:;! n1 for all n E Z+.
=
r
=
9.
10.
np = 3/4 and q = 1 - np = 1/4, so E(X) = np(n
(1/4)n = (3/S)n + (3/8)
(1/4)n = (5/8)n + (3/8).
1)/2 + nq = (3/4)[(n + 1)/2] +
+ 1)], so
Pr(X = i) =
i/[n(n 1)1 =
Pr(X = i) =
(l/fn(n + 1)))
i = (l/fn(n + 1)1)[n(n 1)/2) = 1/2 and q = 1 - (1/2) = 1/2.
(1/2)n =
1)J)
i2
-11 _ 2.!!: +'
\} 6
11.
begin
i:= 2
j:= 1
j<i
aj = aj then location := i
else j := j + 1
t := ~
1
end
end
is
subscript of
that
a previous
array entry; location is 0 if the array contains n distinct integers}
For n
let fen) count the maximum number of times
while loop is
for
value of i, where
executed.
second while loop is executed at most n - 1
2 5 i ::; n. Consequently, fen) = 1 + 2 + 3 +... (n - 1) = (n - 1)(n)/2, which occurs
is an - l and an.
when the array consists of n distinct integers or when the only
2
(n - 1)(n)/2 = (1/2)(n - n) we have f E O(nZ).
12.
a)
procedure Fi'f'stDecreasc (n: positive integer; all a2, a3, ... , an: integers)
begin
location := 0
i:= 2
while i ::; n and location = 0 do
if ai < ai-l then location := i
else i := i + 1
end {location is the subscript of the first array entry that is smaller than its
immediate predecessor; location is 0 if the n integers in the array
are in increasing order}
b) For n 2:: 2, let fen) count the maximum number of comparisons made
the while
loop. This is n - 1, which occurs if the integers in the array are in ascending order or if
a1 < a2 < a3 < ... < a -1 and an < an-I- Consequently, f E O(n) .
11,
.- 0 --
1.
riB and
and
or
x
E
x
xA
result
true.
( x, y) E
x
U
x
:=:} (x 1 Y) E A x B or (x,
E (B x A) :=:}
E AyE
E B and yEA) :=:} (x E A or x E B) and (y E A or y E B) :=:} x, yEA U B :=:}
(x, y) E (A U B) x (A U B).
or
(a) True
(e) False:
(e) False:
(b)
Let A =
Let f: Z -+ Z,f(x) = 2x.
Let A = {1,2}, = {1,2,
g= {(1,1),(2,2),(3,3)},h
(f)
False.
Let
= {(I, I),
= {1,2,
2},B = {x, ,I = {(1,x),(2,y)}.
(d) True.
,C = {1,2,3,4},f =
1),(2,2)},
2),(3,4)}.
4}, B = {5, 5}, Al = {I, 2}, A2 = {2, 3, 4},
f = {(I, 5), (2, 6), (3, 5), (4, 5)}. Then f(Al nA z) = f(2) = {5}, but f(A 1 )nf(A2 ) = {5,5}.
(g)
3.
(a) f(l) = J(1 ·1) = 1 . f(l) + 1· f(l), so f(l) = O.
(b) f(O) = 0
(c) Proof (by Mathematical Induction): When a = 0 the result is true, 80 consider a =I- O.
For n = 1, jean) = f(a) = 1 . aO • f(a) = nan - I f(a), so the result follows in this first
case, and this establishes our basis step. Assume the result true for n = k(2:: 1) - that
is, f(a k ) = ka k - 1 f(a). For n = k + 1 we have f(a H1 ) = f(a· ak ) = af(a lc ) + ak f(a) =
akak-1/(a) ak f( a) = kale I(a) + ak f(a) = (k + l)a k f(a). Consequently, the truth of the
result for n = k + 1 follows from the truth of the result for n = k. So by the Principle of
Mathematical Induction the result is true for aU n E Z+.
4.
21AxBj:= 262,144 ==? IA x BI = 18 =? IAI := 2, IBI := 9 or IAI = 3, IBI = 5.
5.
(x,y) E (A n B) x (C n D) ~ x E An B,y E enD ~ (x E A,y E C) and
(x E B, y ED) ==? (x, Y) E A x C and (x, y) E B x D ~ (x, y) E (A x C) n (B x D)
6.
(a)
7.
0 x < then
= 0
x 2 = 1/2. So x = 1/V2,
If 1 ~ x < 2, then lxJ = 1 and x 2 = 3/2.
x := )3/2.
k E Z+ and k 2:: if k S x < k 1, then lx J == k
if x satisfies the given equation
2
we
x =1:
k2::2we
k>
sok(k-l)2::1>1/2,
- k >
> k
:=:} k >
= x
we
(b)
51
1.
k E Z+
41
with ( a, b) E 'R:
we have
2:: b,
a 1 a it
2a b;
b+ 1) with (a, b) E 'R: Now we find that 2a 2:: b => 2a 2 ~ b + 1 =? 2( a + 1) 2::
(ii) (a
b+ Ii and
(iii) (a 1, b + 2) with (a~
b 2 =? 2( a + 1) ~ b + 2.
Consequently, for
( a, b) E
9.
E 'R.:
we
In this last case it follows
2a 2:: b.
(a) f2(x) = f(f(x» = a(f(x)
- b = a[(a(x + - b) + b] - b = a2 (x + - b
2
f3(x) = f(f'l{x)) = f(a (x b) - b) = a[(a 2 (x + b) - b) b} - b = a3 (x + 11) - b
(b) Conjecture: For n E Z+, rt(x) = aTt(x + b) - b. Proof (by Mathematical Induction):
n =1the
f( x ). Hence we have our basis step.
formula is
Assume the formula true for n = k(2:: 1) - that is, P'(x) = ak(x + b) - b. Now consider
n=k+ Wenndthatfk+l(x)=f(P'(x»=f(ak(x b)-b)=a[(ak(x+b)-b) b]-b=
ak+l(x + b) - h. Since the truth of the formula at n = k implies the truth of the formula
at n = k + 1, it foHows that the formula is valid for all n E Z+ - by the Principle of
Mathematical Induction.
10.
Let n = IAI-jAII. Since IBV~ is the number of ways to extend f to A and IBITt = 6f!. = 216,
then n = 3 and IAI = 8.
11.
(a) (7 x 6 x 5 x 4 X 3)/(75 ) .:... 0.15.
(b) For the computer program the elements of B are replaeed by {1,2,3,4,5,6,7}.
10
20
30
40
50
60
70
Random
Dim F(5)
For I = 1 To 5
F(I) = Int(Rnd*7 1)
Next I
J =2
5
For K = 1 To J - 1
F( J) = F(K)
150
170
5
190
. I'1 "".
F(I)',
"
,
GOrrO 120
200 Next I
210 End
12.
For each subset A of S, let SA denote the sum of the elements of A. Consider only those
A S S : ; 5. There are 27 - 1 - 1- 7 = 1 such subsets and
nonempty
here 1 SA 20 21 22 + 23 24 = 110. The result follows
Pigeonhole Principle
for there are 119 subsets (pigeons)
110 possible sums (pigeonholes).
13.
For 1 s:; i s l e t Xi
the number of letters typed on day i. Then Xl + X2 + Xli + ...
Xg
Xg + XiO = 84, or X3
••• + Xg = 54. Suppose that Xl + X2 + Xa < 25, Xa
X3
X4 <
25, ... , Xa + XIt + XIO < 25. Then Xl 2X2 + 3(X3
Xs)
2X9 Xl{) < 8(25) = 200, or
3(X3
••• + XS) < 160. Consequently, 54 = Xl
••• + Xs < (160)/3 = 53 1/3.
14.
If two elements in {Xl' X:h ..• ,xr} have the same units digit then their difference is divisible by 10. If this does not happen consider the ten possible units digits as follows:
{OJ, {I, 9}, {2, 8}, {3, 7}, {4, 6), {5} -- these are the pigeonholes for the problem. vVhen the
seven pigeons {Xl, X2, ••• ,X7} go to the pigeonholes where their units digits are located,
at least one two-element subset is fined and those two numbers (pigeons) will sum to a
multiple of 10.
15.
For nk:=l(k - ik) to be odd, (k - i k ) must be odd for all 1 S k n, i.e., one of k, i k must be
even and the other odd. Since n is odd, n = 2m + 1 and ill the list 1,2, ... , n, there are m
even integers and m + 1 odd integers. Let 1,3,5, ... , n be the pigeons and iI, is) is, ... ) in
the pigeonhole:s. At most m of the pigeonholes can be even integers, so (k - ik) must be
even for at least o:ne Ie = 1,3,5, ... , n. Co:nseque:ntly, llk::::l(k - i k ) is eve:n.
16.
(a.)
The answer is the number of onto functions f : A ---i' B where IAj = 10 (weekly
chores) and IBI = 3 (for the three young men). There are 3IS(10,3) such fuuctions.
2!8(9,2)
only mows the lawn) +3!S(9, 3) (Thomas does more than just mow
the lawn).
2".---1
'-'v•.a"..". ...."""
+ 5fS(7, 5)
m)
with
no container
and the
one. This can
containers
are two cases.
in (;) ways. The other possibility is
contain two objects and the others one. This happens in
objects
two
(1/2)(;) (n;2) =
(n!)/[2!2!2!(n - 4)1] = 3(~) ways.
21.
Fix m = 1.
n= 1
is
Assume foP: = fk 0 f and consider f 0 plH.
I 0 fHl = I 0 (f 0 fk) = f 0 (fk 0 f) = (f 0 fk) 0 f = J't H 0 I. Hence f 0 fn = p1- 0 I for
Now assume that for t :2: 1, 0 til. =
0
•
0 P" =
0 P) 0 fn =
n E
f 0 (P 0 fn) = f 0 (P" 0 P) = (f 0 p") 0 P =
0 f) 0 P = til. 0 (f 0 P) = In 0 It+l, so
fm 0 j'" = In 0
for
m, n E
r
22.
un
(b) Y E f(rlieIAi) ¢=} Y = I(x), for some x E
Y E f(Ai), for all i E I{::::::=:} Y E niEl f(Ai).
=:}
(c) From part (b), f(rliEI Ai) niEI f(Ai). For the oppositei:ndusion let y E niEI f(Ad.
Then'll E I(Ai) for all i E
80 Y = f(Xi),Xi E Ai, for each i E
Since f is one-to-one,
all of these Xi'S, i E I, yield only one element x E niE1Aj. Hence y = I(x) E l(rlieIAi),
so ~EI I(Ai) /(~El Ai) and the equality follows.
The proof for part (a) is done in a similar way.
23.
Proof: Let a E A. Then
l(a) = g(/(f(a») = f(g(l(/(f(o,»») = I(g 0 /3(0,».
From 1(0,) = g(f(/(o,») we have pea) = (/ 0 /)(0,) = f([I(f(/(a»). So I(a) =
I(g 0 f3(a» = l(g(/(f(f(a»)))) = j2(f(a» = j2(g(f2(a»)) = I(f(g(f(.f(o,»») =
l(g(/(a») = g(a).
Consequently, 1= g.
(a) n(nxn) = n{n2 )
(d) Since IAI = '11, there are n choices for each selection of size k, with repetitions allowed,
of
n.
are r =
25.
o iff x ¢ n
iff x ¢ A or x ¢ B
XA (x) = 0 or
=0
XA • XBeX) = O. Hence
XAnlil = XA . 'Xlii·
(b)
The proof here is similar to that of part (a).
(c) XA:(x) = 1 iff x E A x ¢ A iff XA(X) = o iff - XA)(X) = XA:(x) = 0
x ¢ A
x E A
XA(X) = 1 iff (1 - XA)(X) = O. Hence XA: = 1- XA'
21.
28.
fog = {(x, Z),(YdJ),(z,x)};goj = {(x,x),(y,z),(z,y)};
f- 1 = {(x,z),(y,x),(Z,Y)}i9,-1 = {(x,y),(y,x),(z,z)}i
(90f)-1 = {(x, x), (y,z),(z,y)} = 1-1 09- 1 ;9- 1 oj-1 = ((x,z),(y,y),
x)}.
(a) f-l(8) = {xl5x 3 = 8} = {I}.
(b) Ix 2 3x + 11 = 1 ==> x 2 + 3x + 1 = 1 or x 2
3x + 1 = -1 ==> x 2
3x = 0 or
x 2 + 3x 2 = 0 ==> (x)(x 3) = 0 or (x + 2)(x + 1) = 0 = } x = 0, -3 or x =
,-2.
1
Hence 9- (1) :.: ; : {-3, -2, -1, OJ.
(c) {-8/5,-8/3}
29.
Under these conditions we know that 1- 1 ( {B, 7, 9}) = {2, 4, 5, 6, 9}. Consequently we have
(i) two choices for each of f(l), f(3), and f(7) - namely, 4 or 5;
(ii) two choices for each of f(8) and 1(10) - namely, 8 or 10; aud
(iii) three choices for each of j(2), f(4), 1(5), feB), and 1(9) - namely, 6, 7, or 9.
Therefore, by the rule of product, it follows that the number of functions satisfying these
conditions is 23 .22 .35 = 7776.
30.
Since P = j and (J-l)! = 1-1, the result is true for n = 1. Assume the result for
n = k : (f"')-1 = (I-I )k. For n = k + 1, (IHI t1 = (J 0 f")-l = (1"')-1 0 (I-I) =
(I-l)k 0 (f-1)1 = (/-1)1 0 (I-l)k (by Exerc.ise 21) = (1-1 )k+1. Therefore, by the Principle
of Mathematical Induction, the result is true for all n E Z+.
31.
(a) (11' 0 n)(z) = (0' 0 1I")(x) = x
(b)
= x - nj O'n( x) = x n( n 2::
(c) 1r-n(x) = X n; O'-n(x) = X - n(n 2:: 2).
--
.··(ek
k = 2 : r(2) = 1'(3) :=
=2
k 3 : 1'(22)
1'(52 ) = 3
k=4;
=4
k=5:
=5
k 6 : 1'(12) =
=6
=
=
::::
=
(c)
Cd)
... ,
1)
... ,
are k
t
1) ... (ek
1)][(11 +
+ ... (It
1)] = r(a)r(b).
(a)
are
distinct primes and
subset A "",1""",h,.,,,..
aeter;Olln,es a distribution of
eight distinct objects
3,5,7,1 13,
19} into
four identical containers with no container left empty. There are S(8, 4) such distributions.
(b) S(n,m)
34.
Define f : Z+ -;. R by fen) = lin.
35.
(a)
Let m = 1
k=1.
for all n ?: k,lf(n)!
2 < 3
so
for all n ?: k, i9(n)1 :::; 4 = 4·1
4!f(n)1 = mlfCn)l, so
f E O(g).
(b)
Let m = 4 and k =
9 E OCf).
36.
(a)1 E 0Ud ::=:;. 3m! E R+ 3k1 E Z+ such that If(n)1 S mll/l(n)1 Vn ?: k 1 •
9 E 0(91) ::=:;. 3m2 E R+ 3k 2 E Z+ such that Ig(n)1 S m2Ig1(n)! Vn ?: k 2 •
Let m = max{ml,mZ}' Then for aU n ?: max{kl,kz},IU g)(n)1 = If(n) g(n)1 =
I/(n)1 + Ig(n)1 s mll!t(n)1 + m2Ig1(n)1 S m(lh(n)1
Igl(n)1) = mlh(n) + gl(n)1 =
mlUl gl)(n)l, so (I + 9) E OUI + 91)'
(b)
37.
Let I, /l,g,gl : Z+ -+ R be defined by fen) = n, /1(n) = 1 - n, g(n) = 1,gl(n) = n.
First note that if logo. n = 'i', then n = ar and log" n = 10&(a1') = r lo~ a = (logb a)(loga n).
Now
m = (lo~a) and k = 1. Then for
n?: k,lg(n)1 = logbn = (logba)(lo~n) =
mlf(n)l, so 9 E O(f)·
Finally, with m = (log" a)-1 = lo~ b and k = I, we
that for all n > k,lf(n)1 =
loga n = (lo~ b)(lo~ n) = mlg(n)l· Hence f E O(g).
166
CHAPTER 6
LANGU AG ES: FINITE
Section 6.1
1.
(a) 25;
2.
(a) 44
3.
12
4.
125
(a) 0
(e) 3
4.
(b)
0
(f)
4
(c)
(g)
1
(d) 2
1
(h)
0
i
5.
Ei=15
6.
There are 100 substrings of length 1: Xl, X2," • XlOO; 99 substrings of length
XIX2, X2X3,'" ,XOO X 100j ••• ; 1 substring of length 100:
XIX2'" XlOO' So there are 100
99 + ... + 1 = E:~ i = (100)(101)/2 = 5050 nonempty substrings
total.
1.
(a) {OO,l1, 000,111, OOOO,l111}
(b) {O,l}
(c) E* -{A, 00, 11,000,111,0000,11
j
}
(d) {O,l,OO,lI}
(e) E*
(f) 2:* - {O, 1,00, Il} = {A,OI, 1O} U {will w 112 3}
8.
(a) AB = {WOO,
Ill}
(b) BA == {0010,00l1,
I}
= {101010,101011,
110,1
B2 == {OOOO,OOl,
11}.
xE
=}
x = ae,
111,11101
some a E
cE
=:} x
11
11
11}
c
E
x E A0 = } x = yz, for some yEA, z E 0. But z E 0
Inanner
10.
(b)
= {A,
if
,it
. .. } ==
(a)
(b)
(e)
Yes
Yes
No
(c)
(f)
(a) Here A'"
all
x of even
where if x
A, then x starts
0
and ends with 1, and the symbols (0 and 1) alternate.
(b) In this case
contains precisely those strings
of 3n D's, for n E N.
( c ) Here a string x E A" if (and only if)
(i) x is a
of n
for n E Nj or
(ii) x is a string that starts and ends with 0, and has
least one 1 - but no consecutive
(d) For this last ease A* consists of the following:
(i) Any string of n 1's~ for n E
(ii) Any string that starts with 1 and contains at least one 0, but no consecutive O's.
14.
There are five possible choices:
(1) A = {A}, B = {Ol, 000,0101, 0111, 01000, 010111};
(2) A = {OI,OOO, 0101, 0111, 0l000,OlO111}, B = {A};
(3) A = {O}, B = {I, 00, 101, 111, 1000, 10111};
(4) A = {O,OlO}, B = {l,OO,ll1}; and
(5) A = {A,OI}, = {01,OOO,0111}.
15.
Let E be an alphabet with 0 i A S; E*. If IAI = 1 and x E A, then xx = x since A2 = A.
But II xx 11= 211 x 11=11 x II==:}II x 11= 0 ==:} x = A. IAI > 1, let x E A where II x 1/> 0 but
II x II is minimaL Then x E A2 ==:} X = yz, for some y, z E A. Since II x iI=1I 11 II + Ii z II,
if II y 1/, II z II> 0, then one of y, z is in A with length smaller than II x /I. Consequently,
one of II y II or II z 1\ is 0, so A E A.
16.
(a) Pfw5a, r
(b) r,d are
Pa(E*) = {a}E*;sc;(E*) = E*{a}
(c) r is invertible and 1'-1 = 1'.
(d) 25; 125i 51'/,/2 for n even, 5(n.t1)/2 for n
(e) (d () PI\!)(:C) = x =
0 d (') 7' (')
1
1'- (B) =
5=
nE
17.
some a E A, b E B ,eE ~
... bm)(C1Cl '" en), where (l.i E 1
lj
E
1 S j S m,
n '¢=} x
ai E A, 1
i S i,
j S
Cit E l k
=
a. E
1 :::; i I, hi E
(AB)C = A(Be).
1 S j :::; m, Ck E C,l S k S n ~ x E A(BC). Hence
Theorem 6.2(b): For a E A, a = Aa
A E B*. Hence A ~
Theorem
6.2(a) A* S; A*A*. Conversely, x E A*A* =* x =
where y = a}(1,2'" am, Z = a~a~ ••• a~, with ai, aj E A, for 1 S i :::; m,l S j S n. Hence
x E A*, so
~ A*
the equality
Since (A*)* = U~=o(A*Y" it
that A"' ~ (A"')"'. Conversely, if x E (A"')*, then
x = X1X2." X n , where Xi E 44'*, for 1 Sis n. Each Xi = ail C4i2 ••• aik;, where
E A, 1
j < ki . Hence X E
so (A*)'" ~
and (A"')'" = A*.
j
-
yt ~ U~=o(A*Y' = (A"')'". If x E (A*)*,
= XIXZ ... X fH where
Xi E A*, for
1 SiS n. Then X = al1a12.·.alkla:;naZZ .•. aZk2, .• anl ••• ankn E A* ~
~=l(A*)n = (A*)+, so (A*)* = (A*)+.
Since
A*, (A+)* ~ (A"')"' by part (d) of this theorem. For x E (A*)*, if x = A, then
x E (A+)*, If x
A, then as above x = an au . •. a1k 1 a:n ••. a2k2 • , . anI •• . ank.. E A+ ~
( A +)'" and the result follows.
x
19.
By Definition 6.11 AB = {abja E A, b E B}, and since it is possible to have alb! = a2~J
with al, a2 E A, a1 I- a'll and hI, b2 E B, b1 I- hz, it follows that IABI S IA x BI = IAilBI·
20.
(a) {y }*x{y}*
(b) {y }*x{y }*x{y}*
(d) {x, y}*yxy
(e) (x{x,y}*) U ({x,y}*yxy)
(c) x{x,y}*
(f) [(x{ x, y}*) U ({ x, y}*yxy)] - [xi x, y}*yxy]
21.
(a) The words 001 and 011 have length 3 and are in A. The words 00011 and 00111 have
length 5 and they are also in A.
(b) From step (1) we know that 1 E A. Then by applying step (2) three times we get
(i) 1 E A =* 011 E A;
(ii) 011 E A=?
11 E A;
(iii) 00111 E A =* 0001111 E A.
(c) If
we aee
word
Likewise, 000111
A =? 0011 is
in A.
are no words
2fad,
are no 't1I?n'1!",'1q
are
23.
(a)
1.
2.
3.
Stepii
( ) is in A.
» is in A.
in A.
(C
Part (1) of the recursive definition
Step 1 and
(2(ii)) of
definition
1, 2, and part (2(i)) of the definition
»( )
« » is A.
« »( ) in A.
« »( )( ) is in A.
Reasons
recursive definition
Part (1) of
Step 1 and part (2(ii)) of
definition
Steps 1, 2, and part (2(i) of the definition
Steps 1, 3, and part (2(i» of the definition
Steps
( ) is in A.
( )( ) is in A.
« )( is in
( )( ( )( » is in A.
Reasons
Part (1) of the recursive definition
Step 1 and part (2{i» of the definition
Step 2 and part (2(ii» of the definition
Steps 1, 3, and part (2(i» of the definition
Steps
1. ( ) is in A.
2.
3.
4.
(c)
1.
2.
3.
4.
»
24.
(1)'\ E A and,8 E A for all sEE; and
(2) For each x E A and sEE, the st:ring sxs is also in A.
[No other string from }J* is in A.]
25.
Length 3:
(;) +
G) = 3
Length 5:
(~) + (:)
(~) +
Length 6:
(:) + (!) + (;) + (:) = 13 [Here the summand (:) counts the strings where there
Length 4:
(;) = 5
(D (;) = 8
are no OSj the summand
(D counts the strings where we arrange the symbols 1,1,1,1,00;
the summand (~) is for the arrangements of 1,1,00,00; and the summand (;) counts the
arrangements of 00, 00,
26.
[Here (~) counts the ar:rangements for oue 111
111'8 and
A:
B:
) >. E A.
(2) IfaE
(1)
111 's
15
two
Oal, laO, lal E A.
IE
HaEA,
2S.
eight OO's; (:) counts the arrangements
E
Of
one
6.2
1.
(a) 0010101; 81
2.
first output 1 we must be at
S2 when
is
then forces
first three inputs to be 1,0,1. To get the second output of 1 we must
at state 82
input
the
two inputs to be O,l.
Hence x = 10101.
3.
(a) 010110
(b) 0000000; B1
(b)
001000000j So
5}, where at state Si, the machine remembers the insertion of a total of
z
1= {5f, 10¢ , 25¢, B,W}
0=
(nothing), P(peppermint), S(spearmint), 5¢, 10¢, 15¢, 20¢, 25¢}
80
i
81
82
8a
85
81
81
82
83
84
85
82
82
83
84
85
85
83
Sa
84
. 85
5.
n
n
n
n
n
n
I n 5;
!
85
8S
85
85
85
85
(a) 010000;82
5¢ n
1O¢ n
15r n
20¢
n
84
84
So
1O¢ 25; S
So ! 5¢
(b) (81)100000; 82
n
n
n
n
P
(82 )000000; 82
(83)110010; 82
(c)
v
w
0 1
0 0
0
1
So
80
03 1
81
1$1
82
1
1
0
0
0
0
1
82
Sa
82
8a
So
Sa
84
82
Sa
1
(e) x = 101 (unique)
6.
(a) The machine recognizes (with an output of 1) every 0 (in an input string x) that is
preceded by another o.
(b)
81 relnembers that at
one 0 has been supplied
all input
x.
(c)
= {l}"',B = {OO}
(i)
(8)
0
1 1 0
0
1
0 1
i 80
80
81
81
81
82
82
82
83
Input
1 0
1 1
o 000 0 0 0 0 1 0
W
1.1
i
1 1
83
83
84
84
84
85
85
85
86
0
0
0
0
0
0
0
O·
0
0
0
1
(c) W(X, 80) = 0000001 for x = (1)1111101; (2)1111011; (3)11101 i (4)1101111;
(5)1011111; and (6)0111111
(d) The machine recognizes the occurrence of a. sixth 1, a 12th 1, ... in an input x.
9.
(a.)
(b)
1.1
W
0
1
So
34
81
81
83
82
82
83
82
8a,
83
83
84
S5
83
85
85
83
0 1
0 0
0 0
0 1
0 0
0 0
1 0
are only two possibilities: x = 1111 or x = 0000.
(c) A =
HI
(d) Here A:;
U {OOOHO}*
HI}'" U {OOOOOHO}*.
6.3
2.
(0110)
(1010)
0,0
3.
Start
4.
0110
0,,0
5.
(i)
1
011
Input 1010
same tasks as the one
Fig.
has
6.
Suppose the contrary and let the machine have n states, for some n E Z+. Consider the
input string
expect
here be
. As
O's
are processed we obtain n 1 states 811 82,'" ,Sn,Sn+1 from the function v. Consequently,
the Pigeonhole
are two
Si! Sj where i < j
Si = Sj.
if
the states 8 m ) for i
1:S; m ::; j I are
along with their inputs of 0, then
machine win recognize
sequence on+1-(J-i)l'\
n 1- (j - i) ::; n. But the
on+l-(j-i)l n ¢ A.
1.
transient
are So, 81.
84 is a sink state.
,8:;" 83, S'b
{ 84},
{SZ,S3,SS} (with
corresponding restrictions on the given function v) constitute submachines. The strongly connected submachines are {84} and {82' 83, ss}.
(b) States 32,83 are transient. The only sink state is 84. The set {SO,Sl,SS,S4} provides
the states for a submachine; {80, 81}, {84} provide strongly connected submachines.
(c) Here there are no transient states. State 86 is a sink state. There are three submachines: {S2' 83, 84, 85, ss}, {S31 84, 85, 56}' and {ss}. The only strongly connected sub machine
is {5s}.
8.
Either 110 or 111 provides a transfer sequence from 52 to 85'
Supplementary Exercises
1.
(a) True
(d) True
(b)
(e)
False
True
(c)
(f)
True
True
2.
No. Let x E L with A = {x, xx}, B :::: {x}. Then A* = B* = {xnln 2 a}, but .4 CJ;. B.
3.
Let x E E and A = {x}. Then A2 = {x 2 } and (A2)* = {.\,x 2 ,x\ ... }. However A* =
().,x,x 2 , ... } and (A"')2 = A"', so (A*)2 (A2)*.
4.
(a) A'"
5.
So we can
requires an input of
Hence 0 02 :;;:;
: to} OO}*{O}
B*. [For example, 111 E B* hut 111 ¢ A"'.}
:e
: { OO}"-
Hl,
6.
(a)
v
0
1
So
So
81
81
81
S2
'~2
82
83
Sa
S3
So
~
0 1
0 0
0 0
0 0
0 1
1)
(h) For any input string X, this IDI3t.c,n:me re.c:oglrll
every fourth 1 in x.
occurrence of
(!)
(c) (:)
+ (~) = 72. (The first summand is for the sequence of eight l's, the second
summand for the sequences of four 1'8 and four
and
last summand
the sequence
of eight O's.)
For II X
12, there are G~)
(~2) +
e:) + e:) =
992 such sequences.
1.
( a) By the Pigeonhole Principle there is a first state s that is encountered twice. Let y
be the output string that resulted since s was first encountered until we reach this state a
second time. Then from that point on the output is yyy ....
(b) n
(c) n
8.
x
= 110
9.
1,0
o
1 (and, 1
v
w
0 1
(811
(Sh 83) 1 1
(Sh S4) 0 1
(82,84) 1 1
(S2,83) 1 1
(8,1) 83)
(SOl84)
(SOl
(SOl
(a)
(Bb
83)
11 1
1
(b) W«SOI 83),1101) = 1111; 11.11
12.
0
in state 80 and ~~2
state 84-
The following program determines the output for the input string 1000011000.
10
20
30
40
150
160
170
Dim A(3,2), B(3,2)
Mat Read A,B
Data 2,1,3,1,3,1,0,0,0,0,1,1
Dim P(IOO), S(100)
Read N
For I = 1 to N
Read X
If I <> 1 Then 120
If X = Then pel) = B(I,I) Else pel) = B(1,2)
If X = 0 Then S(l) = A(l,1) Else S(l) = A(1,2)
Go
140
Y=X 1
pel) = B(S(I-l)Y) : SCI) = A(S(l-l),Y)
Next I
Data 10,1,0,0,0,0,1,1,0,0,0
Print
For I = 1 To N-l
190
Next I
50
60
70
80
90
100
110
120
130
140
°
nmv
CHAPTER 7
RELATIONS: THE SECOND TIME AROUND
Section.
1.
(a) {(I,l )j(2,2)~(3,3),( 4,4),( 1,2),(2,1 ),(2,3),(3,2)}
(b) ((1,1),(2,2),(3,3),( 4,4),(1,2)}
(c) {(I,l ),(2,2),(1,2),(2,1))
2.
-9, -2, 5, 12, 19
3.
(a) Let it,/2,/s E with II{n) = n 1, 12(n) = 5n, and 13(n) = 4n
(b) Let 9}' 92, 93 E F with 9l(n) = 3, 92(n) = lIn, and 13(n) = sin n.
4.
(a) The relation R on the set A is
(i) reflexive ifVx E A (x,x) E 1(.
(ii) symmetric ifVx,y E A [(x,y) E 'R = } (y,x) E 'R]
(iii) transitive ifVx,y,z E A [(x,y),(y,z) E R = } (x,z) E 'R]
(iv) antisymmetric if Vx, yEA [(x, y), (y, x) En=} x = y],
(b) The relation 'R on the set A is
(i) not reflexive if 3x E A (x, x) ¢ 'R
(ii) not symmetric if 3x, yEA [(x, y) E n A (y, x) ¢ R]
(iii) not transitive if 3x,y,z E A [(x,y),(y,z) E R/\ (x,z) ¢'R]
(iv) not antisymmetric if
yEA {(x, y),(y,x) E R /\ x "I
5.
(a) reflexive, antisymmetric:, b:ansitive
(b) transitive
(d)
(f)
transitive
a
x E A,
so (x, x) E
n
lin.
ofthese
are
if ,R2 are
(x, y), ('II, E
Rl n'R2 then (x, y), ('II, z) E 'R b 'R.'},1 so (x, z) E 'R(1 ) 'R2 (transitive property) and (x, z) E
'R1 n
for
symmetric and
are oUU.u.c:.w.
8.
(a)
For aU x E A, (x, x) E 'R1 , 'R'}, ~ 'R1 U 'R. 2 , so if either 'Rl or 'R.2 is reflexive, then
U 'R'}, is reflexive.
(b)
(i) H x, 'II E A and (x, 'II) E R.l U'R'},1 assume without loss of generality, that
(x, y) E 'R.t • (x, y) E 'Rl and
symmetric:::::::} (y, x) E 'R1 ::=:} ('II,
E 'Rl U'R 21 80
'R.1 U 'R.'J, is symmetric.
False:
A = {1,2}, R.l = {(I, 1),(1,2)}, R'J = {(2,
(1,
(2,1) E
,
'R.1 U 'R2 and 1 f:. 2, 80 'R.1 U 'R'}, is not antisymmetric.
(iii) False: Let A = {1,2,3},'R1 = {(I, 1), (1,2)},'R'}, = {(2,3)}. Then ,2),(2,3) E
In.
R.l U 'R. Z1 but (1,3) ¢ n 1 U'R.21 so n1 U R'}, is not transitive.
9.
(a)
(b)
False:
(i)
(ii)
(iii)
(c)
(i)
(ii)
(iii)
(iv)
(d)
True
Let A = {1,2} and n = {(1,2),(2, I)}.
Reflexive: True
Symmetric: False. Let A = {I,2}, Rl = {(I, I)}, Rz = {(I, 1), (1, 2)}.
Alltisymmetric & Transitive: False. Let A = {I, 2}, Rl = {(I, 2)},
R2 = {(I, 2), (2, I)}.
Reflexive: False. Let A= {1,2},R1 = {(1,1)},n2 = {(1,1),(2,2)}.
Symmetric: False. Let A = {1,2}, 'Rl = {(I, 2)}, 1(.z = {(I, 2), (2,
Antisymmetric: True
Transitive: False. Let A = {1,2}, 'Rl = {(l, 2), (2, In, 'R2 = {(I, 1),(1,2),
(2,1), (2, 2)}.
In.
10.
(a)
(d) 211
(g)
(b)
(24 )(26 ) = 210
(2 4 )(2 5 ) = 29
(c)
(f)
(i)
26
24 .36
1
2
12.
Since 5880
-
2)(k+l),
we
For n = p~pipg there are (5
so IAI = 168.
13.
1)(3 + 1)(6
There may exist an element a E
such
1) = (6)(4)(7) = 168 positive integer divisors,
b) nor (b, a) E'R·o
for
There are n ordered
of
form (x, x), x E
each of
(n:! - n)/2 sets
{(x, y), (y,x)} of ordered pairs where x, yEA, x =1= y, one element is chosen. This results
in a maximum value of n + (n 2 - n)/2 = (n 2 + n)/2.
2
The number of antisymmetric relations that can have this size is 2{n -n)/2.
15.
r - n counts the elements in 'R of the form ( a, b), a =1= b. Since 'R is symmetric, r - n is
even.
16.
(a) xRy if x < y.
(b) For example, suppose that R satisfies conditions (ii) and (iii). Since R =1= 0, let
(x,y) E R, for X,Y E A. Since R is symmetric, it follows that (y,x) E R. Then by the
transitive property we have (x, x) E R (and (y, y) E R). But if (x, x) E R the relation R
is not i:rreflexive.
(e) 2(n2 -n); 2",2 _ 2(2(n2 -n»
en en
G) (~l) + G) + (~)
G) (~) + G) (;1) + G) (;1)
17.
(a)
(b)
18.
(a) Let Al = I-l(x), A2 = j-l(y), and As = j-l(z). Then R = (AI X AI) U (Az X A 2) u
(As x As), so 11(,1 = 102 102
= 225.
(b) ni + ni + n~ n~
1.
xE
(a, d) E ('R,1 0
,(b,
(a,
E
E
E 'R,3
some c E
E
E R 2 ,(c, d) E
E 'R1 (;I
(;I
for some b E B, c E C ==>
(;I
0
0
(Ka 0
b) E 'R,1I
0
4.
(8.) R 1 o('R2 U
::::::R1 o{(w,
(W,
(x,6),(y,
= {(1,4),(1,5),(3,4),(3, (2,6),(1,6))
(R.! 0 R. 2 ) U CR.! 0 R.a )
= {{I,
(3,
(2,6), (1, 4), (1,
U ,4), (1,5),
= ((1,4), (1,5), (1, 6), (2,6), (3, 4), (3, 5)}
(y,
5)}
(b) R1 0 ('Rz n 1'i3) = R1 ° {( w, 5)} = {(I,
(3,5))
(R1 0 R 2 ) n (R! 0 'R3 ) :::::: {(I, 5), (3, 5), (2, 6), (1, 4), (1, 6)} n {(I, 4), (1, 5), (3,4), (3, 5)} =
{(1,4), 5),(3,5)}.
°
'Rl 0 ('R. a n 'R3 ) = 'Rz {(m,3), (m,
= {(1,3), (1,4)}
CRI ° 'R 2 ) n CR1 o'R.a ) = {(I, 3), (1, 4)} n {(I, 3), (1, 4)} = {(I, 3), (1, 4)}.
6.
(a) (x,z) E 1'il 0 (n 2 U 1'il ) ¢=.} for some y E B,(x,y) E 1'il,(y,z) E n'}, U n3 ¢=.} for
some 11 E B, ({ x, 11) E 1'i1, (y, z) E 'R'},) or «x, y) E 'Rh (y, z) E 'R3) ==> (x, z) E 'Rl ° 'R'}, or
(x, z) E 'R1o'R3 -¢=::} (x, z) E ('R! o'R2 )U('R1 o'Ra), so 'Rl O('R2 0'R3 ) ('R} o'R2 )U('R1 o'Ra).
For the opposite inclusion, (x,z) E CR.l o'R2 )U('R1 oRa):=::} (x,z) E 'R1 oR'). or (x,z) E
'R.l o'R3 • Assume without loss
generality that. (x, z) E RIO 'R 2 • Then there exists an
element y E B 80 that (x, y) E'R1 and (y, z) E R 2 • But (y, z) E Ra :=::} (y, z) E 'R2 U'R31
so (x, z) E Rl ° (Ra U R 3 ), and the result follows.
(b)
proof here is similar to that in part (a). To show that the inclusion can be
proper, let
= B = C = {I, 2, 3} with Rl = {(I, 2), (1, In, R2 = {(2, 3)}, R3 = {(I, a)}.
Then 'R1 ° ('R2 01(.3) = R1 00 = 0, but (Rl ° 1(.a) 0 (n l 0 R 3 ) = {(1,3)}.
1.
This follows by the Pigeonhole Principle. Here the pigeons are the 2,,2 + 1 integers between
o and 2n2 , inclusive, and the pigeonholes are the 2",2 relations on A.
8.
Let S = {(1,1),(1,2),(1,4)} and T == {(2,1),(2,2),(1,4)}.
9.
Here there are two choict'.s for each "ii,l ::; i < 6. For each pair "ij, "ji, 1 ::; i < j
are two choices,
there are (36 = 15 such pairs. Consequently
(26 )(215 ) = 221 such matrices.
10.
For
6,
are
o in E the
G.
)
1
'-'V.llU,l.u,uof
row
is O.
12.
yEA (x,
¢
yEA where xRy. Hence (x, y) E
and R
Men) = 0,
then
n
R = 0.
if
f-
0.
m = I, we have M(nl) = }d(R) = [1\d('R)]1, so the
is true in this
case. Assuming the truth
the statement for m = k we have M(Rk) = {1\d(1<..)]k. Now
consider m = k + 1. M(1<..k+l) = ]1,.f(R 0 Rk) = M('R) . M(1<..It:) (from Exercise 11)
= 1\;[(1<..) . [M(R)]k = [M(R)Jk+l, Consequently this result is true for all m ;;::: 1 by the
Principle Mathematical
13.
(a) n
{:::::;:::} (x, x) E R, for all x E A ~ mxx = 1 in M = (mij)nxro for all
x E A {=} In :5 M.
(b) 'R synunetric '¢::=} [\lx,y E A (x,V) E
M ~ mya: = 1 in M] ~ M= Mt'l".
~ (y,x) E R) -¢=>
[Vx,y E A mx'll = 1 in
14.
10!
20!
40!
50!
60
PROGRAM MAY BE USED
DETERMINE IF' A RELATION
ON A SET OF SIZE N, WHERE N <= 20,
WE ASSUME
OF
EQUIVALENCE
GENERALITY THAT
ELEMENTS
1,2,3, ...
80
90
100
110
120
130
140
150
160
170
180
190
200
210
220
230
240
250
260
270
280
=0
320
330
360
370
NEX.T 1
NOT TRANSITIVE"
+ Z = 3 THEN
184
d
f
c
(b)
16.
(a)
b),
(a,
1vl(R) =:
(c)
.
(e)
(/)
rows
0 1
1 0
0 1
0 1
1 1
0 0
0 0
1 1
0 0
0 0
0 1
0 1.
1 0
1 0
0 0
1 1
0 0
0 0
are
e),
(e,
In
o 1 0 0 o0
0 0 0 0 1 0
0 0 0 0 0 0
o 1 1 0 0 0
0 0 0 o 0 0
0 o 0 o 0 0
M(n) =
n=
a),(a,b),(b,a),(c,
c), (d, e), (e, d),
I), (I, d), (e, 1), (I, e)}
1 1 0 0 0 0
Men) =
(iv) n = ((b,a),(b,c),
1 0 0 0 0 0
0 0 0 1 0 0
0 0 1 0 1 1
0 0 o 1 0 1
000 1 1 0
b), (b, e), (c, d), (c, d)}
o 0 0 0 o0
1 0 1 0 1 0
M(n) =
18.
x),(w,v),
x),
0 1 0 1 0 0
0 0 o 0 0 0
0 0 0 1 0 0
o 0 o 0 o 0
y),(w,z),(x,z),
(b)
x),
v),
7.
b.
2t.
225 ; (25 )(2 10 ) = 21 .5
22.
2211; (25 )(2 10 ) = 215
'R2 :
'Rl :
1 1 0
1 1 0
0 0 1
0 0 1
0 0 0
0 0
0 0
1·0
1 0
0 1
1 1 1
1 1 1
1 1 1
0 0 0
0 0 0
0 0
0 0
0 0
1 1
1 1
(b) Given an equivalence relation n.. on a finite set A, list the elements of A so that
elements in the same cell of the partition (See Section 7.4.) are adjacent.
resulting
relation matrix will then have square blocks of 1's along the diagonal (from upper left to
lower right).
24.
(~); G);
(;)
25.
r
When n = 2, we
x_
<
a
on
where
= m. Let he the directed graph Q"'l!>VIv~a.I11C:U
'R,. .- each component of G is a directed cycle Ci on mi vertices, with 1
z ~ k.
ml m2 + ... + mAl = m.. ) The smallest power
n
loops appea.:r is
, for
t = min{mil1
i ~
( c)
Let
Let s = lcm(ml' m2,'"
= 'R,
all r E
27.
(~) = 703 => n = 38
7.3
j
mk). Then 'R,.'f'S = the identity (equality) relation on A and
smallest power of n
n is 8 1.
/,
6" /9
18
2
3
1 .
3.
For all a E A,bE B,an1a and llR',lb so (a,b)'R(a, b), and'R is reflexive. Next
{a,b)'R.(c,d),(c,d)'R(a,b)::=:;:} a'R1c,c'Rla and b'R2 d,d'R2 b:::=:} a = c,b = d =;. (a,b) =
(c, d), so 'R is antisymmetric. Finally, (a, b)'R,( c, d), (c, d)'R( e, f) ~ a'Rt c, e'R,l e and
b'R2 d, d'R 2 f ::::::::} a'Rl e, b'R'2f :=:} (a, II )n( e, f), and n is transitive. Consequently, n is
a partial order.
4.
No. Let A = B = {1,2} with each of n I l n 2 the usual "is less than or equal to" relation.
Then n is a partial order but it is not a total order for we cannot compare (1,2) and (2,1).
5.
0 < {I} < {2} < {3} < {1,2} < {1,3} < {2,3} < {1,2,3}. (There are other possibilities.)
6.
(a)
(b)
(a)
(b)
(e)
(d)
(e)
(a)
(b)
1
0
1
1
1
0
1
1
M(n):= (e)
(d)
0
0
0
0
0
0
1
0
0
1
1
0
1
1
1
1
1
(b) 3 < 2 < 1 < 4 or 3 < 1 < 2 < 4.
7.
(c) 2
8.
Suppose
X, yEA and
both are least elements.
xRy since X is a least
element, and y'Rx since y is a least element. \Vith'R antisymmetric we have x = y.
9.
Let x, y
be greatest
bounds. Then x'Ry since x
a lower bound
Y IS a
greatest lower bound. By similar reasoning yR.x. Since R. is antisymmetric, x = y. [The
proof
lub is similar.}
10.
U = {1,2,3,4}.
A be the collection of all proper subsets of U, partially ordered
under set inclusion. Then {1,2,3}, {l,2,4}, {1,3,4}, and {2,3,4} are aU maximal elements.
11.
Let U = {l,2}, A = P(U), and 'R the inclusion relation. Then (A, R.)
total order. Let B = {0, {I}}. Then (B x B) n R. is a total order.
12.
For all vertices x,y E A,x =I- y, there is either an edge (x,y) or an edge (y,x), but not
both. In addition, if (x,y),(y,z) are edges in G then (x,z) is an edge in G. Finally, at
every vertex of the graph there is a loop.
13.
n + (;)
14.
15.
a paset but not a
n + (~)
(a) The n elements of A are arranged along a vertical line. For if A = {all a2,'"
where at R.a2 ~ R. ... 'R.an , then the diagram can be drawn as
(b) n!
16.
(a) Let a E A with a minimal. Then for x E A, xRa ::=::? x = a. So if M(R) is the
relation matrix for R, the column under 'a' has all O's except for the one 1 for the ordered
pair (a,a).
(b) Let bE A, with b a greatest element. Then the column under 'b' in M(R) has alII's.
If c E A and c is a least element, then the row of M(R) determined by 'e' has alII's.
17.
(a)
(b)
18.
lub
glb
{1,2}
{1,2,3}
0
0
(c)
(d)
lub
glb
{1,2}
{1,2,3}
0
(e)
lub
glb
{l,2,3}
0
{l}
(a) (i) Only one such upper bound - {1,2,3}. (ii) Here the upper bound has the form
{I, 2, 3, x} where x E U and 4 ::; x < 7. Hence there are four such upper bounds. (iii)
There are (~) upper bounds of B that contain five elements from U.
(b)
(~) + G)
(c)
lubB={1,2,3}
(d)
One - namely 0
glb =0
(e)
(~)
(~) + (1) = 24 = 16
each a E Z it follows
aRa because a - a =
h, c E Z with anb
o1?e
an even nonnegative integer. ....u,,""'""ov
IS
a- b =
some mE
b - c = 2n, for some n E
a - e = (a - b) + (b - c) =
+ n), where m n E
a1?c and n
transitive. Finally,
that a1?a and bRa for some a, b E
Then a - b and b - a
are both nonnegative integers. Since this can only occur for a - b = b - a, we find
[a1?b A bRa] => a = h, so
antisymmetric.
192
'VA04,'Vll 'R is a
order for
it
not a total
example, 2,3 E Z and we have neither 2'R3 nor 3n2, because neither
nor 1, respectively,
a nonnegative even integer.
20.
(8.) For all (a, b) E A, a = a and b 0, so Ca, b)n(a, b) and the relation is reflexive. If
d) E with (a,
d)
(e,
b),
if a e we
that
(a, b)'R(e, d)::} a < c,
( c, d)'R.( u, b) "* c < a,
and we obtain a < a. Henc£) we have a = c.
And now we find that
b)n( c, d) "* b ::; d, and
( c, d)n( u, ::} d $; b,
so b = d. Therefore, (u, b)'R( c, d) and (c, d)'R( a, b) => (a, b) = (e, d), so the relation
antisymmetric. Finally, (',ol1sider (a, b), d), (e, f) E A with (u, b)R( e, d)
(e, d)'R( e, f).
Then
(i) a < c, or (ii) a = c and b ::; d; and
(iY c < e, or (ii)' c = e and d :5 f.
Consequently,
(i)" a < e or {iiY' a = e and b f - so, (a, b)R(e,f) and the relation is transitive.
The preceding shows that R is a partial order on A.
b) & c) There is only one minimal element -- namely, (0,0). This is also the least element
for this partial order.
The element (1,1) is the only maximal element for the partial order. It is also the greatest
element.
d) This partial order is a total order. We find here that
(0, O)'R(O, 1)'R(1, O)'R(l, 1).
::n.
(a) The refiexive, antisymmetric, and transitive properties are established as in the
vious exercise.
e!eDlent (2,2)
(o,O)n(O,l)'R(Oj
22.
O)R(l, 1 )'R(1, 2)'R(2, 0)'R(2, 1 )1(.(2, 2).
1
23.
193
(A,
a total order, then
x, 'II E
xRy oryRx. FbI'
and glb{x, y} = x. Consequently, (A, R) is a lattice.
'II} = y
24.
is finite, A has a maximal element, by
x, 'II (x =f. y) are both
maximal elements, since x, yRlub{x, y}, thell1ub{x, y} must equal either x or y. Assume
lub{x,
= x. Then yRx, 80 'II
a maximal
A
a
maximal element x. Now for each a E A, a ¥ x, if lub{ a, x} f. x, then we contradict x
a E A, so x is the greatest eu:~m.:lnt
being a maximal element. Hence aRx for
[The proof for the least element is similar.]
25.
(a) a
(b) a
(A, R) is a lattice with z
(d) e
(e) z
(f) e
(g) v
greatest (and only maximal) element and a the least (and
(c)
c
only minimal) element.
26.
a)
d)
21.
b) and c) n + 1
e) and f) n + (n - 1) + ... + 2 + 1 = n(n + 1)/2.
5
10
Consider the vertex paqbrc, 0 :$ a < m, 0 :$ b < n, 0 :$ c < k. There are mnk such
vertices; eaeh determines three edges - going to the vertices pa+1qOrC,paqb+1rc,pa(lrc+1.
This accounts for 3mnk edges.
Now consider the vertex pmqbrc, 0 :$ b < n, 0 :$ c < k. There are nk of these vertices;
each detemines two edges - going to the vertices pmqb+l r c, pffi qb r c+l. This accounts for
2nk edges. And similar arguments for the vertices paqflrC(O
a < m,O :$ c < k) and
pCqbrk(O :5 a < m,O :$ b < n) account for 2mk and 2mn edges, respectively.
Finally, each of the k vertices p'mqnrc, 0 S C < k, determines one edge (going to pfflqn r c+1)
and 50 these vertices account for k new edges. Likewise, each of the n vertie.es pm (lrk ,
o :$ b < n, determines one edge (going to pmqHlrk), and so these vertices account for n
new edges. Lastly, each of the m vertices paqnrk, 0 :$ a < m, determines one edge (going
to pl1l+1'i'rk) and these vertie,es account for m new edges.
The preced,mJ;
28.
29.
a)
'''.U.U'-'''''. of edges as (m+n+k )+2( mn+mk+nk )+3mnk.
. 3. There are 4 . 2 = 8
for
totally
are2·7=
all 1
30.
194
)mxn be (O,l)-matrices, with E ~ F and F :5 E. Then,
Now let E = (eij )mlo., F =
all 1 :5i :5 m, 1 ~ j :5 n,
~
lij :5
=}
,so E =
-- and
"precedes" relation is antisymmetric.
that
= (fij)mxnl
G =
are (0,
F and F < G. Then, for all 1 :::; i ~
1 :5 j :::; », eij :5 Iii and lij :5 gij =}
:5 Uij, so E
~. and the "precedes" relation is transitive.
Finally,
with E
In so much as the "precedes" relation is reflexive, antisymmetric, and transitive, it
this
a partial order - making into a poset.
Section 7.4
1.
(a) Here the collection AI, A:h Aa provides a partition of
(b) Although A = Al U A2 U A3 U~, we have At n A2 0, so the collection All A:h A a)
A.t does not provide a partition for A.
2.
(a) There are three choices for placing 8 - in either At, A2! or Aa. Hence there are three
partitions of A for the conditions given.
(b) There are two possibilities with 7 E At, and two others with 8 E AI. Hence there are
four partitions of A under these conditions.
(c) If we place 7,8 in the same cell for a partition we obtain three of the possibilities. If
not, there are three choices of cells for 7 and two choices of cells for 8 - and six more
partitions that satisfy the stat,ed restrictions. In total - by the rules of sum and product
- there are 3 + (3)(2) = 3 + 6 = 9 such partitions.
3.
n = {(I, 1), (1, 2), (2,1), (2,2), (3, 3), (3,4), (4,3), (4,4), (5, 5)}.
4.
(a) [1] = {1,2} == [2]; [3] = {3}
(b)
{1,2} U {3} U {4, U {fi}.
5.
'R, is not transitive since 1R2, 2R3 but
=
6.
Y) E
(Xl, Yl)'R.(X2dJ2) :::::::::;} Xl
).
X
= X2
)n.(X2'
1"1.3.
Xl
+ 111 = Xz + Y2,X2 + 1.12 :::::: X3
reflexive,
Y3, so Xl Yl = X3 + Ys and (xl,lI1}R{x:hY3)' Since'R is
it is an
(b) [(1,3)] = {(1,3),(2,2),(3,1)};
{(2,4)] = {(1,5),(2,4),(3,3),(4,2),(5,1)};
,1)] :::::: {(1,1)}.
(c) A = {(I,
U HI, 2), (2, I)} U {(I, 3), (2,2), (3, l)}U
{(I, 4), (2, 3), (3,2), (4, I)} U {(I, 5), (2,
(3,
(4,2), (5, l)}U
{(2, 5), (3,4), (4,3), (5, 2)} U {(3, 5), (4,4), (5, 3)} U {( 4, (5,
8.
U {(5, 5)}.
(a) For all a E A, a - a = 3 ·0, so 'R is reflexive. Fo:r a, b E
a - b = 3c, for some
c E Z =::} b - a = 3( -c), for -c E Z, so aRb =::} bRa and R is symmetric. If a, 0, c E A
a'Rb, b'Rc, then a - b = 3m, b - c = 3n, for some m, n E Z ::::::::} (a - b) (b - c) ::::::
3m 3n ::::::::} a - c = 3( m + n), so aRc. Consequently, 'R. is transitive.
(b) [1] = [41 = [7J :::::: {1,4, . [2] = (5) = {2, 5};[3} = [6J = {3,6}.
A = {1,4, 7} U {2,5} U {3,6}.
9.
(a) For all (a, b) E A we have ab = ub, so
b)'R.(u, b) and 'R, is reflexive. To see that n is
symmetric, suppose that (u, b), (c, d) E A and that (u, b)'R( c, d). Then (a, b)'R( c, d) =} ad =
be =} CO = da =} (c, d)R( a, b), so 'R. is symmetric. Finally, let ( a, b), (c, d), (1'3., f) E A with
(a, b)'R( c, d) and (c, d)'R( 1'3., f). Then (a, b)'R,( c, d) :::} ad = be and (e, d)n( e, f) =} cf = de,
so adf = bef = bde and since d i= 0, we have af = be. But af = be :::} (a, b)'R.(e,f), and
consequently 'R is transitive.
It follows from the above that 'R is an equivalence relation on A.
(b) [(2, 14)] = {(2, 14)}
[( -3, -9») :=: {( -3, -9), ( -1, -3), (4, 12)}
[(4,8)] = {(-2,-4),(1,2),{3,6),(4,8)}
(c) There are five cells in the partition -
in fact,
A = [( -4, -20)] U [( -3, -9)] U [( -2, -4)] U [( -1, -11)} U [(2, 14)}.
n
(b)
Cd)
11.
(a)
,so
l
8~onefor
{I,
}
the partition {a, 1:1, c} U {d, e, f} that corresponds with an. equhr...Jence relation.. But the
selection {d, e, f} gives us
same '"'=",,.. . ,.-. ......,
(b) (;) [1
= 4(~) After selecting a of the elements we can partition the remaining a
-
(i) 1 way into three equivalence classes of
1; or
(ii) :3 ways into one equivalence class of size 1 and one of size 2.
(c) (:) + = 2(:)
(d) n)(~)
12.
4(~) + 2(:) +
(:)
(a) 210 = 1024
(c) 1024 - 52 = 972
(e) 2:t=:l 8(4, i) = 1 + 7 6 1 = 15
(g) Et:=l S(3, i) = 1 + 3 1=5
(b) E:=1
i) = 1 + 15 +
10 1 =
(d) S(5, = 15
(f) L~=l Sca,i) = 1 a 1 = 5
(h) CEf=:l S(3,
(E~=l S(2, i» = 3
i» -
13.
300
14.
(a) Not possible. With n reflexive, I'R.I ~ 7.
'R. = {(x, x)lx E Z, 1 ::; x :5 7}.
(c) Not possible. With 'R. symmetric, I'RI- 7 must be even.
(d) R = {(x, x)lx E Z, 1 ::; x ::; 7} U {(I, 2), (2, I)}.
(e) 'R = {(x, x)/x E Z,l :5 x $ 7} U {(I, 2), (2, In u {(a, 4), (4, 3)}.
(b)
(f) and (h) Not possible with r - 7 odd.
(g) and (i) Not possible. See the remark at the end of Section 7.4.
15.
Let {~}ieI be a partition of a set A. Define n on A by x'Ry if for some i E I, X, Y E Ai.
For each x E A, x, x E A, for some i E I, 80 xRx and 'R is reflexive. x'Ry ==> x, y E Ai,
for some i E I==> y, x E Ai, for some i E I==> y'Rx, so'R is syxnmetric. If x'Ry and y'Rz,
then X, y E Ai and y, z E Aj
some t,J E
Since
n Aj contains y and {Adi€l is a
partition, from Ai n Aj :;;: 0 it follows
= Aj, so i = j. Hen.ce x, z E Ai, so x'Rz
'R is
16.
Let P =: UieIA i
a
= P, so f
of A. Theu E =
x
yE A
17.
{B 1 ,Bz,Bl}, ... ,Bn }
a
U ... U
1 :5 i < j :5 n,
(Bi)
Part (b)
1.
.t:JACId..... p ...v
7
is a special case
this
(v(st, 1) =
(V(8,*,
83,
0) =
=
so 82~2S3'
(V(85,0) = S3) and {V(S2!
= 83)E1(v(S5,
-
so
and 82E2'~5' it follows that
Hence P'J, is given by P2 :
x = 0,1. Hence 82Ea85 and
{..'la, }, {84}'
},
= 83)
-
Consequently, states 82 and 85 are equivalent.
(b) States 82 and 85 are equivalent.
(c)
2.
States 82 and 8., are equivalent; S3 and S4 are equivalent.
(a)
P4 : {S1
{82,S.,},
{8a,s,d,
{.'Is},
1,0
Pa :
{Sll oS!)},
P2 :
1.0/
{82,S.,},
{Sa,S4},
is:!, 83, 84) 8.,},
0,0
100
o
iss}
1,0
{8e}
0,0
iss}
-,
v
M:
0
1
S
84
81
w
0 1
1 0
81
82
1
1 0
0 0
1 0
I 821
!
83
86
'~1
84
.9S
84
86
82
81
()
Supplementary Exercises
1.
(a) False. Let A = {I, 2}, 1 = {I, 2}, n 1 = {(I,
,n2 = {(2, 2)}. Then UiEI'R.i is reflexive
but neither n 1 nor'R.2 is reflexive. Conversely, however, if ~ is reflexive for aU (actually
at least one) i E 1, then UiElni is reflexive.
(b) True. niEI'R.i reflexive {:=} (a, a) E niEI'R.i for all a E
and all i E 1 {::::::::} n; is reflexive for all i E 1.
2.
{:=}
(a, a) E 'Ri for all a E A
n. Then 'Rl U 1(.2 is symmetric
(i) (a) False. Let A = {I, 2}, 1(.} = HI, 2)}, 'R.2 = {(2, 1
although neither n 1 nor 1(.2 is symmetric.
Conversely, however, if each nil i E 1, is symmetric and (x, y) E UiElnil then (x, 'II) E n i
for some i E 1. Since ni is symmetric, (y,x) E nil so (y,x) E UieiRi and UiEI'R. i is
symmetric.
(b) If (x, y) E nie/R.i' then (x, y) ERr" for all i E I. Since each 'Iii is symmetric,
(y, x) E 'R.il for all i E I, so (y, x) E nie/Ri and niE/Ri is symmetric.
=
n
The converse, however, is false. Let A = {l, 2, 3}, with 'iiI
HI, 2), (2, 1), (1, 3)} and 2 =
{(1,2),(2,1),(3,2)}. Then neither 'R1 nor'R2 is symmetric, but Rl n'R2 = {(1,2),(2,l)}
is symmetric:.
(iii) (a) Let A = {
with Rl = {(1,2)} a.n,d R'J = {(2,
tra.nsitive but Rl U'R2 is not transitive.
not ".. ""£.«>~u
z) E Ri
iE
E nU:.IRi and niEiR-i
some b E
b) E 1i2 , (0, c) E
E
E 'R.11 so
E Rl 0
0 'R.I' (c,
a) E 'R1 , for some d E A. Then
E 'R-a, (at
E Rl
,i E
follows.
(a)
(b)
Equivalence relation. Each equivalence class is
t = "," E R+}. Then T = U"eR+A...
(c) Reflexive, antisymmetric.
Symmetric.
(e)
Equivalence relation. [(1,1)]
E
the area
= {(I, 1), (2,2), (3,3), (4, 4)}j
[(1,2)J = {(I, 2),
1), (2,3), (3, 2), (3,4), (4, 3)}i
[(1,3)1 = {(1,3),(3, (2, (4,2)};f(1,4)] = {(1,4),(4, I)}.
= [(1,1)] U {(I, 2)J U [(1,3)] U [(1,4)}.
5.
(c,a) E ('R.I 01?.2l ¢:=:> (a, c) E n 1 oR'), ¢:=:> (a, b) E nb(b,c) E~, for some b E B ¢:=:>
(c,h) E 'R,21 (b,a) E nIl for some bE B ¢=:} (c,a) E 'R.2 on1,
6.
(a) If P is a partition
A then P $ P, so n is reflexive. For partitions Pi, Pj of A
if Pi $ Pj and Pi
then Pi = Pj and'R. is antisymmetric. Finally, if ~,Pj) PIr: are
partitions of A and Pi npj, Pj'RPr;, then Pi < Pj and Pj $ Ph, so each cell of Pi is contained
in a cell of
Hence
transitive
a partial order.
(b)
n
1.
Let U = {1,2,
= P(U) - {U,0}. Under
relation A is a poset with
the five minimal elelnents {x}, 1 $ x $ 5, but no least element. Also, A has five maximal
elements - the five subsets U of size 4 - but no greatest elemen.t.
8.
(b) [(l,l)J = HI,1)}; [(2,2)] = {(1,4),(2,2),(4,1)};
[(3,2)] = {(1~6),(2,3),(3,2),(6,1)}i {(4,3)] = {(2,6),(3,4),(4,3),(6,2)}.
9.
n = 10
as
by g and fJ UVJ../.LUJ'UIJ";U by h.
S transitive.
part (a), 1 is UOlllllliate~U by II,
(c) Let f,/t,/z E .Fwith fen) = '11,11('11) = '11+3,
and
h ¢ [1], because f is not dominated by
Adjacency
List
1
2
1
1
2
3
3
4:
5
6
7
1
2
3
4:
2
3
5
6
8
4:
5
3
5
...
\}
6
Adjacency
List
"1
2
2
3
(b)
3
1
4:
5
4:
15
= 5,
Index
List
1I 1
2 2
3
4:
5
3
4:
5
6
6
201
(c)
Adjacency
List
1
2
2
3
3
1
4:
4
5
5
1
6
4:
7
Index
List
1
1
2
2
3
4
5
3
6
()
8
7
12.
13.
(a)
each v E V, v = v so vRv. If vRw
there is a path from v to w. Since the
graph G undirected, the path from v to w is also a path from w to v, so wnv and n is
symmetric. Finally, if v'Rw and w'Rx, then a subset of the edges in the paths from v to
w and w to x provide a path from v to x. Hence 'R is transitive and 'R an equivalence
relation.
(b) The cells of the partition are the (connected) components of G.
14.
(a) PI: {8t, 83, 87}, {82' S'h 85, 8e, ss}
P2 : {Sll 83, 87}, {S:h Sli! 8S}, {s,,) .ss}
P3 : {8d, {83,S.r}, {S2,SS,8s}, {S4},{Se}
P4 =P3
v
w
M: 0
1
0
81
83
8a
82
83
83
0
0 0
1 0
0 0
0 0
83
83
82
84
82
83
86
84
8}
(b)
1
1
Hence w( 84,000) = 001 ::f. 000 = w(ss, 000), so 000 is a distinguishing string for 84 and. S615.
One possible order is 10, 3,8, 6, 7,9, 1,4,5,2, where program 10 is run first and program
2
n == 2:
16.
n
= 4::
n=6:
4
r
2
2
1
l
1
(vi) n =
n = 12:
8
8
4
4
4-
2
3
2
1
1
(viii) n =
n==
8
(ix) n ==
15
5
3
rouuce a new
part
.tiX.ample 4.45.
(b)
= HO.3., O.7)}
on
[(0,0.6)} =
0.6), (1, O.6)}
[(1,
= {(O, 0.2), (1, O.2)}
general, if 0 < 6 < 1, then [(6, b)] = {(a, b)}; otherwise, [(0, b)] = {(O, b), (1, b)} =
(c) The lateral surface of a cylinder of height 1 and base radius 1!21r.
18.
U, then 0 :5 ICI:5
ICI = kj each such subset C
C
For 0:5 k :5 3 there are
C
U where
2'"
B ~ C. Hence tbe relation 'R contains
(n 21 + (;)22 + G)23 = (1 + 2)3 = 33 = 27 ordered pairs.
(~)~
(b) For U = {I ~ 2, 3, 4}
(:)2 = (1 +
4
( c)
b)].
number of ordered pairs in n is (~) 2° +
21 + (:)
+ (!)
= 3 = 81.
For U = {1, 2,
4
... , n} 'IV"''''?"''' n ~ 1, there are 3'11, ordered pairs in the relation n,
j
19.
Since lUI = n, IP(UI = 2'11. and so there are (2'11,)(2'11,) = 4n ordered pairs of the form (A, B)
where A, BU. From Exercise 18 (above) there are 3n order pairs of the form (A, B)
where A
B. [Note: If (A, B) E n, then 80 is (B, A).] Hence there are 3n + 3'11, - 21>
ordered pairs (A, B) where either A ~ B or B ~ A, or both. We subtract 2'" because we
have counted the 2n ordered pairs (A, B), where A = B, twice. Therefore the number of
ordered pairs in this relation is 4'11. - (2· 3n - 2Vl) = 411. - 2· an + 2'11,.
20.
(a) There are 2m equivalence classes - one for each subset of B.
(b) 2n - m
21.
(a) (i) B'RARCj
(ii) BRC'RP
B1lARC'RF is a maximal chain. There are six such maximal chains.
(b) Here 11 'R 385 is a maximal chain of length 2, while 2 n 6 n 12 is one of length 3.
The length of a longest chain for this poset is 3.
(c) (i) 0 {I} ~ {1,2} {1,2,3} ~ U;
(ii) 0 {2} ~ {2, ~ {I, 2) 3} ~ U.
There are 4! = 24
(d)
n!
elelmelut a E
The proof for C'Il maximal
}, {{I
(b)
}; 4
(c) Consider
set M of all maximal elt""ments in (A, R).
this
not an antichain
then there are two elements a, b E 1.\1 where aRb or bRa. Assume, without loss of generality,
aRb. If
is so, then a is not a maximal
of
R). Hence CJo.,f, (M x M)n'R)
is an antichain
(A, 'R).
The
25.
for the
of all minimal elements is similar.
If n = 1, then for all X, 11 E A, if x f::. 'II
antichain,
the result follows.
;cRy and y'Rx. Hence (A,'R) is au
Now asS1..1lIle the result true for n = k ~ 1, and let (A, 'R) be a poset where the length of
a longest chain is k + 1. If AI is the
of
maximal elements in (A, 'R), then M 0
and M is an antichain in (A, 'R). Also, by virtue of Exercise 23 above, (A - M, 'R'),
for R' = «A - M) x (A - M» n 'R, is a poset with k the length of a longest chain. So
by the induction hypothe-.sis A - M = C 1 U C'}, u ... U Ck, a partition into k antichains.
Consequently, A = G1 U G'}, U ... U Gk U M, a partition into Ie + 1 antichains.
26.
(a) Since 96 = 25 ·3, there are t(~) = 132 ways to totally order the partial order of
positive integer divisors of 96.
(b) Here we have 96 > 32 and must naw totally
the partial order of 10 positive
integer divisors of 48. TIllS can be done in ~
= 42 ways.
(c) Aside from 1 and 3 there are ten other positive integer divisors of 96. The Hasse
diagram for the partial order of these ten integers - namely, 2,4,6,8,12,16,24,32,48,96 - is
structurally the same as the Hasse diagram for the partial order of positive illteger divisors
of 48. So as in part (b) the answer is 42 ways.
(d) Here there are 14 such total orders.
en
27.
(a) There are n edges - nan1.ely, (0,1),(1,2), (2,3), ... ,(n -l,n).
(b) The number of partitions, as described here, equals the Uih'nber of compositions of n.
the answer is
(c) The number of
and
partitions
of 5(= 12 - 7).
= 64, for
are
of 3
FURTHER TOPICS
IN
ENUMERATION
AND
Section 8.1
1.
xES and let n
the number of
(from among C17 C2, C3, (:4) satisfied
x:
(n = 0): Here x
once in N(C£CSC4) and once
N(CIC'J,CSC4)'
(n = 1):
x satisfies C1 (and not C2, Ca,
x 18
once
and once
in N(CIC2C3C4)'
If x satisfies Cj, for i =I 1, then x is not counted in any of the three terms in the equation.
(n = 2, 3, 4): If x satisfies at least two of the foul' oondi tions, then x is not counted in any
of the three terms in the equation.
The preceding observations show that the two sides of the given equation count the same
elements from S, a.."1d this provides a combinatorial proof for the formula N(c2c3c4.) -
N(CIC2C3C4) + N(CIC2 CSC4)'
2.
Proof (By the Principle of Mathematical Induction):
If t = 1, then we have N = N(c}) = the number of elements in S that do not satisfy
condition Cl = N - N( (1)' This is the basis step for the proof.
Now assume the result true for k conditions, where k (2:: 1) is fixed but arbitrary, and for
any finite
S. That is, N(Cle2c3'" eA:) = N - [N(ed + N(C2} N(C3) + .. , + N(Ck»)
[N( C1 (2) +N (C1 (3) +, .. + N (CI C,,)+ N( C2( 3)+' , . + N( cacl!:)+' . . + N (C3CA:)+' .. +N (Ck-l C1:)J[N(C1CzCS)
N(Ck-2Ck-1Ck)]' ..
+ (-1)& N(CIC2CS'" Ck) •
• u'-n......,~.u',."
hypothesis we have
N(cz) ... N(Ck) N(CkH)] +{N(Ct C2) .. ·+N(C1Ck) N(CICkH)
) N(CkCkH)] - [N(C1 C2C3)
N(Ck_2 Ck_1 Ck)
N(Ck-1 CkCkH)}
(_1)k+1 N(C1C2C3'" CkChH)'
N-[N(Cl)
true for
set Sand
Principle of Mathematical Induction.
So
t
3.
N == 100
N(c1) = 35; N(c2) ==
= 30; N(C4) = 41
N(CICZ) = 9; N(C1C3) = 11; N(C1C4) = 13; N(C2CS) = 10; N(C2C4) = 14; N(CSC4) = 10.
N(CIC2C3) = 0; N(CIC'},C4J = N(CIC3C4) = 6; N(CZC3C4) = 6
N(CIC'lC3C4) = 4
(a) N(CIC'j,CaC4) = N(CIC2C4) - N(CIC'j,CSC4)
N(CIC2C4) = N - [N(Cl) N(cz) N(C4)]
+[N(CICZ) N(C1C4) N(C2C4)] - N(C1C2C4) = 100 - f35
= 100 - 106 + 36 - 6 = 24
N(CIC2C3CJ,,) = 12 (as shown ill Example 8.3)
30 + 41] + [9 + 13 + 14] - 6
SO NCCIC'lCSC4) = 24 - 12 =
Alternately~
N(C1CaC4) = N - [N(el) + N(C2) + N(c4)] + [N(CIC2) + N(CIC4) + N(C2C4)] - N(CIC'j,C4), so
N(CICaC3C4) = N(cs) - [N(CICS) + N(C2C3) + N(C3 C4)] + [N( C1C2(3) N(CIC3C4) + N(C'lC3C4)]
-N(C1CaC3C4) = 30 - [11 + 10 + 10J + [5 + 6 + 6}- 4 = 30 - 31 17 - 4 = 12.
N(C4)] + N(CIC4), so N(CtC2C3C4) = N(C2C3) - (N(CIC2C3) +
N(C2C3C4)] + N(CtC2C3C4) = 10 - [5 + 6] +4 = 3.
(b) N(CIC4) = N - [N(el)
4.
St.aff member brings hot dogs
Staff member brings fried chicken
C3: Staff member brings salads
c,,: Staff member brings desserts
Cl:
N=
= 4; N(Cl(.!ZC4) = 6;
=2.
~- [4+6
5
2=
-[N(C1C3)
N( C1C3(4) +N( CaC3(4)]-2=
N(CIC2 CaC4) =
- flO +
N(CIC2C3C4) = N(C4)-[N(C1C4) N(CaC4)+N(cSC4)] [N(CICZ C4) N(CICaC4) +N(CaC3C4)]- 2 = 8.
N(CIC2CaC4) =
+ 17 14] [6 5 7] - 2 = -40
So
answer is 2 8 6 8 = 24.
(a)
n is divisible by 2
n is divisible by 3
C3:
n
divisible
5
N(Cl) = l2000/2J = 1000, N(ca) = l2000/3j = 666,
N(C3) = l2000/5J = 400, N(C1CZ) = l2000/(2)(3)J =
N(C2CS) = l2000!(3)(5)J = 133, N(CICS) = l2000!(2)(5)J = 200,
N(CIC2Ca) = l2000!(2)(3)(5)J = 66.
N(CICaCs) = 2000 - (1000 + 666 + 400) (333 + 200 + 133) = 534
Cl:
C2:
(b) Let C1!C2, C3 be as in part (a). Let C4 denote the number n is divisible by 7. Then
N(C4) = 285, N(C1C4) = 142, N(C2C.d = 95, N(CaC4) = 57, N(CIC1C4) = 47, N(CIC3C4) =
28, N(C2CaC4) = 19, N(CIC1CSC4) = 9. N(CIC2C3C4) = 2000 - (1000 666 400 + 285)
(333 + 200 + 133 + 142 + 95 57) - (66 + 47 28 19) 9 = 458
(c) 534 - 458 = 76.
6.
Xl
+ Xa + X3 + X4 = 19.
= (22\)
(a) 0 < x· 1 < i < 4. (4+19-1)
19
19
-
(b)
$,
For 1
-
-
i S; 4 let Ci: Xi 2: 8.
N(Ci): Xl+X2
X3
X4=11:
(H~i-l)=G:), lS;i
4
N(c;Cj): Xl + X2 Xa + X4 = 3: ("+:-1) = (~), 1 S; i < j S; 4
N(C1 C2C3C4) := - 51 + 52 =
4(i:) 6(:)
G!) -
-
=1
8
"{
N(CIC2 C3C4) = G~)
1.
- [2e:) + (:) +
8
+ e)]·
Let Cl
condition
an
11
consecutive pair IN. "-''''.".....' "' similar conditions c:"
rences of
consecutive pairs NI, 10,
NO,
ON, respectively.
occur-
N = 80 = 1
N(cd = 9!/(2!)2,
= (~)[9!/(2!)2J;
N(C1C2) = N(CICa) = N(CI C6) = N(C2c'4) = N(C2CS) =
N{CI(4) =
= (6)[71/2IJ; and
83 = S4 = Ss = 8 6 =
Consequently, the number arrangements under the given restrictions is Nec! C2C3C4CSCs) =
So - 8 1 + 8'}, = [11!j(2!)3] - (~) [9!/(21)2] + (6)[71/2!] = 4,989,600 - 544,320 15,120 =
4,460,400.
8.
The number
integer solutions for Xl X2 X3 X4 = 19,
< Xl :::; 10, 1 :::; i :::; 4,
equals
number of integer solutions for 111 + 1/2 + 113 + 114 = 39, 0 'IIi:::; 15.
For 1:::; i :::; 4, let c..;,: l1i ~ 16.
N(c..;,) , 1 :S i S 4: 'Ill
N(CiCj), 1
112 + 113 + 114 = 23: e+~;-l) = (~)
i < j :::; 4: '111
112 + '113 + 1/4
N(CIC~ii3C4) = (~;) - (~) (:) + (~)
9.
c:)
= 7: e+~-l) = e:)
x be written (in base 10) as Xl X2 ••• X..,. Then the answer to the problem is the
number of nonnegative integer solutions to Xl + X2 + . . . X'l = 31, 0 :::; Xi :::; 9 for
1 S i 7.
J <
1
Xl
X2
+ ...
Xl,
>9
••• ,Xi
of
an
Xj ~ 10).
0
,0
Xi
= 11, {}:::;
0
Xi
-
-
-
::£., =
!
11.
For each distribution
15
are 15! arrangements. Consequently, order
to answer this question we need to know the
positive integer solutions for
Xl + X2
X3
X'" + X5 =
where 1 ~ Xi ~ 4 for all 1 SiS
is equal to
numher of nonnegative integer solutions for
Y1 Y2 + 1/3 !l4 Y5 = 10, where 0 ~ Yi S 3 for all 1 < i ~
'!Ii
1 = Xi for all
5.]
1S i
For 1 S i ~ 5 let Ci denote the condition that '!II 112 '113
where
?: 4
j
i. Then N(Cl) is the number of &£UJu.U<;,I;G,,,,,
(or Yi > 3) and Yj ?: 0 for 1 ~ j ~ 5
integer solutions for
Zl
Z2
Z3
Z4 + Zs = 6. [Here ZI + 4 = 'Ill, and Zi = 'IIi for all 2 :s; i
5.] This is
1
e+:- ) = (~O), and so 8 1= (~).
II 1 S i < j S 5, N(CiCj) is the number of nonnegative integer solutions for
'WI
Wa 'Wa W4 'Ws = 2. [Here Wi +4 = '!Ii, Wj 4 = 1/j and WI: = 1/k fur alII < k S 5,
k i,j.]
This is e+;-l) = (~), and so 8 2= (~) (~).
Similar calculations show us that 8 3 = S4 = S5 = 0, and so N(CIC2C3C4CS) = So - S1 + S2 =
(S+~~-l) + (~) (~) = (i~)
(~).
Consequently, Flo can arrange these 25 plants, aceording to the restrictions given, in
(15!)[G~) - (~) (~O) +
ways.
(D
(D e:)
- (:) e:)
e)
e) (:)]
12.
The answer is the number of integer solutions for Xl + Xl + X3 + X4 = 9, 0 < Xi :$ 3, 1 :$
i S 4. For 1:$ i :5 4 let Ci dellote that Xl, X2, X3, Xo{ is a solution with Xi?: 4. Then
N(CI CaC3C4)
13.
Let
C1
= (~) - (~) (!) + (:) (:).
pattern spin. Likewise, let C2, C4
denote that the arrangement contains
for
game, path,
net, reEiOOC'tl
-[3(230+
241} - (2n!
211)
N = 9!1[(3!)3] N(ed = N(c?,) = N(cs) = 7!j[(3J)2]
= (51)/(3!), 1 -::.; i < j :5 3
= 31
N(CIC2C3) = 91/[(31)3] - 3[7!1[(3!)2JJ 3(51/3!) - 3!
18.
we need the number of integer solutions for
where 1 :5 Xi. :5 20 for i = 1, 3,4.
is the same as the number integer solutions for
where 0 :5 '11i :5 19 for i = 1,2,3,4.
Let S be the set of integer solutions for equation (*) where 0 :5 '11i for 1 :5 i :5 4. Then
N = 80= lSI = (4+::-1) = (:). So define conditions c}, Ca, Cs, C4 on the elements of 8 as
follows:
Ci: (111,1,/2,113,'114) E S but !Ii > 19 (2:: 20), i = 1,2,3,4. Then
1
) =
N(Ci) =
1 :5 i 4;
e+::- G:),
N(cicj) = e+:- = (:), 1 :5 i < :5 4;
1
j
)
N(CiCjCk) = 0, 1 :5 i < j < k -::.; 4; and N(CIC2C3C4) = O. Consequently,
N(c} C2C3( 4) = So - 51 + 52 - 53 + 54 = (::) - (~) (;:) +
e) (:)
= 18424 - (4)(3654) + (6)(84) :::: 4312.
So the probability the selection includes at least one boy from each of the four troops is
4312/ (!!) = 4312/18424 ~ 0.234.
19.
Here we need to know the number of integer solutions for
where 1 :5 Xi 6 for 1 :5 i $ 5.
Tllis equal to
intege.r solutions for
Y1
'11i :5 5
i :5 5
Cj ael"l,ote
114
15, where 0 :5
'11;;h
=
i, but
a ~J.·'""""-'U
;::: 6.
nonnegative integel' W,",,"M"~'~""" for
50 =
.J
find
.... "
...,Vv
where Zi =
for i
I,
Consequently,
(~) (~).
N(CIC2): Now we
to count
nonnegative integer
for
where Wi = 'Iii, for i = 3,4,5j!l1 = Wl +6, and Y2 = W2 +6. This number is
and, as a
we ha.ve S'}, =
(n G) .
Since S3 = 8 4 = S5 = 0, it follows that N(CI C2CSC4CS) = So - S1
G) G) = 3876 - (5)(715) (10)(35) = 3876 - 3575 + 350 = 651.
82 =
CS+;-l) = G),
G:) - (~) e:)
The sample space here is S = {(Xl,X2,X3,X4,x5)!1 ~ Xi ~ 6, for 1 ~ i
5}. And since
5
lSI = 6 = 7776, it follows that the probability that the sum of Zachaxy's five rolls is 20
equals 651/7776":" 0.08372.
20.
For 1::;; i ::; 7, let Ci denote the situation where the i-th friend was at lunch with Sharon.
Then N(CIC2 .•. Cr) = 84-(;)(35) G)(16)-G)(8)+G)(4)-G)(2)+(;)(1)-G)(0) =0.
Consequently, Sharon always had company at lunch.
21.
(a)
22.
(a) 5186 = (2){2593), and ¢(5186) = (5186)(1/2)(2592/2593) = 2592.
(b) 5187 = (3)(7)(13)(19), so ¢(5187) = (5187)(2/3)(6/7)(12/13)(18/19) =
(b)
32
96
(c)
3200
(2)(6)(12)(18) = 2592.
(c) 5188 = (22)(1297), and ¢(5188) = (5188)(1/2)(1296/1297) = 2592.
Hence ¢(5186) = ¢(5187) ~(5188).
=
23.
(a) 2n - 1
24.
</J( 11,) odd ===> 11, = 2.
25.
(a) 4>(6000) == 4>(24 .3.53 ) = 6000(1 - (1/2)(1 -- (1/3»(1 - (1/5» = 1600.
-1 (for
-
pit
26.
if
Therefore,
L
II( 1---)=
p
)=
= 16.
28.
have one
II(l- 1
_
(2) n =
29.
one
n is divisible by 8;
(2) n divisible by two
more)
(3) n is divisible by an odd prime p (such as 13,
17) u.rnp'l"P 4 divides p - 1; and
(4) n is divisible by 4 (and not 8)
least one odd ""'"''"'.-.
30.
1 ::;; i
5
condition Ci denote the situation where
family i seated. all together.
answer
seating arrangement
N(c-y,,752caC4C5)'
Here So is the number of ways one can arrange 15 distinct objects around a circular table.
This is (15 - 1)1 = 14!
N(el) = 6(13 - 1)1 = 6(12!), for there are (13 - 1)! = 12! ways to arrange 13 distinct
objects [family 1 (considered as one object) and the other 12 people] and 6 ways to seat
the three members of family 1 so that they are side by side. Consequently, 51 = (~)6(12!).
Similar reasoning leads us to
N(CICaCa) 63 (8!)
S3 = (~)63(8!)
N(C1Ca) = 62 (1O!)
SJ = (~)62(1O!)
5
4
4
N(CICaC3C4C5) = 6 (41)
55 = (!)6 5 ( 4!).
N(CICaC3C4) = 6 (6!)
S4 = (:)6 (6!)
=
Therefore, N(CICaC3C4C5) = So - 8 1 + Sa - S3 + S4 - S5 = Ef"",o( -1)iG)6i (14 - 2i)! =
87,178,291,200 -14,370,048,000 + 1, 306, 368, 000 - 87,091,200 + 4, 665, 600 -186, 624 =
74,031,998,976.
Section 8.2
(20160) = 80640
Ez = S2 -
53
= 5'}, -
S3 +
(ii)
54 = (:)(5040) = 5040
54 = O'ii:'ii:J;CU
S4 =
= 53 - (~)S4 =
~
Cs
~,~'~'
(a) N = (14!)/(2!)5
N(c}) = (13!)/(2t)4; 51 = (D[(131)/(2!)4]
N(CIC2) = (121)/(2!)3; S2:::: (;)[(12!)/(21)3]
N(CICZC3) ::::
= (;)[(111)/(21)2J
S's, respectively.
S4
N(C1C2C3C4) = 101/21;
= (:)(10!/2!)
N(CIC2CaC4Cs)
9!
N(C1C2C3C4CS) = 1,286,046,720
= = S5
(b) E'J, = 52 - (~)S3 + (~)S4 - (;)S5 = 350, 179,200
(c) L3 = S3 - G)S" + e)S5 = 74,753,280
4.
For 1:S; i < 7 let Ci denote the condition that i is not in the :range
J. Then the
number of functions I: A - t B where Il(A)1 = 4 is Es = S3 - (~)S4 (~)S5 - (:)S6+
G)S7 = G)410 - (i) (:)310
(~) (;)210 - (;) (~)llO + G) (;)010 = 28648200.
G) 4!S(10, 4) = 28648200.
L3 = S3 - G)S4 (~)S5 - (~)S6 + (~)Si = (;)410 - G) (~)310 + (~) G) 2
(~) 110.
Note:
Using Stirling m.u:n'bers of the second kind the result is
10
5.
6.
-
(;)
Here A = {1,2,3, ... , 1O}, B = {1,2,3,4}. Using the ideas in the first part (of Exercise
4) for I/(A)I = 2 we find that E2 = 6132. For I/(A)I < 2 we find that L2 = 6136.
l:S;i:5
let Ci
,l:5i
a :replacement where card i is plac-ed
its
,1
l"A..
,~"""..t"
i<j<k
probability thatthe
include at
G)
(b) El = 8 1 + (~) S3 - (!)
of exactly one void is Ell (~;).
one card from each suit
= (1) (~) - 2 (~) (::)
3 (!) (:;) - O. The probability
t'wo voids is Ea/ (~;) .
. The probability of
8.
N(CIC2CaC4)/ (~;).
= Lt Et- 1
(b) E t - 1 = St-l - tSt;
L t - 1 = L t E t - 1 = St St-l - tSt = St-l - (t -l)St = St-l - G::::;)St
(d) Lm = Lm+l Em
(e)
= St
L t - 1 :;::: St-l - (::i)St
A
L k+! = S k+l - (k+l)S
l)t-k-I(t-l)S
l1.SSUme
h
k+2 +(k+2\S
k ) 1.:+3 - • • •
k
t
Lie = Lk+l
et
En = [Sk+! - etl)SH2
(_l)t-k-le~l)St] + [SI.: - C"il)Sk+l
2
)Sk+3 - ... +
e~2)Sk+2 --...
(-.l)t-k(t~k)St]
For l:5r:$t-k,thecoefficientof SHr is (-l)"-le+~-l) (-lye;7')=(-lY-l[(k
r - 1)!/lk!(r -I)!] - (re + r )!/(k!r!)] = (-lY-l[1'( k r 1)! - (k + r )ll/(k!r!) = (_ly-l[k +
r -l)!(-k)lI(k!r!) = (-lYet:~l).
k)Sk+l + (k+1)S
Consequently, L k = SIe - (1.:-1
ie-I
"'+2 -
.•.
(_1)t-k(~-_11)S"
,.'
Section 8.3
1.
For 1 < i < 5 let Cj be the condition that 2i is in position 2i.
10li N(Ci):;::: 9!, 1 :$ i :$ 5; .lV( qCj) = 8!, 1 :5 i < j:5
N(CIC2C3C4Cll) = 10! - (;)9!
e)8! - (!)71 + (:)6! - (:)5!
N =
'"
j .lV(ClC2C3C4CS)
= 5!
23154
- (1/5!)] =
5-1=
3.
1
- (1/3!) + (1/41) ,- (1/5!)
5040 -- 1854 = 3186 nell1DIUa1!;1011S
);
m.
), and 80 d m
8.
(a) (4!)d4 == (4!)2e- 1
9.
{lO!)dlO ':'" (lO!)2( e-"l)
10.
(a)
(ii)
(b)
11.
(iii) 1 - (dn/nl)
(ii)
(1) d"./n!
= 265 =
(iv)
(1/r!)e- 1
(iii) 1. - e- 1
(a) (dw ?
(b) For 1 i .$ 10 let Ci denote that woman. i gets back both of her possessions.
N = (10!)2; N(ci,) = (9!)2, 1 SiS 10; N(CiCj) = (81)2, 1 i < j S 10j etc.
s
N(C1C2'"
CIO)
= (10!)2 - CI0) (9!)2 + C20)(8!)2 - .,. + (-1)10 e:)(O!)2.
12.
(a) (12!)d 12
13.
For each n E Z+, n! counts the total number of permutations of 1,2,3, ... , n. Each
such permutation will have k elements that are deranged (that is, there are k elements
x}, X2) ••• ) XIc in {I, 2, 3, ... ,n} where Xl is not in position Xl, X2 is not in position X2)" 0,
and XI. is not in position XA:) and n - k elements are fixed (that is, the n - k elements
Yh 'II')., •
Yn-k in p, 2, 3, ... ,n} - {Xl, X:h' •• , XI;} are such that YI is in position !I!, !l2 is
in position 'liZ)" OJ and Yn-k is in position 'lin-h).
The n - k fixed elements can be chosen in
ways and the remaining k elements
0
.,
C,:k)
(r.:k)d
can then be permuted (that is, deranged) in d k ways. Hence there are
k = (~)dk
permutations
1,2,3, ... , n with n - k fixed elements (and k deranged elements). As k
o n we count all
1, 3,,, ~,n
elements.
01
GUj;"vU
+
1
i~n-l
1
i<j n-l
1 i<.i<k n-
Pd·p.n:. . . .
(-
('~~l)(n -1)1
) = n! -
N(C1C2'"
(n;l)(n - 2)! - (n;l)(n - 3)! + ...
C'~'l)(n - !e)!
-1»!
(b) dn + dn - 1 = [n! - (;)(n -I)!
(;)(n - 2)1-...
(-lY-(:)(n - n)!J+
(n - 3)! - ... + (-It- 1(:=!)((n - - (71 - I»)!]
The coeffident of (n-h)! in £4.+d n - 1 is (_l)k(~)+(-l)k-l(;=n=
[en - I)! - (n;l)(n - 2)1
2
H(n-l)!![(k--l)!(n-[n!f[!e!(n-k)!]J = (-l)k-l[[k(n- -n!]J[k!(n-k)!J] =
(_1).I:-1(n - l)![(k - n)/[k!(n - k)!H =
l)k(n - I)l/[k!(n - k - I)!] = (_l)At·~l).
(~)(n - 1)1 - (~)(n 16.
(n~l)(O!)
(n - 3)! - ... +
+
(a) (11,088)/(10!) == 0.003
l)n(:)
(b) (13, 264)/(101) -=- 0.004
Sections 8.4 and 8.5
These results follow by counting the possible locations for the desired numbers of rooks on
each chessboard.
2.
Consider a chessboard made up of 10 squares arranged in a diagonal so that in each row
and column there is only one square.
3.
(a)
(:) + (n8x + (~)(8 . 7)x 2 + (!)(8 ·7 . 6)x 3 + (:)(8 . 7·6· 5)x4 + ... + (!)(8!)x 8 =
}1",o (~)P(8, i)xi.
(b) Ei=o (7)P(n, i)xi
r( 0 1 , x) = 1
5.
4x + 3x 2 = r( G'l, x)
(a) (i) (1 + 2x)3
1
blocks,
n
k row i"V'O""~""'J""
for the first row containing a
are n-l
1
l)(n -
- m
(:)(n){n -
e:) (
lxi,
l)x!'l~
row 1 to row 2
row
n colulnn choices. For the sec()n<l
we
==
·.·(n-m
reG, x) =
-1)(n -
1.
C++
(1)
(2)
(3)
(4)
(5)
Jeanne
Charles
Todd
Paul
Sandra
1'(C,x)= +4x+3x 2 )(1+4x 2x 2 )=1+8x 21x2 20x3 + 6x 4
1 :::;: i 5
Ci be the condition that an assignment is
with person (i) assigned
to a language he or she wishes to avoid.
N(CICaC3C4C,,) = 5! - 8(4!) + 21(3!) .- 20(2!) 6(1!) = 20.
8.
The factor (61) is needed because we are counting ordered sequences.
9.
20
10.
1
(b) 3/10
5
Z
4
f.i
J
1
5
r( C, x) = (1 + 4x + 2X2) . (1 + 3x + x 2 ) • (1 + x) =
1 + 8x + 22x2 + 25x3 + 12x4 + 2 X 5.
For 1:S: i :s: 6, let Ci denote the condition where,
having roned the dice six times, all six values occur
on both the red die and green die, but i on the red
die is paired with one of the forbidden numbers on
the green die .
.. . 41) {6! - 25(3!) 12(2!) - 2(11) + O(O!)] = 160.
The probability that every V'dllle came up on both the red die and the green die is
(6!)(160)/I(28Y'}.2- 0.00024.
=
= 1
- 31(3)
9x
12.
Consider the Ch€:8StJ~:'d
of shaded
Here r(C, x) = 1 + 8x + 20x 2 17x3
Cl , C2, Ca, Co( denote the conditions:
C1:
f(l} = v 01' W
C2:
f(2)=uorw
4X4. For any one-to-one fundiou f: A -+ B, let
C3 :
C4:
J(3) = x
J(4)=v,x,ory
The answer to this problem is N(CIC2C3C4) = 61 - 8(5!) + 20( 4!) - 17(31) + 4(21) = 146. So
there are 146 one-to-one functions f : A -+ B where
J(1) "1= v, w
J(3) "1= x
J(2)#u,w
f(4)=f. v ,x,y.
Supplementary Exercises
1.
vVe need only .......
by
..,'OA ...<JL
the divisors 2,3, and 5. Let
Cl
denote divisibility by 2,
C2
l500/3J = 166; N(C3) = l500/5j = 50; N( C2C3) =
= 33;
::;;:: 500 - (250 + 166
nrh""'''''''
1
(83
50 + 33) - 16 = 134.
0 S; 71,i ::; 9 for 1 S; i :5
problem
OS; n7
37).
vv..'.'-.'''Ii.'...... Ci as follows:
71,1) n:'h
<
So= N = (7 +
37
the
of
integer solutiollS for Xl X2 + X3
Xr =
Xl
10 = nl, and Xi = ni for 2 < i 7. So N(cd = C+;~-l) = (~) and S1 = (~) G~).
N(CIC2) is
of nonnegative integer solutions
'112 + '!:Is + ... + '!ir = 17
- here '!it 10 = nI, '!i2 + 10 = n2, and 'Yi = ni: for 3 S; i 7. This is C+~~-l) = (:;), and
N( Cl)
so S2 = (:) (;:).
N (CI CZC3) counts the number of nonnegative integer solutions for Zl Z2 + Za + ... + z., = 7,
wbere Zi + 10 = ni for 1 is; 3, and Zi = ni for 4 S; i S; 7. So N(CI C2Ca) = e+~-l) =
e:)
S3 = (:)e:)·
Since S4 = S" = S6 = 0,
answer to this problem is
C
N(C1C2 a ... (:6) = So - 8 1 + S2 - Sa S4 - S5 + S6 = (:~) - (~) (~) + (~) (~;) - (:)
930,931.
3.
e:)
=
For each distribution of the 24 balls (among the four shelves) there are (24!)/(61)4 possible
arrangements. Hence we need to know how many ways the boys can distribute the balls
for the given restrictions. This is the number of integer solutions for
where 2 ::; Xi ::; 7 for all 1 :; i :; 4.
This equals the number of integer solutions for
'YI
'Y2 + 'JJa + 'JJ4 = 16,
where 0 ::; 'Yi ::; I) for alII:; i ::; 4. [Here 'Yi 2 = Xi
each 1 S; i S; 4.]
For 1 ::; i ::; 4 define Ci to be the condition that Yt, 112,113, 'JJ4 is a solution of
~In ..r ..• 'IIi > I)
2: 6)
2: 0
1 S; j ::; 4, j
numt)e.t' of nonuegative integer solutions for
z.
) is
- (~)
[
]
ways.
8 ::;:::; {1,
... ,
and::;:::; So =
define
elements of S as follows:
C1: n E Sand 11. a perfect squ.are;
C2: 11. E Sand n is a
cube; and
C3: 11. E Sand 11. is a perfect fourth power.
N(C1) : ;: :; 31, N(e2) = 10, N(ca) = 5,
N(C1C2) = 3, N(CICS) = N(C3) = 5, N(C2C3) = and
N(CIC2CS) = N(C2Ca) = 1. Consequently,
N(Cl caca) = So - 8 1 S}, - Sa =
1000 - [31 10 5] + [3 5 1] - 1 = 1000 - 46 9 - 1 = 962.
5.
Let Ci denote the occurrence of the pattern i( i + 1) for 1:S; i :s; 7.
The occurrence of the pattern 81 is denoted by C8.
For 1:S; i ::; 8, N( Ci) = 7!; N( CiCj) = 6!, 1 :s; i < j ::; 8; etc.
N(CIC2'''CS) =
6.
8! - (~)7!
(~)6! - (~)5!
(-lr(~)l! = 14832.
(a) Label the walls of the room (clockwise) as 1,2,3,4, and 5. Let C1 denote that walls 1,2
have the same color. Condition Ca denotes that walls 2,3 have the same color.
a similar
way we define conditions Cs and C4, while C5 denotes that walls 5,1 have the same color.
N = k5 ; N(Ci) = k4, 1 <i ::; 5; N(CiCj) = k3 , 1 :s; i < j ::; 5i
N(CiCjCl) = k 2, 1 :s; i < j < f :s; 5;
1
i < j < l < m :s; 5;
So
k=3
=k.
.IS
k.
)
2
~
Sk =
8.
(lkO) (~~ =~)(1O - k)!
= L:l~o( -l)kekO) (!:=Z) (10 - k)! = 1,
),-m (~) Si - . . . ( -1 )n-7n (::J Sn.
(,.,.?1'. . 2)8m+2 - , ••
St = (7) (:) (8;') ('-;.21') ... (a-(i;l)r) (n_iy-ir , where (7) is
the selection of the i containers (from the n possible distinct containers), each of which will contain exactly r elements.
The product (:) ("~f') ... (a-(i,:-l}r) is for the selection of r distinct objects for each of the
i distinct containers. Finally, (n because
of
remaining s - ir
objects there are n - i containers to select from.
9.
The total number of arrangements is T = (13!)![(2!)5].
(a) 8 3 = (!){(10!)/(2!)2]
8 4 = (!) [(9!)/(2!)]
S5 = (:)(81)
E3 = [83 - (:)S4 + G)Ss]/T
(b)
= [S4 - (D
The answer is
10.
- (E4 + Es)JIT.
(i =
Ci denote
(i =
.!JlJJ.1.U1V
COI1Ull;lOI18 Gill
:5 i :5 5, as
the same
C2: band c have the same color.
band e have the same color.
C4: c and e have
same color.
Cs: C and d have the same color.
Cl:
N = ).5 j N(Ci) =).\ 1:5 i ~ 5;N(CiCj) = ).3,1:5 i < j :5 5;
N(C2CSC4) = )./J.,N(CiCjCk) =,\,2 for all other 1 :5 i < j < k :5 5;
N(C1CaCsC4) = N(CZC3C4CS) = ).2,N(CIC2CaCs) = N(CIC2 C"C a) =
N(CIC3C4CS) = )';N(CI CzCaC4Cs) =)...
N(C1 CZCS C4 CS) = ).5 - 5)..4
+ 10.,\3 - ().3 + 9).2)
(2).2 + 3),,) _).. = )..5 - 5).4
9).3 -7 X~
+ 2)".
For)" = 1,2, this result is O. When).. = 3 the result is positive and 80 the chromatic
number is 3.
(b) Draw a graph with a vertex for each room. IT two rooms share a common doorway
draw an edge connecting their corresponding vertices.
The result is the graph in part (a) and the answer is 6 5 -5(64 )+9(f?) -7(6 2 )+2(6) = 3000.
13.
Consider the derangements of the symbols L,A,P 1 ,T,O,P z. There are ds such arrangements.
Of these there are
(i) d" arrangements where P a is in position 3 and PI is in position 6;
(ii) d5 arrangements where PI is in position 6 and
is not in position 3; and,
(iii) d5 arran,gements where P z is position 3 and PI is not in position 6.
There a.re c4 - 2ds - d4, Buch
of L,A,PhT,O,P2. Hence there are (1/2)[~ 2df) - d4 ]
(1/2)[265 - 2(44) - 9} = 84 ways to arrange the letters
none of
original
P is
=
1/2? .....1D<.,c'*u",""
PI TO.]
<
m > 1. Then
14.
n-l-
15.
17}
< n-
or 1
1 -. a.
(b)
16.
(a.)
= 2 = 4;(6)
151 1 = 6 =
ISat = 6 = ¢t(9)
k E Z+,1 S k
= 1 = <fo(2)
IS61 = 2 = <p(3)
ISu'Il = 1 = 4;(1)
S m. Then gcd(k, m) = d ::5 m, for some d E Dm. If k E
Stl}, Sa2 then d1 = gcd(k, m) = d a. So the collection Sa, d E Dm" provides a. partition of
{ 2,3, ... ,m(b) Recall that gcd(n, m) = d if and only if gcd(n/d, mid) =
so
= l{nlO < n
m and
m) = d}l = I{nlO <
mid and
mid) = I}I = 4>(m/d).
s
Proof:
(a.) If n is even then by
Fundamental Theorem of Arithmetic (Theorem 4.11) we may
write n =
where k ~ 1 and m is odd. Then 2n = 2k+1m and ¢(2n) =
(2kH)(1_ D4>(m) = 2k¢(m) = 2(2k)(~)<p(m) = 2{2k (1 - t)¢(m)l = 2[¢(2 k m)J = 2¢(n).
(b) When n is odd we find that ¢(2n) = (2n)(1- DII(l-!), where the product is taken
pin
P
over all (odd) primes dividing n. (Ifn = 1 then I1(1-!) is 1.) But (2n)(1-lHl(1-!) =
~
p
~
P
1
nIlCl - -) = ¢(n).
p11/'
P
So 4>(ab)¢(c)
=
~1+nlp~2+n2 ••• p~e+nt(l_ :)(1-
!) ... (l- :)]
IT
1<;«t
Pimin{m;,ni}(l - -1 ) and
Pi
min{ me;nt }:FO
miil.{m;,n,}]
Pi
the following cues!
_ .1.).
'Pi
227
.
CHAPTER 9
GENERATING FUNCTIONS
Section 9.1
1.
The number of integer solutions for the given equations is
coefficient of
(a)
in (1+x+x2 ... +X1)4,
20
(b) x in (1 + x + x 2 + , .. + x20r~(1 + x 2 X4 +... x 20 )2 or
(1+x+x 2 + ... )2(1 X2+X4 ••• )2.
(c) x So in (x 2 x 3 + X 4 )(x 3 + X4 + ... XB)4.
(d) x 30 in (1 x + x:! ... + x 30 )3(1 + x:! x4 +... x 30 )·
(x + x 3 + x 5 + ' .. + X 29 ) or
(1 + x + x 2 + ... )3 (1 + x 2 + X4 •• • )(x + x 3 + XiS ••• ).
2.
)5 or (1
x x2 + ... )5
(b) (x + x + ... + x )0 or x 5 (1 + x x 2 ••• )i>
35
x )l' or xlO(l + x
x 2 ... )5
(c) (x 2 + x 3
(d) (1 +x x 2 + ... X 25 )4(X 10 + xU ... + X 35 ) or
(1 + x x 2 ... )'i( x10 XU X 12 + ... )
15 )3 or
X
(e) (x 1O xl!
x 25 )2(1 + x + x 2 + '"
(xl!} + x ll ... )2(1 + x + x 2 ... )3
(a) (1 +x +x 2 + ...
2
3.
X
35
35
+x 3 + ... ),3.
[The number of ways to select
candy bars is the coefficient
XlO in either case.]
2
3
7t
(b) Thegeneratingfunctioniseither(l x+x x
••• +xT') or{1+x
x 2 +x3 + ... )n,
[The number of selections
r
is the coefficient of
either
(a) The generating function is either (1 +x+x 2 x 3 +...
X
10
)6 or (1 +x +x
2
second
(b)
6.
(a)
9.2
( a) (1 + X)8
x)
(b)
(e)
xtl
(c)
2
x /(1 - ax)
(a) -27,
-36,8,0,0,0, ...
(b)
0, 0,1,1,1, 1, ...
3
2
X4 + xtl + ... ] = x
XU
... , so f(x)
(c) f(x) =
- x ) =
generates
0,0,
0,1, 1, 1, ...
""y·,.. .,"'''' the sequence
... ,so J(x)
(d) f(x) = 1/(1 3x) = 1 (-3x)
3
2
1,
3 , _3 , •••
(X/3)2 (X/3)3 ., so
(e) f(x) = 1/(3 - x) = (1/3)[1/(1- (x/3))] = (1/3)[1
f( x)
the sequence
(1/3)2, (1/3)3, (1/3)4, .. .
(£) f(x) = 1/(1 - x) 3x7 = (1 +, x x 2 + x 3 + ... ) +
- 11, so J(x) generates
the sequence Q.o,aha:h"'j whexe ao = -10, ar = 4, and ai = 1 for all i 0,7.
3.
(a) g(x) = lex) - asx 3 3x3
(b) g(x) = f(x) - a3xs + 3.1: 3 - arx7 7x 7
(e) g(x) = 2f(x) - 2alx x - 2a3x3 3x 3
Cd) g(x) = 2f(x) [5/(1- x)] (1 - 2al - 5)x + (3 - 2as - 5)x 3
4.
Cs (3 )(210)
5.
(a) (-~5)(-lr = (-lfC5+;-l)(-lf =
5
1i
)
en
(-.,n)C--lf = (_1)"(n+;-1)(_1)7 = (nis)
(b)
6.
(~6)(_1)8 = (-1)8e+:- 1)( -1Y~ = (x:)
7.
(~:)
- (~) (~5)( -1)5
+
o
G) C~5) = G~) - G) (:) + (~)
(7 - 2a7 '- 5)x'1'
IS
package
answer to
120
6
the coefficient of x
(x
= x24(1 + x
is the same as the coefficient of x 96 in [(1 _. X 35 )/(1 - X)]4 = (1 - x 35 )4(1 - X)-4 =
[1
+6x
(~4)+
(;:)(_x)26+ ... (~:)(_X)61
(;;)(_XY*I+ ... ].
70 _...
...
...
= (:)-4
Consequeutly the answer is (;:) ( -1 )96 - 4 (;:) (
12.
(a) The coefficient of X24 in (X2 + X3 + .. .)1> = xW(1 + x + x.2 + .. .)l' = xlO(l- X)-5 =
Xl0[(-~S)+(-;5)(_X) (-;5)(-x)2+ ... ]
(~!)(_1)14=(_1)14e+;:-1)
)14=
This
is the number of
distribute the 24 bott-les
one type of Boft drink among the
surveyors so that each gets at least two bottles. Since there are two types, the two cases
2
can be distributed according to the given restrictions in
ways.
G:).
in (x 3 + X4 + ... y; is
(b) The coefficient of
13.
en
G!)
and the answer is
G:) (~) .
+X 3 +x4 x 5 x6 )12 = x 12 [(1-x 6 )/(1-x)]12 = X12«1_x)6)12{(-;2) + (-;2) (-x) +
(-:2)( -x)2+ .. .J. The numerator of the answer is the Coefficient of X 18 in (1_X6)12{(;';)
18 + ... + x72J[(-~2) + (-:2)( -x) + ...J and
(-:2)( -x) ...] = [1- x6 + C22)x12 X2
(x
ene12) (-;;2)(
e22) (-;2) {_1)6 - en (-;2) = (;:) _ (;2) (;n +
e,2). The final answer is ohtained by dividing the last result by 612, the size of
this equals (-;~2)( _1)18 -
e22)
en -
C:)X
_1)12 +
the sample space.
14.
x 3 + X4 + x5l~(X5 + x lO )2 = x 26(1 + x x 2 + x3 )')(1 + x 5 )2, so we need the coefficient
of x14 in [(1 - x4 )/{1 - x)J8(1 2x 5 + x 10 ) = (1 - x 4)8(1 - xt8(1 2x 5 + XlO) =
[1 (~)(_X4) (~)(_X4)2 ... +(_x4Y~H(~8)+(~8)(_x) (-;8) (_X)2 ... ](1 2x 5 +X 10 ).
(x 2
2(~8)
(~8)( _1)4] - (~)[(~:) 1)10 2(~8)(_
o ] + (!)[(~8)( _1)6 2{~8)( -1)] - (:)[(-;8) 1)2J =
+ 2(~) (~l)] - (~)[G~)
the
+ 2(~)] IS
by
coetnclent is
[(;:)(-1)1"
[G!)
to
the probability.
... ) -
15.
231
-
X))4,
This is the coefficient of x 5 in (1 - X)-4,
= (:).
... )( 1 + x + x 2 + . . .
_
...],i.e., (~:)(-1)13-(~)(-;4)(-lr+G) 1
=
For the hot dogs we
the coefficient of x Hi
(x3 +
x3(1/(1-x»«1-x6)/(1-x»)3. This is
coefficient
(nx6 G)X12_xlS][(~4)
G~) - G) (~) G)(:)·
(~4)(_X)
,(-;,4)
in (1-x 6)3(1-x)-4 = [1-
By the rule of product the total number of distributions for the prescribed conditions is
(:) [(i~) - (~) ei)
==1
(~) (:)].
(x +x 2
"---.....,....--....,.,
-.. .---v-----'
two rolls
one roll
where the 1 takes care of
18.
(1 - 4x)-1/2 = [(-~2)
(-~2)(_4)n
case where the die is not rolled.
(-~/2)(-4x) + (-;tZ)(-4X)2 + ... J. The coefficient of xl'>
=
« -1/2) - n + 1)« -1/2) - n + 2)··· « -1/2) - 1)( -1/2) (-4t =
n!
(1 + 2n - 2)(1 + 2n - 4)··· (1 + 2)(1) (2t =
n!
(2n - 1)(2n - 3)··· (5)(3)(1) (2)11. =
17,1
[(2n -: 1)(2n - 3)··· (5)(3)(1)J(2n )(nl) = (21£)!. = (21£).
nln!
19.
n!n!
n
1S
12, we find that 2 llO / 2j =
2 t6 /2J = 8 start with 3, and 2l4/2j = 4 start with 4.
start
end) with t is the nmnb~~r of palindromes
21.
number
of n
of n - 2t. This is 2 l(n-2t)/2J.
22.
Suppose a palindrome
n has an even number, say 2k,
of the last k summands.
n = 2s,
n
23.
Let n = 2k. The palindromes
n with an even number of summands have a plus sign at
the center and their number is
number of composit.ions of k - namely, 2"'-1 = 2(n/2)-1.
n 2
Since there are 2 / palindromes
total, the number with an odd nm.nber of summands
is 2n / 2 - 2{n/2)-1 = 21">/2(1 = 2n/2G) = 2(10/2)-1.
start
Let .9 be the sum
<> ••u ........... UY"-U>.
i)
24.
(a)
number of palindromes of 10, where all summands are even, equals the number
of palindromes of 5, which is 2 t5 /2J = 4.
(b) 2t6 / 2J = 8
(c) 2 Ln/4j
25.
(a) Prey = y) = (~)jJ-l(~), Y = 1,2,3, ... ,
(b) aIld (c) Using the general formulas at the end of Example 9.18, with p =
q = 1 - p = ~, it follows that
E(Y) = = ~ = 6, and
1
and
6
!
= JVar (Y) = Jq/p2 = .j(~)/(~)2 = .j(V(36) = v'3O ~ 5.477226.
Here we wantEi:l Prey = 2i).
2:~1 Prey = 2i) = Ei:l(~)2i-l(!) = (~) E~1(~)2i-l = (~)[(~) + (~)3 + (~)5 ...J =
(1)(~){1 + (~? (!)'l ... J = (~)l-tfrj = (~)l-hi) = (is)
= (is)(i~) = 151'
qy
26.
27.
Let the {liscrete random variable Y eouui the number of tosses Leroy makes until he gets
y 1
first tail.
Prey = =
- (3), y = 1,2,3, .. ,.
(V
Here we are interested in P1'(Y = 1)
... =
(,
Prey = 3)
... J =
Pr'(Y = 5)
'" =
- (1)(§!) - ~
-
3
5
-
<l) + (;)2(l) +
S'
lR P =
(~)
••• J =
(
(d) P7'(Y ~ 5!Y ?:
- -'-----------'- = Prey
5)1 Prey ;;:: 3) =
(ir l(iY~
t
=
(~)2.
;;::
~ 4) = (
= (~)2.
(f) Var (Y) = qlp'l, where q = 1- p = ~. So Vax (Y) = (~)/(~)2 = (~)/(
Consequently, cry =
1.763834.
J28i9 . :.
29.
(a) The differences are 3 - 1,6 - 3,8- 6,15 - 8,
where 2 + 3 2 + 7 + 0 = 14.
(b) {3,5,8,15}
(c) {I
30.
a,l + a + b, 1
a +b
15-
= (~)(~) =
~ that is 2,3,2,7, and 0,
c,l + a + b e d }
Using the ideas developed
Example 9.17, we consider one such subset: 1 ~ 1 < 3 <
6 < 10 < 15 < 30 < 42 ~ 50. This subset determ.ines the differences 0,2,3,4,5,15,12,8,
which sum to 49.
A second sum subset is 1 S 7 < 9 < 15 < 21 < 32 < 43 < 50 ~ 50, which provides the
differences 6,2,6,6,11,11,7, 0, which also sum to 49.
These observations suggest a one-to-one correspondence between the subsets and the in••• + Cg = 49
where Cl, Cg ?: 0 and Ci;;:: 2 for
teger solutions of Cl + C2 + C3
2 ~ i ~ 7. The number of these solutions is the coefficient of x 49 in the generating function (1+x+.'E 2 •• • )(X 2+x3 .. . )6(1+x+x 2 ... ) = [1/(1-x)2][x 12 /(1-xY>J = X12/(1-xY~.
The answer then is the coefficient
( -1 )37 e+~~-l)( -1 )37 = e~).
i(P-2ki
-I}.
32.
(a)
a1
Cz =
Cl ;::;;::;
(b) (i) en =: n
en =: 1
(:2
==
= 2; Cz =
=
CJ = 4:
Cs
1
2+
Co = 1; C1 =
= 2ft+! - 1
C2 == 3; en =
n 2: 3.
T '..
00
+
... ) =
1 + 2 = 3, C3 = 1 + 2
cn=n (n-I)
Cz =
3 = 6, C4 = 1
(n - 3) +
2 +3
4=
- 4) = 5n -- 10
and
all n ;:::
- X
(b)
- x x 2 - x3
x 2 - x 3 ••• ) = (1::1:)2 = (1 + X)-2, the generating
lUU.ctlC)u for the sequence (~2) (~2), (-;2)) (-;2)).... Hence
convolution of the given
pair of sequences is co) Cl, C21 ... , where
COl =
=
1)ne+:- =
== (-1)ft(n I), n E N.
[Tills is the alternating
I, -2,3, -4,5, -6, 7, ... oJ
j
1)
(:2)
(n!l)
1.
7;6+1;5+2;5+1+1;4+3;4+2+1;4+1+1+1;3+3+1;3+2+2;3+2+1
3+1+1+1+1;2+2+2+1;2+2+1+1+1;2+1+1+1+1+1; 1+1+1+1+1+1+1
2.
(a) I(x) = [1/(1 - x 2 )]fl/(1 - x4)][1/(1- X)6J ... = n~l[l/(l - x 2i )]
(b) $l(x) = (1 + x 2 )(1 + x 4 )(l + x6 ) ••• = n~l(l x 2i )
(c) h(x)=(1+x)(1+x 3 )(1 x5)···=n~1(1+x2i-l)
3.
The number of partitions of 6 into l's, 2'8, and 3's is 7.
4.
(a) (l/(l - t 2 )]{1/(1 - t 3 )][1/(1 - t 5 )][1/(1 - t 7 )]
(b) [1/(1 - fl)][t 12 /(1 - t 3 )][t 20 1(1 - t 5 )J[t 35 /(1 - t 7 )]
5.
(a)and(b) (1+x 2 +x 4
=
6.
X4+X8+ ... )(1+x6+x12+ ... ) ...
nco
1
1=1 1-.1,.2.
(a) l(x)=(1
(b)
X+X2
••• +x
5
)(1+X 2 +X 4 + ... +X lO ) ••• =
[(1 _ x 6i )/(1 _ xi)]
+ ... + x 5i ) =
n~!l(l + xi + X2i +... x 5i ) = n!!l[(l - x 6i )/{1 -- xi)]
xi
8.
x6 + ... )(1
X2i
g(x) =
... = f(x)
result follows from
one-to-one correspondence between the "'A"~~'~
summands
not exceeding m and the transpose graphs (also Ferrel's graphs) that
have m summands (rows).
10.
Consider the Ferren; graph for a. partition
2n
n
(rows). Remove
the first column of dots and the result is a Ferrel'S graph for a. partition of n. This
correspondence is one-to-one, from which the result follows.
Section 9.4
1.
2.
(b)
(e)
(a)
(d)
(a.)
f(x) = 3e 3.v = 3Ei:o t3~)i, so I(x) is the exponential generating function for the
sequence 3,32 , 33 , •••
(b) f(x) = 6e51lJ - 313 23: = 6 E~o (5~)i - 3 I:~o (2~)i, so I(x) is the exponential generating
function for the sequence 3,24,138, ... ,6(5") -- 3(2"), ...
(c) 1,1,3,1,1,1,1, ...
(d) 1,9,14, -10,2\ 2'\ 26 , ...
(e) f(x) = 1 + x + x 2 x 3 + ... ::: 2:~oi! (::), so f(x) is the exponential generating
function for the sequence 01,11, 2!, 3!, .. .
2x + (2x)2 + (2x? + ...}+
,so f(x) is the exponential generating
the sequence 4,7,25,145, ... , (311,!)2n
(f) f(x) = 3[1
function
g(x) =
+
(c)
(d)
+
(eX - 1)4 is
coefficient
... ,12}
x+
+ ..
B=
(1/4)(e 2W )(e 2X + 2
2e 2x + 1),
e- 2$) = (1/4)(e4X
this
(1/4)[412 +
is even and the number of black Hags is even.
the
of
signals where the number of
g(x) = (1 x+(x 2j2!) .. . )2(X+(x 3/3!) ... )2 = e2X (ell:_e-Il:)/2j2 = (1/4)(e 4X -2e 2z +
coefficient of x 12 /(12!) in g(x) is (1/4)[412 - 2(212)], and this counts the signals
where the num.bers of
and black flags are both odd.
Consequently, the number of signals where the total number of blue and black Hags is even
(1/4)[412 2(212)] (1/4)[412 - 2(212)] = (1/2)(4 12 ).
5.
We find that
_1_
1-.:;
= 1 x + xl + XS + ...
= (O!) ~~ (1!)~~ + (2!);~ + (3!)~~
so 1/(1 - x) is the exponential generating function for the sequence O!, I!, 2!, 31, " ..
6.
(a) (i) (1 + x)2(1
(x 2 /2!))2
x
(ii) (1 x )(1 + x + (x 2/2!»(1 + x + (x 2/2!) + (x 3/3!) + (x 4/4!»)2
(iii) (1 +x)3(1 x + (x 2 /2!))4
(b) (1 +x)· (1 +x+(x 2 /2!))·(1
x (x 2 /2!) +( x3 /3!)+(x 4 /4!»·«x 2 /2!) +( x S /3!) +(x4/4!»).
.. ,+ ~O!HI)4 •
7•
The answer is the coefficient of ;:~ in (;~
8.
hex) = f(x)g(x) = Co CIX + C2(x 2/2!) + C3(x 3/3!) + .. " where
cn(xn/n!) == U""o(aixi/i!)(bn_iXn-i/(n - i)!) =
[Ei=o(aibn-i)/(i!(n - i)!)]x" =
[Ei.::o[n!/(i!(n - i)!)Jaibn_iJ(X n In!) = [Ei:=o e:)a.bn_i](x n In!)
9.
(a)
1J/(320 )
20
(c) (1/2)[3 - IJ/(3~W)
(b) (1/4){3 20 3J1(320 )
(d) (1/2)[3 20 - 1)/(32°)
... ) .
\
-~ 1)(e3 1!; -
= (1
x
(x 3 /3!) +
= (1/2)(e 4X _
e)e (x3/6)2 - (!)
/6)3 (x /6)4,
of x zo/(20!) in hex) is
(i) (lj6){317)(20)(19)(18) + (~) (1/6)2(214)(20)(19)(18)(17)(16)(15) - (:) (1/6)3[(20!)/(11!)].
3
2X
x 2°/(20!)
(d) The coefficient
(1/2)(318 )(20)(19).
Section 9.5
1.
(a) 1
so (1
x
Xl is the generating function
2
x + x )/(1 -
1,1 + 1,1
1 + 1, 1
the sequence 1,1,0,0,0,.
is the generating function for the
1 + 1 + 0, ... - that is, the sequence 1,2,3,3, ....
"j
(b) 1 + x + x 2 x 3 is the generating function for the sequence 1,1,1,1,0,0,0, ... , 80
+ X x 2 + x3 ) / (1 - x) is the generating function for the sequence 1, 1 + 1, 1 1 1, 1
1+1+1,11+1+10,1 1+1 1+0+0, ... -thatis, the sequence 1,2,3,4,4,4, ....
(c) 1 2x is the generating fu.nction for the sequence 2,0,0,0,0, .. so (1 2x )/(1 - x)
is the geneI'ating function for the sequence 1.11 + 2, 1 + 2 + 0, 1 + 2 + + 0, . ,. - that
the sequence 1,3,3,3, .... Consequently, (1/(1- x»[(l 2x)/(1 - x)] = (1 + 2x)/(1- X)2
is the generating function for the sequence 1,1 + 3, 1 + 3 3,1 3 3 3, ... - that is,
the sequence 1,4,7,10, ....
'j
2.
(a) (i) x
(ii) x/(1 - x)
2:k""l Ie (b)
-
_
_
°
(iii) x/(1-x)2
(iv) xl(l - X)3
the coefficient of xn in x/(l- x)3
the coefficient of xn in x(1 - x 3
the coefficient of x n - 1 in (1 - x t 3
~n~lj( _1)'11,-1 = (_1)n-l(3+~:=g-1)( _1)'11,-1
r.+1 = l(n +
t
.-1
2
.
: : : : lx(1+x)J/(1-x)3 generates 02 , ,2 2 ,3 2 , ••• ; [x(l
·x 3 ... ; (djdx)[(x
+ ·x
. xl
(d/dx)[(x x 2)/(1- X)3] =
BI.:!(jUe!J.ce (10,
(12
ao
+a3
0,2
generating function
5.
(l-x)f(x) = (1-x)(aO+alx+a2x2+aax3+ ... ) = aO+(al-ao)x+(a2-a:dx2+(aa-a2)x3+
... , so (1- x )1(x) is the generating function for the sequence
at a2 aI, as - aa, ...
6.
[I(x) - f(l)]/(x - 1) =
- l)[(ao - ao) (alx (a2x2 - a2) + ..
For n ~
n
n
n 2
2
0, (anxn -- an)/(x - 1) = an(x - l)/(x - 1) = an(x - + x - + '" + x + X + 1), so
[I(x) - /(1»)/(x - 1) = al a2(x + 1) + a3(x2 + x 1) a4(x 3 + x 2 x 1) +... Hence
the coefficient of x'\
n ~ 0, is
ai.
Since e V is
generating function for 1,1, 1/2!, 1/3!, ... , it follows that eX /(1 - x)
generates the sequence aO,aI,a2,'''' where an = E~o(1/i!).
8.
6
(a)
= 1 + x + x2 +
+ ... is the generating function for the sequence 1, 1, 1, 1, ....
Applying the summation operator, we
learn that (1~$)2 is the generating function for
the sequence 1,1+ 1,1 1 + 1,1 + 1 1 + 1, ... - that is, the sequence 1, 2, 3, 4, ....
Consequently, x/{l- x)Z is the generating function for 0, 1,2,3,4, ... and x/(l- xl the
generatillg functioll for 0,0+ 1,0 1 + 2,0 + 1 + 2 + 3, 0 + 1 2 3 + 4, ... - that is, the
sequence 0, 1, 3, 6, 10, ... (where 1, 3, 0, 10, ... are the triangular numbers).
(b) The sum of the first n triangular numbers is the coefficient of xn in the generating
function x/(l- x)4 = x(l- X)-4 = X[(~4) + (~4)( -x)
-xr~ + ... J. SO the answer is
(-;4)(
the coefficient of x n - 1 in (1-x)-4 and this is (n:~)(_l)n-l = (_l)n-l e+~:=~~-l)( _l)n-l
(:~;) = (l/o)n(n + l)(n
=
2), as we learned in Example 4.5.
Supplementary Exercises
1.
(a) 6/(1- x) 1/(1- X)2
1/[1 - (1 + a)xJ
(b) l/(l-ax)
(d) 1/(1- x) 1/(1- ax)
+X12)tO. The coefficient of X 83
_ x15)10(1 _
=(1
x+
... ) ... ::::;:
...)
5.
I{x) he the generating function for the number partitions of n where no even
summand is
an
summand may be
g( x) is
generating
function for the number of partitions of n in wh.ich no summand occurs more than three
~,uu,:;;o. Then g(x) = (1+x+x2+x3)(1+x2+x4+x6)(1+x3+x6+x9) ... = [(1+x)(1+x 2 )][(1+
x 2 )(1 x 4 )][(1 x 3 )(1 + x a)] ... = [(1 - x 2 )/(l - x )J(1 + x 2 )[(1 - x 4 )/(1 -~ x 2 )J(1 x 4 )[(1 x6 )/(1-x 3 )J(1+x 6 ) ••• ::;:; [l/(l-x)](l
x+x 2
. . . )(1
6.
x 2 )Il/(1-x 3 )1(1+x'-'){1/(1-x 5 )J(1+x6 ) ••• = (1+
.. )(1
+X5+xlO+XlS .. . )(l+x S ) . . . = I(x).
This result is the coefficient of XIO /(1Ol)
(1 +(x2 /2!)+(x 3/3!)+ ... )4 = (e:l' __ X)4 =
(~)xe3~ + e)x 2e2X - (~)x3eZ:
X4. This coefficient
410 - (~)(10)(39) (~)(10)(9)(28)-
is
(!)(10)(9)(8).
(a) (1 - 2X)-5/2 = 1 + E~l (-5/2l{{-5/z1-1)«-~f2)-2} ... (-5/2)-1'+11( -2x)"
= 1
(5H7H9}~
.. (3+2r) x r , so g( x) is the exponential generating function for 1,5,5(7),
r.
5(7)(9), ...
(b) (1 - ax)b = 1 + E~l {b}{b-'1}(b-r~)···(b-"'+11 (-ax Y = 1 - abx + b(b - 1)a2X2 /2! + ...
Consequently, by comparing coefficients of like powers of x, we have -ab = 7, b( b - 1 )a2 =
7 . 11 and a = 4, b = -7/4.
8.
For each pa,rtition of 71, place a row of 71 + k dots above the top :row in its Ferrel'S graph
and the result is a Ferrern graph for a partition of 2n + Ie where n + k is the largest
summand. This one-to-one correspondence yields PI::;:; P'J,. Taking the transpose of a
Ferrel'S graph for a partition
Pz yields the Ferrel'S graph for a partition in Pa, and vice
versa. The result now follows from these two observations.
9.
FnreochnE
,(l+x)n= (~)
of
we find
(;)X+(;)X2+(~)X3
+
)=
10.
+
+
(:) XU.
Taking the derivative
1
(a)
xS
(x x 2 •• .)12 =
••• )12=(1-x)-12,and
IS
x 20
(1+x+x 2
coefficient of X10 in (x + x 2
(b)
The probabilHy
12.
•••
)6 = x S (1
x +x2
...
)1' is (:6)(_1)4 = (:).
this type of distribution is (:) (:) / (~).
Fix m, 0 :::; m .:::; n.
m objects can be arranged in (k)(k + 1)··· (k + m - 1) ways.
Since there are (;:) ways to
m of these objects, there are (;:) (k )( k 1) ... (k
m -1) ways to select m
the n objects and place them the containers as prescribed.
ere /(1-
xY' = [1 + x + (x 2J2!) + (x 3/31) + ...J. [(~k) + (~k)
eiIJ(l-x)-k is (~k)
coefficient of x71'1nf
(~k)(-l)(n)
(-;1»( _X)2
(-;k)
1)2(n)(n-
+ ...
(n-!l) (-l)n-l(n!/l!) + (:k)C -l)n(nl/o!) = E~=o (~k)( _l)m (n~~)! =
(m+~-l) (n~~)P
and
(n)('k)(k
+ 1) ... (k + m -1) -- L.,..m:::::O
""n
_--11:.l-(mH-l)l - ""n
(mH-lll-11L..m
m!(n-m)! (k-l)!
- L.,..m:::O m!(k-l}! (n-m)!
= E~=o (m+~-l)
13.
(a) We start with a+ (d- a)x, the generating function for the sequence a, d-a, 0, 0,0, ....
Then [a (d - a)xJ/(l - x) is the generating function for the sequence a, a + (d - a), a +
(d - a) 0, a + (d - a) + 0 + 0, ... - that is, the sequence a, d, d, d, . ... Consequently,
[a + (d - a)xJ/(1 -- X)2 generates the sequence a, a + d, a + d d, a + d + d d, .. . - that
is, the sequence a, a + d, a + 2d, a 3d, . . .. [Note: Part (c) of Exercise 1 for Section 9.5 is
a special case of this result: Let a = 1, d = 3.J
(b) Here we need the coefficient of x n - 1 in (1/(1 - x»[a + (d - a)x}/(l - X)2 = La +
(d - a)xJJ{l - x)3
[a + (d - a)xJ(l - xta. This coefficient is aC:::.31)(-1)n-l + (d-
=
a) (n-:.32) (-1 )n-2 = a( -1 )n-l e+(:=~)-l)( -1 )n-1+( d-'a)( -1 )n-2 (3+(:=;>-1)( -1 )1&-2 = a(:~D
(d--a)(n~2) = a(V(n l)(n)+(d-a)(i)(n)(n-l) = a(!)(n)[(n+l)-(n-l)]+d(i)(n)(n1) == na
witb the result
reader may
Chapter 4.1
1=
k(
1
<4
(b) Pr(X = = (3/4)(1/4)3 = 3/256
Pr(X ::; 3) = E!",o P,'(X =
= (3/4)[1 (1/4) (1/4)2 (1/4fl = (3/4)(85/64) =
255/256
Pr(X > 3) = 1 - Pr(X 3) = 1- (255/256) =::
[Alternately,
> 3) = Pr(X ~
Pr(X = x) = (3/4)[(1/4)4 + (1/4)5
(1/4)6 + ...J
= (3/4)(1/4)4[1 + (1/4) (1/4)2 .. oJ = (3/4)(1/4)4[1_<;/4)1 = (1/4)4 = 1/256.]
Pr(X ~ 2) E:'2 Pr(X x) = E:: 2 (3/4)(1!4Y
= (3/4)(1/4)2[1 (1/4)
.,.J = (3/64)[1_(;/4)] = (1/4)2 = 1/
(c) For n E Z+, Pr(X ?::
= E:"n Pr(X = x) = E:='I1.(3!4)(1!4)X =
(3/4)(1/4)'11. E:o(1/4)i = (1/4)".
Pr(X > 4
X> 2)
Consequerltly, Pr(x ?::
?:: 2) =
P~'(X ~ 2) = Pr(X ?:: 4)/ Pr(X ?:: 2) =
=
=
(1/4)4/(1/4)2 = (1/4)2, Likewise Pr(X ?:: 104!X ?:: 102) = (1/4)2,
17.
For Ie E Z+, Ie fixed, we find that Pr(Y?:: Ie) =:: E~kqg-lp (where q = 1- p)
= qk-l p qh) + qkHp + ... = qk-lp[l + q + q2 + ...]
= qk-lp~ = qk-l p (!) = (/-1. Consequently, Prey ?:: mlY ;?: n) = Prey ;?: m and Y ~
n)/Pr(Y?:: n) = PreY ~ m)/Pr(Y ;?: n) = qm-l/qn-l = qm-n. [This property is the
reason why a geometric random variable is said to
memorgles.s. In fad, the geometric
random variable is the only discrete random variable with this property.]
18.
(n,) The car travels the first mile in one hour, the second mile in 1/2 hour, the third mile in
1/4 [= (1/2)2] hour, and the fourth mile in 1/8 (1/2)3] hour. Consequently, the average
velocity fox the first four miles is 4/[1+(1/2)+(1/2)2+(1/2)3] = 4/[[1-(1/2),'11[1-(1/2)]] =
4/[2(15/16)] = 32/15 = 2121) miles per hour.
(b) The average velocity for the first n miles is n/[1 + (1/2) + (1/2)2 + ... + (1/2)n-l] =
»/[[1 - (1/2)n]f[1 - (1/2)]] = n/[2[(2n - 1)/2nlJ = n(2n - 1 )/[2n - 1] miles per hour.
(c) For n =
the average velocity is 4980736/524287 ...:.. 9.500018120 miles per hour. For
n = 20
average velocity is 2097152/209715 ...:.. 1O.OOn009M miles per hour. Hence the
smallest value n
which the
for
first n nriles
10
per
hour is n 20.
=
CHAPTER 10
RECURRENCE ...
LJU ........
an = 5an- I , n :2: 1, ao = 2
(b) an = -3a,.._1l n :2:
ao = 6
2.
(a)
= 1.5a,0 an = (1.5)nao, n
(b) 4an = 5a,.-1, an = (1.25)nao, n > O.
(c) 3an +! = 4a'H 3al = 15 = 4ao1 ao = 15/4, so
an = (4/3)nao = (4/3)n(15/4) = 5(4/3)n-1, n:2: O.
(d)
an = (3/2)a n - l ' an = (3/2)nao1 81 = a4 = (3/2)4 ao so
(16)(3/2)\ n :2: O.
aD
= 16 and an =
3.
an +l - dan = 0, n 2: 0, so an = dnao. 153/49 = a3 = d3a01 1377/2401 = as = d5ao ::::::::::}
a5/a3 = rP = 9/49 and d = ±3/7.
4.
art+! an 2.5an, n 2: O.
an = (3.5)nao = (3.5)"(1000). For n = 12, an
5.
Pn = 100(1 0.015)n, Po = 100
200 = 100(1.015)''' ==> 2 = (1.015)'"
(1.015)46 . .:. . 1.9835 and (1.015)47 . .:. . 2.0133.
Henc,e Laura must wait (47)(3) = 141 months for he:r money to double.
6.
Pn = Po(1.02)n
7218.27 =
=
19+
9 8
1
SO
= (3.5)12(1000) . .:. . 3,379,220,508.
= (7218.27)(1
= 145
= $2200.00
243
(b) The input for the following procedure is an array A of n real numbers. The output
is the
array
with
$;
$; ... $;
P:mcedure BuhhleSort2( var
array; n: integer)i
Val'
boolean;
{The value of Switch is true if}
{an interchange actually takes plac,e.}
ij: integer;
Begin
Switch := true;
While 8wi teh do
Begin
Switch := false;
For i := 1 to n-I do
For j := n downto
1 do
If Am < AU-l} then
Begin
temp ;= AU-I];
AU-I] := Am;
Am := tempi
Switch := true
End {if}
End {while}
End. {procedure}
(c) The best case occurs when the array A is already in nondecreasing order. When
this happens the procedure is only processed for
1 and j:= n down to 2. This results
n- 1
80 the
complexity
O(n).
The worst case occurs when a Switch
made for aU i := 1 to n - 1. This results
so the
18
ill
(c ) The value of
1 or
IS
(d) Let Pl,]>z,P3, •.. , he an
permutation of 1,2,3, ... , n. Then PI is either 1 or
n. P1 = 1, then P2 - 1, P3 - 1, ... ,Pn, - 1 is an
permutation of 2,3, ... , n - 1.
For PI = n we find that
... , is an orderly permutation of 1, 2, 3, ... , n - 1. Since
these two ca....es are exhaustive
ha'lt'e nothing in common we may write
an = 2an-l, n;? 3, a2 = 2.
Hence, aa = 2a2 = 2 . 2 = 2 2 ,
a4 = 2a3 = 2 . 22 =
and, in general,
an = 2ft - I ,
n?:
Section 10.2
(a) an = 5an -l + 6an -:h n;? 2, ao = 1, al = 3.
Let an = ern 1 e, 'I' =f. O. Then the characteristic equation is '1'2 - 51' - 6 = 0 = (r - 6)(1'
so r = -1,6 are the characteristic roots.
an = A( _1)'11, B(6Y'
1),
l=ao=A+B
3 = al =
+ 6B, so B = 4/7 and A = 3/7.
an = (3/7)( _l)n + (4/7)(6)\ n ?: O.
(b) an = 4(1/2)n - 2(5)n 1 n?: O.
(c) a n +2+ a n=O,n;?O,ao=O, (11=3.
With an = ern 1 e, r f= 0, the characteristic equation r2 + 1 = 0 yields the characteristic
roots ±i. Hence an = A(i)n+B(-i)n = A(oos(-;r/2) + isinCrr/2)Y* + B(cos(1r/2) +
isin(--1r/2)Y' == Ccos(n-;r/2) + Dsin(n-;r/2).
0::::::: ao::::::: C, 3 == al = Dsin(n/2) = D, so an = 3sin(n7r!2), n?: O.
Let
- 61' + 9 = () =
- 3)2, so the
J2[(-1/J2) + B(l/v'2)], so 3 = -1 + B, B = 4
an = (J2yt[cos(31in/4) 4
n 2: 0
2.
Example 10.14: an =
= 0, r = '5
an = A(5 6V6t + B(5 - 6V6Y~
al = 100:::::: 10a1 + 29ao =
29ao, so ao = 0
0= (to = A + B, so
= -A.
an = A.[(5 6V6)n - (5 - 6J6)n}
10 = a1 = A[5 6V6 - 5 + 6V6] = 12v'6A, A = 5/6V6.
an = (5/6v'6)[(5 6\116)10 - (5 - 6V6)"'j, n 2: o.
Example 10.23: an = cl(2n ) + c2n(2n), n 2: 0, ao = 1, a1 = 3.
aQ = 1 = Cli a1 = 3 = 2 + ca(2), C2 = 1/2.
an = (2n)[1 + (n/2)], n 2: O.
Example 10.16: an = a n -1 + (tn-2, n 2: 2, ao = 1, a1 = 2.
1 = 0, r = (1 ";5)/2.
ao::::::l=A+B
a1 = 2 = Al(l + .;5)/2]
B[(l - V5)/2]
4= A(1+v'5)+B(1-V5):::::: (A+B)+v'5(A-B) = l+V5(A-B), so 3 = V5(A-B)
(b)
1'2 -
r -
and A - B = 3/../5.
2A = (A B) + (A - B) = 1 + 3/../5 = (3 + V5)/../5, A = (3 + -15)/2-../5; B = 1 - A =
(V5 - 3)/2../5,
an = [( v'5 3)/2V5]{(l V5)/2]n + [{../5 - 3)/2../5][(1 - \I"5)/2Jn, n ~ 0
3.
= 0): 62
cao = 0 = 4 + b(l)
c(O), so b = -4.
(n=!): aa-4a2+cal=0=37·-4(4) C,so c=-21.
an +2 - 4a,.+1 - 21an = 0
r2 - 21 = 0 ::;;; (r - 7)(1' 3), r = 7,-3
an = A(1Y~ + B( _3)n
(n
bal
O=aQ=
~B=
1 = al 1A-3B = lOA, so A = 1/10, B = --1/10 and an = (1/lO)[(7Y'
=
an
=
-3)10], n ~ O.
an'~l
1 = 0, r =
a,. := A«1 ..;g)/2Y~ +
ao a]. 1 ~
-~ " -
= =
aw,=
n 8uac:es
further, there are G n -2 ways to fill the n spaces when a compact car occupies positions
n - 1 and n. These two C8.-'<eS are exhaustive and have nothing in common, 80
Let an = ern, c ::/:: 0, r
and an = Cl(1 + V2)n
Upon substitution we have 1'2 - 1 = 0, so l' = 1
c2(1 - J2)n, n ~ O. From 1 = ao = Cl
C2
2 = al Cl(1 J2) c2(1 - V2), we have Cl =
and C2 =
lln = v'2 + 2)/4)(1 +
"Ii)n «2 - .;2)/4)(1 - V2)n = (1/2\1'2)[(1 + Vi)nH - - Vi)nH], n ~ O.
(b) Here 1'10 = 1 and I'll = 1. For n ~ 2, consider the nth space. This space can
occupied
by a motorcycle
one way and accounts for a n -1 of the an ways to fill the n spaces. H
a compact ear occupies the (n - l)st and nth spaces, then we have the remaining 3an -2
ways to :fill n spaces. So here an = tin-l 3a n-2, n :::::: 2, ao = 1, a1 = 1.
a.¥
«
Let an = ern, C =1= 0, r =1= O. Upon substitution we have
- r - 3 = 0, so r =
(1 vi:f3)/2, and an = cI[(l V13)/2]n + c2[(1 - v13)/2]n I n 2:: 0, From 1 = Go = Cl + C2
and 1 = a1 = cI[(l v'i3)/21 c2[(1 - v'i3)/2j, we find that C1 = [(1 .Ji3)/2V13] and
C2 = [(-1 + 03)/203]. So an = (l/vU){(l + vU)/2]nH - (1/~)[(1- Ji3)/2]n+l,
n:::::: O.
(c) Comparable to parts (a) and (b), here we have an = 2a n - l
3a n -2, n ;:::: 2, ao = 1,
al = 2. Substituting an = cr,n, c I- 0, r I- 0, into the recurrence relation, we find that
1'2-21'-3 = 0 so (1'-3)(1' 1) = 0 and r = 3, r = -1. Consequently, an = cl(3 n )+C2( -1)'\
n?: O. From 1 = ao = Cl + C;z and 2 = al = 3Cl - C2, we learn thai Cl = 3/4 and C2 = 1/4.
Therefore, an = (3/4)(3 n ) + (1/4)( _l)n, n ~ 0.
6.
For all three parts, let b'H n
0, count the number of ways to :fill the n spaces under the
condition(s) specified _. including the condition allowing empty spaces.
(a) Here bo =
hI = 3, and hI!, = 3bn _ 1 bn - 2 , n > 2. This recurrence relation leads us to
the characteristic equation 1,2 - 3r' - 1 = 0, and the eharacteristic roots l' = (3 ..ff3)/2.
Consequently, bn = c:tf(3 v"13)/2]fl ca[(3 -- M)/2]n, n
O. From 1 = be = Cl C2
3 = hI = cl[(3 .jl3)/2] c2[(3 - v13)/21, we find that Cl = (3 Vi3)/2.;i3 and
C2 == (-3
03)/2Ji3. So bn = (1/v'I3)[(3 Ji3)/2]nH - (1/.;13)[(3 - v1.3)/2]n+l,
n;:::: O.
7.
(a)
Fl
-
F3
Fs
-
F'J, - Fo
F4- F'J
F2n _. 1 -
F'ln-
Conjecture:
n E Z+, Fl
+
... + F2n - 1 =
- Fo = F zn •
Proof: (By the Principle of Ma,thematical Induction).
For n = 1 we have Fl =
and this is
since Fl = 1 = Fa. Consequently, the result is
true in tIris first case (aud this establishes the basis step for the proof).
Next we assume the result true for n = k
1) - that is, we assume
find that
Whenn=k+l we
Fl
F3 + Fi) + ... + F2k-l + F2(kH)-1 =
(Fl F3 + Fs
F2k- 1 ) + F2kH = F2k + FU:+1 = F2k+2 = F2(kH)'
Therefore the truth for 11. = k implies the truth at n = k + 1, so by the Principle of
Mathematical Induction it follows that for all n E Z+
(b)
F2
F4
F6
- F3- Fl
-
Fs-F3
F.,-Fs
- F
F2n- 1
··,+F'},'fi. = Fo +F2
2nH -
ConJecture: For
-1.
nE
+
F4
n = k
1
) -1 =
by the Principle of Mathematical Induction.
8.
.
~ _ .
(l/V5)Hi!±V5l!J)n+l_((1_{§1I 2 ),,+11
(a) hmn-+oo F" - lim n..... oo (1/-fi)[((HJ5)/2)"'-W-...!5)/2)n]
- r n oo [(Hv'sU2)"'+1-«1-~)f2)nHl
- uu .....
[((H.,;5)/2)"-((1--.ft)/2)"j
'l
01,,+1_13»+1 ( h
l±.i! a l-:I§)~
= lmn .....oo an-f3"
w ere a = 2 ' P = 2
Since IPI < 1 and lal > 1, it fonows that I~I < 1
=a=
(0)
(i)
ACjAX =
AXCI
ACX = sinl08°/sin36° = 2
36°cos36"/sin36° =
2 cos 36°
(ii) cos 18° =
72° = 2 sin 36° cos 36° =
2(2 sin 18° C{)S 18°)(1 - 2 sin2 18°) ==> 1 =
4 sin 18"(1 - 2 sin2 18°) = 4sin 18" - 8sin3 18°.
0= 8 sin3 18° - 4 sin 18° + 1, so sin 18° is a fOQt of 8x 3 - 4x
8x3 - 4x + 1 = (2x -1)(4x 2 2.1: -1) = O.
The roots of 4x 2 + 2x -1 = 0 are (-1 ± V5)/4.
Since 0 < sin 18° < sin30° = 1/2, sin 18" = (-1 + V5)/4,
(c) (1/2)(AC/AX) = 00836" = 1 - 2sin 2 18" = 1 - 2[( -1
ACIAX = 2(1 ..;5)/4 = (1 + .../5)/2.
9.
,
an = an-l
1 = O.
y'5)/4J2 = (1 + V5)/4.
a n -2, n 2:: 0, au ;:::;; 0.1 = 1
~
(Append '+1')
(Append '+2')
an = A[(1 y'5)/2J1t B[(1 - v!5)/2Y'
1 = ao = A+.8; 1 = a1 = A(l + J5)/2
2 = (A + B)
1=A +
(1/../5){(1
B(l- ../5)/2 or
v'5(A - B) = 1 + v'5(A - B) and
- B = 1/..;5.
l/-If> = A-===? A =
-/5)/2-./5, B = hIS - 1)/2v'5 and an =
-I5)/2)nH - ((1 - \1"5)/2)1$+1], n >
n 2:: 3, we t~a.n
an-I strings of .."",u."-",,,,,,
n - 2. These two cases have nothing in common and cover
possibilities, so
We
that an =
= (an +2 -(3'lt+2)j(a-fJ) where (Jt =
../5)/2 and f3 = -../5)/2.
b)
hI :::.:: 1 since 0 is the only string of length 1 that satisfies both conditions.
n =
there are
strings: 00, 10, and 01 - so b2 = 3. For n
3, consider
bit in
the nth position of such a binary string length n.
(1) If the nth bit is a 0, then there are a"-l possibilities for the remaining n - 1 bits.
(2)
the
bit is a 1, then
(n - 1)st and 1st bits are 0,
so
are an -3
possibilities for
remaining n - 3 bits.
, from part (a).
The characteristic equation x 2 - x - 1 = 0 has characteristic roots (Jt = (1 V5)/2 and
j3 :::.:: (1 - ../5)/2, so bn = Clan + c2j3n. From 1 == b1 = CIa c'J,j3 and S = b2 = cIa2 + C2j32
we learn that Cl = C2 = 1. Hence bn = an + j3n = L n ) the nth Lucas number. [Recall that
in Example 4.20 we showed that Ln = Fn+1 + F n - 1 .)
12.
Let an count the number of ways to arrange n such chips with no consecutive blue chips.
Let On equal the number of arrangements counted in an that end in blue; C n = an - bn .
3a
Then an +1 = 3bn + 4c n = 3(bn + cn ) + Cn = 3an
n -I.
Hence aYi +1 - San - 3an _1 = 0, n
1, ao = 1, al = 4. This recurrence relation has
characteristic roots r = (3 .;21)/2 and an, = A«3 + .;21)/2)"" B«3 - ffi)/2)n.
ao = 1, a1 == 4 =:;.. A = (5 + v'2I)/2v'zl, B = (v'zl- 5)/2-121 and
an = [(5
13.
vI2I)/(2v'2I)Jl(3
,J2i)/2]n - [(5 - v21)/(2v'2T)J{(3 - v'2T)/2jn, n ~ O.
For n ~ 0, let an count the number of W01'ds of length n in E* where there are no consecutive
alphabetic characters. Let a~)
those words that end
a numeric character, while
a!?) couuts those that end with an alphabetic character.
an = a~l} a~2).
For
n;::: 1,
=2
1 = ao =? 1 11 = al => 11 -
-
so A = (9
B,
and
+ 4V21 + B[2 - 4V2]
A[2 + 4V2]
- A)[2 - 4V2J
[2 - 4V2] + A[2 4V2 - 2 + 4V2]
[2 - 4v'21
8V2A,
4..;2)/{SV2) = (8 + 9V2)/16, and B = 1 - A =
- 9..;2)/16.
Consequently,
14.
Using the ideas developed in the prior exercise we find that 7k = 63, or k = 9.
15.
Here we find that
ao = 1,
and, in general, an = 2F "" where
is the nth Fibonacci number for n ;::: O.
16.
= 1. For n;::: 4, let n = Xl X2 + ... + Xt, where Xi;::: 2 for 1 SiS t,
and 1 S t s In/2J. If Xl = 2, then X2 ••• + Xt is counted in a n -2- If Xl 1= 2, then
Xl > 2 and (Xl - 1) + X2 + .. Xt is counted in an-I- Hence an = a"-1 + a n -2, n ~ 3,
and an = F,~-b the (n - l)-st Fibonacci number.
11.
(a) From the previous exercise the munber of compositions of n 3 with no Is as summands
is Fn +2 •
(b) (i) The number that start with 2 is the number of compositions of n + 1 with no 18 as
summands. This is Fn.
(ii)
(iii) The number that start with k, for 2 S k S n 1, is the number of compositions of
+ -k
-_.
2 ksn
If
a1 = 0, 0,2 = 1, a3
)
- x-I
(1
= 0, so x = (1
-
and
1, or
19.
1 + ~ = x we learn that x 1 = , or x 2 -x-l = O.
x = (1 ..;5)/2 and the points
-15)/2) =
a) and «1-V5)/2, (1-V5)/2) = (/3,/3).
ofintersection are «1 V5)/2,
20.
(a) a 2 = [(1+v'5)/2J2 = {1+2.;5+5)!4 =
= (3+.;5)/2 = [(l+v5)/2]+(2!2) =
a+1.
(By Mathematical
n = 1, we
= a 1 = a = a· 1 0 =
aFt + Fo = aFn + Fn - 1l so the result is true in this case. This f'...stablishes the basis step.
Now we assume for an arbitra;ry (but fixed) positive
k that (i~ = aFir;
. This
is our inductive step. Considering n = k + at this time, we find
-
a( a k ) = a(aFk Fk-l] (by the inductive step)
a 2F]k aFk-l
(a + l)Fk + aFk _ 1 [by part (a)]
a(Fk + Fh- I ) + Fi:
aF.lc+l + Fl;.
Since the given result is true for n = 1 and the truth for n = k 1 follows from that for
n = k, it follows by the Principle of Mathematical Induction that an = aFn + F:.-l for all
n E Z+.
21.
Proof (By the Alternative Form of the Principle of Mathematical Induction):
(a) F3 = 2 = (1 + -./9)/2 > (1 + ..(5)/2 = 0: = a 3 - \
F." = 3 = (3 + ..;9)/2 > (3 -./5)/2 = a 2 =: 0:4 - 2 ,
so the result is true for these first two cases (where n = 4). This establishes the basis
step. Assuming the truth of the statement for n = 3,4,5, ... ,k(~ 4), where k is a fixed
(hut arbitrary) integer, we continue now with n = k + 1:
Fk+l
--
Fit
Fk-l
>
a k- 2
+ a(k-l).-2
_
ak- 2
a k- 3 =
. a 2 = a k- 1 =
all n ~ 3 - by the Alternative Fm:m of the Principle of
n == 3,
1)
basis
22.
n is counted by
= d(2n), n > 1 and b:t = 1 :::;.. 2d = 1 ::::} d = 1/2,
= 2bn1 n 1,
so
. Hence, for n
21(nH)/21-1 = 2(1'6-1)/2 = 2ln/2J.
234
= 2 ::::} c = 1,
(a) Since an+! = 2an we have an = c(2)>), n ~ l.
SO an = 2ft. Consequently,
n even, the number of Da.Jlm.::l.rolm(~
n 2j
n 2
Gn /2 = 2 / = 2l / •
the number of palindromes of n is counted by
-
Here we shall use
For n 2:: 1, let a~) count the number of ternary
of length n where there are uo consecutive 1s and no consecutive 28 and the nth
symbol is O. We define a~) and a~2) analogously. Then
I.ln
+ ail) + a~2)
-
a~O)
-
an-l
-
2an - l [a n - l - a~~l - a~21J
(0)
2an -1 + an -1 = 2an -1 + an -2
(l)]
[a n ...... 1 -- a n
-l
+ [a n "".. 1 - a n(2)-l ]
Letting an = cr'''', c 0, r -=f 0, we find that 1'2 - 2r _.- 1 = 0, so the characteristic roots
are 1 -J2. Consequently, an = Cl (1 \1'2)10 c2(1 - \1'2)". Here al = 3, for the three
one-symbol ternary strings 0, 1, and 2. Since we cannot use the two-symbol ternary strings
11 and 22, we have a2 = 32 - 2 = 7. Extending the recurrence relation so that we can use
n 0, we have a2 = 2a1 + ao so ao = a2 - 2a1 = 7 - 2 . 3 1. With
=
=
=
+ C:h and
= Cl(1 + -/2) + c2(1 --/2)
= (Cl C2) + ¥'2(Cl - (2),
we now have 1 = C1 + C2 and -/2 = Cl - CZ so C1 = (1 + J2)/2 and C2 = (1 - ../2)/2.
1 ao
3 = at
-
Cl
l
Consequently,
24.
an = (1/2)(1 + v'2t+ 1 + (1/2)(1 - v2)n+l,
Here a1 = 1, for the case of Olle
(ii)
horizontal dominoes; ox
0.2
n:2:: o.
= 3 --- use (i) one square tile ; or
For n :2:: 3
the nth
nth
cases account for all
<.~f'A3 have auythin.g
common so
x1=
as the
+
c2(2) and 3 = a2 = Ct( -1)2 + c2(2)2 we learn that Cx = 1/3, Cll = 2/3. So an = (1!3)[2 n +1
n :2:: 1. [The
1, 5,11,21, .. "
113
as
sequen<-"e. ]
25.
an count
of ways one can tile a 2 x n
using
2
2
dominoes and square tiles. Here ell = 4, a2 = 4 + 4 + 5 = 37, and, for n
3, an =
2
4an -l + 16an _2
= 4an -l +21an _2' The characteristic equation is x -4x-21 = 0 and
this gives x = 7, x
-3 as
characteristic roots. Consequently, a,;. = c1(7)'I1. + ca( -3)"",
n:2::1.
=
Here Cl(l = (1/21)( a2 - 4(1) = 1 can be introduced to simplify the calculations for ell C2'
From 1 = ao = C1 + Ca and 4 =
"- 3C2 we learn that Cl = 7/10, Ca = 3/10, so
an = (7/10)(7)n + (3/10)( _3)71,) n ?::
When n = 10 we find that the 2 x 10 chessboard can be tiled in (7/10)(7)10 +(3/10)( -3)10 =
197,750,389 ways.
26.
Here al = 1 (for the string 0) and a2 = 3 (for the strings 00, 01 and 11). For n ?:: 3, there
are three cases to consider:
(1) The nth symbol is 0: There are an-l such strings.
(2) The (n - 1)st and nth symbols are 0, respectively: There are an -2 such strings.
(3) The (n - l)st and nth symbols are.both 1: Here there are also a»-2 strings.
These three cases include all possibilities and no two cases have anything in common.
Consequently,
The characteristic equation, 1'2 - r - 2 = 0, yields the characte:ristic roots 2 and -1, so
an, = cl(2)'n
ca( _l)n. From 1 = al = 2Cl - C2 and 3 = a2 = 4Cl + Ca, we learn that
Cl = 2/3 and C2 = 1/3. So
an = (2/3)(2)71,
[So here we find
(1/3)(-1)\ n?: 1.
occurrence of the Jacobsthal numbers.]
= 2, as = 5.
C08(2;)
Z
we have
1 = 0.1 = 2C1 - C2/2 cs( .;3/2)
2 = 0.2 = 4Cl - C2/2 - es( .;3/2)
we
5 = 0.3 = 8Cl + C2,
that Cl = 4/7, C2 =
an = (4/7)(2y>
and C3 = ...(3/21, so
(3/7) cos(2mr /3)
(J3/21) sin(2mi/3), n
1.
[Note that an also counts the number of ways oue can tile a 1 x n chessboard using 1 x 1
square tiles of one color, 1 x 2 rectangular tiles of one (,-<lior, and 1 x 3 rectangular tiles
that come in two colors.]
28.
Here 0.1 = 1 (for 0), 0.2 = 2 (for 00,01), as = 4 (for 000, O~, 010, 011), 0.4 :::::: 9 (for 0000,
0001,0010,0100, 0011, 0110, 0111, 1111,0101), and for n ~ 5
The characteristic equation r.4 - '1'3 _,..2 - r - 2 = 0 tells us that (1' - 2)(1' + 1)(r2 + 1) = 0,
so the characteristic roots are 2, -1, ±i. Consequently,
From
1 = 0.1 = 2C l
-C2
+C4
2 ::::: 0.2 = 4C1
+C2
-C3
4 ::::: as = 8Cl
-C2
-C4
9 = 0.4 = 16cl
+C3
we
that C1 = 8/15, C2 = 1/6, C3 = 3/10,
(1/6)( _1)n + (3/10) cos(mr /2) + (1/10) sin(nu/2), n
1X 1
x,. = ,
-1), so Xf~ :::::
xo=l=A B
Xl = 5=
4: = ..4,
255
C4 ;:
1.
1/10, so an = (8/15)(2Y' +
30.
Expanding by row 1, Dn:::::: 2Dn- 1 - D, where D is an (n -1) by (n -1) determinant
whose V"cUue, upou expansion by its first column, is D n - 2 - Hence Dn = 2Dn- 1 - D n- 2 •
This recurrence relation determines the characteristic roots r = 1, 1 so the value of
Dn = A(l)" Bn(l)'fI = A + Bn.
!11 2 I= 4 - 1 = 3
D2 = 2 1
2 = Dl = A
Bi 3 = D2 =
+
bn = a~, bo = 16, ill =
This yields the linear relation bn+2 - 5bnH
r = 4,
80 bn = A(l)'"
B(4)n.
=1
4010 = 0 with characteristic roots
bo = 16, bt = 169 ==> A = -35, B = 51
J51(4)n - 35, n > O.
and
On = 51(4)n - 35. Hence a n -
32.
c2(7)n, n 2:: 0, is the solution of a"+2 + ban+! + can = 0, so ,.2 + br c = 0 is
the characteristic equation and (r - 1)( r - 7) = (,.2 - 8r 7) = r2 + br + c. Consequently,
b = -8 and c = 7.
33.
Since gcd(FhFo) = 1 = gcd(F"J"Fx), consider n 2:: 2. Then
Fs=F2+FI 1)
F4=F3 F2
aT' = Cl
F5 =F4+ 1i3
FnH = F.,. + Fn- 1 •
Reversing the order of these equations we have the steps in the Euclidean Algorithm for
computing the gcd of .Fn+l and Fn ) for n ? 2. Since the last nonzero remainder is FI =
it follows that gcd(Fn+h Fn) = 1 for all n ? 2.
34.
output);
to
app.!'opriat.e. ')
= 0 then
U<4"U'-""''''
(,Your ~LUl!.U'-""'L 18
if number = 1 then
Wdteln ('Your ~"U.Jl"""J"''' 18
. . . UO" .....
..,'L'..
')
>
Begin
Fibonacci [1] := 1;
Fibonacci [2} ;;;.;;: 1;
:= 1;
i := 3;
While number> current do
Begin
Fibonacci [i] := Fibonacci [i-I] + Fibonacci [i-2]j
current := Fibonacci [i];
If number < current then
Writeln (,Your number is not a Fibonacci number.')
Else if number = current then
Writeln ('Your number is the " i:O, '-th Fibonacci number.')
1 {number > count}
Else i : = i
End {while}
End {else}
End.
Section 10.3
(a) an +l - an = 2»
al
ao 0 3.
ila = ltl
2
=
3, n 2:
ao = 1
=1
-
, n>
,n
0
ao = 1
=1
an = +
2
n
(d) a n =2 +
+
), n ~ O.
't'2 •
= a""
+lr~, n;:::O, ao=O.
an+! - an = (n
lY' = n 2
aUt) = A. a ip ) = Bn
,n
f!,
2n + 1
Cn 2
B(n 1) C(n
+
1)3 =
Cn 2 + Dn 3
2n I::::.=:>
3
2
Bn B Cn'}, 2Cn C + Dn
3Dn
3Dn + D == Bn + On? Dn3 + n 2 + 2n + l.
By comparing coefficients on like powers of n we find that C + 3D = C + 1, so D = 1/3.
Also B 2C + 3D = B +2, so C = 1/2. Finally, B + C D = 1 ==> B = 1/6.
So an = A (1/6)n + (1/2)n 2 (1/3)n 3 • 'Vith ao = 0, it follows that A = 0 and
an = (1/6)(n)[1 3n + 2n2] = (1/6)(n)(n 1)(2n + 1), n > O.
3.
(a) Let an = the number of regions detennined by the n lines under the conditions
specified. vVhen the n-th line is drawn there are n - 1 points of intersection and n
segments are formed on the line. Each of these segments divides a region into two regions
and this increases the number of previously existing regions, namely an-I, by n.
an = an-l + n, n ;::: 1, ao = 1.
a(h) = A, a(p) = Bn
Cn 2
n
n
En + Cn? = B(n -1) + C(n _1)2 + n
Bn Cn 2 - Bn + B - Cn'}. + 2Cn - C = n.
By comparing the coefficients on like powers of n we have B = C = 1/2 and an =
A + (1/2)n (1/2)n 2 •
1 = ao = A so an:::;;; 1 (1/2)(n)(n + 1), n ~ O.
(b) Let On = the number of infinite regions that result for n such lines. When the nih
line is drawn it is divided iuto n segments. The first and nth segments each create a new
infinite region. Hence bn =
+2, n ;::: 2, bI = 2. The solution of this recurrence relation
bo = 1.
is On = 2n, rt
4.
n
started.
-1.005ptl. =
C - 1.0050 == 200 =:::::} C =
000
- 40,000:::;;; 1000, so A =
000
000 = $11,
1
5.
= (1.005)P47 = $11,890.05
(a) an+2 + 3an H
2an = 3n, n ;?: 0, ao = 0, a1 = 1.
With an = ern, c,r::/= 0,
characteristic equation '1'2 3'1' 2 = 0 = (1' + 2)(r + 1)
yields
r =
Hence a~h) =
1)" + B( _2)1\ while a!i) = C(3)"'.
C(3)91,H 30(3)tt+1 20(3)'10 = 3'10 :=:.=:} 90 + 90 + 20 = 1 :=:.=:} C = 1/20.
an = A( _1)" + B( _2)n + (1/20)(3)"
0= Go = A B + (1/20)
,
1 = a1 = -A - 2B (3/20)
1 = ao al =
and B = -4/5. Then A =
an = (3/4)( _1)14 + (-4/5)( _2)71. (1/20)(3Y', n ~ 0
(b) an = (2/9)( -2Y' - (5/6)(n)( _2)'10 + (7/9), n ~ 0
6.
6a n H + 9a n = 3(2)n
a~h) = A(3)'" + Bn(3)'"
anH -
- (1/20) =
7(3)n, n ;?: 0, ao = 1, a1 = 4.
a!!')
= 0(2)'10 + Dn2(3)n.
Substitutillg a!r) into the given recurrence relation, by comparison of coefficients we find
that C = 3, D = 7/18.
an = A(3)'" + Bn(a)n + 3(2Y~ + (7/18)n 2 (3)n
1 = 110,4 = a1 ==} A = -l,B = 17/18, so
an = ( -2)(3)'· + (17/1S)n(3)n (7/1S)n2(3)n + 3(2Y', n > O.
1.
1'3 - 31'2 + 31' - 1 = 0 = (r - 1)3, so r = 1,1,1 and
3 + En4.
A Bn + Cn 2 , a(p)
=
Dn
n
D(n+3)3+E(n+3Y4 - 3D(n+2)3 -3E(n +2)4 +3D(n 1)3 +3E(n 1)4 -Dn 3 - En4 =
3+5n ~ D = -3/4, E = 5/24.
an = A + Bn + Cn 2 - (3/4)n 3 + (5/24)n 4 , n ~ O.
Here the characteristic equation is
a(k) =
n
8.
anH = 3a n
that end in
3n , 110 = 1, a1 = 4. The term an accounts for the sequences of length n
3an accounts for those sequences of length n that end in 0, or 2.
a(h)
'10 == Aan •, a(~)
'10 =
B(n + 1)3""+1 = 3(Bn3n )
1 = Go = A, so a~t =
+
3t1.:=:.=:} 3B(n
1) = 3Bn + 1 => 3B = 1 ===> B = 1/3
, n ;?:
9.
-(
10.
=6
7
-7] -
a(p) = n n
1)-
+
(a) Let a! = o,.,n ~ 0
bn +2 -- 50n +1 + 6b,. = 7n
b~) = A(sn) + B{2n ), of;> = Cn + D .
C(n 2)
-5[C(n+1)+DJ+6(Cn+
=7n=>
D = 21/4
1>,. = A(3") B(2n) (7n/2) (21/4)
-- a02 -- 1, b1 -_...,2
- 1
'~1 1 = bo = A + B + 21/4
1 = hI = SA 2B+ 7/2 21/4
3A 2B = -31/3
2A 2B = -34/4
A = 3/4, B =-5
an = [(3/4)(3)'" - 5(2Y' (7n/2) + (21/4)1112, n ~ 0
(b) Q,! - 2an - l = 0, n ~ 1, ao = 2
a! = 2an -1
log2 a! = log2(2an -d = log:! 2 log2 an - l
21og2 an = 1 + log2 a n -1
Let bn = log2 an'
The solution of the recurrence relation 2bn = 1 + bn - 1 is On = A(1/2)n + 1.
bo = loga ao = log22 = 1, so 1 = bo = A 1 and A = O.
Consequently, bn = 1, n ~ 0, and an = 2, n 2 o.
12.
Consider the nth symbol for the strings counted by an' For n ~ 2, we consider two cases:
(1) If this symbol is 0, 2, or 3, then the preceding n -1 symbols provide a string of length
n - 1 counted by Gin-t.
(2) If this symbol is 1, then the preceding n - 1 symbols contain an even number of Isthel'e are 4n - 1 - a n -1 such strings of length n - 1.
two cases are exhaustive and have
in common we
n
2~
)=
al = 1 = c(2)
c=
1.
We can
Here an is
n
13.
(a) Consider the 2n binary strings of length n. Half of these strings (2n-l) end in 0 and
other half
) in 1. For the
bimuy
length
are
runs, When we append 0 to each of these strillgs we get t n - 1 + (l)(2 n - 1 ) runs, where the
additional ( )(210 - 1 ) runs
when we
0 to the )(2,,-1) strings of length (n - 1)
that
in 1. Upon appending 1 to
of
211.-1 binary strings of length n -1, we get
the remaining t n - 1 (V(2 n - 1 ) runs. Consequently we find that
(t
!
so t!!) = An(2n). Substituting t!f) into the recurrence relation we have
Here t{k)
n --
An(2n)
_
_
2A(n _1)2n - 1 + 2n - 1
An(2n) - A(2U) + 2n - 1
By comparison of coefficients for 211. and n2n we learn that A = ~. Consequently, in =
n
t!!) = c(2n) n(2n- 1), and 2 = tl = c(2) + 1 => c =
80 tn = C!)(2 )
n(2n-l) =
(n + 1)(2)>-1), n ~ 1.
(b) Here there are 4n quaternary strings of length 11, and 411,-1 of these end in each of the
one symbol suffices 0,1,2, and 3. In this case
!,
tih)
in = 4[t n_ 1
(~)4n-l] = 4t n- 1 + 3(4 ,-1), n?: 2, t1 = 4.
11
Comparable to the solution
part (a), here t~h) = c(4n) and t!f) = An(4n). So An4fl. =
4A(n - 1)411.-1 + (3)(41'1.-1) = An4 n - A(4n) (~)4n, and A =~. Consequently, tn =
c(41t) + (~)n4n and 4 = tt = 4c + G)(4) => c = ~, so in = (D4 n + (~)n4n = 4 n- 1 (1 + 3n),
n > 1.
(c) For an alphabet E, where lEI = r ~ 1, there axe 1'11. strings of length n and these rn
strings determine a total of 1' n - 1 [1 + (r -l)n] runs. [Note: TIns formula includes the case
where l' = 1.J
2)/2
- Dn ;;;;: (1/2)(n 2
(b)
810,000,000
~""' .....,.. -
s:w.,ooo :;:
x
2)=*
15.
Var
disks}
Procedure Move_The_Disks (n: integer; start, inter, finishj char);
{This procedure will move n
fmm
start peg to the finish peg 'lwing inter as
the intermediary
}
Begin
If
then
Writeln (,Move disk from ',start, 4 to ',finish, '.')
Else
Begin
Move_The_Disks
1, start, finish, inter);
Move_The_Disks (1, start, ' " finish)j
Move_The_Disks (n-I, inter, start, finish)
End {else}
End; {procedure}
Begin {main program}
Write ('How many disks are there? ');
Readln (number);
If number < 1 then
Writeln (,Your input is not appropriate. ')
Else
End.
00
00
-3
00-
2
=0
00
00
00
=0
(J(x)-l-
-3x(/(x)-
l(x)(1-3x
= 1 6x-
= 1
Consequently,
lex) == _ _ _ _ _ _ _ _
5_
and an = 5(211,) - 4, n 2: 0.
(d) 0,11.+2 - 2an +1 +ar. = 2n , n 2: 0, Go = 1, a1 = 2
E:'o an +2 xn+2 - 22::'0 an +1 Xn +2 2::::::0 an x n +2 = E:=o 2n x n +2
Let lex) = r::=oanx n • Then
[/(x) - 0,0 - a1xj - 2x[j(x) - 0,0] x 2 /(x) == x 2 E:""o(2x)n
I(x) -1- 2x - 2xl(x) 2x + x 2 1(x) = x 2 /(1- 2x)
(x 2 - 2x + l)/(x) = 1 + x 2 /(1- 2x) => I(x) = 1/(1 - x)2+
x 2 /«1-2x)(1-x)2) = (1-2x+x 2 )j«1-x)2(1-2x» = 1/(1-2x) = 1+2x+(2x)2+ ... ,
so an = 217., n 2: 0.
2.
a(n, r) = a(n - 1, l' -1) + a(n - 1, r), 1" 2:
E~l a(n, r)x 7 = E~l a(n - 1, r - l)x1'
E~l a(n - l,r)x7
a(n, 0) = 1, n 2: 0; 0,(0, 1") = 0, r > O.
Let In = E~Oa(n,r)xr.
In - a(n, 0) = Xfn-l + In-l - a(n - 1,0)
In == (1 + X)/",-l and fn = (1 x)nfo
lo==E~oa(O,r)x1'=
0)+
l)x a(0)2)x2
a(n, r), r ~
... =0,(0,
=1,80 fn=
_
.L.-_ _-:........_~
- 6x)(1- 2Xt2 = (1-
1-
+ 1
2
(_2X)2
... J
- 2n), n ~, 0
.
1
f(x)(-4x) (1- 6x)g(x) = 0 ===> g(x) = (4x)f(x)ll - 6xt ===> g(x) =
4x(1 - 2x )-2 and bn = 4(,.:21)
= n(2 n ·tl ), n 2:: O.
an =
(:2)
- 6(~~21)
0
= (1 -
1=
(b) an = (-3/4) + (1/2)(n + 1) + (1/4)(3 n ), n 2:: 0
On = (3/4)
), n 2:: 0
Section 10.5
1.
04 = bobs + 01 02 ~bl babo = 2(5 2) =
= [(2n)!/«n + l)!{n!))], b4 = 81/(5!4!) = 14
1» n 12) =
5.
(c)
tn+!:
n
VI, V~h'"
be the
of a convex (n + I)-gon.
the side
each partition of this polygon into triangles, with no diagonals
Vl'Vn +! is
of one
these triangles. The triangle is given
, 2 ::; i ::; n.
For each 2::; i ::; n, once the triangle VIViVn+l is drawn, we consider the resulting
polygon on Vi, V2, ••• ,Vi and the other polygon on Vi, Vi+! , .•• ,Vn +!- The former polygon
can be partitioned into triangles, with no intersecting diagonals, in ti ways; the laUer
polygon in t n +1 -Hl = t n +2-i ways. This results in a total of ti' t n +2-i triangular
partitions with no overlapping diagonals. As i varies from 2 to n we have
t'Jtn t3 t n-l + ... + tn-Its tnt2 = Ei::::2 iji n +2-i.
(a)
j
(b) From Example 10.36, in = bn - 2 , n 2: 2. With hn = (2n)!j[(n
tn = (2n - 4)lj[(n - l)!(n - 2)!], n 2.
7.
l)!n!] we have
(a)
5
t'l (b(
))
...._ _ _ _--.;.5
a«
)d)
3111-_ _ _ _....,..5
(a (
)
2(2n 1)
- (n 2) bn •
In
10.23 note how vertex 1 is always
with an even numbered vertex.
the case
n 2
otherwise we end up with intersecting chords.
For each n
1, let 1:5 k
n, so that 2
2k:5 2n. Drawing the chord connecting
vertex 1 with
2k I we divide the circumference of the circle into two segments - one
containing the vertices
3, ... ,2k - 1, and the other containing the vertices 2k 1, 2k +
2, ... ,2n. These vertices can he connected by nouintersecting chords in ak-l an"-k ways,
so
an
= aoan-l + alan -2 + 0.20.11,-3
•••
+ 0.11,-20.1 + an-laO·
Since ao = 1,0.1 = 1,0.2 = 2, and 0.3 = 5, we find
10.
11-
= brn the nth Catalan number.
an
Consider, for example, the second mountain range
Fig. 10.24. This path is made up
from the moves N S N N S S. Replace each 'N' by a '1' and each'S' by a '0' to get 10 110
0- a sequence of three 1's and three 0'13, where the number of D's never exceeds the number
of 1'13 as the sequence is read from left to right. We know that the number of such sequences
is 5(= b3 ). In general, for n E N 1 there are bn such sequence.':; and, consequently, bn such
mountain ranges. [Note: the above argument could also be established by replacing 'N'
by 'push' and'S' by 'pop', setting up a one~to-one ('.orrespondence between the mountain
:ranges and the permutations obtained with the stack]
(a) x
fI(x)
1
1
2
2
3
3
(b) The
hex) hex) hex) fs(x)
3
3
3
2
2
3
1
2
3
3
3
3
(0,
in part (a) correspond
y
y
/
/
/'
/'
/
/
/'
/
/
/
/
/
/
/
-----"~
x
12.
(a)
x
1
2
3
1
2
3
93(X)
1
2
2
1
1
1
95(X)
1
1
:3
1
1
2
Ii [in part (a)
<
the previous exercise]
correspondence for i = 1,2, and 4.
(i =
x
1
2
3
hex)
91(X)
1
2
1
2
3
3
(i =
x
1
2
3
with fJi. We
(i =
!-.l(x)
3
3
3
fJ2(X)
1
1
1
x
14(x)
94(X)
1
2
3
2
3
3
1
1
2
Consider the column for any Ii- In that column replace each entry Ie by 3 - (Ie - 1): so
1 's and 3's are interchanged while 2'8 remain as 2'8. Then reverse the order of this new
column. The result is the column for 9i. [In order to generalize this to the case where the
domain and codomain are {1,2,3, ... ,n}, n E Z+, we write down two columns - one for
1,2,3, ... ,n and another listing Ii(l), li(2), li(3), . .. ,Ii(n). Each entry k (in the column
for Ii] is replaced by n - (Ie - 1). Then the order of the column is reversed, giving us the
image under the corresponding function gi-
( c) For n E Z+, the number of monotone increasing functions 9 : {I, 2, 3, ... , n} -+
{1,2,:3, ... ,71}, where g(i)::; i for alII::; i::; 71, is (l/(n +
= bn, the n~th Catalan
lnC:)
number.
13.
For n E N, let an count the number of these arrangements for a row of n contiguous
pennies. Here ao = 1, al = 1, a2 ::::: 2, and a3 = 5. For the general situation, let n E N
and consider a contiguous row of n + 1 pennies. These n + 1 pennies provide n possible
locations for placing a penny on the second level. There are two cases to consider:
(1) The first location (a.~ the second level is scanned from left· to right) that is empty is
at position i,
1 :5 i ::; 11..
there are i - 1
(above
i pennies
the
bottom contiguous I'O'\.v) ill the positions to the left of position i. These i - 1 contiguous
n - [(i 1J == n - i ""...<'L ....""U.P
UCI,cn.llU1CU by a row of n-i+ 1 . . ~..&~~l',""..,.
== aoan
al lkn-l
Catalan
+
14.
(a) 82 = 6
(b) 82 = (~)~
(c) 83 = (~)b3
(:)
(~)b2
e)bo
(:)01 + (~) bo = 22.
there are r
rrlr.~r.<"Q
Consider those paths from (0,0) to
o 5 r 5 n. How can one generate such a path'? It
contain (n - r) R's
(n - r)
U's and
2(n - r) letters provide 1 (location at the start) +2(n - r) - 1 [locations
between the (n - r) R's and (n - r) U's] +1 (location at the end) = 2( n - r) + 1 locations
in total, for inserting the r
Further, these 2( n - r) + 1 locations are selected with
repetitions allowed. So there are
= nr-I")bn-'I' paths with 'r D's, n '- r
R's, and n - r U's (with the path never crossing the
y = x). Sunurnng over r we have
UU"l'.,""U<;;W.
e
e<n-r};-Hr-l)
Sn
= L~=o enr-r)bn _ r •
15.
(a) 1 32,2 3 1: E3=2
(b) 13254
14253
14352
15243
15342
34152
34251
35142
3524 1
23154
24153
24351
25143
25341
45132
45231
E5 = 16
(c) For each rise/fall permutation, n cannot be in
position (unless n = 1); n is the
second
of a
8'U("h a
""' ................ , Consequently, n
be
2
or 4 ... or 2ln/2j.
nelrm1ltttl~tlCm XIX2XS ••• Xn-1X n
• , ., n.
are 2i - 1 ......J,.................
The (n - 1) - (2i- 1) = n -
to En-<Ji
n~
m
1
a permutation. Therefore, 1 must
remaining (n - 1) - 2i = n - 2i - 1 nu:mo;ers
to
rise/fall
n
(g)
(
(
.n.UU.Ullji;
)
n-l
)
2t{n-l)/2j
(n;l)
-
these
n-1
2ln/2J-l
-1
or
-
(1/2) Ef;J (n;1)E i E n_i _ 1 •
Ee -
(1/2)
-
(h)
E1 -
e) E2 E
(~)EoE5
--
3
(~) EsEa
·1·2 10·2·1
(1/2)[1· 1 . 16 5·1·5
(1/2)[16 + 25 20 + 20 + 25 16] = 61
(1/2) Ef=o (nEiEs-i
(1/2)[1·1·61 6·1·16 + 15 . 1 ·5
272
(!) E4 E l +
5·5·1
20·2·2
15·5·1
1·
(3·16· 1 + 1 ·61· 1]
(i) Consider the Maclaurin series expansions
secx = 1 x 2 /2! 5x4 /4! + 6Ix o/6! ... and
tanx = x + 2x3 /3! + 16x5 /51 + 272x1/7! + ...
finds that sec x + tan x is the exponential generating function of the sequence
1,1,1,2,5,16,61,272, ... - namely, the sequence of Euler numbers.
Section 10.6
(a) fen) = (5/3)(4nleS3 4 - 1) and f E O(nloSa4) for n E {3 i li E N}
fen):
n
fE
f E
on
E
IE
on
E
=
c= 3
d ;:::;
a : 2, b
n)
nE
E
fen) = 3[n1og,,2 - 1]
f E o(n1<>fi,Sc 2)
(b) d = a = 1, b = 2, c = 2
fen) = 1 + 2
n
f E O(logl) n)
(a) 1(1)=0
fen) = 2f(n/2) + 1
From Exercise 2(b),
= n - 1.
(b) The equation fen) = f(n/2) + (n/2) arises as follows: There are (n/2) matches
played in the first round. Then there are (n/2) players remaining, so we need f(n/2)
additional matches to determine the
6.
(i) Corollary 10.1: From Theorem to.l
(1) f(n)=c(lo&,n+l) for n=1,b,b2 , ••• ,when a=1. HencefE O(logbn) on
S = {bkik EN}.
(2) fen) = [c/(a - l)][anlo~a - 1] for n = b,02, ... , when a 2: 2. Therefore
1 E O(nlogoa) on S = {bklk EN}.
(ii) Theorem lO.2(b): Since f E 0(9) on S, and 9 E O(nlogn), it follows that
f E 0 (n log n) on S. So by Definition 10.1 we know that there exist constants mER+
and s E Z+ such that fen) = If(n)! :::; minlog
= mnlogn for all n E S where
n 2: s. We need to find constants ME R+ and 8. E Z+ so that fen) < Mnlogn for
all n 2: 81 - not just those n E S.
Choose t E Z+ so that s < bk < t < bk+l (and 10gB > 1). Since f is l'llOnotone
increasing and positive,
J(t)
f(lf+!)
< mbk+!log(bk +!)
m bk +! [log lf log b]
-
m bk +! log f} + m bk+l log b
-
m b[ti'(log
10gb)]
m bff} log
log b)]
m
+ b)(ll~logbk)
m b(l +
b)t
t
<
-
<
=11i
t)
(Cue 1:
It),
8} =
f(t):::; M(tlogt),
we
+
we
f E O(nlogn»).
tE
2 = f(2k) = fen),
f(k).
2:
k
1 is even) Now we write n
Then fen
= f(21') = l(r) + fer) 21(1')
/('1') ~ fer - 1) by the induction hypothesis.
'11,
1 < n,
induction
1 =
where r E Z+ (a.n.d r 2:: 3).
fer -1) 2 = f(2r -1) = fen),
Therefore f is a monotone increasing function ..
(b)
From part (a),
all nE Z+.
TheoreI.ll10.2 (c) it
10.48,
f E 0('11,) for
(a)
fen)
a/en/b)
2
a f(n/b 2 )
as f(n/b 3 )
ak - 1 f(n/li- 1 )
<
<
<
<
+ en
+ ac(n/b)
2
2
af(n/b)
a2 f(n/b 2 )
a3 f(n/li)
a4 f(n/b 4 )
+
< akf(n/bk )
a c(n/b )
a3 c(n/b3 )
ak - 1 c(n/li- 1 )
fen) 5 ak f(n/b)" + cn[l + (a/b) + (a/b)2 + ... (a/b)k-l] = ale f(l) cn{l +
(a/b) + (a/b)?' + ... + (a/b)k-l], since '11, = bk • Since f(l) c and (n/hle ) = 1, we have
fen) 5 cn[l + (a/b) + (a/b)2 ... (a/b)k-l + (a/b)kJ = (en) Ef=o(a/b)i.
Hence
(b) When a = b, fen) 5 (en) E::=G Ii = (en)(k + 1), where '11, = bk , or k = 10gb n. Hence
fen) 5 (en)(logb n + 1) so f E 0('11, 10gb '11,) = O(n logn), for any base greater than 1.
(c)
k
,
For a i= 9, en ~(a/b)$ = en
=
=c
C
[1 1-_(a/b)kH]
(a/b) .
Ii -
------'---'--'-
1-
=c
hen) = n E O(niog;!
for € = 1.
by case (i) for the Master
we have f E
(b) a = 2, b =
=
= n
hen) = 1 E o (n1ogJ 2 - f) for f =
By case (i) for the Ma.'!iier Theorem it follows that f E
'" -- 1 b -....
- .
1
...,
,.logl> a
- _ I -- h( n) = 1 E 9( n10g3/2 1 )
Here case (ii) for
Master Theorem applies and we find that f E 6(nlogb<l\ 10g2 n) =
9(10g2 n).
(d) a = 2, b =
n!O{!;b lt = n lois2 :... nO. 531
hen) = n E O(nlogg 2+t:) where f"':" 0.369.
sufficiently
n, a hen/b) = 2h(n/3) = 2(n/3) = (2/3)n ::; (3/4)n =
Further, for
c hen), for 0 < c = 3/4 < L Thus, case (iii) ofthe Master Theorem tells us that f E 6(n).
(e) a = 4, b = 2, nlogbGi = nlos24 = n 2
hen) = n 2 E 9(nlog24)
From case (ii) of the Master Theorem we have f E 8(nlog241og2 n) = 8(n 2 1ogz n).
GI
)
Supplementary Exercises
1.
2.
n ) =(k+l)!(n-k-l)!=(k
nl
(n - k)l)k!(n-k)!=
n!
(nk+l
- k) (n)k
(k+l
(a) Conside,;t· the element n 1 in S = {1,2,3, ... ,n,n+ I}. For each partition of S
we consider the size of the subset containing n + 1. If the size is 1, then n 1 is by
itself and there are Bn partitions where this happens. If the size is 2, there are (~) = n
ways this can occur, and B n - 1 ways to partition the other n - 1 integers. Tllis results
in
B n - 1 partitions of S. In general, if n 1 in a subset of size i + 1, 0 z n,
(7)
there are (~) ways this can occur with (~)
of sum Bn+! = Ei=o (7)B n- i
(b)
= Ei;:;o (n:~)Bn-i =
1
resulting partitions of S. By the rule
(7)Bi •
For n ~ 0, Bn = E?"'oS(n,i). [S(O, 0) =
az = 1.
v.rrite n =
1
n
t,
an
If
Xl
= 1,
3 and n - 2 = (Xl - 2) X2 + ... + Xt, a summation counted in a,..-2. Consequently,
an = (£'1'1.-1 0.,..-2 for all '11;::: 3,
an =
number,
n ~ 1.
Xl
5.
(a)
A2 =
[~ ~ 1= [Fs FI 1 AS = [~ i 1= [F Fzl'
j
3
=[~ ~]=[F4 ~:l·
(b) Conjecture: For n E Z+,
where Fn denotes the nth Fihonacci number.
Proof: For '11 = 1,
l
A =A =
[~ ~] = [~ ~: 1,so the result is true in
result true for n = k
this first case. Assume
1, i.e.,
Consequently, the result is true for all n E Z+ by the Principle of Mathematical Induction.
1vf=[~ ~ ],
=[~ ~l'
=[~ !j=[F,
]
_- [58 8 ] = [Fe FrJ
We
that
1)
n E Z+,
= [
=
[5 8]
8 13 ) M
4
=
[13 34]'
21
=[~ ;]=[F4
]
] = [Fr F91
M4 = [
1.
n = 1
as,
"..",,"'' ' ,.,.,'' when n :::;: k +
Mn=[i ;r=[! !][; !r=[i ill
]
=[
1
F2k+2
=[
)
F:u;;+2
=[
F2k+1
1
] = [ F2k+1
1
F2k+2
= [ F'ln-l
It follows
Fan+! ].
the Principle
Mathematical Induction
all n 2: 1.
1.
F'romx 2 -1 = 1+~wefindthatx3-x =x+l,orx3 -2x-l =0. Since (-1)3-2(-1)-1 =
-1 + 2 -1 = 0, it follows that
is a root of x 3 - 2x -1. Consequently, x - (-1) = x + 1
is a factor and we have x 3 - 2x - 1 = (x 1)(x 2 - x-I). So the roots of x 3 - 2x - 1 are
+ ..(5)/2, and (1 - ~5)/2.
For x =
11 = (_l)a - 1 = O.
For x = (1 + ~)/2, 11 = [(1 + ..(5)/2]2 - 1 = (1/4)(6 2v'5) - 1 = [(3 + ..(5)/21 - 1 =
(1
.;5)/2.
For x = (1 - V5)/2, 11 = [(1 - vrs)/2]2 - 1 = (1/4)(6 - 2V5) - 1 = [(3 - v'5)/2] - 1 =
(1- ../5)/2.
So the points ofintersection are (-1,0), «1 +v'5)/2, (1 +\1'5)/2) = (a, a), and «1-v'5)/2,
(1 - -15)/2) = (fJ, f3).
8.
(1 + v'5? /4 = (6 + 2../5)/4 = (3 -15)/2
a 1 = (1 ../5)/2 1 = (3 v'5)/2
f32 == (1 - ../5)2/4 = - 2~5)/4 = .- ../5)/2
fJ 1 = (1 - v'5)/2 1 = - -15)/2
(a)
0'2 =
=2
=2- fj)
= [1/(a - P)J
(:)2 k a/' - ~k={) (~)2kPk]
= [1/(a-
(:)(2a)k -
(~)(2p)k]
-(1
2/3)"] = [1/(a-/1)1[a 3n -
= [l/(a-j3)][(l+
9.
(a) Since 0 2 = 0 + 1, it follows t.hat a 2 + 1 = 2
a and (2
a)2 = 4:
4a + a 2 =
2
2
4(1 +a) +a = 5a •
(b) Since;92 = P+1 we find that /1 2 + 1 = ,8+2 and (2 /1)2 = 4+4/1+/1 2 = 4(1 +;9) ;92 =
5f32.
= (I/(a - p»
[E (2;)
(a 2)kam -
E(2kn)C{32)k,a
m
]
= (I/(a - p»fam(l + a2r~n - pm(1 + p2?n]
= (1/(a - p»)[am(2 + a?n - fJm(2
= (I/(a - fJ))[a m«2
fJ)2n]
a)2)'" - pm«2 + p)2r~]
= (I/(a - p»fa (5a 2 t - pm(5p2tJ
7n
= 5,Tj,(1!(a - fJ»[a 2n+m -- p2n+m] = 5np2n+m'
10.
(a) Let po = $4000, the price first set by Renu, and let PI = $3000, the first offer made
by Narmada. For n ~ 0, we have
us
characteristic equation 2x2 - x - 1 :;::: 0; the characteristic roots are 1
,ncO.
10th
n =
n U1C17eas'es the
$3333.33.
-
+
= $3200. So 4000 = PQ =3200
Pn=
so
occurs
-1
11~
Consider the case where n
is similar.) For the fence
= {al,a2, ••. ,an },
are (:'11,-1
f:
-+ {
f(a n ) = 2. [Note that ({I, 2},:5) is the same partial order as .1"2.] \Vhen such a function
satisfies J(an ) = 1,
we
have f(an-d =
and
are C,.-2 of
preserving functions. Consequently, since these two cases
nothing
common and
cover
possibilities, we find that
So Cn = Fn+2, the (n
12.
13.
2)nd Fibonacci number.
This combinatorial identity follows by observing
l(n+l)/2J, each count the
of subsets of {I,
integers.
Fn+2
and
3, ... , n}
'11,-1<:+1) ,lor
.t
m =
( k
contain no consecutive
(a) For n > 1, let an count the number of ways one can tile a 1 x n chessboard using the
1 X 1 white tiles and 1 x 2 blue tiles. Then al = 1 and 0,2 = 2.
For n ~ 3, consider the nth square (at the right end)
the 1 x n chessboard. Two
situations are possible here:
(1) This square is covered by a 1 x 1 white tile, so the preceding n -1 squares (of the 1 x n
chessboard) can be covered in 0,.,.-1 ways;
(2) This square and the prer.eding
l)st) square are both covered by a 1 x 2 blue
tile, so the preceding n - 2 squares (of the 1 X n chessboard) can be covered in a n -2 ways.
These two situations cover all possibilities and are disjoint, 80 we have
«n -
Consequently, an = Fn +b the (n + l)st Fibonacci muuber.
(b) (i) There is only 1 = (~) = C,n.::;~o) way to tile the 1 x n chessboard using all white
squares.
(H) Consider
equation Xl + X2 + ... Xn-l = n-l,
~
n-L We
Xi = 1
·
(71.
..
C&l. select one of the Xi,
1 a n - ,1l n 1::::::: no-2
ways. Increase
of
Xi
2
we
1)
(n-1) =
ones.
(v)
ones.
(}
k
In/2j,
n-
(c) Fr.H =
Et/;J ('fl.;:") = 2:1:(;.1
n-k
•
[Compare this result with the formula presented
the
c2 :;:: 1
-c-l=O
c:::::: a or c
= /3.
c> 0 it follows that c = a = (1
15.
(a) For
position i, 2 :$ i ::; n. Two
derangement, 1 is placed
then occur.
Case 1: (i is in position 1) the other n - 2 integers are deranged
With n - 1 choices for i this results in ('1'1 - 1)dn _ 2 such derangements.
d n - 2 ways.
i».
Here we consider 1 as the new natural
Case 2: (i is not in position 1 (01' position
position for i, so there are n . - 1 elements to derange. \Vith n"~ 1 choices for i
we have ('1'1 - 1)dn - 1 derangements. Since the two cases are exhaustive and disjoint, the
result follows from
of sum.
(c) dn - nd"._l = d".-2 - (n - 2)dn - 3
(b) do = 1
(d) dn - ndn - 1 = (-l)'[dn-i - ('1'1 - i)dn - i - 1 ]
Let i = '1'1 - 2.
dn - ndn_1 = (-lY·-2(d 2 - 2d1 ] = (_1)n-2 = (_l)n
(e) d". - ndn - 1 :;:: (_l)n
(d". - ndn_l)(Xn In!) :;:: (-lY-(x nIn!)
2::::2(d". - ndn _ 1 )(x n /n!) :;:: ~2( _x)'fI In! = e- X - 1 + x
E:"",,2dnx'TI./n! - xE~2 dn _ 1 x n - 1 /('1'1 -1)!:;:: e-a: -1 x
[f(x) - d1x - doJ- x[J(x) - dol:;:: e-$ - 1 + x
lex) -1 - x/ex) + x =
-1 + x and /(x):;:: e- X /(1- x)
16.
Drawing the (n 1)st oval, '1'1 ~ 0, we get 2'1'1 new points of intersection which split
the perimeter of tllis oval into 2'1'1 segments. Each segment takes an existing region and
divides it into two regions. So
= an
an.H
2'1'1, '1'1
a~)::::::
-
a1 = 2.
+
= n(Bn
C)
:=::}
= 0 =:} B = 1,
an:::::: n - n + 2:::::: 2{n(n 2.
-
B(n 2
c = Bn 2 Cn 2n =::.?
2'1'1+1)
= -1, so
-
'n.
2 = ell =
2
-1
:::::: s( t - 1) :::::: 2 -
00
S =
t=
-~
be
2n times with the sequence of H's and T's counted
an. For
1 i 5
is a
i
equals the number
for
first time after 2i tosses. This sequence of 2i tosses is counted in hi; the given sequence
tosses counted
Since bo = as ~
from 0 to n,
an =
aibn-i.
Let a
Let g(x) = i:~=o bnx'\ f(x) =
anx n = (1 - 4xt 1 / 2 •
~x"' ::::: i:~""'l(aobn
anbo)xfl. => f(x) - ao = f(x)g(x) or g(x)
a1 bn - 1
= 1 - [1/ f(x)} = 1- (1- 4X)1/2.
(1-4x)I/2 == [C~2)
ei
Z
)(-4x)
The coefficient of x"' in (1·- 4X)1/2 is e~2)( _4)"' =
(1/2)«1/2) -1)«1/~) - 2)·· ·«1/2) - n + 1)(_4)71 = (-1)(1)(3)(5)·· ·(2n - 3)(2'1'1,) =
n!
n!
(-1)(1)(3)···(2n-3)(2)(4)···(2n-2)(2n)
n!n!
(-1)
(2n)1
= (2n - 1) ~!n! =
n).
1/(2n-l»)(2
n
Consequently, the coefficient of x'11 in g(x) is bn = [1!(2n - l)]C:), n ~ 1, bo = O.
18
IPI::::: 11-)1]1
= 1&-1
< 1' so 2:k=O pk = _1_
. l-~ = ~
=
:I
2
I--a = l - (1¥ ) = .....l....¥ = ~
Hva 1- 5
1-5
00
•
~ _ _ (~) _ _ R
Z
2
-
-
fJ·
Since 0: + f3 = (¥) +
'1\'00
e-?) = 1, it follows that
0: -1
= -{3.
IRlk -- L...k=O:2
(iA-l)k -- 1-( b,- - _
1
L ~ - w.i§ - !!.:tM - (1)(3+ 15)
(¥)
-- 3-VS"
i+J5' - 9-5 4
2
VO ,
L...h=O fJ
,,\,,00
2- ) --
and 0: 2 = (¥)2 = (6+:1&) = n)(3 + J5).
x, fj,z E
f(f(x, y), z) = f(a+bxy
=a ac
+
+
z) = a+b[Ca+bxy+c(x+y»z]+c[(a+bxy+c(x+y»)+z)]
ely +
cz,
-cc=aorc=
a-fJ=(
-
--1-)2 _ H2.fi
- ( H,;5
4
-
4_
6-215 __
-
6+2l5 -
4
-
(!)(3 + v'5 - 3 + v'5) = J5.
{3-2 - (32 = (i-~r;j - e-~,(frY;~ =
_3-15_~_!i~_
2
4
2-
+ y'5 - 3 v'5) = J5.
(b) Using the Binet form we have
_ F2
F.2
= (ClI R+1_l!n+l)2 _ (an-1_rr -1)2
n+l
n-l
a-fi
Ot-fJ
_ a 2,,+2+p2n+2 _2(Oitl),,+1_a: 2,.-2_l;12n-2+2(O'@ln-l
-
-
(0'-13)2
a
2n
(a
2
2
-a-:l-fi?"W-(t2) ( •
(ClI-;3}2
smce a 'f3 -- -1)
= (0'2» - {:J2'f1.)/(a - (:J) [from
results in part (a») = F 2n •
(c) Here the base angles are 60° and the altitude is (1/2)(y'3)Fn • Consequently, the area
of Tis (1/2)( v'3/2}Fn [Fn - 1 FnH ] = (v'3/4)Fnf Fn-l Fn+d.
Returning to part (b) we find that F2n = F:+1 - F:_l = (Fn+1 - Fn- 1 )(Fn+1 + Fn- 1 ) =
FnFn+l + FnFn- 1 • Consequently, the area of T = (v'3/4)Fln •
21.
Since An B = 0, Pf'(S) = Pr(A U B) = Pr(A) + Pr(B). Consequently, we have 1 =
p + p2, so p2 P - 1 = 0
p = (-·1 ../5)/2. Since (
- ../5)/2 < 0 it follows that
p = (-1
V5)/2 = -{:J.
22.
The probability that Sandm wins is p+(1-p)(1-pYlp+(1-p)(1-p)2(1-p)(1-,p)'J.p +··· =
pI1 + (1 - p)3 + -- p)6 (1- p)9 ...J = p{l/[l - p)31J.
For the game to be fair we must have 1/2 = prI/[1 - (1 - p)3JJ, so
p -
(1/2)[1 - (1 - p)3]
2p - [1 - (1 - p)3J = 1 - (1 -, Sp + 3p2 - y)
2p -- Sp - Sp:! p3, and
o _ p3 -- Sp2 P = p(p2 - Sp 1).
Since p > 0, it follows that p2 - 3p
1 = 0, or p = (3 ± V5)/2. Since p < 1, we find that
p=
have u'.n,;'UJ..l.r::;. ln common
+
,n?:: 1, and 00 we have another instance where the
Here an -
24.
2
Here:to = a, Xl = h, X2 = XIXO =
X3 = X2Xl = 1;2 a, X4 = X3X2 = iJ3a , and Xl) =
3
Fn
X4X3 = 1/'a •
results suggest that Xo = a and, for n
1, Xn = bF"a - 1 , where
denotes the nth Fibonacci number (for n ?:
To establish this
general we proceed by
mathematical
The result is true for n =
as
as
n = 2,3,4,5.
... ,k k, where k is a fixed (but arbitrary)
Fk
2
Xk-l =
a and XI:; = bF"aFk - 1 , so Xk+! = XkXk-l =
(lI'ha Fk - 1)(bFk - 1a1<1.-2) = l/'1<+Fk-laFk-l+Fk-2 = bFh +1a F", by
recursuve definition of the
Fibonacci numbers. Consequently, by the alternative form the Principle of Mathematical
Induction
result is true for (n = 0 and) all n ;:: 1.
Assume the
positive integer.
(Second Solution). For 11, ?:: 0 let Yn = log X n . Theu Yo = log a, Yl = log h, and y".. tin+!
tln-:'h n ?:: 2. So Yn = Clan + czlr, where a = (1
V5)/2 and f3 = (1 - V5)/2.
log a = Cl + Ca, log b = C1 a + c'),{J ::::}
c2=(-1/v'5)logb [(1 v'5)/2v'5Jloga,
C1 = (1/.;5) log b + [(-1 + V5) /2V5]log a,
where the base for the log function is 10 (although any positive real number, other than
1, may be used here for the base).
Consequently,
tin
Xn
c2{3n
-
Clan
-
[(1/-15) log b [(-1 + v'5)/2V5] loga}a71
+[( -1/ \1'5) log b [(1 + J5)/2V5] log a}f3n ,
-
10C1 O:"+C2P"
10[[( -Hv's/Zv's] log a+(l/v's) log bJa" •
1O[(Hv'S)/2.fi)log a+{ -ltv'S) !ogb]P"
-
a(Il<,,-1_p.. - 1}/.fi
b{ 0/" - f3n )/.,f5
-
-
Fl - FoF] - FJ = 12 -. 0 . 1 - 02 = 1
-
i!:"2_.
·K'2
~-
- F; =
·....·1·1-1.°'"
- 2 . 3 .-
--1
-
0,
={
3,
neven
n
so
k odd. We shall establish
result fo!' k even, the
us that
- FkFHl n =k
that Ft+2 - Fk+JFkH - F1+1 = (FkH + FIc? - FkH(Fk+l +Fk ) - FiH = F;+1 2FH1 F..., +
- F:+l - Fk+1Fk F~ - Pi+1 = - F1J = -1.
result follows for all n EN, by the Principle of Mathematical Induction.
26.
21.
The answer is the number subsets of {1,2, 3, ... ,
We learned in Section 10.2 that this is FnH ,
(a) r(G1 ,x) = 1 x
r(G2 ) = 1 + 2x
1'( Gs , x) = 1 3x + Xl
which contain no consecutive entries.
2)nd Fibonacci number.
r(G4 ,x) = 1 + 4x + 3.x 2
r(Gs ,x) = 1
+ 6x 2 + x 3
r(G.. , x) = 1 + 6x 10x 2 + 4X 3
In general,
n;?: 3, r-(Gn,x) = r(Gn-x,x) + xr(Gn _ 2 ,x).
(b) r(Gl? 1) = 2
r(Gs , 1) = 5
r(Gs ) = 13
r(G2 , 1) = 3.
r(G41 1) = 8
r(G6 , 1) = 21
[Note: For 1 i ::; n, if one "stra-ightens out" the chessboard Gi in Fig. 10.28, the result
is a 1 x i chessboard - like those studied in the previous exercise.]
28.
For 0 S; n
18, let Pn be the probability that Jill bankrupts Cathy when Jill has
n quarters. Then Po = 0 and PIS = 1 and the answer to this problem is PIG. For
0< n < 18, if Jill has n quarters, then after playing .another game of checkers,
Pn = (1/2)Pn-1 + (1/2)Pn+1
--...--- --...---
Jill h8..'3 lost
the game
Jill wins
the game
Pn+l - 2Pn + Pn-l = 0 has characteristic roots r = 1,1, so Pn = A + Bn. Po = 0 ==:} A =
0, 1 = Pta :} B = 1/18, so pn = n118. Hence Jill has probability 10/18 = 5/9 of
bankrupting Cathy.
29.
(a)
The partitions counted
m
occur more than once.
f ('(1" m) fall into two categories:
a summand.
are
fen -
rn
integer;
Function
If n=O
f:= 1
Else if
<
or (m < 1) then
°
f:=
f := f(n,m-l)
End; {of function f}
f(n-m,m)
Begin
Writeln (,What is the value of n?')j
Readln (n);
Writeln ('What is the value of m?');
Readln (m);
Write (,There are " f(n,m):O,' partitions of ');
Write (n:O, , where " m:O, , is the largest ');
Writeln (,summand possible. ')
End.
(c)
Program Parlitions(input,output);
Var
n: integer;
FUnction f( n,m: integer): integer;
Begin
If n=O then
f:::::::: 1
if < 0) or
< 1)
f:= 0
f:= {(n,m-I) f(n-rn,m)
tunc:tlOn f}
'1V'riieln (,What
value of
'. ')
30.
f(a) : : : : b
aE
For m
1, nffl = the total number of functions f : A --l> B, If 1:5 i :5 n - 1,
there are
i)
g with ....
a
of
t.
Furthermore, any function h: A --t B that is not onto found amOl1g these fm:,tctions g.
Consequently, a( Tn,
- Ei,;-"l
a( m,
V ..,... UA,J,&
31.
The following program will print out the units digit of the
- F 129•
Program Units(input,
Vat
FibUnit: array[0 .. 129}
iJ: integer;
integer;
Begin
FibUnit[O] := 0;
FibUnit[lJ := 1;
For i := 2 to 129 do
FibUnit[i] := (FibUnit[i-l] + FibUnit[i-2]) Mod 10;
For i :=0 to 12 do
For j := 0 to 9 do
IT j < 9 then
vVrite (FibUnit[lO * i + jJ: 4)
Else {j = 9}
Writelu (FibUnit[lO >I: i + 9J: 4)
End.
130 Fibonacci numbers:
3
GRAPH THEORY
AND
APPLICATIONS
CHAPTER 11
INTRODUCTION TO
THEORY
Section 11.1
1.
represent
air routes traveled a.r.nong a certain set of cities by a particular airline.
(b) To represent an electrical network. Here the vertices can represent switches, transistors,
and an edge (x, y) indicates the existence of a wire connecting x to y.
( c)
the vertices represent a set of job applicants and a set of open positions in a
corporation. Draw an edge (A,b) to denote that applicant A is qualified for position b.
Then all open positions can be filled if the resulting graph provides a matching between
the applicants and open positions.
2.
(a) {b,e},{e,j}, {/,g}, {g,e},{e,b}, {b,c},{c,d}
(b) {b,e},{e,j},{f,g},{g,e},{e,d}
(c) {b,e},{e,d}
(d) {b,e},{e,f},{/,g},{g,e},{e,b}
(e) {b,e},{e,f},{/,g},{g,e},{e,d},{d,c},{c,b}
(f) {b, a}, {a, c}, {c, b}
3.
6
4.
We claim that x( G) = 2. To verify this consider the following:
(1) Let Ct be the set of all vertices v E V where the binary label of v has an even number
of Is. This includes the vertex z whose binary label is the n-tuple of all Os. For any
Vo E C1, where Va i= z! we can find a path from Vo to z as follows. Suppose that the binary
Vo has 2m 18, where 2
2m ~ n. Change the first two Is in the binary
label
Vo to Os and call the resulting vertex Vi' Then VI E C 1 and {vo,vd E E. Now change the
for VI
v2·
the vertex Vm = Z
C1 connected
to z.
upon changing
vertex W2 E C 2 with {WI!
E E.
vertex Wm = z*'
with W m -l E C'), and {Wm-l, w m } E E. Consequently each vertex in C 2 - {z*} whose binary
label
1 is connected to z*.
(ii) There are 2m+ 1 Is in the binary label for Wo, with 3:5 2m 1 n, and the
the label for WI} is O. Change
first entry in the binary label for Wo to 1
1 in the binary label for Wo to O. This results in the vertex WI E OJ. with {wo, WI} E
Upon changing the second and
18
the binary label for WI to Os we obtain the
u",..
W2 E O2 with {Wb
E
Continuing this process we reach the vertex Wm+1 =
with {wm' W m+1} E E.
shows that each vertex in C'J, whose binary label starts with 0
is also connected to z*.
WI
a 1
'!:&>.".
(3) We claim that the components of G are the graphs determined by C 1 and C2 • Can
there exist an edge {x, y} E E where x E Ct, 11 E C2 ? Here the binary label for x has an
even number of 11'1 while the label for y has an odd number of 18. This contradicts the
definition of E - for if {a, b} E E then the total number of Is in the binary labels for a, b
18 even.
5.
Each path from a to h must include the edge {b,g}. There are three paths (in G) from a
to b and three paths (in G) from g to h. Consequently, there are nine paths from a to h
in G.
There is only one path of length 3, two of length 4, three of length 5, two of length 6, and
one of length 7.
6.
7.
c:
1
i: 4
e:
j:
1-22
f:
1
3
k:
b
1
2
g:
2
1: 3
1-22
h: 3
m:
3
;;:
{(g, d), (d,e),(e,a)}j
{(g, b), (b, c), (c, d), (d, e), (e, a)}.
Two: One of {(b, e), (e,
and one
of {(b, j), (J,g),(g,d)}.
(e, a), (a,
8.
3-
9.
a cycle, then its renlOvai disconnects a
a
b
P,
{a, b},
o}
x
a
exist
11
y EV
of a
C, then the edges
(P - {e} ) U (C - {e}) would
a second
connecting x to
11.
12.
(c)
n-1
(a) In a loop-free undirected graph (that is not a mw.tigraph) the maximum numher of
edges is
e::.s;
= v(v -1)/2, so
: .s; v 2 - v.
a loop-free directed graph (that is not a multigraph), e::.s; v 2 -
(b)
V.
13.
This relation reflexive, symmetric and transitive, so it is an equivalence relation. The
partition of V induced by 'R.
the (connected) compollents
G.
14.
(a) There are three cycles of length 4 in W 3 , :five cycles of length 4 in W 4 , and five such
cycles in Ws.
(h) Denote the cOllsecutive cyde (rim) vertices of Wn hy 'Vb V2," • 'Vn and the additional
(central) vertex by vn +!-
(i) For n f:: 4, there are n cycles of length 4:
(1) VI -+ V2 -+ V3 -+ V n +! --+ VI;
(2) V2 -+ V3 -+ V4 -+ V n +! -+ V2;
" ... ,
(n - 1) Vn-l -+ Vn -+ VI --+ V n +l -+ Vn-I; and
(n) Vn -+ VI -+ V2 -+ V n +l -+ Vn'
When n = 4 the vertices VI, V2, VS, V4 provide a cycle. The othel' four cycles of length 4
consist of vertex Vs and three of the four vertices Vb Va, Vs, V4.
(ii) There are n 1 cycles of length n in Wn:
(1) VI -+ Va -+ Vs -+ ..• -+ Vn-l -+ Vn -+ VI;
(2) VI -+ V n +l -+ V3 -+ V4 .--t ••• -+ Vn-l -+ Vn -+ VI;
(3) V2 -+
-+ V4 -+ Vs -+ ... -+ Vn-l -+ Vn -+ VI --.; V2;
.. .
"
($1) V .. _1 -+ V n +! -+ VI -+ 'Vjj -+ •.• -+ V n -3 -+ V n -2 -+ V n -l; and
+
Vn -+
$1
1,
we allow
3
16.
-+ V2 -+ Us -+ ... -+ V Y._3 -+ Vn-l -+ Vn'
--.. 1
'If
.
If
0
. •
",
1 0
1 0
Ii
iIIP
WI
W2
. ••
0
!lIP
1
1
Ij!II
@)
0
1
010101
b) For four unit intervals there are 14 uui.t-inteval graphs.
n n!l (:n)
n
For n > 1, there are b =
unit-interval graphs for n unit intervals. Here b is
the nth Catalan number. The billary representations set up a one-to-one correspondence
with the situations in Exaluple 1.40 - in particular) change 0 to 1 and 1 to 0 in part (b) of
Example 1.40 to obtain the binary representations of the 14 unit-interval graphs on four
unit intervals.
Section 11.2
1..
(a)
{b,a}, {a,e}, {e,d}, {d,al
(2) {f,e},{c,a},{a,d},{d,e}
(3) {i,d},{d,c},{c,a},{a,d}
Three: (1)
(b) G1
the 8ubgraph .............v~"c. by U = {a, 0, d, I, [J, h, i, j}
G1 =0-
G2
i,j}
b--j
j
290
a, bE
rt E1 ·
but
Then G - e is a subgraph of G but it is not an induced subgraph.
(b)
3.
(a) There are
= 512 spanning subgraphs.
(b) Four of the spanning subgraphs in part
(c)
is only one - the graph G itself.
4.
5.
are connected.
G is (or is isomorphic to) the complete graph Kn? where n =
6.
(2)
"
.---•.
undirected graphs with four verlices.
v.
z
I I
•
ol
(11)
1.
(b)
R
B
'f
V
W
w
'III
Fl0B 'f~}~w
B
Y
B
W
R
No solution.
(a,b,g,h)
a cycle.
is not so for
graphs are
the first graph the vertex d is incident with four edges.
graph has this property, so
graphs are not isomorphic.
G has 'IJ vertices and e edges, then by the definition
10.
vertex
the SC<:OD,Q
G, there are (~) - e
in G since there are (;)
11.
If G1 = (Vl, E 1 ) and G'}, = (Vi, E 2 ) are isomorphic, then
is a function
f : Vi --+ Vi that is
and onto and prt',senres adjacencies. If x, y E Vi and
{x, y} Ell then
f(y)} ¢ E 2 • Hence
same function f preserves adjacencies
and can be used to define an isomorphism for G 1 , G2 • The converse follows in
for G},
a similar way.
'
(a)
(b) They are not isomorphic. The complement of the graph containing vertex a is a
cycle of length 8. The complement of the other graph is the disjoint union of two cycles
of length 4.
12.
(a) Let el be the number of edges in G and e'}, the number in G. For any (loopfree) undirected graph G, el + (:>z = (;), the number of edges in ](n' Since G is
self-complementary, el = el, so el
(b) Four vertices:
= (1/2)(;) = n(n - 1)/4.
ClUb
c
G:
d
or
b
e
G:
n
d
n-l
some k E
G is
with
{a,
,{e,d},
e}
{e, a}, then G is the cycle with edges {a, c},
, {e,b}, {b,d}, {d,a} . Hence, G and G are
isomorphic. Conversely, if G is a cycle on n vertices
and G, G are isomorphic, then n = (1/2)(;), or n =
(1/4)(n)(n -
and n =
(a) All of the examples in Exercise 12 above satisfy
(b) Since G is not connected, there
vertices x, y and no path in G connecting
these vertices.
{ x, 11} is an edge in
For
vertex a in G, a =I x, y, either
{a,x} or {a,y} is in G. If not, both {a,x},{a,y} arein G and {x,a},{a,y} provide
b, c E V. If {b, x},
al':e both in G,
a path in G connecting x and y.
there is a path connecting b, c: namely, {o, x}, {X 1 c}. The same is true if {b, y}, {c, y}
both occur in G. If neit,her of these 8i tuations occurs we have { ii, x}, {c, 11 } in G ( or
{b,y},{c,x}) and then the edges {b,x},{x,y},{y,c} provide a path (',onnecting band
c.
15.
(a)
Here 1 must also maintain directions. So if (a, b) EEl, then (l(a), feb»~ E E 2 •
(b) They are not isomorphic. Consider vertex a in the first graph. It is incident to
one vertex and incident {rorn two other vertices. No vertex in the other graph has this
property.
16.
17.
(:)(2(D)
(a) (:)(23 ) = (:)(2(D)
(b)
(c) E~=l (:)(2e»
Cd) Ek:l (:)(2(~)
There are two cases to consider:
Case 1:
Ifj
Ii
V
W
Case
.
•
x
w
--·------4.l1li'------••
41>'4)-----·----4
..........
Y
V
z
w
are n - :2 ""....~....'A"'"
ro':)le~:lS for z -- u,,"'.....'-"."~
are
3
tJ
=6
(b)
IVI = 1 or 2 or 3 or 5 or 6 or
first
to w.
2=: 31vl, so
3.
Since 38 = 21EI =
IVI is 11.
E deg(v) 2=: 41Vj, the largest possible value foOr IVI is
We can have
vEV
degree 4
two
or (li) eight vertices of degree 4
one
graph in part (a) of the figure is an eXMu.ple for case (i)i an example foOl'
4.
We :must note here tha,t G need no~ be connected. Up to isomorphlm:.n G either a
vertices or (a disjoint union of) two cycles, each on three vertices.
b)
either a cycle oOn seven vertices or (a disjoint union oOf) two cycles - one on
three vertices and the other on four.
c) FoOl' Much a graph GlI 0'1 is one of the graphs in part (a.). Hence there are two such
graphs G1 •
d) Here G1 is one of the graphs in part (b). There are two such graphs G1 (up toO
isomorphism ).
e)
G 1 ::::;; (Vi,
a
(n with
= n.
toO
isomoOrphism the Ulunber of
graphs G 1 is the number of partitioOns of n into summands
<"1"",ttt·.,·.,. 4 -
294
observe
G 1 and Ga are
is to consider once again the
degree 4
G1
vertices induce a
subgraph
consisting of the two edges {b,c} and {f,g}. The four vertices of degree 4 in graph G2
n u n . . , ... a
that has
edges - every
o",,-,vu,U
way
f
....----.e
a --------
...----.e
(ii i)
(11 )
1.
19
b)
~(~)
=
I.
a) There are 8 . 27 1024 edges in Qs.
b) The maximum distance between pairs of vertices is 8. For example, the distance
between 00000000 and 11111111 is 8.
c) A longest path in Qs contains all of the vertices
Qs. Such a path has length
8
2 -1 = 255.
9.
a) n· 2»-1 = 524,288 =} n == 16
b) n ·2»-1 = 4,980,736 ==> n = 19, so there are 219 = 524,288 vertices in this hypercube.
10.
The
path of
2 uses two edges
vertex b as any vertex of Qw., so there are 2'1l choices
IS
n
ways.
11.
are (;)2'1l path."S
{a, b}, {b, c}.
can select
b. The vertex b (labeled by a
we can
(;)
2in Qn.
edges in Kn is (~) = n( n - 1)/2. If the
Kn can be partitioned
4:
or
v
same
position i-hence, 1l is reflexive.
v, w E V and vRw
v, w have
same
position k and
same bit
position f.. Hence w, v have the smne bit
poSli;lon
k and the same bit in position e. So wRv and R is symmetric. Finally, suppose that
v,w,x E with vRw
v,w have
same bit
Ie
the same
position i, and w, it
same
in position Ie a:nd the same bit
l.
Consequently, v,x have the same bit in position k and the same bit in position i, so v'Rx
- and 'R transitive.
80 much as 'R is reflexive, symmetric and transitive, it follows
that 'R is an
relation.
There are four blocks for (the partition induced by) this equivalence relation. Each block
contains 2",-2 vertices; the vertices in each such block induce a subgraph isomorphic to
Qn-2.
(b) For n ~ 1 let V denote the vertices in Qn. For 1 :$ kl < k2 < ... < k t < n and
V define the relation 'R on V by w'Rx if w, x have the same bit in position kl' the
same bit in position
... , and the same bit in position kt. Then 'R is an equivalence
relation
V and it partitions V into 2t blocks. Each block contains 2 n - i vertices and the
vertices in each such block induce a subg;raph of Qn isomorphic to Qn-t.
W, x E
13.
SIVI
LilEV deg( v) :$ AIVI. Since 2!EI = EVEV deg( v), it follows that
SIVI :$ 21EI :$ AlViso 6:$ 2(e/n) :$ A.
14.
(a) f- 1
one-to-one and onto. Let x, 'II E V' and {x, 'II} E E'. Then J one-to-one
and onto =::} there exist unique a, bE V with f(a) = x, feb) = y. If {a, b} ¢. E, then
{f(a), feb)} ¢ E'.
(b) If deg(a) :;:::: n, then there exist Xl) Xl,"" xn. E V and {a, Xi} E E,l :$ i ::5 n. Hence,
the edge {f(a), f(Xi)} E E! for all 1:$ i ::5 n, so deg(f(a))
n. If deg (f(a» > n,
let y E V' such that y
f(Xi) for all 1 ::5 i ::5 n, and y = f(x). Since 1-1 is an
isomorphism by part (a), {a, x} E E and deg(a) > n.
degf(a) = n.
15.
Proof: Start with a
1 VkH}, {Va) Vk+2},"
Vt -+ V2 -+ V3 -+ ... -+ V2.1:-1 -+ V2k -+ VI-
.,
The resulting
Vi+I.J, ' .. ! { 'Jin
ae2::ree 3.
16.
(By the Alternative
The :result is true fur n = 1 (for
Now
296
Tben draw the k
has 2k vertices
Mathematical Induction.
17.
(Corollary 11.1) Let V = Vi u V2 where Vi (V2) contains all vertices of odd (even)
2jEI deg( v) is an even
For
I odd,
degree.
deg(v)
(Corollary 11.2) For
converse let G = (V, E) ha;lr'e au Euler trail with a, b as the
starting and terminating
respectively. Add the edge {a, b} to G to form the
graph G' = (V, E'), where G' has an Euler circuit. Hence
is connected and each
..,,,,,,,,1",,,.... has even degree. Removing edge {u, b} the verti('.-es in G will have the same even
degree
for 0" b.
= degG,(a) - degG(b) = degG' (b) -1,00 the
a, b
have odd degree in G.
since the edges in G form an Euler trail, G is connected.
18.
Select VI, V2 E V where {VI, V2} E E. SUdl an edge must exist since V
0 and
deg( v) ~ k ~ 1
all V E V. If k = 1 the reBul t follows. If k > 1, suppose that we have
selected VhV2, ... ,V,. E V with {Vt,V2},{V2,'03}, ... ,{Vk-hVk} E E. Since deg(vk) k,
there exists Vk+l E Vi where 'Ok+! 1= Vi for 1 =:; i ::; k - 1, and {'OJ;, vk+d E E. Then
{VI, V2}, {V:h va}, ... , {'Ok-I, VI:}, {VI., Vk+1} provides a path of length k.
19.
(a) Let a,b, X,Y E V with deg(o,) = deg(b) = deg(c) = 1, deg(x) = 5, and deg(y) = 7.
Since deg(y) = 7, y is adjacent to all of the other (seven) vertices in V. Therefore vertex
x is not adjacent to any of the vertices a, b, and c. Since x cannot be adjacent to itself,
it follows that deg(a:) 4, and we
a graph for the given
unless we
conditions.
(b)
a~b~c~g~.~j~g~b~f~j~i~f~e~i~h~d~e~b~
d~o,
~
d~o,~b~d~h~i~e~f~i~j~f~b~c~g~k~j~g~.~
e
n=2
23.
Yes.
~odel the
SUll'ro'unl!llIltllr
corridor,
1;
24.
We
that
L id(v) = e =
od(v}.
fJEV
25.
(a) (i) Let the vertices of
be Vl,V2,V3,V4,
Va, where deg(vi) = 5 for alII < i:5 6.
vO:IlBl~(leI' the subgraph S
He
(from K 6 ) by deleting
{V2'
and
{V3)
S is connected with deg(vl) =
= 5,
deg(vi) = 4 for i E
{2,3,5,6}. Hence S has an Euler trail that starts at VI (or V4) and terminates at V4 (or
vd. This Euler trail in S is then a trail of
length in K(;h and its length
(~) - (1/2)[6 - 2] = 15 - 2 = 13.
(ii) (~) - (1/2}{8 - 2] = 28 - 3 = 25
- (1/2)[lO - 2] = 45 - 4 = 41
e;) - (1/2)[2n - 2] = n(2n - 1) - (n - 1) = 2n
(iii) e~)
(iv)
2
-
2n + 1.
(b) (i) Label the vertices of Ke as in section (i) of part (a) above. Now consider the
subgraph T of Ke. obtained (from Ke.) by deleting the edges {VI! 7)41, {V2' vol, and {V3' V6}'
T is connected with deg(vi) = 4 for all 1 :5 i < n. Hence T has an Euler circuit
and this Euler circuit for T is then a circuit of maximum length in K 6 • The length of the
circuit is (~) - (1/2)(6) = 15 - 3 =
(ii) (:) - (1/2)(8) = 28 - 4 = 24
(iii)
(iv)
26.
e:) - (1/2)(10) = 45 - 5 = 40
e;) - (lj2)(2n) = n(2n - 1) - n = 2n
2
-
2n = 2n(n - 1).
(a) If G = (V, E) has a directed Euler circuit, then for all X, y E V there is a directed
trail from x to 11 (that part of the directed Euler circuit from x to y). This results in a
directed path from X to y, as well as one from y to X • Hence G is connected (in fact,
G is strongly connected as defined in part (b) of this exercise). Let s be the starting
vertex (and termiual vertex) of the directed Euler circuit. For every v E V, v f.. 8, each
time the circuit comes upon vertex v it must also leave the vertex, so ode v) = ide v). In
the case of s the last edge of the circuit is different from the first
and odC s)::;:: id( s ).
cmnpOIMem; has a vertex on
a we travel on
the component C1
we
VV.U.~AA4U.JtU):. the process,
V2 on COlnDOll'ent
OU1;Wll a ntF'{'!~l::P'(1 Euler circuit for G.
---4---.. . . d
If G = (V, E) is a
graph with a directed E-uler
for all x, y E V, :;r. :F y, there is a directed path
:;r.
y ,
y to
graph is
strongly connected.
false. The
directed graph shown
strongly
HO¥rever,
od(b)
id(b) the graph does not have a
circuit.
tI:Y1i'!'t"r.ll~ 24 we see
[ ode v) - ide v)1 = O. For each v E V, ode v)+
c ......
= nso 0 = (n -1) ·0= EVEV(n -1)[ od(v) - id(v)] = LUEV[ od(v)+
ide v )][ ode v) - ide v)} = EVEV [( ode v))2 - ( ide v) )2J, and the result follows.
28.
Let G he a directed graph satisfying the three conditions. Add the edge (x, y). Then
by part (a) of Exercise 26 the resulting graph has a directed Euler circuit C. Removing
(x, y) from C yields a directed Euler trail for the given graph G. (This trail starts at
y and terminates at :;r..) In a similar manner we find that if a directed graph G has a
directed Euler trail then it satisfies the three conditions.
29.
(a) and
'VI
V2
V3
't'4
VI
0
1
0
V2
A= Vs
1
1
0
V4
0
VI)
1
1
1
1
1
1
1
e1
e2
1=
1
~~0 01
Va
v, ~
V4
0
Vs
0
0
0
1
1
e3
1
1
0
0
0
1
1
0
1
€4
0
1
0
0
1
e5
0
1
1
0
0
e6
(;.,
€g
0
0
0
1
0
0
0
0
0
1
0
1
0
1
1
e9
0
0
0
1
1
€lO
0
0
1
0
1
(b) If there is a walk of length two between Vi and V j, denote this by {Vi, VA:}, {VI:, Vj}.
Then aik = (Lkj = 1 in A and the (i, j )-entry in A 2 is 1. Conversely, if the (i,j)-entry
of A2 is 1 then there is at least one value of k, 1 S k < n, such that (LiAr = akj = 1, and
this indicates the existence of a walk {Vi) Vk}, {VI., Vj} between the ith and jth vertices of
V.
(c) For all 1 S i,j S n, the (i,j)-entry of A~ counts the number of distinct walks of
length two between the ith and jth vertices of V.
(d) For v at the top of the column,
column sum is the degree of V, if there is no
loop at v. Otherwise, deg( v) = [(column sum for v) - 11 2 (number of loops at v).
(e) For
column of [ the column sum is 1 for a loop and 2 for an edge that is not a
loop.
33.
(a) Label the rows and columns of the first matrix with a, b, c.
the gl'aph for this
adjacency matrix is a path of two edges where deg(a) = deg(b) = 1 and deg(c) = 2.
Now label the rows and oolunlll8
is a path of two
the second matrix with x, '$I, z. The graph for this
deg(x) ;;;::;
where deg(y) = deg( z) = 1
by f(a) =
Alternatively, if we start with
,!,u,,"".. ",-,_co;:,,,, OO~,WJ:UlS 1
3
a
on
f(11) = z,
= x.
(b) The graphs
are not isomorphic. The graph for the first incidence matrix is a cycle
of
3 with the
(remaining) edge incident with one
the
second graph is a cycle on four vertices.
(c)
35.
No. Let each person represent a vertex for a graph.
v 1 w represent two of these people,
draw the edge {v,
if
shake hands.
were possible, then we would
have a graph with 15 vertices, each of degree 3. So the sum of the degrees of the vertkes
would
45, an odd
contradicts Theorem 11.2.
36.
Define the function f from
domain A x B (or
set of processors of the grid) to the
OOITefij)ondm,e: vertices of Qr, as follows:
f «o,b, cde) = abcde, where ab E A, cde E B,and a, h, c, d, e E {O, 1}.
If f«ab, cde» = j(al bll Cldl cd), then abede = alblCldl€ll so a = 0,11 b = b,}, c = C:t, d = dll
e = ell and (o,b,cde) = (o,tbt,c1d1el), making j one-to-one. Since IA x BI = 15 = the
number of vertices (of Q5) in the codomain of j, it follows from Theorem 5.11 that j is
also onto.
Now let {( o,b, cde), (vw, xyz)} be an edge in the 3 x 5 grid. Then either o,b = vw and
cde, xyz differ in (exactly) one component or cde = xyz and o,b, vw differ in (exactly) one
component. Suppose that o,b = vw (so a = v, b = w) and C = X, d = 1/, but e f. z. Then
{abcde, vwxyz} is an edge in Qs. [The other four cases follow in a similar way.] Conversely,
suppose that {f(a 1 b1 , elatel), f(VIWh Xlf11Z1)} is an edge in the subgraph of QI) induced by
the codomain of f. Then alo1cldlel and VIW1XIYIZ1 differ in (exactly) one component - say
the last. Then in the 3 x 5 grid, there is an edge for the vertices (albl,cldlO), (a1b1c1d1l).
[Similar arguments can be given for allY of the other first four components.] Consequently,
f provides an isomorphism between the 3 X 5 grid and a subgraph of Q5'
[Note that the 3 x 5 grid has 22 edges while Qs has 5.24 = 80 edges.]
31.
Assign the Gray
{DO, 01,11, IO} to the four horizontal levels: top - 00; second (from
the top) second from the bottom - 11;
- 10. Likewise, assign the same
the four vertical levels:
(or, first) - 00; second - 01; third - 11; right (or, fourth) 00),
10), and
Conversely, ",..
differ
",.fi..,.'"",
,Cl
for
other three components.
three-by-three grid and a subgraph of Q4.
[Note:
gri d has
d1 ),
are
f establishes an isomorphism between the
edges
Section 11.4
In this situation vertex b is in the :region formed
by
edges {a,d},
e is
outside of this region. Consequently the edge
{b,e} will cross one of the edges {a,d}, {d,c},
{a,e} (as shown).
demonstrate
2.
3.
(a)
N umber of vertices
Number of edges
11
28
18
m+n
77
mn
m=6
4.
Let G = (V, E) be bipartite with V partitioned as Vi U V;, so that each edge in E is
the form {a, b} where a E Vi, b E V2 • If H is a subgraph of G let W denote the set
of vertices for H. Then W = W n V = W n (Vi U Vi) = (W n Vi) U (W n Vz), where
(W n Vi) n (W n V2) = 0. If {x, y} is all edge in H then {x, y} is an edge in G - where,
say, x E 1'; and y E V2. Hence x E WI, Y E W 2 and H is a bipartite graph.
5.
(a) Let Vi = {a,a,e,h} and V2 = {o,e,j,g}. Then every vertex of G is in Vi U Vz and
Vi n Va = 0. Also every edge in G may be written as {x,V} where x E Vi and y EVa.
Consequently, the graph G in part (a) of the figure is bipartite.
(b) Let V{ = {a,b,g,h} and V; = {c,a,e,j}. Then every vertex of G1 is in V;U V; and
Vi n V; = 0. Since every edge of G' may be written as {x,y}, with x E Vi and y E V;,
it
that this graph bipartite.
fact G1 is (isomorphic to)
complete bipartite
graph K 4 ,4'
(c)
graph
not biprui;ite.
= (V", Eft) were bipartite,
U
edge in GU is of
{x,
V{'. Now
the
0, c,
we
haveb,c
an
E E" => e E V;"
E Ell ::;:;} e E VI',
6.
Alternately,
the vertex of ue2:ree :3
](1,3 -
303
can
done in n ways. Then
-
this can
to
- 1)(n - 2)(n - 3)/6 = (4)[(n)(n --l)(n - 2)(n - 3)/24] =
'1.
4(~).
The vertices in Km,n may be partitioned as 'Vi U Va where IVi I = m, !V2 ! = n, and each
edge of the graph has the form {x, y} where x E Vi and y E 1tJ.
In
to obtain a
of length four we
to select two vertices from each of
in (;) (;) ways - eadl resulting in a distinct cycle of length
'Vi and Va- This can be
four.
[Note: Say we select vertices a, b from Vi and vertices c, d from Va. We do not distinguish
cycles a ....... c ....... b ....... d ....... a and a ....... d ....... b ....... c ....... a. J
(b) For a path of length two there is one vertex of (path) degree 2 and two vertices of
(path) degree 1. If the vertex of (path) degree 2 is in lIi then there are m(;) such paths.
There are
n(';) sut'll paths when the vertex of (path) degree 2 is in Vz. Hence there are
m(~) + n(~) = (1/2)(mn)[m + n - 2J paths of length 2 in Km,n'
( c) Here a path of length '3 has the form a ....... b -+ c -+ d where (1" cElli and b, d E V2. By
the rule of product there are (m)(n)(m - l)(n -1) =
such paths in Km,n'
4{r;) (;)
(c) 14 (= 2(7»
8.
(a) 2
9.
(a) 6
(b) (1/2)(7)(3)(6)(2)(5)(1)(4) = 2520
(c) 50,295,168,000
(d) (1/2)(n)(m)(n -l)(m - l)(n - 2)··· (2)(n - (m l»(l)(n - m)
10.
Let G = (V, E) be bipartite with V = Vi U V2 , Vi n V2 = 0. If G has a cycle of odd
length then there is an edge in the cycle of the form {x, y} with x, y E 'Vi (or x, y E V2).
This contradicts the definition of a bipartite graph.
11.
Partition V a.<J VI U V'2 with
(b) 6
maximizes m(v -m) = [(v+
= (v 2 _
(a)
(i)
n E Z+, n
2(3))
(d) 2m
jV2 1 = v - m.
G is bipartite, then the
- m) = - (v/2)]2 (v/2)2,
m = v/2
a:
f:
b: {3,4}
c: {1,5}
d: {2,4}
e: {S,5}
g: {2,5}
{2,3}
i: {1,3}
j: {I
b
(b) G is (isomorphic to) the
f
c
Graph
that the first graph contains a subgrnph homeomorphic to Ks,a, so it is
not planar. The second graph is planar and isomorphic to the second graph of the exercise.
Tbe third graph provides a subgraph homeomorphic to Ka,a so the third graph given here
is not planar. Graph (6) is not
it contaitlS a
to'
Kfj.
b
--------..;.ub
b
If both m, n are
J(m,n has an
edges and cannot
decomposed
into
isomorphic subgraphs - since each such subgraph has the same number edges
as the other.
16.
how the vertices of
Petersen
are labeled in Fig. 11.52(a).
correspondence of vertices provides an isomorphism for
two
a~s
b~v
c~z
d~y
e~t
f~'U
g~'r
h~w
z~x
)~q
following
17.
(a)
are 17 vertices, 34 edges and 19 regions and v - e r = 17 - 34 + 19 = 2.
(b) Here we find 10 vertices, 24
and 16 regions and v - e + r = 10 + 16 = 2.
18.
Proof: Since each region has at least five edges in its boundary, 21EI > 5(53), or lEI 2':
(1/2)(5)(53). And from Theorem 11.6 we have rlll = IEI-53+2 = lEi-51 2': (1/2)(5)(53)51 = (265/2) - 51 = 8q. Hence IV! 2': 82.
19.
10
20.
(a) For each component Ci := (Vi, E i ), 1 ~ i ~ n, of G, if ei = IEil and Vi = IViI then
e.. -v.+2=ri. Summing as i goes from 1 to n we have e-v+2n=r+(n-1) because
the infinite region is counted n = K(G) times. Hence e-v+n 1 = r = e-v [K,(G) + 1].
Using the same notation as in part (a) we have 3ri < 2eil 1 ~ i
'n, so 3r ~
Ei=::1(3ri) E~12ei
Also, Ci :s; 3V i -6, 1 ::s; i :s; 'n, so e = 2:i=1 Ci l::'i::::l(3vi -6)
3v - on :s; 3v - o.
(b)
21.
=
=
not, deg( v)
() for all 'V E V. Then 2e = Lvev deg( v)
contradicting e::; 31VI- 6 (Corollary 11.3.)
G = (V, E) with
61VI, so e > 31V!,
= 11. Then
= (V, E 1 ) where {a, b} E El
lEI, Cl = IEll. If both G and G are planar, then by
of Exercise 20, if necessary), e
- 6 ::=
- 6 =
=1
are:<l
,so
;:: 28. Hence, Olle of
22.
If G=
f ....-->11111:
of
011
v
kr=
connected, it
(d)
e-
=?
v =
[kl(k-
>
= 8 < 9 = e.
be nonplanar.
k == 5, v =
= (40/3) <
= e.
are no
But
does
contradict
because the loops contain other vertices and edges of the graph.
25.
(a) The dual for the tetrahedron (Fig. 1l.59(b» is the graph itself. For the graph (cube)
Fig. 11.59( d) the dual is the octahedron, and vice versa. Likewise, the dual of the
dodecahedron is the icosahedron, and vice versa.
(b) For n E Z+, n ~
dual of the wheel graph Wn is Wn itself.
26.
(a) The COl'l·~,pornd,en~~ a.......; v, b - t W, C - t y, d - t Z, e .......; x providE'S an isomolphism.
(b)
(d)
(1)
>"
ill,
{ (/.ll}.
"
} ; {{ p,r',
}
21.
~
b'
28.
The number of vertices in Gel, the dual of G) is r, the UW,UIJ'II:I.!. of regions in a planar depiction
Since G is
to Gil it
r = n.
IVi - IE! r =
2 ~ n - lEI + n = 2 :::} lEI = 2n - 2.
[The converse follows in a "'UJ"~""" manner.]
11 .. 5
2.
The graph
a path (cycle).
8.
(a) Hamilton cycle: a -+ 9 -+ k -+ i -+ h -+ b -+ c -+ d -+ j -+ f -+ e -+ a
(b) Hamilton cycle: a -+ d -+ b -+ e -+ 9 -+ j -+ i -+ f -+ h -+ c -+ a
(c) Hamilton cycle: a -+ h -+ e -+ f -+ 9 -+ i -+ d -+ c --+ b -+ a
(d) The edges {a,e}, {c,d}, {d,b}, {b,e}, {e,f}' {f,g} provide a Hamilton path for the
given graph. However, there is no Hamilton cycle, for such a cycle would have to include
the edges {o, d}, {o, e}, {a, c}, {a, e}, {g, f}, and {g, e} - and, consequently, the vertex e
will have degree greater than 2.
(e) The path a -+ b -+ c -+ d -+ e -+ j -+ i -+ h -+ 9 -+ f -+ k -+ 1 -+ m -+ n -+ 0 is
one possible Hamilton path for this graph. Another possibility is the path a -+ b -+ c -+
d -+ i -+ h -+ 9 -+ f -+ k -+ I -+ m -+ n -+ 0 -+ j -+ e. However, there is no Hamilton
cycle, For if we try to construct a Hamilton cycle we must includ.e the edges {a, b}, {a, f},
{I,k}, {k,l}, id,e}, {e,j}, {j,o} and {n,o}. This then forces us to eliminate the edges
{f,y} and {i,i} from further consideration. Now consider the vertex i. If we use edges
{d,i} and {i,n}, then we have a cycle on the vertices d,e,j,o,n and i-and we cannot
get a Hamilton cycle for the given graph. Hence we nU18t use only one of the edges {d, i}
and ii,
Because
symmetry ill
graph let us
i} - and
{ h, i} 80 that vertex i will have degree 2 ill the Hamilton cycle we are trying to construct.
{d,
{ d,
are now being used, we
d}
{b, c} and
h} in our construction.
ffl~
~
a-+b~c-+d~e-+j-+i-+h-+g-+
I ~ m --+ n -~ () -+ t -+ S -+ '{' -+ q -+ p --+ k -+ f -+ a,
4.
go to
c or
e. (i) we go
vertex c we
edge
{e, a}, and this forces the
from oonsideration, but we must now include edge.s {e, j}
elimination of edge {a, b}. Now we must consider vertex h, for by eliminating edge {a, b}
we are now
to
edges {b,g} and {b,c} in
us to
remove edge
h}
further consideration.
we have now removed
if, h}
and {c, h} and there is ouly one other edge that is incident with h, so no Hamilton
can be obtained. (ii) Selecting vertex: e after d, we remove edge {d, c} and include
{c,
and {b,e}. Having removed {fiji} we must include {g,b} and {g,i}. Twsforce8
the elimination of
b},
inclusion
e} (and the elimination of {e, j}). We now
have a cycle containing a, f, i, d, e, hence
method has also failed.
However, this graph does have a Hamilton path: a -+ b -+ e -+ d -+ e -+ j -+ h -+ f -+
z -+ g.
(b) For example, remove vertex j and the edges {e, j} {g, j} {h, j}. Then e -+ a -+
f -+ h -+ c -+ b -+ 9 -+ i -+ d -+ e provides a Hamilton cycle for tills subgraph.
j
j
5.
(a) H we remove anyone of the vertices a, b or g, the resulting subgraph has a. Hamilton
cycle. For example, upon removing vertex a, we find the Hamilton cycle b -+ d -+ c -+
f -+ 9 -+ e -+ b.
(b) The fonowing Hamilton cycle exists if we remove vertex g: a -+ b -+ c --+ d -+ e -+
j -+ 0 -+ n -+ i -+ h -+ m -+ 1 --+ k -+ f -+ a. A symmetric situation results upon
removing: vertex i.
6.
Let the vertices on the cycle (rim) of Wn be consecutively denoted by VI, V2, ••• ,V'll.' and
let V n +1 denote the additional (central) vertex of W n . Then the following cycles provide n
Hamilton cycles for the wheel graph W 1".
(1) VI -+ Vn+l -+ Va -+ V3 -+ V4 -+ ... -+ V n -l -+ Vn -+ VI;
(2) VI -+ Vz -+ V n +! -+ Vs -+ Vti -+ ... -+ Vn-l -+ Vn -+ VI;
(3) Vi -+ V2 -+ Va -+ V n +! -? 1)4 -+ ... -+ V n -l -+ V'll. -+ VI;
en - 1)
VI -+ V2 -+ V3 -? V4 -+ ... -+ Vn-l
-? V n +I
-+ Vn -+ VI; and
VI -+ Vz -+ Va -+ V4 -,,) ••• '--7 Vn-l -+ Vn -+ 1Jn +1 -+ Vl'
10
8.
9
such
part (a).)
n=l ..~"u,.,.""" sense
this part
not
= (V,
a loop-free undirected graph with no
asSun::l.e that
is connected - otherwise, we work with the components of G. Select any vertex x in V
Vi = {v E Vld(x, v), the
of a shortest path between x
v, is
Va = {w E Vld(x, w), the length a shortest path between x and w, is even}. Note
that
x E Vz; (ii) V =
U 1/2; and (iii) Vi n 1/2 = 0. We claim that each edge {Il, b}
in E has one vertex in Vi and the other vertex: in V;z.
suppose that e = {(1., b} E E with
b E Vi.
proof for (1., b E
is similar.)
, x}} be the m edges in a shortest path from 11
Let Ea = {{ 11, vd, {V}, V2}, ... ,
to x, and let Eo = {{ h, tiD, {v~,
I ••• ,
, x}}
be
n edges
a shortest path
from b to x. Note that
n are both odd. If {Vb V2, ... , vm-d n {V~, V~, ... , v~_d = 0,
then the
of edges E' = {{a, b}} U EfJ. U Eb provides an odd cycle
G. Otherwise,
let w( i= x) be the first vertex where the paths come together, and let E" = {{ a, b}} U
Ha, vd, {VI, vz}' ... , {Vi, w}} u Ub, vB, {v~, v~}, . .. , {vj, w H, for some 1 < i m-l and
1 :$ j
n - 1. Then either E" provides an odd cycle for G or e - E" contains an
odd cycle for G.
vn
10.
(a) Suppose that G has a Hamilton cycle C. Then C contains IVl edges and the
vertices on C must alternate between vertices in Vi and those in V2 because G is
bipartite. This forces IVI to be even and IVi 1= 1V2!.
(b) In a similar way, if G has a Hamilton path P, then P has IV! - 1 edges and
the vertices on P must alternate between the vertices in VI and those in V2 • Since
IViI =F 1V21, it follows that IViI-IV21 = ±1.
(c) Let V = {a, h, c, d, e} with Vi = {a, b}, Vi = {c, d, e} and E = {{a, c} l { a, d}, {a, e},
{b,c}}.
311
C
od(a) = 3 id(6) == ()
OO(b) = 2
od(c) = 0
id(b) = 1
idee) = 3
od(d) = 1 id(d) = 2
.----111---... b
od(6) == 3
od(b) = 1
od(c) = 1
od(d) = 1
idea) = 0
ideo) = 2
id(c) = 2
idea) = 2
=0
=2
=2
=2
=3
..,..--'/11"---. b
2
=1
=1
=2
=2
oded) = 2
=1
=1
1:::::X
=1
=1
=1
Qn+l that
with 1. Each of Qn,(})
more than one but we
to
same cycle in
in
the cycles is the first bit in the vertices of an edge - that is, if {Ox, Oy} is an edge in the
(where 11 are
strings of length 11. that differ in only one
position), then {lx,1y} is the corresponding edge in the Hamilton cycle for Qn,l') Seled
edges {Ov, Ow} and {lv, lw} from the Hamilton cycles for Qn,Q and Qn,l' respectively.
Remove these edges and replace them
edges {Ov, Iv}, {Ow,lw} (in Qn+l)'
result is a Hamilton cycle for Qn+!'
It now follows from the Principle of Mathematical Induction that Qn
aU 11.~
13.
Proof: If not, there exists a vertex x such. that (v, x) f/- E and,
all 11 E V, 11 ::f:. v, x, if
(v,y) E E then (y,x) rt E. Since (v,x) rt E, we have (x,v) E E, as T is a tournament.
Also, for each 11 mentioned earlier, we also have (x, 11) E E. COllsequently,
x) ~
v) 1
- contradicting
v) being a maximum!
ode
ode
14.
a Hamilton cycle
Let G
with more
any
ode
vertices.
the multi graph in the given figure, IVI = 4 and
deg(a) = deg(c) = deg(d) = 2 and deg(b) = 6. Hence
dege x)
deg(y) ~ 4 > 3 = 4 - 1 for all nonadjacent
y E V, but the multigraph has no Hammon path.
b
16.
Proof:
11 E V, deg(x)
deg(y) ~ 2[Cn - 1)/2] = n -
80 the
result follows from Theorem 11.S.
Corollary
a, b E where {a, b} rt
(n/2) = 11., so the result follows from Theorem 11.9.
17.
en = (V E) <1eUCtte
j
but
all v E V,
(n/2)
cycle on n
=2<
18.
x,y E ,
wennd
21.
When n = 5
vertices.
Theorem 1
22.
graphs Cs
both are Hamilton
are isomorphic,
on five
u, v denote nonadjacent vertices
. Since deg( u) = deg( v) = n - 3
+ deg( 'I) = 2n - 6. Also, 2n - 6 ?: n ~ n?: so it
that the cocycle
contain.s a Hamilton cycle when n 6.
en
(a) If x v and y v,
deg(x) = deg(y) = n - 2, and deg(x) + deg(y) = 2n - 4 ?: n,
for n ?: 4.
one of x, y is v, say X, then deg(x) = 2 and deg(y) = n - 2, and deg(x) + deg(y) = n.
(b) From part
it follows
deg(x) + deg(y)
n for all nonadjacent x, y
V.
'1''Ihere{o:re
has a Hamilton cycle - by virtue of Theorem 11.9.
(c) Here lEI =
-1 +2, where we subtract 1 for the edge {ih,V2}, and add 2 for the
an
(n;l)
pair of edges {Vb V} and {V, vz}. Consequently, iEI = (1'1.;1) 1.
(d) The results in parts (b) and (c) do not contradict Corollary 11.6. They show that the
converse of this corollary is false - as is its inverse.
23.
(a) The path V -+ VI -+ 1)2 --+ V3 -". ••• - t '1)71,-1 provides a Hamilton path for Hn. Since
deg( v) = 1 the graph cannot have a Hamilton cycle.
+ 1. (So the number of edges required in Coronary 11.6 cannot be
(b) Here lEI =
decreased. )
(";1)
24.
(a)
Since the given graph has a Hamilton path
we use this path
provide the following
Gray code for 1, 2, 3, ... ) 8.
1:
9:
000
11
1010
0101
010
3:
100
111
7:
001
6:
1011
11:
1
0111
1100
1001
1000
1101
(i)
25.
(c)
(i)
(d)
I,
w,
(ii) j3(G) = 3
j3(G) =
(li) 3
(iii) 3
(iv) 4
3
n.
m
The maximum
complete
on II! vertk.-es.
In
on 'V
(b)
(v) {)
a,bE I.
contradicts
nlust have v
I = {(1, b, C, d, f}, as shown in the figure.
v=l e=18,
e-Eve1deg(v) 2IIj=18-(4
4: 3 4 + 3) 2(5) = 10 < 11, so by part (b), the
Herschel graph has no Hamilton cycle.
Draw a vertex for each species of fish. If two species x, y must
kept in separate aquaria,
draw the edge {x, y}. The smallest number of aquaria needed is then the chromatic number
of the resulting graph.
2.
Draw a vertex for each committee. If someone serves on two committees Ci, Ci draw the
edge joining the vertices for Ci and
. Then the least number of meeting times is the
chromatic mnnber of the graph.
3.
We can model this problem with graphs. For either part of the problem draw the undirected
graph G = (V, E) where V = {1, 2, 3, 4, 5,6, 7} and {i, j} E E when chemicals i and j
separate storage compartments. For part (a), the graph (in part (a) of the figure)
cm:'O:l!llat:lC number
1-....____•
a)
the
the
K 5 • Consequently,
Jeannette will need five separate storage compartments to store these seven chemicals
safely.
G
a cycle on n
1""·,,,,,, where n is odd
'{T", ...
5.
(a) P(G,)..) = )..(A _1)3
(b)
G=
we
that P( G, )..) =
X(K1 ,n) = 2.
6.
(i) Here we have).. choices for vertex
1 choice
vertex b
same choice as
that for vertex a), and)" - 1. choices for each of vertices x, 11, z. Consequently, there are
)..().. _1)3 proper colorings of K'},,3 where vertices a and b are colored the sam.e.
(ii) Now we have).. choices for vertex a, ).. - 1 choices for vertex b, and), - 2 choices for
each of the vertices x, y, and z. And here there are ..\(..\ - 1)().. - 2)3 proper colorings.
(b) Since the two cases ill part (a) are exhaustive and mutually exclusive, the chromatic
polynomial for ](2,3 is
..\(A - 1)3 + ,,\(,,\ - 1)().. - 2)3 = A().. - 1)()..3 - 5,,\2
10)" - 7).
X(K'},,3) = 2.
(c) P(K2 ,n,)..) = ,,\()" - 1)1'1
..\(..\ - 1)(..\ - 2)1'1
X(K2 ,n) = 2.
1.
(h) 2 (n even); 3 (n odd)
(a) 2
Figure 11.59(d): 2; Fig. 11.62(a): 3; Fig. 11.85(i); 2; Fig. 11.85(ii): 3 (d) 2
G = (V,
IS
then V = lt1 u V2 where Vx n V2 = 0 and each edge
form {x, y} where x E ,11 E V2. Color all the vertices in V'i with one color and those
Vi.! with a second color.
X(G) =
if
one
with
= 0 and
9.
-1
(3) 3
(c) (1) 720;
10.
(a)
f
(2)
1020j
(3) 420
graph
axe two cases
the
(b)
Case (i): Vertices f and k have the same color: Here there are JI.(JI. properly color
vertices.
(ii):
f
Ie axe colored
colors: Here
JI.(A - 2)2(). - 3y~ ways.
properly colored
By the rule of sum, P(G,,x) = ,x{>. - 1)2(,x - 2)2 ),(). - 1)(..\ - 2f"(..\ - 3)2 =
)'(..\ - 1)2(A - 2)2(..\2 - 5), + 8).
ways to
Using the same type of argument, with the two cases for vertices u and z, the chromatic
polynomial for
second graph is also
to be A(). - 1)2(>. - 2)2(),2 - 5,x + 8).
(c)
Gt G 2 are two graphs with P( GIl>') = P( G 2 ! >'), it need not be the case that G 1
and G'}; are isomorphic.
j
11.
Let e = {v, w} be the deleted edge. There are ,x(1)(,x - 1)(>. - 2)··· (>. - (n - 2)) proper
colorings of Gn where v, w share the same color and ),( A-I )(>. - 2) ... (,x - (n - 1))
proper colorings where v, w are colored with different colors. In total there axe P(Gn,,x) =
,x(,x -1) ... (,x - n + 2) + A(,x -1)···(,\ - n + 1) = ,x(). -1)· .. (). - n + 3)(>' - n + 2)2
proper colorings for Gn •
Here X(G n ) = n -1.
12.
a) Here (;) + (~) = (~) + g) = 15 + 3 =
and (r~g) = (~) = 36. So there are 18 edges
that are red or green, and 18 blue edges.
(~) = (1/2)(r~9) *> (1/2)r(r - 1) + (1/2)g(g - 1) = (1/4)(1' + g)(r 9 - 1) *>
21'(1' - 1) + 2g(g - 1) = (r + g)(1' + 9 -1) *> 1'2 - r + g2 - 9 = 21'9 *> (1' - g)2 = f' + g.
Let l' = g+k, k > O. Then [(r-g):! = P :;::; 1'+g = 2g+k] # [g = (1/2)(k 2 -k) = (1/2)k(k1) = tk-l and r = g+k = (1/2)k(k-l)+k::: (1/2)k[(k-l)+2k) = (1/2)k(k+l) = tkJ *> 1',g
b) (;)
are two consecutive t:riangular numbers.
1S.
(a) IVI =
IEf = (1/2) Lv€V deg( v) = (1/2)[4(2)
3n - 2, n;;::: 1.
n = 1, we find
result is
here
(2n - 4)(3)J = (1/2)[8
6n - 12] =
vertices in l,l have
(b)
colored.
nE
A(>' - l)(A -
from
PnY'''''rn
11.10
(c) Follows by the :mle of product.
(d)
P(Cn,
-
-
P(~.-b
- P(Cn-l, >.) = A(>' - 1)11.-1 - P(Cn- h A)
[(A - 1) IJ(A - P(Cn - b A)
(>. - l)n + (.X _1),..-1 - P(Cn- I , A):::;;:::;}
P(Cn, - (A _l)n = (A - - P(Cn-hA).
Replacing n by n - 1 yields
Hence
(e) Continuing from part (d),
P(CM >.) = (A - I)" + (-1)n-3[p(C3 ! A) - (A - 1)3J
= (A - l)n (_l)n-l [A(>' - l)(A - 2) - (A _1)3]
= (A _l)n + (-l)n(A -1).
16.
(a) X(Wn ) = x(Cn ) + 1. [Cn has n vertices; Wn has n
(b) peW?'!, A) = AP(Cn, A-I) = A[(A - 2)n
1 vertices.]
(-l)tt(A - 2)}.
>.) =
_2)5
,\(,\ -2) - For k
we
P(W1"),
(-l)1'jk(k - 2) = k(k - 2){(k _. 2}4 - 1] proper colorings, whenever k ~ 4.
11.
IVI -
!V! _.
= 1:
-
=0 :
= 1:
lEI =2:
= 3;
-
-
-A
A
-
-
(b) Let G = (V, E) be a loop-free undirected graph where IVI = n 4 and lEI = k ;2;: 1.
k =
P( G,).. = An
the
11.10, peG,
P(Ge )") - P(G~,)..) where e = {a, b} is an edge in G. Since G e has n vertices but
k -1
by the
where k -- 1, Cn-z, Cn-a, ... ,Cx ;2;: O. (When a coefficient
this list is zero, all successive
coefficients are
Likewise, since G~ has n - 1 vertices, by the induction
P(G~,)..) =
where bn - z , bn - 3 , •••
j
"V~ - (k).,\r.-J
19.
bn-2 A\n-2 + bn-3 A\11-3 -- •••
(-
bt ;2;: O.
Then P(G,).) = P(Gej )..) -
( c)
-
P(G~, A) =
+ (C n -2
bn _ z ».rI-2 + ... + (_It-I(C} + b1)..\.
This was shown in part (b).
(a) For n E Z+, n ;2;: 3, let en denote the cycle on n vertices.
If n is odd then x( en) = 3. But for each v in en, the subgraph Cn - v is a path with n - 1
vertices and x( Cn - v) = 2. So for n odd en is color-critical.
en.,
However, when n is even we have X(Cn ) = 2, and for each v in
the subgraph en - v
is still a path with n - 1 vertices and X(Cn - v) = 2. Consequently, cycles \\rith an even
number of vertices are not color-critical.
(b) For every complete graph Knl where n ~ 2, we have X(Kn) = n, and for each vertex v
in K", Kn -v is (isornorpbic to) K n- h so X(Kn -v) = n-1. Consequently, every complete
graph with at least one edge color-critical.
not (::onnected.
G1 be a component of G where X(G 1 ) = X(G),
(c) Suppose that
and
G2 be
other component G. Then xC G 1 )
x( G2 ) and for all v G a we find
)=
so
(d) If not,
Supplementary Exercises
1.
(;) = 56 + 80 = 136 => n(n - 1) = 272 => n = 17.
n ;::::: 1,
Cn count
number
cycles
in Qn' Then Cl = 0 and
C2 = 1. Recall
recursive construction of Qn+I fl'om Qn - given in Sedion 11.3.
Vn.~l denote all the vertices Qn+l that start with 0, and Vn~~ those vertices in Qn+1 that
start
1. [Each of the subgraphs of QnH induced by V~~l and V,!~l is isomorphic to
Qn.} Let 'Vl -» 'V2 -+ V3 -;. 'V4 -» VI denote a cyde of
four
Qn+l. There are three
cases to consider.
V~~l: Here there are en such cycles;
(2) Vl,'V:!,VS,'V4 E V~~l: Here there are also Cn such cyclesj and,
(3) one edge of the cycle (call it the first) is in (V~~l) and another edge (namely, the
third) is in (V~~l): Here the other two edges are eoch adjacent to a vertex V~~l and one
in Vn'!>l' [Let {VI' Va} E (V~~l)' then {V3' V4} E (V~!~) and the binary labels on VI and Vol
(1)
Vb V2,Vs, V4 E
differ only in the first (left-most) position, while the binary labels on V2 and V3 also differ
only in the first (left-most) position.} Since there are n2n - 1 possible choices (the number
edges in Qn) for the so called "first" edge, here we find n2,,-1 new cycles of length four.
The preceding discussion gives us
Cn +! = 2cn + n2'~-1 = 2en
(1/2)n2'1l
cih ) = .A211., c!r) = nCB + Cn)2?t
n;::::: 1, Cl = 0, Cz = 1.
(71 + l)(B + C(n + 1»2"+1 = 2n(B Cn)2?t + n2 n - 1
=> [B(n + 1) + C(n 1)2J2n+1 = [Bn Cn2]2n+l + (n/4)2n+l
=> 2C = 1/4, B + C = 0 => C = 1/8, B = -1/8.
So c!r) = (1/8)(11. 2 - n)2n.
0= CI = c~h,) + c~p) = 2.A + 0 => A = 0, so
en = (1/8)(71 2 - 11.)271 = (~)
n 2: 1.
j
Alternate Solution: Let VI -;. '02 -;. Vs -;. V4 --+ Vl be a cycle of length
urn",,...,. 1
i ~
1
in Qno Say VI, V2
j ::; n,
i
J.
VI there are 211,
POfU tli:mS 't, J .
til .....;. V4 -+ V3 -;.
(~)2n ""IU"t_
same cycle as VI -» Va --+ Va -;. Vol -» VI'] SO
*~ -+ 'Os --+ V4
-+ V1 "m+ 'V2, ~hl --+ ~J4 -+ Vi -~ V" --%
-;. Va -;. V.q
-» VI j
n;:::::1.
3.
If two.
are friends
a.
respective venices. The result then fo.llo.WS from part (a.).
pe(]~p!e a.'l
4.
(a)
= (1/2)(~)
(ii) For any undirected
G ) if G is
so G is connected.
connected then
is connected. In
(b) Proof:
n = 1 we have K 1 •
n = 4 the path on four vertices is an example of
a self-complementary graph.
cycle on
provides an example for n = 5.
a self-complementary graph
= (V, E). Construct the graph
Now suppose we
G 1 = ("Vi, E 1 ) where 'Vi = V U {a, b, c, d} (so none of a, b, c, d is
V) and El = E U
{{a,b}, {b,c}, {c,dH U {{v,a}/v E V} U {{v,d}lv E V}. Then
self-compleDnentary
1"Vi1 =
5.
(a) We can redraw
(b) 72
tt_------i/III b
6.
Only the graph for the cube is bipartite as seen
in part (b) of the given figure. In any of the other
four graphs (See Fig. 11.59(b) and Fig. 11.60)
there are cycles of odd length, 80 these graphs
cannot be bipartite.
1.
321
dllF-----~
(b)
V; as
are (1/2)(3)(7)(2)(6)(1) paths of &~U""~~
4
and end
a vertex
Vi,
are also (1/2)(7)(3)(6)(2)(5)
length 4 that start and end with a vertex in V2 • Consequently, there are 126 630::::: 756
of
4
(c) (Case 1: p odd, p = 2k + 1 for kEN). Here there are mn paths of length p = 1
(when k = 0) and (m)(n)(m -l)(n - 1)··· (m - k)(n - k) paths of length p = 2k + 1 ~
(Case 2: p is even, p = 2k for k E Z+). When p < 2m (i.e., k < m) the numher of paths
of length p (1/2)(m)(n)(m -1)·,· (n - (k -l»(m - k) + (1/2)(n)(m)(n -l)(m1)··· (m - (k -l»(n - k). For p = 2m we find (1/2)(n)(m)(n - l)(m '" (m - (ml»(n - m) paths of (longest) length 2m.
8.
(a) (n = 2): X::::: {l, 2} and G consists of the single vertex v that corresponds to X.
(n = 3): X = {I, 2, 3}. Here G if> made up of three isolated vertices.
(n = 4): X = {I 2, 3~ 4}.
has six vertices and is drawn as follows:
a: {1,2}
d: {2,4}
j
b: {3,4}
c: {I,3}
e: {1,4}
f: {2,3}
v({a,b}) and w({z,
two vertices of G. If {a,b} n {x,y} = 0, the edge
{v, w} is in G. If {a, b} n {z, y}
assume without loss of generality that a = x but
b f:. y. Hence a, b, yare three distinct elements of X and since IXI ~ 5, let c, d E X
with c d and c, d ¢ {a, b, y}. Then there exist edges from {a, b} to {c, d} and from
{c,d} to {x(=a),y},since {a,b}n{c,d}=0={c,d}n{x,y}. Hence G isoonnected.
e,
(c) For
n 2: 6
n:::::: 5
G is (isomorphic to) the Petersen graph, which is nonplanar. For
G contains a subgI"&ph isomorphic
the Petersen graph
consequently G is
a nor b is
Vit, b E
maep€:nGEmt. Conversely, if I
V with V - I a
Me
yEI
e
a t:>'.r~'".'·",,," ....."....'u.u..... UJlUeiJCli,ut:ll
<
10.
VP,'l"t,~'lt m
v-
D - {x} and
- {x}
a
I is independent but not ..u ...""-.~.... InClependent,
is a vertex v E
v is not
I and is not adjacent
vertex in I. But this contradicts I being a dominating set. Conversely, if I is maximal
vertex
V is
I or is
to a vertex
Hence I is
dominating.
(b)
<U.
(c) 1'(G) S; f3( G) follows from part (b). For the other condition,
x( G) = m. We
can partition the vertices of G into m cells 'Vi, 1
i < m, where two vertices are in
same cell if they
same color in
Each of these cells is an independent set
00 IViI (J(G), for all 1 S; i S; m. Since IVI = L~l lVii, IVI S; 2:~1 (J(G) = mf3(G) =
f3(G)X(G).
11.
Since we are selecting n edges and no two have a common vertex, the selection of n edges
will include exactly one occurrence of every vertex. vVe consider two mutually disjoint and
exhaustive cases:
(1) The edge {xn, Yn} is in the selection: Then {X n-l, xn} and {Yn- b Yn} are not in the
selection and we must select the remaining n - 1 edges from the resulting subgraph (a
ladder graph with n - 1 rungs) in an-l ways.
(2) The edge {xn) Yn} is not in tile selection: Then in order to have Xn and Yn appear in
the selection we must include edges {Xn-b Xn} and {Yn-l dlrJ. Consequently, we must now
select the other n - 2 edges from the resulting subg:raph (a ladder graph with n - 2
rungs) in a n -2 ways.
Hence an = a n -l +a n -2, ito = 1, at = 1, and an = Fn+ll the (n+ l)st Fibonacci number.
12.
There are two cases to consider:
(1)
vertex Un is not used. Then there are an-l independent subsets that contain Xn ,
and another (1.,.-1 such subsets that do not contain x n •
'Vertex '!in is included the independent subset. Now we
use either of the
vertices Xn or !In-t. Consequently, there are «$1-2 such subsets for
of the following
(i) X n --1 is
(ii) X n -l
These ('A)llsiderations give
to
recurrence
,rtr,',l""".,."..
,,,,<» ~ ;= 1.)
O.
-2r-2 =
and the
=(A1 +
+
to
V3(A 1 - A z ), so 2/../3 = (AI - A 2 ).
= (.;3 2)/2-/3, A2 = (J3 - 2)/2../3,
+ 2)/2.;3J(1 .;3)n [(.;3 - 2)/2.;3](1- V3)f6, n ~ 0 (or n ~ 1).
13.
If the vertex '!In is included the independent subset then we cannot use any the vertices
'!In-I, X n -1, or XII,' There are an -2 such
another an -2 independent subsets
where Xn, is included. In addition,
are an -l independent subsets when both Xn and
'!In are excluded. This leads us to the recurrence relation
with initial
To solve this recurrence relation let an = Arn I where A
characteristic equation
1'2 -
r -
0, r =f: 0. This leads to the
2 = 0,
and the characteristic roots -1 and 2. Therefore, an = A l ( _1)1'1. + A 2 (2 n), where All A2
are constants.
al = 3, a2 = 5 ::::} 2ao = 5 - 3 =} ao = 1.
1 = ao = At A 2 •
3 = al = -AI + 2Az = -(1 - A 2) + 2A2 = -1 + 3A z , 80 A2 = 4/3, and Al =
1- A2 = -1/3.
Consequently, an = (-1/3)(-1)1'1. (4/3)(2n), n > 0 (or n ~ 1).
14.
ao = Gl = 0
For n > 2, an = (;) = (1/2)n(n - 1) > O.
1/(1 - x) = 1 + X x 2 + x 3 + ...
(djdz)[l/(l - x)] = 1 + 2x + 3x 2 4X 3
(d/dx)[l/(l- x)] (dldx)[(l] (-1)(1- x)-2(-1) = (1- xt2
(1 = 1 + 2x + 3x2 4X 3
(d/dx)(1-x)-2J = (-2)(1-x)-3(-1) =
-xt3, so 2(1 ~·xt3 = 2
5. 4x3 + ...
=
=
3·
4·
00
n(n -
=L
VI=!)
IS
15.
:=
2; ,8(0) = 3;
(b) 0
16.
nor an Euler
m=
function for the ...... ,' ...""""....
n=8
(ii) Tn = n = 4
a Ha.IDilton
(b) (i) Km,m
m. n, has an Euler circuit but not a ..........
even and m:f:. n.
$ ...... ,,'-'..,.
cycle if m. and n are
(ii) When m, n are both even au.d m = '11;, then Km,n has both an Euler circuit
cycle.
17.
(a) X(G)
18.
(a)
w(G)
are equal.
(i) Here vertex 1 is
{b,c}, and 4 for {c,d}.
(ii) Here the co:rre8pondem~~ hpt..~,rt¥''I'l V,p''1'T.,r,f''Jil;
and edges in G is given by
1: {y,Z}i 2: {x,z}; 3: {w,x}j
4: {w,y}; 5: {u,y}; 6: {u,x}
(b) Let v E V with deg( v) = k. Then th(~re af«':l Ie
G of
{ Vi, v}, 1 :$ i :$ Ie. Any two of these edges are adjacent at
v and give rise to an
edge in L(G). Hence v brings about ee~(v)) edges in L(G). In total, L(G) hM
EvEV (de~(v}) = (1/2) Lvev deg( v )[deg( v) - 1] = (1/2) Evev deg( v)2 - (1/2) Lvev deg( v) =
(1/2) EVEV deg(V)2 - e edges.
(c) First we shall prove that L( G) is connected. Let (;1, e2 be two vertices
L( G)
where el arises from edge {a, b} and ea from edge {x, y} in G. Since G is connected
there a path in G
b
x: b ..-~ '/)1 - t Va ................. Vk ....... x and a path trom a
1.1: a - t b - t VI -l> ••• -l> Vk -l> x -l> y. These vertices and edges thel1 determine a path
L(G)
el to ez, so
e
L(G) , let
b}
be
edge in G that
:::::; (deg(a)- +
l1elClCe by
11
( e) Suppose that G = (V, E)
Vl --+ V:a --+ Va --+ ••• --+ Vn -" VI and
let el =
}, 1 ::; i ::; n en = {Vn , Vi}' Then
it cycle
on
the vertires (;i, 1 :$ i
n. If lEI = n, then this cycle is a Hamilton cycle. If lEI > n,
e E E, where e =/: ei, 1 :::; i :$
e = {Vi,V;}, 1
i < j
n.
also
takes care of the case where G
a rnultigraph.) In L( G) there are edges { eli_I? e},
where ei-l = en if i =
and
ei}, and we can extend
cycle
L(G) by replacing
{e.-I, c.;} by the edges
,e} and {e, ei}. Since lEI finite, as we continue enlarging
our present cycle in this way, we obtain a Hamilton cycle for L( G).
(f) The graph Fig. 11.99(b) has no Hamilton cycle, but its line graph, as seen in part
(a), has a Hamilton cycle.
(g) For G = KSl L(G) has 10 vertices and 30
Since G
connected, L(G) is
connected. But since 30 > 3(10) - 6, it follows by Corollary 11.3 that L( G) is nonplanar.
For G = K 3 ,3 we number the edges as shown in the first figure. Then in L( G) we find
the submph sho,,'1l in the second figure, so L( G) is nonplanar.
1
G be the graph shown here with six vertices (five pendant
one of degree 5). Then in L(G) there are five vertices each
four,
L( G) =
a nOllplan.ar graph.
" . ."<I'rfi'p
not O. This ('.antradiets Theorem 11.1
1.
m ..."_,,,.,. "'·A .. ..,
11.6.
2 ....._"_ _.....
J--.---......
= xy(x + y) is always even.
units digit either
follows. Also, if
y have the same units digit, then
x or y is 0 or 5, then. the
a multiple of 10 and so is x 3y - xy3. In all other cases we have three positive
x-y
units
the
= {1,2, 4,6,7,8,
By the
integers x, z with
V~,ii:."',""UU'''''''V principle two of these integers,
x and '!II
same component
(K4) of G. Since the component is complete, {x, y} is an edge, so either x+y or x-y
is divisible by 5. Hence x 3 y - xy3 is divisible by 10.
21.
(a) 0,1 =
0,2 = 3. For n;::: 3 label
edges are {Vb V2},
va}, ... , {Vn-b V n }.
Pn we consider two cases:
Pn as VI, V:;!,
••• , Un
the
constructing an independent subset S from
¢ S: Then S is an independent subset of Pr.-l and there are 0,'11,-1 such subsets.
(1)
Vn
(2)
Vn E S:
Then V n -l ¢ Sand S - {v n } is one of the an -2 independent subsets of
Pn -2.
Hence an = a n -1 +an -2, n 3, 0,1 = 2, a2 = 3, or an = 0,,..-1 +an -2, n
2. So an = FnH , the (n + 2)nd Fibonacci number.
2, 0,0 = 1, 0,1 =
(b) Consider the subgraph of 0 1 induced by the vertices 1,2,3,4. From part (a) we know
that this subg:raph determines 8
Fs ) the sixth (nonzero) Fibonacci number) independent
subsets of {l,2,3,4}. Therefore, the graph G1 has 1 + Fa independent subsets of vertices.
Likewise the graph G 2 has 1 F., independent subsets (of vertices), and the graph G n
determines 1 + Fn+2 such subsets.
(c) HI: 3 + F6 = (22 -1) + Fa
H'2 : 3 + F., = (22 - 1) + F.,
Hs : 3 + F~+2 = (22 - 1) + Fni-2
(d) There are
22.
- 1
vertices for graph 0' = (VI, E ' ).
m independent subsets
Proof: First we prove that G is connected.
and
with VI a vertex
= deg(Vl)
not, let 01) C 2 be two of tbe components of
Ca.
(n2-
Vz a
<
18 "'VJu..u<~'-
deg(v) = (i)(50) = 25.
-6=
G were planar,
1.
CHAPTER 12
TREES
1
1.
2.
IEII = 17 ==:::} !V2! = 18. 1V21 = 21Vi =: a6 =:.} IEzl = 35.
3.
(a) Let el, ez) •• • ,e7 denote the numbers of edges for the seven trees, and let 'VI, Vz, ••• V7,
respectively, denote the numbers of vertices. Then Vi :::: ei
1, for alII:::; i :::; 7, and
IViI = VI V2 + ... + V7:::: (el + e2 + ... + e7) + 7 = 40 7 = 47.
(b) Let n denote the number of trees in F2 • Then if ei, Vi, 1 :5 i :5
denote the numbers of
edges and vertices, respectively, in these trees, it foHows that Vi = ei 1, for all1 ::; i ::; n,
and 62 = Vl+V2 " .+v" = (el+l)+(ez+l)+ ... +(e n +l) = (el+e2+' .. +en)+n = 51+n,
so n = 62 - 51 = 11 trees in F2 •
4.
e= V -
K,
5.
A
is a tree with only two pendant vertices.
6.
(a)
no
have a
either Ks
or K 3 ,3'
(b)
T = (V,E)
'h"" ......P1i'n
11.6,
By
b
7.
.d
8.
(a) Let x be the number ofpendallt vertices. Then 21EI = EvEydeg(v) = x
1(3) + 2(4) 1(5) and lEI = IVI- 1 = x 4 1 2 + 1 - 1 = x + 7.
So 2(x 7) = x 24 and x = 10.
4(2) +
(b) 21EI = LVEY deg(v) = VI '02(2) + '03(3) + ... + tJm(m)
lEI = IVI - 1 = (VI + '02 + ... 'Om) - 1
2('01 '02+'"
Vm -l)=Vl+ 2v Z+."
mvm,so v1=v3+2v4+3v5+ ... (m-2)v m +2,
and IVI = VI '02 ... + 'Om = ['03 + 2V4 ... + (m - 2)vm + 2] V2 + Va + ... + Vm =
V2 + 2'03 + 3V4 + ... + (m - l)v m 2, lEI = IVI- 1 = '02 + 2'03 + ... + (m -I}vm + 1.
9.
If there is a (unique) path between each pair of vertices in G then G is connected. If
G contains a cycle then there is a pair of vertices x, 'II with two distinct paths connecting
x and y. Hence, G is a loop-free connected undirected graph with no cycles, so G is a
tree.
10.
31
11.
Since T is a tree, there is a unique path connecting any two distinct vertices of T. Hence
there are (;) distinct paths in T.
12.
If G contains no cycles
vertices. This
G is a tree. But then G must have at least two pendant
o:ne i!eJrlUi;W"
c
d
14.
d
spanning trees for
IS
.uu.uJ......;·... is L( n + 1)/2J.
U.UU'~...
L ....
of partitions
n into two (nonzero)
1
six spanning trees for Cs
(a) 6: Anyone of
the path connecting f to k.
(b) 6 . 6 = 36
16.
(1) This graph has 9 = 3·4 - 3 :::::: 3 3{ 4 - 2) vertices, so any spanning tree for it will have
= a . 4 edges
total) so we shall remove four edges. Two
eight edges. There are
must he removed from one 4-cycle (a cycle on four vertices) and one edge from each of
the other two 4-cydes. When two edges from a 4-cycle are removed one must be from the
a-cycle (induced by a, b, and c) - otherwise, we
a disconnected subg:raph. There are
three ways to select the 4-cyde for removing two edges and three ways to select the edge
not on the 3-cycle. We then select one edge from each of the remaining 4-cycles in 4 . 4
ways. So the number of nonidentical spanning trees for this graph is 3(4 -1)(42) = 144.
on six vertices) ,,'-","""'"'''u'....... with
(2) Here the graph has 8 = 4·3 - 4 = 4 + 4(3 - 2) vertices and 12 = 4·3 edges. There are
4(3 - 1)(33 ) = 216 nonidentical spanning trees.
(3) This graph has 16 = 4 . 5 - 4 = 4 4(5 - 2) vertices and 20 = 4·5 edges. There are
4(5 - 1)(53 ) = 2000 nouideutical spanning trees.
17.
(a) n ~ m
1
(b) Let Ie be the uumber of pendant vertices T. From Theorem 11.2 and Theorem 12.3
we have
2(n -1) = 2lEI =
deg(v) ~ Ie + men - k).
L
vEV
Consequently, {2(n - 1) ~ Ie + men - k)] =? [2n - 2 ~ k mn - mk] =? (k(m - 1) ~
2-2n mn = 2+(m-2)n ~ 2+(m-2)(m+l) = 2 m 2 -m-2 = m 2 -m = m(m-l)],
so k ~ m.
= 2(1Vj _.
20.
= 2(999) =
G', Pu {{a, b}} is a cycle. If G' contains a second cycle Cb then C1
{a, b}.
a
second cycle
PI is a path in G
P. This contradicts
==? (a):
G is not
let Ct,
Then adding
edge {a b} to G would not
with no cycles, so G is a tree.
components
a E
b E C 2•
in a cycle. Consequently, G is connected
j
21.
(a) (i) 3,4,6,3,8,4
(ii)
U
3,4,6,6,8,4
(b) No pendant vertex of the given tree appears in the sequence so the result is true
these vertices. When an
{ x 1 y} is
and y is a pendant vertex (of the
or one of the resulting subtrees),
decreased by 1
x is placed in
the sequence. As the process continues either
\"ertex x becomes a pendant vertex
in a, subtree and is removed but not recorded
ooquence, or (ii) the vertex x is left
as one of the last
of an edge. In
case x has been listed in the sequence
( deg( x) - 1) times.
(c)
(d) Input: The given
code Xl) X2," .,
Output: The unique tree T with n vertices labeled with 1,2, ... , n. (This tree T has
the Priifer code XI, X2, ••. , X n -2')
C := [Xl, Xlb .•. ,Xn -2]
L := {I, 2, ... , n]
{Initializes C as a list (ordered set).}
{Initializes L as a list (ordered set).}
. _ fA
. - VI
for i := 1 to n - 2 do
v := !\!nlLw.t~t> e!e:ll.U~nt
L
w:=
oecu,rrence
z are
23.
If the tree contains n + 1
it, is (isomorphic to) the complete bipartite
graph K1,n called the star
(b) If the tree contains n vertices then it is (isomorphic to) a path onn vertices.
24.
the
codes for
labeled trees on n
a given labeled
the Pl"ufer
tree, the pendant vertices (of degree 1) have the labels which do not appear
code for
If
are k pendant vertices, then there are k labels missing from
the code and these can be selected in (:) w.ays. That leaves n - k labels that must a.ll be
of
Priifer code. This can be counted as the number of
placed in the n - 2.
onto functions from the set of n - 2 positions to the set of n - k labels - that is,
(n - k)! Sen - 2,n - k).
result then follows by the rule of product.
25.
Let El = {{a, b}, {b, c}, {e) d}, {d, e}, {b, h}, {d; i}, if, i}, {g, iH and
Ea = {{a, h}, {o, i}, {h, i}, {g, h}, {i,g}, {e, i}, {d, I}, {e, i}}.
Section 12.2
1-
(g)
2.
(b)
(e)
(a) f,h,k,p,q,s,t
(d) e,f,j,q,s,t
(c)
(f)
a
q,t
k,p,q,s,t
a
Vertex
p
s
Level Number
35
36
36
t
37
38
v
W
x
'$I
z
(b)
u has 37
'$I
1
4.
w-xY*'#ftz3
2.1.3
5
1
x
4
2,1.1,
d
2
5.
l'j,h,g,e,d,b,a.,c,f,i,k,m,p,s,n,q,t,v,w,u .
InordeI': h,e,a,b,d,c,.g,fj,i,r ,m,s,p,k,n, v,t,w ,q,u
Postorder: a,b,c,d,e,f,g,h,ij,s,p,m,v,w,t,u,q,n,k,r
Preorde:r: l,2,5,9,14,15,10J6,17,3,6,4,7,8,11,12,13
Postorder: 14,15,9,16,17,10,5,2,6,3,7,11 ,12,13,8,4,1
7.
(a)
(i) & (iii)
(b)
(i)
(ii)
(ii)
8.
f
f
b
9.
(h)
11.
-1
( c) i = (m - l)i + 1 ==} i = (f n = mt 1 ==} i = (n -
- 1)
(Corollary 12.
is
<£S
h 1
m
< i S m ==} logm(m - ) < lo~(£) S
(h- <logmiSh==}h=
~l
h 1
-
12.
h
Theorem 12.6
we
(a) (£ - l)/(m - 1) = (n - l)/m ==} (n -l)(m - 1) = met' - 1) =>
n --I = (mi - m)j{m ==} n == [emf - m)/(m 1=
[(ml- m) (m - l)]/(m - 1) = (mf - l)/(m - 1).
(b) (i - l)/(m - 1) = (n - l)/m ==} i - I = (m - 1)(n - l)/m = }
i = [em - l)(n -1) mJ/m = [em - l)n 11jm.
13.
(a) From part (a) of Theorem 12.6 we have IVI = number of vertices
T = 3i 1 =
3(34) 1 = 103. So T has 103 - 1 = 102 edges. From part (b) of the same theorem we find
that the number of leaves in T is (3 -1)(34) + 1 = 69. [We can also obtain the number of
leaves as IVI- i = 103 - 34 = 69.J
(b) It follows from part (c) of Theoreln 12.6 that the given tree has (817 - 1)/(5 - 1) =
816/4 = 204 internal vertices.
14..
[)
(c), with l = 25, m = 2, it
16.
that i = (25-1)/(2 -
= 24.
lj"U,"'~O balls are opened and 24 mt:~tcJl1e8 are played.
17.
21845; 1
18.
2[5
52 +
rn.
= (mil. - l)/(m54 + 55 + 56 + 57 J; 2(5 5 + + 51 1
J
19.
{1,2,3,4}-'
10, 11,
- {S, 6, 7, 8}
{1, 2} - {3, 4}
{s, 5} - {7, s}
{5} {7} - {a}
1o
{12}
20.
Tbe number of vertices at level h - 1 is m h - 1 • Among these we find m h - 1 leaves of T. Each of the bh - 1 branch nodes account for m leaves (at level
1 = m h - 1 - bh - 1 mbh _ 1 = m h -- 1 + (m - 1)bh _ 1•
21.
Let T be a complete binary tree with 31 vertices. The left and right subtrees of T are then
complete binary tree8 on 2k 1 and 30 - (2k 1) vertices, respectively, with 0 ::; k ::; 14.
The number of ways the left subtree can have 11 (= 2· 5
19( = 2·9
the I
Therefore,
bh-l
1) vertices is (i) (~). This leaves
1) vertices for the right subtree where there are Uo) (~8) possibilities. So by the
rule of product there are (~)eIO)( 11o )e98) = 204,204 complete binary trees on 31 vertices
with 11 vertices
the left subtree of the root. A similar argument tells us that there are
U'l)(i~)(l)(:) 235,144 complete binary trees on 31 vertices with 21
in the right
subtree tl~e
=
24.
(a) 11,12,13,14,5,2,6,7
(b)
the
11
~
~
2.
5~
]
-2,]
-',0.2
-1,0
-1
/ 0
-
I
...
n-I
... ,ra
,
-2 \
2
\ -2I
2
'.-3,15
'.-3 /s
3
\
6 ..I -3
3
'/\3
....1/\0
.-
\5
1 ,1 13
~
1
\
1
-
II
"
&3
11,5. -8
~11,5
11
L5
1\5
-8
\
-8
11
-----~---------------------.----------
f-1.~-U6.-'Ji.1.4)
/'
'\. (ua.U,1.4}
(b)
4.
this result we use mathematical induction (the alternative form). We
g(
g(2):::; g(3):::; g(4). So we assume that for all i,j E {1,2, .,. ,n},
i < j -=} g( i) :::; g(j). Considering the case for n 1 we have two results
examine.
If n + 1 is
then n
1 = 2k + 1 for some k E
$1(21: l)=g(k)+g(k l)+fk+(k+l)-lJ=g(k) g(k
g(2k) = g(n),
g(k
2:: g(k) by
induction
.
9 IS a ~JAi!,UV"'U'''''O:;; ~U<"'''' 'l:4,""U,1};
338
case, g( n
1) =
g(k)+(2k--l) =
Section
.4
(b) tatener
(c)
1
a:
b: 110101
e:
f:
h:
01
g:
L
(a) tear
2.
x=y=z=l
3.
c:
10
0111
11011
1:
J:
010
00
110100
0001
(b) 27
~
(d) 2 10
(c)
4.
(a)
5.
Since the tree has m 7 = 279,936 leaves, it follows that m = 6, From partee)
Theorem
= (279,935)/5 = 55,987 internal vertices.
6. v = 1+m+m2+ .. ·+mh = (1-mh+ 1 )/(1-m) = (rn/t+1-l)/(m-l), sov(m-l)+l = m h +!,
h 1 = logm
1]
l~ = logm[v(m - 1) + 1] - L
12.6 we find that there are (m 7 - l)/(m -
30
30
10
2
2
3
1
filii! follows.
(a)
there are
2)
trees with smallest root weights w and
(of
most
at
in
total,
W')
Merge L2 and L'h then merge
the resulting list (for L 1 ) L 2 ) L 4 )
compansons.
resulting list (for L 2 , L 4 ) with Lh and finally merge
requires at most a total of 89 164+274 =
(d) In
to minimize the numher of comparisons in the sorting process oonstnict an
.... ".........~.. tree with the weights Wi, 1 < i ~ n, given by Wi = ILd.
Section 1
The a..rticttlation points are Il, (:;, j, h,j, k. The biconnected
are
B 1 : {{a,b}}; B'J.: {{d,e}};
B3: {{b,c},{cJ},{j,e},{e,b}}; B 4 : {{f,g},{g,h},{hJ}};
Bo: Hh,i},{i,j},U,h}}; B6: {{j,k}};
B r : {{k,p}, {p, n}, {n, m}, {m, k}, {p, m
n.
2.
If every path from x to y contains the vertex z, then splitting the vertex z will result
in at least two components GX ) Gy where x E Gre , y E G'II' If not, there is a path that
still connects x and y and this path does not include vertex z. Conversely, if z is an
articulation point of G then the splitting of z results in at least two components C1 , C2
for G. Select x E G11 Y E C2 • Since G is connected there is at least one path from x to
y, but since x and y become separated upon the splitting of z, every path connecting x
and y in G contains the vertex z.
3.
(a) T can have as few as one or as many as n - 2 articulation points. If T contains a
vertex of degree (n - 1), then this vertex is the only articulation point. If T is a path
with n vertices and n - 1 edges, then the n - 2 vertices of degree 2 are all articulation
points.
(b)
all cases, a tree on n vertices has n - 1 biconnected components. Each edge is a
biconnected component.
4.
( a) From Exercise 2, if v is an articulation
~"n"""'''' every path
.'l to y includes ..,."" ..1'""..,.
deg( v) > 1, let (I"b E II
that
v}, {v, b} E
Cb
5.
t
k}.
there are vertices x, y
deg( v) > .1. Conversely, if
splitting VP1I·t,PY
nl . 11,2 ••• ns distinct spanning trees.
6.
The graph G
7.
Proof: Suppose that has a pendant vertex, say and that {w, x} is
.,U,~,.u."",,,,, with x. Since IVI ;;:: 3 we know that deg(w) ~ 2 and
2>1=
wisan
(unique)
8.
e(;1J
d.(I;I)
Itt I)
3((3/5)
a('(,1)
(a)
the depth~flrst
tree T for G with e as the root.
(b) The second tree provides (low'( v ), lowe v» for each vertex v of G (and T). These results
follow from step (2) of the algorithm.
For the third tree we find (cin( v ), lowe v» for each vertex v. Applying step (3) of the
algorithm we:find the articulation points d, I, a.nd g, and the four biconnected components.
9.
r
C(l)
C(2'1)
f(3}
f\
b(8)
}j \
j lam
g(4
e(6)
f
d(1,1)
Jd(2)
f(2.2)
g(3, 3)
h(4, 4)
\
I)
J ~
b(1, 1)
e(2, 2)
a(3. 3)
(a) The first tree provides the depth-first spanning tree T for G where the order prescribed
for the vertices is reverse alphabetical and the root is c.
(b) The second tree provides (low'(v),low(v» for each vertex v of G (and T). These results
follow from step (2) of the algorithm.
For the third tree we find (dfi(v),low(v» for each vertex v. Applying step (3) of the
algorithm we find the articulation points d, j, and g, and the four biconnected components.
10.
The ordered pair next to each vertex v in the figure provides (dfi( v), lowe v)). Following
step (3) of the algorithm for determining the articulation points of G we see here that
this graph has four articulation points - namely, c, c, j, and h. There are five biconnected
components.
components - the figure shows the spanning trees
i(8; z)
11.
No! Fot
graph G = {V,
'~1h.""",,,,, IVl 2 2, we have
low(xt} :; low(x2) = 1. (Note: Vertices Xl and Xl are always on
same biconnected
component. )
12.
(a) The vertex set for each graph is V - {v}. If e = {x,y} is an edge in
- v then
e is not in G - v, and since X, y '# v, e is an edge in G - v. For the opposite inclusion
if e = {x) y} is an edge in G - v, theu x, y i= v and e is not an edge in G, nor the
subgrnph G - v. Here e is an edge in G - v.
Since
- v and G - v have the same vertex and edge sets, these graphs are equal.
> x;( G), so G - v
not couneded.
,-=-"""""-, ::::: 1 ::; x;( G), and consequently v
13.
graph and
E and {c,d} rJ El, but cis
!:-,~LUU.,L5 tree shown
an ancestor nor a descendant of d in the tree
d} E
Supplementary Exercises
1.
G is a tree, consider G as a rooted tree.
the root of G and (.:\ choices for coloring
there are .:\ chokes
coloring
descendants. The result
follows by the rule of product.
Conversely, if peG,
= ..\(..\ -l)n-t,
since the factor ..\ only occurs
the graph
G is con.nected. P(G,..\) = ..\()' _1)1'0-1 =..\'1'1 - (n _1»),,,,-1 +...
1)n-1), =::} G has
n vertices and (n - 1) edges. Therefore by part (d) of Theorem 12.5, G is a tree.
2.
Model the problem with a complete quaternary tree rooted at the president.
(a) SinC',c there are 125 executives (vertices) there are 124 edges (phone calls).
(b) The total number of executives making calls is the number internal vertices. From
Theorem 12.6 (c), i = (125 - 1)/4 = 61. So 60 executives, in addition to the president,
make
3.
(a)
(b) (i)
a preen-def
of every complete binary
{4o,9} {6,15}
{2,7} {-1O,35}
{-lO,4,9,35} {-2,5,6,15}
From
part oithe
of
the last two
metric property let x1~:JJ and yRx for x, y E V. xRy =::} x is on· the path from r to y. If
x
y
x
y as we
r
y. Hence
by the uniqueness of such a path we cannot have yRx.
(:cRy 1\ yRx) =::} x = y.
Lastly, let x, y, z E V with xRy and yRz. Then x on
unique path from r to y
y
is on the
path from l' to z. Since these
are unique the path
r
z must
include x so x'Rz and n. is transitive.
i«
1.
number of vertices v where deg(v) = i. Then
Xl + X2
I, so 21El = 2( -1 + Xl X7, ••• + xn-d. But
Lvev deg( v) == (Xl 2X2 + 3X3 + ... + (n ). Solving 2(-1 Xl + Xli + . . . X»-l) =
Xl +2xa
.. . +(n-l)xn._l for Xl! we :find that Xl = 2+xa+2x4+3x5 ... +(n--3)xn._l =
2 + Ueg(1J;)2 3 [deg(vi)-
8.
(a) For all e E E, e = e, so e'Re and R is reflexive.
If eh ez E E with el
e2 and e1'Re:h then e1 and e2 are edges of a cycle C of G.
Hence e2 and el are edges of the cycle C, so ez'R.el and R is symmetric.
For
1 ~
n), let Xi =
Xn-l = IV! = lEI
Let el, ea) (33 be three distinct edges with e1'Rez and e2 'Rea. Let C1 be a cycle of G
containing eh e2 and let C 2 be a cycle of G containing ea, ea- If C 1 f. C2 ) let C be
the cycle of G made up from the edges of C1 , C'},) where common edges are removed. (In
tenus of edges, C = Cl~C2') Since el, C3 are on C we have e 1'Re3, and 'R is transitive.
(b) The partition of E induced by n provides the biconnected components of G.
9.
(a) G'Z is isomorphic to K 5 •
(b) G2 is isomorphic to K 4 •
(c) G'}, is isomorphic to K n +h so the number of new edges is
(ntl) - n = (;).
( d) If CP has an articulation point x, then there exists U I'VE V such that every
path (in G2) from u to 'V pa.~ses through x. (This follows from Exercise 2 of Section
12.5.) Since G is connected, there
a patll P (in G) from u to v. If
:t
on
in a~),
we
:t being an
articulation point in
(in G) passes through x, and we can write
: u -+ 'Ul -+ ••. --+ Un-l --+ Un -+ X -+ Vm -+ 'Om-I -+ ... -+ Vi -+ 'V.
theu in
we
the edge
'Om}, #lud
path pi (in a~)
by pi: u--+
Un -+ Vm -+ Vm-l -+ ... --+ Vl --;. 'V
point
G2,
x.
X
18
no articulation points.
nu,llll1!lmn case i
=
-1)
m=
m h - 1 + (m - 1) and IVI = [m/(m - 1)J[mh - 1
we
i =
1=
-m 1
11.
(rn - 1) - 1] + 1.
= [rn/(m-
the maximum case
-m]
(a) in = l,n-l l",,-2,
n ?: 3 and l1 = £2 = 1. Since this is precisely the Fioonacci
In =
the nth
,
n 1.
recurrence relation, we
(h) in = i n - 1 + i n - 2 1, 11. ?: 3, it:::::
= 0, The summand "+1" arises when we count
the root, an internal vertex.
(Homogeneous part of solution):
;(11,) =
,;(h)..., > 3
"n
·71.-2, •• i~h) = Aa n + BPn, where a = (1 + -15)/2 alld P = (1 - -J5)/2.
(Particular part of solution):
i!r) = 0, a constant
Upon substitution into the recurrence relation Zn-2
1, n ?: 3, we find that
C=0+C+1,
so 0 = -1,
and in = Aa'" BfJn -1.
With il = i2 = 0 we have
0= it = AD! BfJ - 1
o ::::: i2 = Aa2 + B fJ2 - 1,
and consequently,
B = (a - l)/l/1{a - /1)] = r«l + .;5)/2) - 11/[«1 - vI5)/2)( J5») =
[1 .j5 - 2}1[(1 - .;5)( -J5)] = -1/..;5, and A = [1 - BfJlIa = 1/-J5. Therefore,
in ~ (l/v'5)an - (l/v'5)f3n - 1 = Fro - 1,
where Fn denotes the nth Fibonacci number, for n ?: 1.
(c) V'I'/, = in + in,
all n E Z+. Consequently, Vn = Fn + Fn - 1 = 2Fn - 1, where, as
parts (a) and (b),
denotes the nth Fibonacci number.
(3) For the graph Gs
Fig. 12.48 (d) there are 12 nonidentical spanning trees in total.
the ~Taph
spanning
f:rom the
cases any t-wu of which are mutually exclusive.
(1)
{a, n
'V'V....,,,......"'..
Gn ,
(1),
(Homogeneous Solution):
tn+! = 2tn
a
Substituting t!r)
B(n
equation (*') we :find that
1)(2'n+l) = 2Bn(2tl ) + 2n
B(2n+1) = Bn(2n+l) +
Bn(2n+1)
Consequently, 2B(2n) = 2n and 2B == 1, or B = 1/2. Therefore, in = A(2f!.)+(1/2)n(2n ) =
A(2n) + n2n- 1 •
Since il = 1 = A(2) + 1, A = 0 and tn = n2n-\ n:?: 1.
13.
(a.) For the spanning tre,es of G there are two mutually exclusive and exhaustive cases:
(i) The edge {Xl, Yd is in the spanning tree: These spanning trees are counted in bn .
(ii) The edge {XI, yd is not in the spanning tree: In this case the edges {XI, X2}, {Yb Y2}
are both in the spanning tree. Upon removing the edges {Xl,X2},{Yby:d, and {xbyd,
from the original ladder graph, we now need a spanning tree for the resulting smaller ladder
graph with n - 1 rungs. There are an-l spanning trees in this case.
(b)
Here there are three mutually exclusive and exllaustive cases:
The edges {XbX2} and {Yt,Y2} are both in the spruming tree: Delete {Xt,X2},
hll, Y2}, and {XIl ytl from the gntph. Then 6'11.-1 counts those spanning trees for ladders
vI,ritb n -1 rungs where {Xl,Y2} is included. For each of these delete {xz,yz} and add
{Xt,Xl}' {Yh Y2} and {Xl, yd·
(ii) The edge {Xl,X2} is
the spanning tree but the edge {Yt,'Y2}
not: NOW'the
removal of the edges {Xt,lh},{Xl,X2}, and {~hd/2} from G results in a subgraph tha,t
is a ladder
on n - 1 rungs.
subg,Taph
an-I spanning trees.
(i)
(iii) Here the edge {yI, Y2} is in the sparuJ.ing
CMC (ii)
preceding ar,!l~Ul~neIlt we
+
¢In-2
=0
,.:l -4,.+ 1;:::: 0
r=
edge
,X2}
are ",..-1
On
an. -
but
±
~-..,.,
=2
B(2-
=o==>
, n
not:
al =
and
1=
=-
o.
Therefore an =
14.
For n
= '11,/2
1
= rn/21
For n odd, i1 = i2 1
(b) Label the vertex of degree 1 with the
1.
the
n vertic.es
vertex
per
with the labels 2,3, ... , n, n +
(c)
IVI = 4 the only trees are a path of
3 and K 1,3' These are handled by
parts ( a) and (b)! respectively.
For IVI = 5 there are three trees: (1) A path of length 4; (2) K 1 ,4i and (3) The tree
with a vertex of degree 3. Trees (1) and (2) are handled by parts (a) and (b), respectively.
The third tree may be labeled as follows.
=
For IVI 6, there are six trees. The path of length 5 and K 1 ,5 are dealt with by parts
(a) and (b), respectively. The other four trees may he labeled as follows.
348
15.
(i) 3
(ii) 5
(b) a fl, = ~h-.:t
I
I/l.
!
..J
1.,.
coosists of
(a)
3) vertices of the spine - ordered from left to right as Vb V:'h""
(b) deg(vi), in the caterpillar, for alII::;; i < k; and
(c) n, the number of vertices in the caterpillar, with n;::: 3.
Vk;
If k = 3, the caterpillar is the complete bipartite graph (Of, star) K 1 ,n-1l for some
n 2:: 3. We label VI with 1 and the remaining vertices with 2,3, ... 1 n. This provides the
edge labels (the absolute value of
difference of the vertex labels) 1,2,3, ... , n - 1 - a
k > 3 we ,,",VLJl"'lUt:::.I.
£:= 2
{£
h:= n-1
349
end
else
begin
if
Vi has unlabeled leaves
are
on
assign the deg(vi) - 2 labels from h - [deg(vj) - 3]
to h
of Vi
assign
label h - deg( Vi) 2 to Vi+l
h := h - deg(vj} 1
end
18.
(a) Fig. 12.50
10001000010001
Fig. 12.51
100001000010000100001
(b) Yes, when the caterpillar is a path.
(c) Yes, when the caterpillar is the complete bipartite graph (or, star) K I ,n-I, where n ~ 3.
(d)
1111
I
1011
1001
.
-~,-""""",~~-.~~
T
l
•
....... .,
I
."
·4
. iii
..
!if
11111
10111
11011
i
,iIID
"'011111
I i"
",a
10101
10001
.\1.
10011
There are six nonisomorphic caterpillars on six vertices. Four of the corresponding binary
strings are palindromes.
(f) Since the caterpillar ha.'3 n vertices it ha.,; n - 1 edges, and its binary string has n - 1
bits, where
first
last
are 113. For each
n - 3 bits there are
choices - 0 or 1. This gives us
binary strings. However, for each binary string s that
(a) 1,
I,
-1
1
1, --I, 1, ---1,
-1
(b)
I'A !\ A\
~r
\
In
\.
on
are
( c) This is another example where the Catalan numbers
rooted trees on n 1 vertices.
20.
are (n! 1 )
e:) ordered
(a) Consider the case for n = 4, shown in part (a) of the figure. The five spanning
subgraphs in parts (b )-(f) of the figure provide pairwise mutually exclusive situations that
account for all the spanning trees of the graph given in part (a). As we scan the figure
from left to right we find that
3
t4=t3
t 3 +t 2
tl+tO=t3
L:tio
i=O
This result generalizes to provide t n+} = tn
~
I
Ei::o ti-
in+!
=
- tn~l' n ~
O,r # O.
-
tl = l.
Let tn::::::: Ar n , A
1 =0
,,2 _
3'1"
'I" =
(3 ± ,;5)/2
So til = B[(3 + V5)/2jn C[(3 - V5)j2]n.
Since 1 i 1 = B[(3 + VS)/2] C[(3 - -/5)/2) and
3 = t2 = B[(3 + v'5)/2)]2 C[(3 - .J5)/21 2 ,
we find that
B = 1/-/5,C = -1/.;5.
Consequently, = (1/J5)[(3 + J5)/2Y' - (1/J5)[(3 - J5)/2]n, n
=
= 1.
Recall that the nth Fibonacci number Fn is given by
Fn = (1/.;5)[(1
V5)/2]n - (1/v5)[(1 - v5)/2]n, n > O.
For n ~ 1, F2n = (1/-.15)[(1 + V5)/2]2n -- (1/v'5)[(1 - v'5)/2r~n = (1/v'5)[(1
(1/J5)[(1 - v'5Y' /4Jft = (1/-/5)[(3 + V5)/2]n - (1/V5)[(3 - .;5)/2]n = tn.
21.
v'5)2/4]?I-
(a) There are (!) -2 = 8 nonidentical (though some are isomorphic) spanning trees for the
kite induced by a, 0, c, d. Since there are four vertices, a spanning tree h8.'9 three edges and
the only selections of three edges that do not provide a spanning tree are {a, c}, { b, c}, {a, b}
and {a, b}, {a, dl, {b, d}.
(b) There are 8 . 1 . 8 . 1 . 8 . 1 . 8 = 84 nonidentical (though some are isomorphic) spanning
trees of G that do not contain edge {c, h}. These spanning trees must include the edges
{g, k}, {l,p}, and {d,a}, and there are eight nonidentical (though some are isomorphic)
spanning trees for each of the four subgraphs that are kites.
(c) Consider the kite induced by a, b, c,d. There are eight two·tree forests for this kite
that have no path between c and d. These forests can be obtained from the five edges of
the kite by removing three edges at a time, as follows:
(i)
(iii)
(v)
(vii)
{a, b}, {a, c}, {b, c}
{a,c},{a,d},{b,c}
{a,d}, {b,c}, {b,d}
{a,
(ii)
(iv)
(vi)
{a,c},{b,
{a,b},{a,d},{b,d}
{a,c},{a,d},
c}, {o,
c}
u",..
(vii),
(viii).
,{bid}
ms(~onnec:tect 'VU.lltC"'. {«, c},
f."".,... d
{o, d}
d}
13
AND MATCHING
Section 13.1
1.
(a) If
Vi E S, where 1 $ i $ m
i
Bubscript. Then
d( Vo, Vi) < d( Vo, V m +d, and we contradict the choice of Vm+l as a vertex v in S for
which d( Vo, v) is a minimum.
(b) Suppose there is a shorter directed path (in G) from Vo to Vk. If this path passes
through a vertex in S, then from part (a) we have a contradiction. Otherwise, we have a
shorter directed path P" from Vo to Vk and p il only passes through vertices in S.
But then P" U {(Vk, Vk+1), (Vk+h VkH),' .• ,(Vm-I, v m ), (Vm, VmH)} is a directed path (in
G) from Vo to Vm +l, and it is shorter than path P.
2.
(a)
Initialization:
(Counter = 0) a = Vo, So = {a}. Label a with (0, -) and
the other six vertices with (00, -).
First Iteration:
So = {b,c,f,g,h,i}
L(h} = 14, L(g) = 10, L(h) = 17.
So we have the labels: g: (10, a); b: (14, a); h : (17, a).
L(v) = 00 for v = c,f, and i. Hence V1 = g, SI = {a,g}
and the counter is increased to 1.
Second Iteration:
8 1 = {b, c,/, h, i}
L(b) =
= L(g) wt(g, b) < 14, so b now
(13,g).
L(h) = 16 = L(g)+wt(y, h) < 17, so h is now labeled (16,9).
-
+
c, f
we
< 00, so
are still .tac.e!t;~a
8 2 :;;::::: {a,
~
now
label
we find
= {e,l,
= L(b) + wt(b, c) and c
L(f) = 23 = L(o) wt(b, f) and f
L(c) =
(22, b).
(23, b).
= 16
h is labeled (16,g).
i is labeled (14, g).
Now we have Vs = i with S3 = {a, g, Ii,
increased
3.
Fourth Iteration:
the
VVU.U"'C./.
IS
8 3 = {c,
L(e) = 22 and c is labeled (22, b).
L(h) = 15 = L(i) + wt(i,h) <
h
labeled (15,i).
L(I) = 21 = L( i) wt(i, f) < 23 and f is labeled (21, i).
we have V4 = h, S4 = {a, g, 0, i, h} and counter is now
assigned the value 4.
Fifth Iteration:
S", = {c, I}
L( c) = 22 and e is labeled (22, b).
L(/) = 21 and f is labeled (21, i).
Now Vs = j, S5 = {a, g, b, i, h, f} and the counter is increased
to 5.
Sixth Iteration:
S5 = {c}
L{ c) = 22 and c is labeled (22, b).
Here Vs = c, S6 = {a,g,b,i,h,f,c}, and now counter::;;::: 6 =
7 - 1 = IV I - 1, so the algorithm terminates.
(b) c: {a,g},{g,b},{b,c}
3.
(a)
(b)
dCa, b) = 5; dCa, f: {( a, c), ( c,
h:
In
f:
{a,
d(a, I) =
,{g, i}, {i, f}
h].
Order
Lo we
a ""til""""""....
b}
[a, 0,
the
{a, g}, {g, i}
d(a, g) = 16; d(a, h) = 12
g: {( a, b), (b,
Ha, 11), (b, h)}
i:
(h, g)}
00,00); So = {a}
L1 :
00, OOt 10,
S1 =
:
L 2 [0,13,00,00,
16, 14]; S2 == {OJ g,
La: {O, 13,22,23,10,
14]; Sa = {a, g, b, i}
L,;,: [0,13,22,21,10,15,14]; S4 = {a, g, OJ i, h}
[0,13,22,21,10,
14h S5 = {OJ g, b, i, h, f}
[0,13,22,21,10,15,
= {o, 0, i, h, c}
Lo: [0,00,00,00,
5.
weighted
Section 13.2
1.
Kruskal's Algorithm generates the following sequence (of forests) which terminates in a
minimal spanning tree T of weight 18:
(1) FI == { h}},
(3) Fa = F2 U {{b,e}},
(5) Fa = F,U He,f}},
(7) F1 = Fs U {{d,g}},
Note:
answer
(2) F'J. = FI U {{a, b}},
(4) F4 = Fa U {{d,
(6) Fs = Fi) U {{a,e}},
(8) Fs = T = uHf,
en,
in.
0._-----...
0
4
3
b
.:l.
e
d
d __- - - - -..e
4
-----=----"'b
a
b
2
3
2
2
.....- - . : : : - . - - . b
d
d
2
e
3.
v -_ _ _ _
-.:~
IN'
Here V = {v,x,w}, E = {
x}, {x,w}}.
4.
Gary - Soutl1 Bend (58); South Bend - Fort Wayne (79); Fort Wayne - Indianapolis
(121); hldianapolis - Bloomington (151); Bloomington - Terre Haute (58); Terre HauteEvansville (113).
5.
(a)
(168); uU;'VJ..lI.uU:.I>VJ.J
(58)i Terre Haute - Bloomington (58);
Lu~u<:tJ..mi.J~uo.u~ (51); South Bend - Gary
Jl.U'-U<3.U4IftUlI;g",
1.
Wayne (79); Indianapolis (168); Bloomington - in<:m\llap0i16
Indianapolis -- Fort Wayne
two occurrences
Evansville (277); Fod Wayne- Eva:nsville (168).
8.
Haute (201); Gary - Bloomington (198); Indianapolis
proof for Prim's Algorithm is similar to that of Kruskal's Algorithm.
IVI = '11,
let T be a
G
by
Algorithm.
The edges ill T are labeled as el, e:<h ... , €n-l, where the subtree Si of T, obtained after
the ith iteration of the algorithm, contains
edges €}, e2,"" e,i, for some 1 S i ::; n-1.
For each optimal tree T' of G define d(T') , as
the proof of Theorem 13.1. Let Tl
an optimal tree
G where d(TI) = r is maximaL We
prove that
= Ti.
If not, then r < 11, - 1, an.d there exists an edge €r = {x, y} with e,. E T, el' ¢ TI Since
Tl is a spanning tree for G, however, there is a unique path P
x and y
in T1 • Assume that x E 5,.-1, y ¢ 5'1'_1. Select an edge e~ in
which joins a vertex
5 r - 1 with a vertex that is
S1'-l' By the minimality condition
Step 2 of Prim's
Algorithm wt( e~) ~ wt( e,.). Adding the edge er to Til er together with the edges in
P form a cycle. Deleting edge e~, the cycle becomes a path and a new subgraph of G
is obtained. Since this subgraph is connected with 11, vertices and n - 1 edges, it is a
tree T 2 , where wt(Tz) = wt(Td + wt(e r ) - wt(e~). With wt(e~) ~ wt(c r ), we find that
wt(T2 ) ::; wt(Td, and since Tl is optimal it follows that wt(T2) == wt(Tt). But then T2
is an optimal tree for the graph G with d(T2) > ", and this contradicts the choice of Tl
(where d(Tl) is maximal).
o
9.
'When the weights of the edges are all distinct, in each step of Kruskal's Algorithm a unique
edge is selected.
Section 13.3
(a) 8=2; t=4j w=5i x =9; y=4
(b)
18
(i) P = {a, 6, h, d, g, .
P
b, h, g};
P = {a,k};
== {6,d,g,i,z}
(c)
(ii)
(iii)
=
2.
If
o
= c(P,
'IJ E
= c(e)
y),
e == (x,
xE
yE
and
e=
=0
,w), v E
if
wE
= I.:
to) = L c(x, y) ._- 0 =
$EP
wEP
lIE.P
tJEP
z
z
ma:nwaJ flow is
is c(P,
for
= {a,b,d,g,h} and P = {i,z}.
4.
(Example 13.12)
wbich is c( P,
P = {a} and
= {b,g,i,j,d,h,k,z}.
(Example 13.14) Four messengers should be sent out - one for each of the IO!10Vlrl
are mutually
in pairs).
(1) a --+ b -+ h -'> P -'> z
(2) a ~ d -+ i -+ m ~ q ~ z
(3) a -+ f -+ j --+ n -+ r -'> z
(4) a -+ 9 -+ k -'> S -'> Z
paths
5.
Here c(e)
a positive
each e E
and the initial flow is defined as fee) = 0
for all e E
The result follows because .6.p is a positive integer for E'..ach application of
the Edwards-Karp algorithm and, in the Ford-FUlkerson algorithm, fee) - t::..p
not
negati ve for a back\~l'(i
6.
(a)
(b)
Section 13.4
1.
5/ (:) = 1/14
2.
(a), (b),
f
The edges
R}, {Ja~D},
{N,
ing which pairs Janice with Dennis and Nettie with Frank.
(d) No. Every complete matching must include {N,F}.
3.
Let the committees be represented as el, C:;;, •• • , C6, according to the way they are listed
in the exercise.
(a) Select the members as foHows: Cl - A; C2 - G; C3 - M; C4 - N; Cs - 1(; ct; - R.
(b) Select the nonmembers as follows:
C1 -
1(; C2 -
A; C3 - Gj c'" - Jj C5 - AI; C6 - P.
4.
(a) (4)(3) = 12
(b) (4)(3)(2)(1) = 4! = 24
(c) (9)(8)(7)(6)(5) = 9!/4! = P(9, 5)
(d) (n)(n -l)(n - 2)··· (n - rn + 1) = n!/(n - m)! = pen, m).
5.
(a) A one-factor for a graph G = (V, E) consists of edges which have no common vertex.
So the one-factor contains an even number of vertices, and since it spans G we must have
IVI even.
Consider
Petersen graph as shown in Figure 11.52 (a) of the text. The edges
{I,
(1: 'II.
We find
n
-""n-
= (2n-,
= (2)> - 1)(2n-
6.
- 5)···
(2)>)(2>> -1)(2n - 2)(211 -- 3)··· (4)(3)(2)(1)
=(2)>)(211, - 2)··· (4)(2)
(Coronary
~ X.
deg( x) ~ k
x EX, there are at least
klAI edges that are incident from the vertices in A.
edges are incident to IR(A)I
deg(y) S k for all y E Y, it follows that klAl:::; kIR(A)I, 80 we have
IAI
IR(A)I, and there is a complete matching of X into Y (by virtue of Theorem
13.7).
s
7.
such an assignment can be made by Fritz. Let X be the set of student applicants and
Y the set of part-time jobs. Then for all x EX, Y E Y, draw the edge (x, y) if applicant
x is qualified for part-time job y. Then deg(x) ~ 4 ~ deg(y) for all x E X, Y E Y, and
the result follows from Corollary 13.6.
8.
(a.) 4 E
,3 E A2 , 1 E As, 2 E A4 •
(b) 2EAt,4EA2 ,5EA3 , lEA,,,3EAs.
( c) Since I Uf::l Ai I := 4 < 5, there is no system of distinct repre,sentatives.
9.
(a) (1) Select i from Ai, for 1 SiS 4.
(2) Select i + 1 from Ai, for 1 SiS 3, and 1 from A 4 •
(h) 2
10.
(a) If there is a system of distinct representatives then I U~l Ail > »,
k > », since
I Ui,..l Ail = lAd = k, for all 1::; i S n. Conversely, if there is no system of distinct
representatives, then for some 1 < i ::; 'I'l, the union of i of the sets AI, A 2 , ••• , An
contains less than i elements. Henc,e k < i ::; n, or k < n.
(b) P(k,
11.
Proof: For
U
a E A.
and
6(G) := max{6(A)IA
(6
be the subgraph of G induced by
GAl
e > 41AI u.:;;""'~U"1C
......... g,u.,,"" deg( b) ::; 5
bE
<
vertices in
tie
(b) If
G) = 0, there is a complete matching of X into Y, and {3( G) = !Yj,
or
= {3(G) - S(G). fi(G) = k > let
X
IAj- IR(A)! = k.
AU(Y -R(A» is a largest maximal independent set in G and {3(G) = IAI+IY -R(A)I =
(lAI-IR(A)l) = iYI 5(G), so
= f3(G) - S(G).
(c) Fig'ure 13.30
Figure 13.32:
14.
Y:;h Y4d/s}i
{X3' X4, YZ,1J3, Y4}'
{XIl X:h
Proof (By Mathematical Induction):
The hypercube Q2 has vertex set V = {00,
1 and edge set
= {{OO, Ol}, {Ol, 11},
{ll, IO}, {IO,OO}}. Here
are t'Wu perfect mat("hings: {{10, 00}, {11,Ol}} and
{{IO, 1 ,{OO,OI}}. So the result is true in this first case, where n =
Assume the result is true for n = k (2:': 2) - that is, that Qk
at least 2(21<-2) perfect
matchings. Now consider the case for n = k + 1. In dealing with the hypercube Qk+b
consider the subgraphs induced by the two sets of vertices V(i) = {vlv is a vertex in Qk+l
with first component i}, i = 0,1. The subgraph of Qk+l induced by V(Q) is (isomorphic
to) Qk - likewise, for the suhgraph induced by V(l). From the induction hypothesis each
of these subgraphs has at least 2(2"-2) perfect ma,tchings. Since the two subgraphs have no
common edges, it follows from the rule of product that Qk+1 has at least 2(21.-2) ·2(2"'-2) =
2{2 k - 2 +2"-2) = 2[2(2"-2)1 = 2 2(k+I)-2 perfect matchings.
The result now follows for all n 2:': 2 by the Principle of Mathematical Induction.
Supplementary Exercises
1.
d(a, b) ==
d(a, c) = 11; d(a, d) = 7; d(a, e) = 8;
d(a,/) == 19; d(a,g) = 9; d(a, h) = 14
(Note that the loop at vertex 9 and the edges (c, a) of weight 9 and (/, e) of weight 5
are of no significance.)
2.
The algorithm is
The
Again
Step (2) u.u...",,,,a
The following weig;ht("Al di'fected graph provides a cmmterex-
of
Algorithm.
the
edges eb e2 form a circuit and G is a multigraph.
(a) In applying Kruskal's Algorithm, the only way we would have to consider edge el
as our last choice is if there is a vertex v in the graph
el = {w, v} and 'v is a
of G. This cannot
here since (;1 is part
a cycle.
.. '"" ..... 0 .....
graph K3 where the edges are assigned the VY""~,,...t.j\l''''
(b) This result is false. Let G be
= 3,
=L
-
t:r&1~po:d
fb:&-t
is determined by
the in degrees of
vertices for the capacities of the edges tenninating at the sink z; the out degrees of
vertices are used for the capacities of the edges that originate at the source a.
6.
(a) One possible selection is qs: qj
tq: tj
ut: Uj
pqr: Pi
srt: r.
(b) There are nine selections that each determine a system of distinct representatives.
Consequently, the probability that the selection yields a system distinct representatives
9/[(23 )(:r')].
n!
(b)
the sum
row or C01.U,ruw
1.
0] (0.3) 00 01 01]
[100
[~ o
o1 0
1
r rows
sum
each :row
sums
o 0
I ~]
1 0
o1
we
s
s. Consequently we
columns, we get a result of IS if we are dealing with all of the
the
sum to a
less
or
to
both 1" > s
r ~ 8, a contradiction.
18 a
of X into Y we
n edges of
{Xi, Yj},
where each of Xi and Y;, 1 ::::; i,j n,
exactly once. These edges are determined
by the n nonnegative numbers vii' where no two of
numbers are in
same row
or column
B. Writing B = clP} +
Cl is the smallest entry in B
PI
is an n X n permutatioll matrix, the sums of the entries
each row and column of B1
is 1 - Ch where 0 1 - Cl < 1.
(e) We now repeat the argument in part (d) for the ma.trix B1 and get
where the sum
the
each :row and column of B2 is 1 - Cl - C:h
1 - Cl - C2 < 1 - Ct. This process is continued until we obtaill B = C I P1 +
where 0
e2P" ... + CkPk Bk where all entries
are O.
9.
The vertices (in the line graph L( G)) determined by E' form a maximal independent
set.
PART 4
MODERN
APPLIED
ALGEBRA
CHAPTER 14
AND MODULAR ARITHMETIC
Section
(Example
(Example 14.6):
2.
-a = a,
-8
= S,
= e, - C = d, = C, -e = b
= y, -v = x, -w = w, -x = v, -y = t
(a.) This set is not a ring under ordinary addition and multiplication hecause there are
uo additive inverses.
(c) alld (d) These sets are rings under ordinary addition and multiplication.
(d)
3.
This set is not a ring because it is not dosed under multiplication.
(a)
b) + c -
(a
-
(h)
d
a(b + c)
-
(c)
c(d+b)+
-
(d) a(bc)
(ab)d
-
-
(b a) + c
b (a + c)
b+ (c a)
d + (ab + ac)
(d+ab)+ac
(ab d) + ac
ab + (d + ac)
ab + c(d + b)
ab + (cd + co)
ab + (cb + cd)
(ab cb)
(a + c)b
(ab)c + (ab)d
+d)
Commutative Law of
Associative Law of +
Commutative Law of +
Distributive Law of· over +
Associative Law of +
Commutative Law of
Associative Law of
Commutative Law of
Distributive Law of . o.ver
Commutative Law of +
Associative Law of +
Distributive Law of . o.ver
Associate Law of .
4.
ueIlmtlo.n of +, lU11:neJ:V 0, there are no
11'1 an
No..
cE Zwe
c=(a+b-1)
a
=a
c-
= a
369
(0
c-l=a
b+c-
c -1) -1 = a b c -
(ii) For
closed binary operation (0 and all a, b, c E Z, we have
(0,0 0c = (a+
ab)(0c = + ab)+c-(a+ ab)c =
aca + b + c - ab - ac - be + abc; and
a0(b0 (b c-bc) = a
c-a(b
= a+
c- -ababc =
o. + b c - o.b - ac - be abc.
Consequently, this dosed binary operation is also associative.
(iii) Given
integers a, b, c, we find that
(b $ c) 8 a = (b + c 8 a (b c - 1) a - (b + c - l)a = b + c - 1 a - ba - ca + a =
20. b c - 1 - ba - co" and
(b 0 a) ED (c (0 a) = (0 + 0.- bo.) 61 (c a - Co.) = (b + a - loa) + (c a - ca) - 1 =
20, + b + c - 1 - ba - ca.
second distributive
holds. (The proof
the first distributive
IS
Therefore
similar.)
=
(b)
all a, b E Z I
b - ab = b
a 0 b= a
a - ba = b 0 a,
because both ordinary addition and ordina.ry multiplication are commutative operations
for Z. Hence (Z, 61, 0) is a commutative ring.
(c) Aside from 0 the only other unit is 2, since 282 = 2 + 2 - (2·2) = 0, the unity for
(Z, $,8).
(d) This ring is an integral domain, but !lot a field. For all 0.,0 E Z we see that
a 8 b = 1 (the zero element) ::.} a + b - ab = 1 :;::;} a(l - b) = (1 - b) :;::;} (a -1)(1 - b) =
O::.} a = 1 or b = 1, so there are no proper divisors of zero in (Z, Efj, 0).
6.
The trouble
is with the Distributive Laws. For a, b, c E Z we find that
a 8 (b ffi c) -
while
b)
+ c - 7) = a
(b + c - 7) - 3a(b
a + b c ,- 3ab - 3ac 21a - 7
22a b c - 3ab - 3ac - 7,
0.8 (0
(a
- (a b-- (a + b - 3ab) + (a
+b c-
+ c -7)
c - 3ae)
-7
c).
we
Since
(x
z-
-
+y-
on k,m
lS ('n:t:D.m,ut.~tnre
+z-
o.n
-k
=* x + 11
z - Ie - mxy - mxz mkx = x + y - mxy + x z - mxz - k
::::;. mkJ: = J: ::::;} mk = 1 =* m = k = 1 or m = k = -1,
m,kEZ.
8.
(b) - 8 = t, -t = s, -x =
--y = Y
(d) Yes, the ring commutative.
(f) The
s, yare a pair
of (proper) zero divisors.
(a) x
(c)
(e)
9.
t(s xy) = y
No, there is no
(a) We shall verify one of the distributive laws. If a, b,e E
a 8 (b €1 c) -
-
while
a 8 (b + c + 7)
a (b + e + 7) + [a(h + c 7»)/7
a + b + c + 7 (ab/7) (ae/7) + a,
(a0 b)ffi(a0e) -
Also, the rational number
number a is -14 - a.
(b) Since a0 b = a
is commutative.
then
(a0h) (a0e)+7
a + b + (ab/7) a + c + (ae/7) 7
a b e + 7 + (ab/7) (ae/7) + a.
is the zero element, and the additive inverse of each rational
b (ab/7) = b+ a
(ba/7) = 00 a for aU a, bE Q, the ring (Q, €1, 0)
=
(c) For each a E Q, a = a 0 u = a + u + (au/7) ::::;} u[1 + (a/7)] 0 =* u = 0, because a is
arbitrary. Hence the rational number 0 is the unity for this ring.
Now let a E Q, where a 'f; -7, the zero element of the ring. Can we find b E Q so that
a0 b
0 - that is, so that a b + (ab/7) = O? It follows that a + b + (ab/7) = 0 =*
0(1 (a/7» = -a::::} b = (-a)/[l (a/7)]. Hence every rational number1 other than
is a unit.
=
10..
6E&(
11.
(a) For example, (a hi)+(c+di) = (o,+c)+(b+d)i = (c+ +(d+b)i = (c+di) (o,+bi),
Z is commutative.
of
other nr{"}npl"tu,,>;:
to be a cOIruTIutative ring with uuity follow from the co:rresponding property of (Z~ +, .).
with respect to divisors
if
bi)(c di) = (ac-bd) (bc+ad)i = 0 and
=1=
then at least one
a, b
nonzero. Assu:m.e, without loss of generality, that
a O. ac - bd = 0 => c = bd/a; be ad = 0 => d = -he/a. cd == (bd/a)(-bc/a) =
(-b 2 /a 2 )(cd) => cd(l + (b2 /a 2 » = 0 => cd(a 2 b2 ) = 0 => c = 0 or d = 0, since
a,b,c,dEZ and a O. e=O, d=-be/a=>d=O. Also d=O, c=bd/a=>c=
Hence c + = 0 and
is an integral domain.
(b) a + hi
a unit in R if there is an element c + di E R with (a bi)(c di) = 1.
1 = (a + bi)(c +
= Cae - bd) + (be + ad)i => ac - bd = be ad = 0 => c =
aj(a 2 +b2 ), d=-b/(a'1. /)2). c,dEZ=>a'J.+b2 =1=>a=
b=Oj a=O, b=
Hence, the units of R are 1, -1, i, and -i.
12.
(a)
[~ ~ 1[: ! 1= [~ ~ 1=> a + 2c = 1, 3a + 7e = 0, b 2d = 0, 3b + 7d = 1 =>
a= 7, b= -2, c= -3, d= 1.
(b)
[! ;
r
= [ (-;/2)
(0~) 1E M,( Q) but this matrix is not in M (Z).
2
13.
a b
[
14.
15.
c d]
-1
= (l/(ad - be»
d -b
-c a '
[ ]
whell ad - be =1= o.
Let U = {1,2,3} and R = P(U). Then (R,A,n) is a ring with eight elements. To
obtain a ring with 16 elements consider U = {I, 2, 3, 4}. In general, for each n E Z+, if
U =
... , n} and
= P(U), then (R,.6., n) is a ring
IRI = 2ft ,
xx=x(t+y)=xi+xy=i+y=x
= (x
= xt
= t t = s
tl1J=
+x)=
=8
s=s
x=x
ty
=
1.
(Theorem 14.5
Suppose tha.t.
U2 E R
tha.t Ul, U2 are both unity ""..Iv"...."'A~'"''
Then Ul = UfU2 = U2. The first equality holds because U2 18 a
element; the second
Ul is a unity vA....<.U'<'••L.
(Theorem 14.5 (b» Let Y1d12 E R with XYI = '!ilX = U = XY2 = '!i2X~ where u is the
unity of R. Then Yl = UYI = (Y2 X )Yl = Y2(XYl) = Y2 U ::;;:: Y2(b» If 5 is a subring of
b E S -> a b,
E S. L.lOl[lVerSel
5 = {Xl, X2,'" ,xn }.
= {;1:i + xlil ::; i ::; n} ~
Xi + Xl =
+ Xl ::::::::::} Xi = Xj,
so ITI = n and T = S. Hence Xi + Xl = Xl for some 1 i ::; n, and Xi = z, the zero
e.l.elmelilt of R. For each xES, X + S = {x
xiiI::; i:5 n} = S. With z E 5, x + Xj = Z
for some Xj E 5, so Xj = -x E S. Consequently, by Theorem 14.9 S is a subring of R.
(Theorem
2.
(a) a(b - c) = a[b + (-c)] ::;;:: ab + a( -e) = ab + [a( -c)] = ab
(b) This part is verified ill a similar way.
3.
(a) (ab)(b- l a- 1 ) = aua"-l = aa- 1 = u and (b- 1 a,-1)(ab) = b--lub = b-- 1 b = U, 80 ab is a
unit. Since the multiplicative inverse of a unit is unique, it follows that (ab)-l = [)-l a-1.
= [
(b)
2 -7]
-1
4'
16
(BA)-l = [ -9
4.
22
'
1-2]
-2
B-1 A- 1 = [
5 '
4 -15]
-9
34
( AB)-l = [
4-15]
-9
34
'
.
Let u be the unity of R and let x be a unit. Hence there is an element y E R with
xy = yx
'U. If xw = z, the 7.ero of R, then y(xw) = YZ = z and y(xw) = (yx)w =
'Uw
w. Hence x is not a zero divisor.
=
=
5.
-39]
B- 1 = [
(-ae) = ab - (ae).
(_41)-1 =_(a-1 )
s=
$
8' S =.9
S=$
lJ+w=~n+s=w
w
~-IJ
w=s
s,w=w's=s
UJ"W=W
= -w = w
'hel:l:reltU 14.9 that
.).
-
.) is a
r E R~ r s = sr = s and rw = 'Wf" ::;;;; S
v,
.).
+ 8
V
X
s
v
X
.Il
S
V
X
8
S
S
V
V
x
.s
S
x
v
.3
X
S
V
t'
X
-8
== 5, -'I} = X, -x = V.
It follows from Theorem 14.9 that (T, +,.) is a subring of (R, +, .).
Also, for all r E R we have rs, sr, rv, V1', rx, and XI'" in T, so (T,
.) is an ideal
(R, +, .).
7.
z E S, ==> z E S n ==> S n T
0.
b E S n T ==> b E S and
a h, ao E S and a + b, ao E T => a+ 0,0.1> E S n T.
a E S n T => a E S and a E T ==> -a E S and -a E T =:} -a E S n T.
So S n
is a subring of R.
8.
For z = 11 = 0 we
bET ==>
[~ ~ 1E S so S is not empty.
j
Now consider two elements of S [
that is, two matrices of the form
X
X- Y ]
X-v
y
and
V
[
v-w
v-w]
w
'
where x, y, v, w E Z. Then
(i)
x-v
(x-y)- -w)]_
X X-y] [ v v-w]_ [
[ x-y
w
(x-y)-(v-w)
y-w
y
v-w
X -
[
(x - v) - (y - w)
v
ex - v) - (y - w)
y- w
1'
an element of S; and
(ii)
v - w ] = [ xv + (x - 11)( v -x- y ][ v
[ x-y
y
v--w w
(x -- y)v
-
-w)
xv-xw
xv -
1=
X
Xl'
[
xv - yv - XW
xv-
yw
,
(x 1I1JJ
xv-
j ==[
b=xv-
yw-
(1
a -b
a-b}
b
this result
it
1=
9.
there
0" bE S with a E Tb a r¢ T'}" and b E T1;, b ¢ T 1 • Since S is a subring
R, it follows that abE S.
a +b E
or a + b E T 2 •
If
Assume without loss of generality that a + b E TI • Since a E Tl we have -a E T1 , so by
closure
we now
that
(a + b) =
a) b = bE
a contradiction.
10.
(a)
r is a proper divisor of zero we are
Otherwise, consider the function
f: R ~ R where 1(0,) = aT, for all a E R. This function I is one-to-one - if not, we
f(o,l) = !(al)
distinct elements aX, a2 in
But f(al) = f(a2) ::::;. aIr = azr =}
(0,1 - a'},)1'
z, the zero element of R. And since 0,1 - a" 1= z and r =I- z we find that r is a
proper divisor of zero. Furthermore, with R finite it follows from Theorem 5.11 that f is
also an onto function. Consequently, there is an element s in R such that sr = f( s) = 'U,
and since R is commutative we have 1'8 = 'U. With '('8 = 'U = s'(' we find that r is a unit of
=
R.
(b) The result in part (a) is not valid when R is illfinite. Consider the commutative
(Z, +,.) with unity 1. For any integer n, if n =f.
0,1, then n is neither a proper divisor
of zero nor a unit.
11.
(a) Follows by Theorem 14.9.
(b)
[~ ~]
[~ ~ 1
( c)
n
(d) S is an integral domain while R is a noncommutative ring with unity.
(e) S i. not an ideal of R
for example,
[i i][ ~ ~ 1 [i
=
not in S.
12.
(a) Let
A=[: ~l' =[~ ~
[ a+d 0
bee
o,+d, b
f ],
e, c + f,
[-a-b -c0] eS
a
S
lIowever, S
[i
n
ES
B=
Then A
] E
~f] with
=[
ce
cf E Z.
So
+
[-a ~c] =
Mz(Z),
of A/2(Z), We
[! i 1
E
14.9.
E S.
and this result i.
[~~][~ ~]=[~ ~]~
= [;: 2b], B = [
(b)
2b
20
+ = [ 2c+
2h ]
AB = [ 40e + 4bg 40f
4ce +
so
ET
A+
.
2f ] E T.
Also [
e)
4bh] = [ 2(20e + 2bg) 2(201 + 2bh) ]
4dh
2(2ce 2dg) 2(2cf 2dh)
-2b ] = [
-2c -2d
CA = [;
If C = [ ; : ] E M2(Z)
then
2aw + 2cx 2bw + 2dx ]
[ 201/ + 2ez 2by + 2dz
[2(OW
2aw
[ 2cw
+
2( -c) 2( -d)
]
and it is in T. So by Theorem 14.9 T is a sub:dng of M2(Z).
is the additive inverse
AC =
= [2(0
:] [~: ~!] =
ex) 2(bw + dX)]
2(oy + cz) 2(by + (lz)
and
[~: ~!] [; :] =
2111/ 2ax + 2hz
2dy 2cx + 2dz
1= [ 2(ow + by) 2(ax + bz) ]
dy) 2(cx + dz)
2(cw
and CA, AC E T, so T is an ideal of Mz(Z).
13.
za = z, it
Z E N(o)
=z - z =
so ~'l (1'8}a ::.;:: (81')a = s(f'a) = sz =
rla - f'2a
r' E
1'U
N(a)
r:b 7'2 E N(o),
'1"2 E N(o).
if r E
so 'f'S, 81" E N(o). Hence N(o)
= E
l'
1=
so
xEI
xEI=:.!>
-
=uel'
:::::: h,
aE
uE
= 'U.
yE
x+by=z==}x=
so u = 11 b( -by) = 11 - ay = 11 +
11 =
= b- 1 = a . .... "",... ""- x = -by = -ba = -u = u.
17.
(a)
= ('It
a)y = by and
a = au EaR, so aR::F 0. If art, a'f'',l EaR, then art - ar2 = a(r! - 1'2) E
Also,
art EaR, r E R, 1'( arl) = (art)1' =
r) EaR.
is an ideal
R.
Let a E R, a ::F z. Then a = au E
so aR = R. Since 'It E R = aR, u = ar
some r E R, and r = a-I, Hence R is a field.
If Zs, ZT
the zero
S, T, respectively,
for
s E S, t E
(s,t)EB(ZS,ZT)=(S+Zs,t ZT)=
+S,ZT t) = (ZS,ZT)EB(S,t), so (zs,
the zero element for
(s,t),(,Sl,t 1),(S2,t2) E R, (s,t) 8 [(Sl,t 1) (S2,t Z)] =
t) (81 S',htl +'
. (31 + 82),t,l (tl +' t 2) = (8' 81 8 ' 82,t·' tl +' t.' t 2) =
(8' 81, t,l tt) 9 (s· S:h t.l t 2 ) = «8, t) (S1, t 1 » «s, t) 8 (82, t2»' Hence this distributive
(a)
law follows from the corresponding law in each of the rings S, T. In the same way one
finds that the remaining ring properties are also satisfied by (R, 9,8).
(b)
For all
(821 t 2 ) 8
(c)
(S11 t 1), (S2' t 2) E R, (Sh t 1) 8
(S2, t2) = (81 • 82, t1 .' t 2) = (82 • 81,12 ,f i1) =
(81, i 1 ).
UR = (US,UT)
(d) No. Let S, T both be the field of rational numbers. In S xT there is no multiplicative
inverse for (2,0). (Also, (2,0) and (0,2) are proper divisors of zero = (0,0).)
(:)(49)
(b)
(d)
Yes, the element (u, u, u, u).
19.
(a)
20.
(a) By the given recursive definition the result is true
all m E Z+ and n = 1. Assume
the result for all m E Z+ and n = k (~ 1). Now consider 111, E Z+ and 11 = k l.
(m + n)6 = (m + (k+ 1 »a = «m + 1) k)a = (m+ l)a + ka (by the induction hypothesis)
= (ma+a)+ka (by the defiuitioll given in the exercise) = ma+(ka+a) = m€I+[(k+l)aJ =
ma + na. Hence the result is true for all m, n E Z+.
m or n is 0 the result remaius
If m,11 are both
we have m = -mll n =
111,1, nl E Z+
and
(111,+n)(€I) (-ml - 711)(a) = (m} nt)( -a) ml( -a) +
)€I +( -nl)(a) =
mn <0.
ma+na.
tn = s
nl
m = nl
nl -- nla = rna na.
=
=
=ie-a) = t(-a)+m(-a)-
is similar
= 'fruJ, na.
m<O
377
case urh"""'''''
(b ),( d), and (e). The n .."".n1'<.l for these parts are done in a similar way.
21.
(a)
each m E , (am}(a 1 ) = alna = am+! so the
true
n = 1.
the result for mE Z+ and n = k (;?::
For mE Z+, n = k+ 1, (0,11'.)(0,"') = (am)(a k +!) =
( am )( all: a) =
- ) ( a) =
- . Consequently, by
Principle of Mathematical Illduction the result is true for all m, n E Z+.
result for
In like manner, (am)''' = "Inn for aU m E Z+ and n = 1. Assuxning
m E Z+
n = k
1), we consider the ease for m E Z+ alld n = k + 1. Then
(a ln )(k+l) = (a'm)k(a m ) = (amk)(a m ) = amk+m (from the first
= "m(k+l) = amn and
the result is true for all m, n E Z+ by
Principle of Mathematical Induction.
(b)
If R has a unity tt, define aO = tt, for a E
as (a-1)n, for n E Z+.
a =F z. If a is a unit of R, define
Section 14.3
1.
(a) (i) 118 - 62 = 56 = 7(8), so 118 = 62 (mod 8)
(ii) -237 - (-43) = -194, but 8 does not divide -194, so -237 and -43 are fwt congruent
modulo 8.
Also, -43 5 (mod 8) while -237 == 3 (mod 8), so -237 and -43 are not congruent
modulo 8.
(iii) 230 - (-90) = 320 = 40(8), so 230 -90 (mod 8).
Also, 230 = 28(8) () and -90 = -12(8) () 80 230 == () == -90 (mod 8).
=
=
b) (i) 243 -76 == 167 = 18(9) + 5 so 243 and 77 are not congruent modulo 9.
Also, 243 = 27(9) 0 while 76 = 8(9) + 4. Since the remainders for 243 and 76 are
different for division by 9, it follows that 243 and 76 are not congruent modulo 9.
(ii) 700 - (-137) = 837 = 93(9), so 700 and -137 are congruent modulo 9.
(iii) 056 - (-1199) := 1143 = 127(9), so -56 and -1199 are congruent modulo 9.
2.
(a)
=
- 6 = 22, so n(> 1) is a divisor of
With 22 2 ·11, there are four divisors of 22
possible values {OJ:
, and 22.
possible valut>B for
24
-7,
(mod
5.
Proof:
we may
a=
;;::} b == c
27,
some m E Z =>
for some k E Z. And
some e E
6.
a= b
=
Proof: If (1, b
a - b = en, for some
kn = b+
m).
a-=b
m),
a - b = km, for some k E Z. Likewise, a = b (mod
eE Z. With = a - b = in, it follows that nlkm.
::::}-
Now gcd(m, n) = 1 ::::}- mx + ny =
for some x, y E Z. Consequently, k =
and since nlkmx (because nlkm) and
we
nlk.
k =
kl E Z,
a - b = km = kl(mn). Hence, a == b (mod mn).
=
=
Conversely, suppose that a b
mn). Then a - b = tmn, for some t E Z. Consequently, a - b = (tm)n => a b (mod 71,), and a - b = (tn)m ::::}- a b (mod m).
[Note that this result does not require gcd(m, n) = 1.J
=
=
7.
Let a = 8, b = 2, m = 6, and n = 2. Then gcd(m,n) = gcd(6,2) = 2> 1, a b (mod m)
and a
b (mod n).
a - b = 8 - 2 = 6 f=. k(12) = k(mn), for some k E
Hence
a 1; b (mod mn).
8.
Proof: If 31n then 71,
0 (mod 3) and 271,
and 271, - 1 2 (mod 3).
=
=
=
= 0 (mod 3). Hence 271, + 1 = 1 (mod 3)
3 J 71" then exactly one of the following occurs:
=2 (mod 3), and so 2n + 1 == 0 (mod 3), while 2n -1 =2
271, -= 1 (mod 3), and so 271, -1 = 0 (mod 3), while 2n + 1 =2
(a) 71, == 1 (mod 3)==> 271,
(mod 3), 80 31(2n + 1).
=
(b) n 2 (mod 3) = }
(mod 3), so 31(2n - 1).
9.
Proof: For n odd consider the n - 1 numbers 1,2,3, ... ,71, - 3,71, - 2, n - 1 as (n - 1)/2
pairs: 1 and (n - 1), 2 and (n - 2), 3 and (n - 3), ... , n - (n~l) - 1 and n - C·;l). The
I:i =0 (mod n).
n-l
sum of each pair is n which is congruent to 0 modulo n. Hence
i=l
When n is even we consider the n - 1 numbers 1,2,3, ... ,(71,/2) - I, (nI2), (nI2) 1, ... ,
n - 3, n - 2, n -1 as (n/2) -1
- namely, 1 and n -1,2 and n - 2,3 and n -'3, ... ,
(nI2) - 1 and (n/2) + 1 - and the sillgle number (nI2). For
pair the sum is n, or 0
,.-1
10"
= (1'1,/2)
n).
14.11) For
If Ci, bE Z,
Z =::;:;} b E a
n)
Then a - b:=
+ (b-c) = a-c=(k+1n)n, so GEe
(Theorem
b+c)]
••
c;,u~.:; the QU'LU",iV.u
379
n) and the
kEZ--=>
(Zn' .) to be a commutative ring with unity
properties of the ring (Z, +, .).
follow
corresponding
(8,) For each (l E Z+ rea) = rea), so the relation is reflexive. If a, b E Z+, rea) =
reb) =.:::} reb) = rea) so
relation is symmetric. Finally,
a, b, c E
rea) = reb)
and r(b) = r(c) ~ rea) = rCc) 80 the relation is transitive.
(b)
12.
No, 2R3, 3R5 but 5,*8. Also, 2R3, 2R5 but 41/-15.
Z11:
= [I}, [2]-1 = [5J, [3t1 = [4], [4}-1 := {3], [5J-l = [9}, [6}-1 = [2},
[7t1 = [8], [8J-1 = [7J, (9)-1 = (5), [10]-1 = (10).
Z13:
[2]-1 = [7], [3t1 = [9J, [4J-1 = [10], [5t1 = (8), [6t1 = [11J,
[7tl = [2}, [8J-1 = [5], [9]-1 = [3], [lOt1 = [4], [11t1 := [61, [12]-1 = [121.
Z17: (1)-1 = [1], [2J--1 = (9), [3]-1 = [6], {4]-1 = [13J, (5)--1 = [7], [6]-1 = [3],
[7tl = (5), [8tl = {15]' [9J-1 = [2}, [10)-1 = (12), [11}-1 = [14], [12]-1 = [10],
[13j-l = [4], [14]-1 = [11], [15t 1 = [8}, [16t1 = [16].
13.
(a)
0<6<17
1009 = 59(17) 6
17 = 2(6)+5
0<5<6
6 = 1(5) + 1
0 < 1 < 5,
so 1 = 6 - 5 = 6 - [17 - 2(6)] = 3(6) -17 = 3[1009 - 59(17)] -17 = 3(1009) -178(17).
Hence 1
(b)
14.
= (-178)(17) (mod 1009), so f17J-l = [-178] = [-178 + 1009J = [831].
[lOOt! = [Ill}
(a) Z12:
(c)
[777]-1 = [735}.
{OJ, {O,6}, {O,4,8}, {O,3,6,9}, {O,2,4,fi,8,lO,12}, Z12.
Z18:
{OJ, {O,9}, {O, 6, 12}, {O, 3,6,9,12, 15}, {O, 2, 4, 6, ... , I6}, ZU3'
Z24:
{O}, {OJ
,{O,8,16},
6,12, 18}, {O,4,8,12,16,20},
... , 21}, {O, 2, 4, 6,. .. , 20, 22}, Z24'
{O,3,
4
(0,3,6,9}
3
{0,2 ,4,6,8, 10)
{O,6}
(o,4,S}
{OJ
{O,2,4 •... ,16}
(0,6,12J
2
lO,2.4,6""'I20~22}
{O,3,6, ... ,,21}
8
(0,9J
(0,4,8, ... ,20J
{O,8,16}
3
divisors
o proper zero divisors
o
zero
.,,n...,,
(b)
72 units; 44 proper zero divisOl's
.rln'·•
16.
a list of n consecutive "U!>'"!::."'''
:retllWUOeI upon
hi =
= {a, 1,2,. , ., n- ~ so hi = 0
some 1
a,
1. For 1:::; i
n,
bi
n), 0 :5 bi ::;
Then
!
n. hi = 0 ~-:::;. lli := 0
m=
c
= d (mod n)
(ck)(c)
case of m = k+1. em = ck +1 =
(mod n),
consider
(ck ) := (4*)
n).
e
d (mod
Mathematical Induction the result
(mod n). By the Principle
? ck =:; dk
=
= am
c!"" = ekH ==
=
all
(b)
9, (a)(10")
Xn
19.
X n --l
=1
=a
+ ... +
by
=
0,
(mod 9), 10 k
lit: = 1 (mod 9) for aJJ k
for
0
... + Xl ·10
9). COllsequently, Xn • Ion + X»-1 • 10'11.-1
Xl
Xo
(mod 9).
a:$
Xo
=
=
(a)
n = 0 we have 10° = 1 = 1(-1)° so 100
(_1)0 (mod 11). [Since
1
10 - (-1) = 11, 10
(-1) (mod 1
or 10
(_1)1 (mod 11). Hence the result
true for n = 0,
the result true for n = k
1 and consider the case
k
for k + 1. Then since 10
(_l)k (mod 11) and 10
(-1) (mod 11), we have
k
101:+1 = 10 ·10 (_l)k( -1) =
l)k+l (mod 11). The result now follows for all n E N
by the Principle of Mathematical Induction.
=
=
=
=
=
If X n X n - l ' " XaXIXO = Xn • Ion X n - l ' 1071.-1 + ... + Xl' 102 + Xl' 10 + Xo denotes
an (n+l)-st digit integer, then
X n X n-l ••. X2XIXO = (--1)nxn + (_1)n-l xn _1 + ... + X2 - Xl + Xo
(mod 11).
Proof; XnXn-l •• ' X2X1XO = Xn ·Ion + X n -l ·10n - 1 + ... + X2 .102 + Xl ·10 Xo == xn( _1)n +
Xn-l( _I)n-l
... + X2( _1)2 + XI( -1) + Xo =
l)nxn + (_1)n-1 Xn _1 +... X2 - Xl + Xo
(mod 11).
(b)
20.
If a2 = a in Zp, then a2 == a (mod p), and it follows that pl(a 2 - a). But pl(a2 - a) :::?
pla(a - 1) => pia or pl(a - 1), because p is prime. V{ith 0 :$ a < p, pia => a = 0 and
pl( a-I) :;::;;}- a = 1. So the only elements in Zp that satisfy a2 = a are a = 0,1. [Or a = 0, 1
are the only idempotent elements under m.uitiplication in Zp.]
21.
Let 9::::: gcd(a,n), h = gcd(b,n).
91b, hla. gth,9ln ::::::::::} gilt;
22.
(a) 1 = 1; ~ =
-
1-I
= 7(9)
= 7(2232) + 1;
=
= 729 = 7(104) + 1; 46 = 4096 = 7(585)
1.
46656 =
7) = 1, then n == i
7) ¢::;::;;.
23.
a
0
3
f.
11
=b
(mod n):::=::} a = b 1m, for some k E Z:::=::}
hla, hln = : } hlg· Since 9, h > 0, 9 = h.
a
I!
9
14
6
9
J
(mod 7),
a
0
3
u
X
1:5 i
6
8
d
:3
6
f.
~
11
8
L
1)
6
=
=1
v
g
d
8
11
8
11
:3
L
L
n
:-
I}
e
4:
7
(J
3
6
t
n
t
14
8
11
L
t
0
16
22
17
h
7
r
10
20
U
W
W
e
4
7
e
4
7
H
1I
a
15 0
18 3
S D
p
t
r
20
s
22
W
For each e in row (2), the corresponding result below it in row (3) is e+ 3
frequently
Since the
plaintext letter e with the
we have (a) ,., = 12; (b) E(9)
=
(mod 26).
letter in
English alphabet is (;., we correspond
leU.er Q. As Q is 12 letters after e in
alphabet
9+
26) and D(a) a - 12 (mod 26).
=
For part (c) consider the following:
T
Ciphertext
(2)
(3)
5
(4) Plaintext
t
M K I
19 16
12 10 8
4 22 0 24 22
7
h
e w
a
y
w
Q
I
8
16
4
e
I
8
22
w
Q
16
4
e
Q
3
or
16
4
e
Here the results in row (3) are obtained from those in row (2) by applying the decryption
function D.
The plaintext reveals the original message as 'The Way We Were'. [This is the title of
an Acadelny award winning song sung by Barbra Streisand, as well as the title of a film
starring Barbra St:reisand and Robert Redford.]
25.
From part (c) of Example 14.15 we know that fo:r an alphabet of n letters there are n· <p(n)
affine ciphers. Here we have:
(a) 24tfo(24) = (24)[24(1 - ~)(1 - !)] = (24)(8) = 192
(b) 25<p(25) (25)125(1 -i)] = (25)(20) = 500
(c) 274>(27) (27)[27(1 -1)] = (27)(18) = 486
(d) 304>(30) = (30)[30(1 - V{l -1)(1 -1)] = (30)(8) = 240.
=
=
26.
nonnegative lUL1B,e;e:I'8 that correspond
the given plaintext
are as T..... II!n121''''W:22
t: 19
X: 23
e:4
encryption function
E(4) E
(mod
~ Z26 is given
by
E(19) =:
26), and 0: E
. 7 == 105 = 1 + 104 = 1
20
(m.od 26).
=
ciphertext letters
W
7
4:
J
9
17
h
e
r
(1) Ciphertext
(3)
( 4:) Plaintext
E
M
Y
4 12
20 10
k
u
W
22
4:
e
T
0 0 M
14 14
18 11 14 14 10
1!.
s
0
0
k
Q
Y
H
8
7
13
n
~
U X
10 20 23
6 0 19
t
a
9
J(
G
6
24
y
o
o
P
15
8
z
3
d
So the original message is
'[Spoken by Humphrey Bogart to Ingrid Bergman in the
'Here's looking at you,
Academy award winning film CMablanca.]
21.
(a) Xo = 10
Xl
== 5(xo) + 3 (mod 19) = (mod 19) = 15 (mod 19), so Xl = 15.
=5(xd + 3 (mod 19) =78 (mod 19) =2 (mod 19), so = 2.
=
3 (mod 19) == 13 (mod 19), so
= 13.
=
+ 3 (mod 19) =68 (mod 19) =11 (mod 19), so
II.
FUrther computation tells us that
= 1,
8,
= 5, = 9, and
10, the sood.
X2
X2
X3
5(X2)
X4
5(X3}
X3
X4 :::
Xs :::
X5
X1
X9 =
X8
So this linearcongruential generator produces nine distinct terms.
(b) 10,15,2,13,11,1,9,5,9,10,15,2, ....
Xo =
28.
1
Xl = 28
=28 + 1 (mod =29 (mod so = 29
=
29 28 (mod 37) =57 (mod 37), so
= 20
Further computation leads to
= 12,
= 32,
= 7,
= 2,
= 9, and = 11.
== Xl + Xo (mod 37)
X3 == X2 + Xl
(mod 37)
37)
X2
XI)
PI'oof: (By Mathematical Induction)
n 2:: (an - l)/(a ::: an - 1
[Note
the
(Z, +,' }.)
we
Xa
X3
X4
29.
37),
X6
X.,
X8
Xg
1, which can be
pre"\'iOtl8
we have
_ a4xo + c[(a4 - 1)/(0. -l)J (mod m)
_ a4xo + c(aS a'l a
m)
4
2
_ 7 Xo +
+ 7 7 + 1) (mod 9)
_ 7Xo +
4 7+
(mod 9)
_ 7xo + 4(13) (mod 9)
_ txo + 4(4) (mod
7xo + 7 (mod 9)
With X4 =
it follows from 1 := txo 7 (mod 9) that 3
(mod 9), we have 12 == (mod 9), so XI) = 3, the seed.
X4
=
31.
Since 7- 1
=4
1, andn+2be three consecutive integers. Thenn3 (n 1
(n+2)3=
3
2
3
S
3
2
n + (n + 3n:!
3n + 1) + (n
6n + 12n
8) = (3n + 15n) 9(n
1). So we
3
2
consider 3n
15n = an( n
5). If 31n, then we are finish~d. If not, then n
1
2
(mod 3) or n
2 (mod 3). If n := 1 (mod 3), then n + 5 1 + 5 == 0 (mod 3), so
31(n 2 5).
n
2 (mod 3), then n 2 + 5 9 0 (mod 3), and 31(n 2 5). All cases are
now covered, so we have 31[n(n 2 5)]. Hence 91[3n(n2 + 5)1 and, consequently, 9 divides
(3n 3 + 15n) + 9(n 2 1) = 11.3 (n + I? + (n + 2)3.
Proof:
n,n
=
=
32.
=7xo (lllod
Since 55
=
==
= 32 + 16 + 4
2
)2, we have 355 = 332.316.34.32.31,
1 = (1101
=
=
=
=
Now, 31
3 (mod 10) and 3 2
9 (mod 10), 80 3 2 • 3 1
7 (mod 10). Further,
4
2
1
4
4
3 == 81 == 1 (mod 10) 80 3 .3 , 3
7 (mod 10). With 3
1 (mod 10) it follows
16
8
32
that 3 == 1 (mod 10), 3
1 (mod 10) and 3 == 1 (mod 10). Consequently,
=
=
=
355 = 332 • 316 • 34 .32 .31 == 1·1 ·7
so the last digit (that is, the units digit)
33.
=7 (mod 10),
355 is 7.
From the presentation given in Exaruple 14.18 it follows that for n E Z+,
1)~ n, n) = n 1
(2n) the nth Catalan number.
+1 n
-II:
_
j
n)
1 (mod n)
k - 2k + 1
1
o 4=4=1
n
= 112-
h(112 -
- 8295) ::
2
2
7
:> = 16 = 2
36.
Val'
8snum: array[1..9]
i, a, b, c, result: integer;
Begin
Writeln ('Input the social security number, "
'without hyphens, one digit at a time. ')j
"V:riteln ('Input the lst
and then type a
');
Read (ssnum[l]);
Writeln (,The 1st digit is " ssnum[l]:O)j
Writeln ('Input the 2nd digit and then type a return. ')i
Read (ssnum[2Dj
Writeln (,The 2nd digit is " ssnum[2J:O);
Writeln ('Input the 3rd digit and then type a return. ');
Read (ssnum{3J)j
Writeln (,The 31'<1 digit is " ssnum[3}:O);
For i := 4 to 9 do
Begin
Writeln ('Input the', i:O, '-th digit and then type a return. ');
Read (ssnum[i]);
Writeln {'The" i:O, '-th digit is " ssnum[i]:O)
a :=
ssnum[2} ssnum[3]) Mod 5j
b := (ssnum{4] +
Mod 3;
c := (ssnum[6} + ssnum.[7J + ssnum[B)
.lO*b c;
ssnum[9]) Mod
31.
that
it is
and 3+ 1;:;:: 4.
I,
or
38.
=
(a) 3x 7 (mod 31)
Since ged(3,
in Z:n.
10(3) + 1, so 1 = 31-10(3) and
.....",........'" 3x 7 (mod
21(3x) 21(7)
(mod 31).
=
=
"*
Euclidean
we
10J = [21]. (Note: 3·21 = 63 = 2(31)+1) .
(mod 31)
x
(mod 31)
x
23
"* =
"* =
(b) 5x = 8 (mod
vVith ged(5, 37) = 1, we use the Euclidean algorithm to determine
37 = 7(5) 2,
0<2<5
5 = 2(2) +
0<1<2
So 1 = 5 - 2(2) = 5 - 2[37 - 7(5)J = 5 - 2(37) + 14(15) = 37( -2)
= [5][15]
Z37
5- 1 = [5tl =
= (nlOd 37) "* 15(5x)
Therefore, 5x
8
x
9 (mod 37).
=
:= 15(8)
{mod
in ZS1.
5(15). Consequently,
"* x =120 (mod 137) "*
==
(e) 6x 97 (mod 125)
Since 6 2·3 and 125 = 53, it follows that gcd(6, 125) = 1. Using the Euclidean algorithm
we learn that
125 = 20(6) 5,
0<5<6
6
1(5) + 1,
0<1<5
Consequently, 1 = 6 - 5 = 6 - (125 - 20(6)] = 6 - 125 + 20(6) = 21(6) + 125(-1) =
6(21)+ 125( -1) and [1] = [6][21] in Z125' So 6- 1 = [6tI = {21J and 6x :; 97 (mod 125) =}
x 21 ·97 (mod 125)
x 2037 (lIlOd 125) =} x 37 (mod 125).
=
"* =
=
=
Section 14.4
1.
s -+ 0, t -+
2.
(Theorem 14.15 (d»
n = k L
the :result follows for
eon.s!aj~
v -+ 2, w -+ 3, x -+ 4, y -+ 5
The result is true for n = 1. Assume the result for n = k and
f(a H !) = f(aka) = f(ak)f(a) = [J(a)Jk f(a) =
n E Z+ by the Principle of Mathematical Induction.
(a»
oS E
there
r ER
f(r) =
r :: UR1" = r'UR, so 8 = J(r)
f(u}?r) :::::: J(tt}?)f(r) = J(u'l')$ and .3 =
=
J
onto.
= !(run) =
3.
(R,+,'),(S,EfJ,0),(T, ,J)betherings. For all a,bER, (go/)(o,+b)=g(/(o,+b» =
= g(/(o,)
= (g 0 1)(0,)
(g (/)(0). Also, (g 0 f)(a . g(/(a· b» = g(f(o,) I(b» = g(/(a» J g(/(b» = (g 0 n(a).' (g 0 neb). Hence, go 1 is a
°
ring hOIlilOU10tI>hlSill
4.
Define
I: R -+ S by 1('r) = [~ ~
a one,-to-one function
and
f(r's) = [r;
l!
for each r E R. Then
1 is
R onto S. For all 'I",S E R,
r'~ 1= [~ ~ 1[~ ~ 1= I(r)f(s).
So f is a ring isomorphism and R is isomorphic to S.
5.
(a) Since l(zR) = zs, it follows that ZR E K and K
0. If x, y E K, then
f(x - y) = I(x + (-y» = f(x) ffi f(-y) = f(x) e fey) = Zs e Zs = Zs, so x - Y E K.
Finally, if x E K and 'I" E R, then f(rx) = 1(7')
f(x) = fer) 0 Zs = Zs, and
f(xr) = J(x) fer) = Zs fer) = Zs, so "X,Xf' E K. Consequently, K is an ideal of R.
°
(b) The kernel
°
°
{6nln E Z}.
(c) If f is one-to-one, then for each x E K, [f(x) = Zs = J(zn)] :::::::::;. [x = znj,
so K = {zn}. Conversely, if K = {ZR}, let x, y E R with f(x) = fey); Then
Zs = f(x) e fey) = f(x - v), 80 x - Y E K = {ZR}. Consequently, x - Y = ZR ==? X = y,
and J is one-to-one.
6.
(a) fI(12)(23) + 18} = J(13)· f(23) + J(18) = (1,1,3)· (1,2,3) + (0,0,3) = (1,2,4) +
(0,0,3) = (1, 2) = f(17), so (13)(23) 18 = 17 in Zao(b) 1[(11)(21) - 201 = J{ll)· f(21) - 1(20) = (1, 1)· ,0,1) - (0,
(0, 0):= (I,
1) = J{I), so
- 20 :::;; 1 Ul
24
0) =
0,1)-
7.
x (in Zzo) f(x)
~
0
(0,0)
1
(1,1)
2
3
(2,2)
(3,3)
(0,4)
(1,0)
4
5
LL
(b)
1-
x Z5)
(i)
x (in
10
11
13
14
15
16
17
18
19
(3,2)
(0,3)
(1,4)
(12)(14» =
1«17)(19)
,2)(3,4)
(0,2)(2,4) = (3,3) + (0,3) = (3,1) and
1 (3,1) = 11.
(ii) J«18)(11) -- (9)(15» = (2,3)(3,1) - (1,4)(3,0) = (2,3) - (3,0) = (3,3) and
1- (3,3) = 3.
1
8.
l(ma
to) = mJ(a) + tl(b) = m(l, 0)
9.
(a) 4
(b)
10.
(a) There are ~(15) = 15(2/3)(4/5) = 8 units in both Zu; and Z3 x Zo.
teO, 1) = (m, t)
(c)
1
No
(b) Yes. Define 1: Z15 -+ Z3 X Zs by f(O) = (0,0); J(l) = (1,1); 1(2) = (2,2); 1(3) =
(0,3); /(4) = (1,4); /(5) = (2,O); /(6) = (O,l)i J(7) = (1,2); 1(8) = (2,3); J(9) =
(0,4); /(10)=(1,0); 1(11)=(2,1); J(12) =(0,2); 1(13)=(1,3); 1(14)=(2,4). In
general, J(x) = (a, b), where 0:::; x ~ 14, and x a (mod 3), x
b (mod 5), for
o a ~ 2,
b~
=
°: :;
11.
No,
12.
Since J
has
units,
0, j-I{J)
-
ring in . . . .""<-••,.uj.'·...., 14.4
=
one unit.
0. If at,a2 E /-l(J) then j(ad,j(a2) E J. Since J is an ideal
+
E
and ala2 E j-l(J). Finally, a E
80
-aE
so al
0',2 E
==> j(a) E J ==>
Now
E
= [1]-1 =
in
= [71]
Z~n
·1
·8·
1
)j ( ael) =
E
E J ==> f( -a) E J =;To>
So the smallest positive solution is 397 and all other solutions are congruent to 397 modulo
Check: 397 = 48(8) + 3 = 4(81)
a
73.
solution for
x 3 (mod 17)
x
(mod 16)
x 0 (mod 15).
£hU<:iVU.",
congruences
=
=
=
So at = 2; a2 = 10; aa = 0; ml::;:;:
m2 = 16; m3 = 15; m = mlm2m3::;:;: 17·16·15 = 4080;
Ml = m/mt = 240; M2 = m/m'}, = 255; and Ms= m/ma = 272.
[xll = [MI]-t : : : :
= [14(17) + 2J-l : : : : [2}-1 = [9} in ZIT
[X2J = [M2]-1 = [255J-1 = (15(16) 15t I = [151- 1 = [15} in Z16
[X3J ::;:;: [M.s]-l = f272J-l = [18(15) +
= [2]-1 = [8] in Z15
x = 3·9·240 10 ·15·255 +0·8·272 = 44730 3930 (mod 4080).
=
So the smallest number of (identical) gold coins that could have been in the treasure chest
is 3930. Any other solution congruent to 3930 modulo 4080.
Check: 3930 = 231(17) + 3 = 245(16) 10 = 262(15).
15.
Here al = 1; a2 = 2; as = 3; a" = 5; m! = 2; m'}, = 3; ffia = 5; ffi4 = 7; ffi = mlffi2ffiSffi4 =
2 . 3 . 5 . 7 = 210; Ml = m/ffi1 = 105; 1vI2 = ffi/m2 = 70; Ms = m/ma = 42 and
M4 m/m4 = 30.
[xl1 = {M1J- 1 = [105}-1 : : : : (52(2) I)-I = {I)-l = [I} in Z2
[X2J : : : : [M2]-1 = [70]-1 = [23(3)
= [l}-l : : : : [1] in Z3
1
[X3] = [MS]-l = 142]-1 = [8(5) + 2J- = [2]-1 = [3J in ZS
[X4] = [M4J- 1 :; {30]-1 :; [4(7) + 2]-1 = [2]-1 = [4] ill Z7
x == 1· 105·1 2·70·1 + 3·42·3 + 5·30·4 = 1223 == 173 (mod 210).
=
So x == 173 is the slua11est positive simultaneous solution for the four congruences. Any
""-"...... "....'... would
to 173 modulo 210.
Check: 173 = 86(2) + 1 =: 57(3) + 2 =: 34(5) + 3 = 24(7) + 5.
s=
S=
E
S =: {[
~ ~] I a E Z} .
(d) True.
(f)
«(J,
2.
.R commutative ¢:::::::} ba = ab
all a , b E R {::::::::} (a b)2 = a2
3.
(b)
0
2
all a , bE{=} a 2 + aD
2ab + b2 for all a, b E R.
+2ab+
for
+ a) =;} a + a = = z.
For each a E R, a a = z =;} a = -a. For 0, b E
+ b) = + b)2 =
2
ab + ba b = a + 0,0 + ba + b :::::::::> ab + ba = z ::::::::::} ab = - ba = 00, so R is
co:mnmtative.
4.
a
hi = c + di {=} a = c, b = d
~ [_~ !] = [_~ ~],
so f is a one-to-one function. It is also onto. (Why?)
Further,
and
f«a
bi)(x
yi»
-
f«ax - by)
(bx + ay)i)
- [_(b: -a~ !: ~ ~ ]- [-~ !] [-: ! 1
X
-
J(a + bi)J(x + yi),
so f is a ring isomorphism.
5.
Since oz = z = za for all a E R, we have z E C and C # 0. If x, yEO, then
(x + y)a = xa + yo = ax + ay = a(x y), (xy)a = x(yo) = x(ay) = (xu)y = (ax)y,
(-x)a = -(xa) = -(ax) = a( -x), for all a E R, so x '11, xy, and -;x E C. Consequently,
a
6.
(a)
(b)
- 1)(22 - 2) = (3)(2) = 6
1.
, it fonows
so
a2 =
a b.
=
-
we have
391
m, n are
we ca."'} write 1 = ms
s, t E
vVith m, n > 0
that one of s,t must
positive,
the other negative . .n>.D<' ....
(without any
loss of generality) that s is negative so that 1 - ms = nt > O.
Then a'''' = bn =:;.. (an)' = (bn)t ==> ant = bnt ::=}
= bI - mll ==> a(arn )(-6) = b(bm
it
.<..LG
But with - 8 > 0 and am = b'ffl, we
(am)(-II) =
)<-s>. Consequently,
since we may use the Cancellation Law of Multiplication in an integral domain.
8.
(a)
is dosed under
and 0. For all a, 0, c E
a$b = ab = ba = bffiai aEB(bEBc) =
aEB(bc) = a(bc) = (ab)c = (ab)E&c = (a$o)
and affil = lffia = a, so EB is commutative
and associative with additive identity 1. Also, for each a E R+, a-I E R+, and a- 1 is
the ( additive) inverse of a.
Now consider 0. For a, b,e E R+, a 0 (b 0 c) = a 0 (yog2 e) = alog2(bloIl2C) = a{log2 c){los2 o)
and (a 0 b) 0 c = (a l01b /}) 0 c = a(log2 b}(loIJ2 c), so 0
associative. Also, for a, b E
R +, log" b 10g2 a = log2 a log;] b ==> 10g2 [a101;2 bJ = loga Wai2 a] :::::::::} alog2 /; = yog2 a =:}
a 0 b = b G a, so 0 commutative. In addition, a G 2 = alo~::I =
= a for all a E R+
80 2 is the multiplicative identity. Finally, a 0 (b ffi c) = a 0 (be) = a1ogz(bc) = a10g2 H1012c =
(alog2b)(alot;:<c) = (a 0 Ii) f:B (a G c), so the distributive law holds and (R+,ffi,G) is a
commutative ring with unity.
(b) For each a E R+, a 1= 1, we find that a 0 21oS;la = a10g2(21<'1:2") = a(log,,2){log2 1. ) =
a iag" 2 = 2, the unity of the ring. So (R+ ) ffi, 0) is a field.
9.
= a1 bt, y = ~ hal for all E A, bbb2 E B. Then x-y = (al-az)+(b1 -b2) E
B. If r E R, and a + b E A + B, with a E A, b E B, then ra E A, rb E B and
b) E A + B. Similarly, (a + b)r E A, + B, and A B is an ideal of R.
Lei x
A
rea
O<k< (~)=(p!)/[k!(p-k)!J=P[(l)-l)!/(k!(p-l)!j(k!(p-k)!)]
an integer because for any 0 < k < p, none of 2,3, ... , max { k, p - k} divides p when
(a)
pIS "'''A''''''''''
. By
11.
(a)
=0 (mod
p) for 0 < k <
it is dosed
set Sis
tEN,
(a) For
=3
3
= 9
4tH
3
=7
4H4
3
= 1
=7
74tH = 9 (mod 10)
4t
7 +3 = 3 (mod 10)
74tH = 1 (mod 10)
4tH
(mod
(mod 10)
(mod 10)
(mod 10)
=
So
order
units digit of 7m + 3n as 8 we must have (i) m
1 (mod 4) and
n
0 (mod 4), or (ii) m
0 (mod 4) and n == 3 (mod 4), or (iii) m == 2 (mod 4)
and n 2 (mod 4).
==
=
For case (i) there are 25 choices for m (namely, 1,5,9, ... ,93,97) and 25 choices for n
(nam.ely, 4,8,12, ... ,96,100) - a total of 25l = 625 choices for the pair. There are also
625 choices for the pair each of ca.'3eS (ii) and (iii). Consequently, in total, there are 625
+ 625 625 = 1875 ways to make the selection for m, n.
(b) For ca.'3e (i) there are 32 choices for m and 31 choices for n, and 32 x 31 = 992 choices
for the pair. There are 31 choices for each of m, n, resulting in 31 2 = 961 possible pairs,
for case (li) and case (iii). Therefore we can select m,n in this situation in 992 + 961 +
961 = 2914 ""'''ays.
(c) There are (100)2 = 10,000 ways in which one can select the pair m,n.
Here we consider three cases:
(i) m 2 (mod 4) and n 1 (mod 4);
(ii) m 3 (mod 4)
n E 2 (mod 4); and
(iii) m 0 (mod 4) and n 0 (mod 4).
For each case there are (25)2 = 625 ways to select the pair m, n. Therefore, we have 1875
ways
total.
Consequently, the probability for the problem posed is 1~~~ = 0.1875 = 3[( l~)( l~)] =
3/16.
=
=
=
=
=
n :."~ 2
k = 1 we
-- 1) =
- l)(k
n=2k
80
=k
n).
(b) \Vhen n 4k it follows tha.t (2k)3 (4k)(2k2)
=
=
n > 2
=
n).
n
(-2
[l+(n--l
-l)+(n-
_
0 (mod
_
0
n)
=
n
n
?:i = <-2")3
= (2') (mod n),
71.-1
Hence
3
n even with n/2
- by virtue of part (a).
a"",l
(ii)
n is even and divisible by 4, then by an argument similar to
in part (i) we
n-l
I)3 = (;)3 = 0 (mod n) - because of part (b).
i::::l
(iii) Finally, consider the case where n is odd. By an argument similar to the one in part
(i) we have
n-l
(n-1)/2
i::d
i=l
{n-I)/:!
I>3 = L: [i 3 + (n - i)3J = L (i + (n - i»[i2 - i(n - i) + (n - ira], where
each summand has t.he factor n -
i::::::l
making it congruent to 0 modulo n. Consequently,
n-l
2>3 = 0 (mod n).
i=1
15.
=
=
Proof: For all n E Z we find that n 2 0 (mod 5) -- when 51n - or n 2
1 (mod 5) or
n 2 == 4 (mod 5). Suppo.'le that 5 does not divide any of a, h, or c. Then
(i) 07' + b2 + (;2 :: 3 (mod 5) - when a 2 b2 c2 == 1 (mod 5);
(li) a 2 + b2 (;2 1 (mod 5) - when each of two of a 2 , b2 , (;2 is congruent to 1 modulo 5
and the other square is congruent to 4 modulo 5;
(iii) a2 b2 + (;2
4 (mod 5) - when one of a'2, 1;2, (;2 is congruent to 1 modulo 1) and
each of the other two squares is congruent to 4 modulo 5j or,
(iv) a 2 1;2
2 (mod
- when a 2
(;2:: 4 (mod 5).
=
=
=
==
= =
16.
Meta
End;
Writeln
End.
17.
From Section 4.5 we know that a - b has (1ft 1)(e2 + 1) ... ( eli: 1) positive integer divisors.
are «(;1 1)( e2 1) ... (ek + - 1 possible values for n which will make
Consequently,
a b (mod n) true.
=
18.
We use the Chinese Remainder Theorem to :find a simultaneous solution for the system of
three cougruenoos:
x 3
8)
x
4 (mod 11)
x:::: 5 (mod 15).
=
=
Here 0,1 = 3; 0,2 = 4; aa = 5; ml = 8; m2 = 11 j m3 = 15; m = ml mama = 8 . 11 . 15 = 1320;
Ml = mimI:::: 165; M2 = m/m2 = 120; and lv/a = mlm3 == 88.
[Xl.] = [M1]-1 = [165J·-1 = [20(8) + 5]-1 = (5)-1 = [5] in ZS
[X2) = [M2tl = [120]-1 = [10(11) 10J- 1 = [lotI = [10] in Zll
[xa] = [MatI = [88]-1 = [5(15) 13J- 1 = [13]-1 = [7] in ZUi
x = 3·165·240+4·120·10+5·88·7 = 10355 = 7(1320)+ 1115 1115 (mod 1320).
=
So x = 1115 is the smallest number of freshman that Jerina alld Noor could be trying to
organize for the pregame presentation.
Check: 1115 = 139(8) + 3 = 101(11) + 4 = 74(15) + 5.
CHAPTER 15
BOOLEAN ALGEBRA AND SWITCHING FUNCTIONS
Sectioll 15.1
(b)
1
(d)
1
1.
(a) 1
2.
(a) Since x has value 1, x xy + w has value 1 regardless of the values of y and w.
(b) Three assignments: (1) y: 1, w: 1; (2) y: 0, w : 1; and (3) y: 1, w : O.
(c) Two assignments: (1) y: I, w: 1; (2) y: 0, w: 1.
(d) Two assignments: (1) y: I, w : 1; (2) y: 0, w : 1.
3.
(a) 2n
(b)
4.
a)
b)
(c)
2(2 n )
(i)
(iii)
w xyz
wxyz
(ii)
(iv)
wxyz
(i)
(iii)
w+x+v Z
w x !7+z
(ii)
(iv)
w+x+V+z
w+x y+z
5.
x y z x+y xz (x+y)+(xz)
0
0
0 0
0 1
0
1
0'
1
1
0 0
1 0 1,
i 1 1 0
1.1 1
0
1
i
6.
1
1
1
0
0
0
0
0
0
0
1
0 ,
1
0
o.
oI
oI
0
0
1
0
1
1
1
1
wxyz
the d.nJ.
9 we know
binary labels for the above minterms are
wxyz: 11
15)
wxyz:
wxyz: 011
7)
wx fjz:
for 9
a product of 12 ma.xtt~:rm
1
13)
The
100I( = 9)
Con.sequently we have the ma.xterms
0000(= 0)
0110(= 3)
0110(= 6)
z
w
z
w +x + y +Z
w +X + y + z
0001(= 1)
0100(=
1000(= 8)
0010(=2)
0101(= 5)
1010(= 10)
w
w
z
x
Z
the c.n.!. for 9 is the product of these 12 maxterms.
b) 9 = Em(7, 9, 13, 15) = nAl(O, 1,2,3,4,5,6,8, 10, 11, 12, 14)
7.
(a) 264
8.
(a) few, X, y, z) = wxyz + wxyz + wx y z + wxyz
(b) f(w,x,y,z) = tv xyz
wxyz
wxyz
tvxyz
wxyz+wxyz+wxyz+wxyz+wxyz
9.
m + k = 2n
10.
If x = 0, then x + y + z = xyz =::::} x + y + z = 0 =::::} Y = z = O.
If x = 1, then x + y + z = xyz =::::} 1 = xyz ===> y = z = 1.
11.
(a) xy+(x+y)z
(b) x+y+
y
y=y(x
=x
1)
(x
y)z=y+xz+yz=y(l
y+(xyz)=x(l+yz)
(c) yz+wz
12.
z+[wz(xy+wz)]=z(y+l)+wx
wz=z{l+w) wx=z wx.
z
wx
x
= 0 =::::} x = 0 =
=::::} x = Y = 0;
x=y=z=O=:>w=1.
z)+xz=y+xz.
y=x+y.
wxyz+wz=z+wx(l+yz)+wz=
x = 'II = 0 =::::} z = 0; 7£y
(a)
!
1
9
0
0
0
0
0
0
1
It.
0
1
0
1
1
1
1
1
0 0
0 1
1 0
1
1
1
fg
0
0
0
0
0
0
1
1
lh
0
1
0
1
0
0
0
0
gh
0
0
0
fg+lh+gh
0
1
0
fg + fh
0
1
1
1
0
0
0
0
0
0
0
1
1
1
1
I
.
1
0
1
I
Alternately, fg 1h = (fg + 1)(Jg + h) = (f 1)(g + ])(J9 + h) = l(g + 1)(1g + h) =
fgg+gh tfg+th=fg+gh Og 1h=fg gh th.
(ii) fg + IIi + 19 + 1Ii = I(g + Ii) + f(g + Ii) = j·1 +]·1 = 1+7 = 1
(b)
14.
(J g)(J + h)(g + h) = (J + 9)(1 + h)
(ii) (f + g)(j + 9)(1 + 9)(1 + Ii) = 0
(i)
(a) For any f E Fn , f has value 1 whenever f has value 1 so the relation is reflexive.
If j,9 E Fn and j ~ 9 and 9 ~ j, then if f ha:" value 1
a certain assignment of
Boolean values to its n variables, g also has value 1 since f ~ g. Likewise, when 9
bas value 1, f does also, since 9 ~ j. So j and 9 have the value 1 simultaneously
and j = g, making the relation antisymmetric. Finally, if j,g, hE Fn with f
9 and
9 $ h, then if f has the value 1 so does 9 (sinc,e f:::; g) and so does h (since g:::; h).
Hence 1 <h and the relation is transitive.
(b) 19 bas the value 1 iff
9 both have value 1 so f9:::; f. When j has the value 1
so does f + g, 60 1 ~ 1 + g.
, --
f1(X, y):;;;; XY
f:kr:, y):;;;; xy
::;:::;xY
~XY
!n(X,y)-::;:!X Y
!12(X,y)::::::X y
!13{X, y):;;;; X + y
!li X,y):::::: X ii
f&\X, y) y, f6('X 1 Y) x,
h(x, y) y, fs(x, y) x,
h(X 1 y) XY + X 11
ftO(X, y) xy + xii
-::;:!
-::;:!
-::;:!
-::;:!
-::;:!
-::;:!
=/e
(vi)
-
g=
[f ffi 9 = f ffi h] =>
IO ffi 9 = 0 ffi => [g =
(vii)
+fg=je
e (f e g) = f e (f ffi
=? [(f ffi j)
!J =
15.2
x
1.
(b)
y=
+
xy
x
y
x
y
2.
(a)
(b)
0---"" xy
4.
(a)
y
....
z
z
f
x
x
(b)
y + i
'1
z
x
x + z
-
x
x, ii,
9
(a)
c '"' xy
x ______~--------------------~
7.
b)
(x fJ + y) whieh simpH~
is
'II
8.
= x + y. This accounts
xy+y=
in part
9.
(a)
w
xy
00
0
1
01
10
11
!(w,x,y)=?iy
@
(b)
f(w,x,y)=x
(c)
wx
yz
00
01
xfi
10
11
!(w,x,y,z) = xz +?iz
00
01
11
10
(d)
!(w,x,y,z) = xz + xz+ wyz or xz + xz + wxy
(e)
wx
yz
00
01
11
10
00
01
11
10
few, x, y,z) = wy Z + xyz + wyz + xyz.
(f)
wx
yz
00
01
11
10
wx
yz
01
11
10
00
00
01
10
01
11
10
(v = 0)
(v = 1) i
11
W,x,
00
=v
z
vwxz
V
wxz
vyz
10.
11
10
00
01
10
few, x, y,z) = (w + y)(x + y)(w + y + z)(x 17+ z)
11.
(a)
2
(b)
3
12.
(a)
64
(b)
32
(a.)
(c)
(e)
If-1 (0)1 = If- 1 (1)1 = 8
If- 1(0)1 = 14, If- 1 (1)1 = 2
If- 1 (0)1 = 6, If- 1 (1)1 = 10
13.
(c)
(c)
k+1
(d)
4
16
(d)
(b)
(d)
(f)
8
If-l(O)1 = 12, If- 1 (1)1 = 4
If- 1 (0)1 = 4, 1f-1(1)1 = 12
If- 1 (0)1 = 7, If- 1 (1)1 = 9
Section 15.3
1.
fCu,v,w,x,y,z)=(v+w+x+y)(u w)(v+z)(u y+z)=
(uv + uw + ux + uy + vw + w + wx + wy)(v + z)(u + y + z) =
(uv + ux + uy + (u + v + 1 + x + y)w)( v + z)( u + y z) =
(uv + ux + uy + w)(uv + vy + vz + uz + yz + z) =
(uv + ux + uy + w)(uv + vy + z) =
(uv + uvx + uvy + uvw + uvy + uvxy + uvy + wvy + uvz + uxz + uyz + wz) =
uv + wvy + uxz + uyz + wz
2.
Due to the size of this table we show only two of the simplifications.
cd ef
00 01
11
10
cd\ ef 100 01
10
00
0
o
00
0
1
01
0
1
01
0
1
11
10
0
0
11
0
10
0
1
1
(a = O,b = 0)
cd
ef
00
01
11
10
00
(a = 0, b = 1)
01
00
01
1
0;:--""""--'""---:;
(a = 1,b= 0)
g( a, b, c, d, e, f) = bf + be + ad + df.
3.
11
(a)
wx
yz
00
01
11
10
01
11
10
00
01
11
10
f(w,x,y,z) = z
(b)
wx
yz
00
00
01
11
10
x,
=xfjz
de + cf + ace
(c)
wx
yz I 00
01
11
10
wx
00 01
yz
11
10
==--::::;::=:r======~
00
01
11
10
(v = 0)
f(v,w,x,y,z)=vyz+wxyz
4.
vwz+vxy
f(a, b, c,e) = abce (2) + abce (3) + (thee (5) + a:bce (7) + abce (11) + abce (13)
(a)
(b) f = Em(2,3,5, 7,11,13)
ab\ce
00
01
11
10
00
01
abc
abe
f
11
bee
10
liCe
( c) f = E m(2, 3, 5, 7) + d(10, 11, 12, 13, 14, 15)
ab
00
01
11
10
ce
00 01
11
10
b
c
b
e
f
b+c
d
g)(d + e
(a+b+c f) (a d e g)(b+c
(a+b c df+ef)(a+d e
c f+g)(d e f
g) =
(a+b+c+df ef)(d+e+g+af)(b+c+ f+g)
df+ef)(f +g)]( d+e+g
c+af +df ag+dfg+efg](d+e
c+af df+ef+ag](d+e 9
=
bd+cd+adf +df +de1 +adg+be+ce+aef +def +ef
acf af + adf aef afg = bd cd df + ag
Section 15.4
(The second Distributive La.w). Let x = 2k1 3 k'.l5 k \ y = 2m1 3 m2 5'ms, z = 2n13n25n3 where
for 1 is 3,0 S ki,mi,ni S l.
gcd(y,z) = 281 382
where 8i = min{mi,ni},l:5 i:5 3. lcm(x,gcd(y,z» = 2t13t25t3
where ti = max{kil Si}, 1 SiS 3. Also, lcm(x,1,I) = 21£131>2 5'1,1.3, 1cm(x, z) = 2V1 3V2 5'1.13
where 'Ui = max{k j , mil, Vi = max{ki ! nil, 1 S i
3, and gcd(lcm(x, y), lcm(x, z» =
W1
2 3'W25W3 where Wi = nlll1{Uil Vi}, 1 :5 i :$ 3.
prove that lcm(x, gcd(y,
=
gcd(lcm(x, y),lcm(x, z» we need to show that ti = Wi, 1 SiS 3. If ki = 0, then
ti = Si, 'Ui = mi, Vi = ni and Wi = min{Ui, Vi} = min{mi,n.i} = Si = ti. If k i = 1, then
ti = 1 = Ui = Vi = Wi·
z»
(The Identity Laws) x 0 = the lem of x aud 1 (the zero element)
gcd of x and 30 (the one element) = x, since x is a divisor of 30.
= Xi x . 1 = the
(The Inverse Laws) x 7i = the lem of x and 30/ x = 30 (the one element of this
Boolean algebra); xx = the gcd of x and 30/x = 1 (the zero element of the Boolean
algebra).
2.
(by
x
= x" 1 + xy
xy
= x(1 + y)
= x·1
=x
(b)
Def. 15.5 (c)'
Def. 15.5 (b)
Th. 15.3 (a)'
Def. 15.5 (c)'
Follows by duality.
= x+x = 1
(b.) () ==
Follows by
0=
=0
=O=?x=x·l==
1
1) = y·1 = y
y ::;:; fj = } x fj = fj = } fi ::;:; "ff.
= w.
w
O::;:;;} w,O
1
x=*l·x=l,
y$z::;:}
-
6.
Proof:
(a) w x::::} wx = w, and y :5 z::::} yz = y. Consequently, (wx)(yz) = wy, and we
that wy = (wx)(yz) = (wy)(xz) ::::} wy :5 xz.
(b)
in part (a), w 5 x
wx = w,
y :::; z ::::} yz = y. Therefore, (w
y)(x z) =
wx + wz + yx yz = (w wz) + (y yx) = w y, by the Absorption Law
(Theorem 15.3 (by). But (w y)(x z) = w + y ::::} w y:5 x z.
*
7.
x:::; y -¢=::} xy = x. The dual of xy = x is x + y = x.
x + y = x ::::::::::.:} xy = (x y)y = xy + y = y(x
Consequently, the dual of x :5 y is y:5 x.
1) = y . 1 = y, and xy = y -¢=::} Y :5 x.
8.
2ft
9.
From Theorem 15.5(a), with
XZXl = Xl, a contradiction.
10.
If 0 and 0' are both zero elements of B then 0 = 0 + 0' = 0', In a similar way, if 1 and I'
are both one elements of B then 1 = 1 . I' = 1',
11.
(d) Since x is an atom of HI, X 0 so I(x) '# O. Let y E Bz with 0 y and y :5 I(x).
With I an isomorphism there exists z E HI with I(z) = y. Also, 1-1: Hz --+ HI is
an isomorphism so f(z):5 f(x) ==> z :5 x. With x an atom and 0 < z :5 x we have
z = x so I(z) = y = I(x), and f(x) is an atom.
12.
(a) 1(35) = f(5 7) = 1(5) U f(7) = {c} U {d} = {c,d}
f(110) = f(2 + 5 + 11) = {a,c,e}
f(210) = /(2 + 3 + 5 + 7) = {a, b,c,d}
1(330) = f(2 + 3 + 5 11) = {a, b, c, e}
distinct atoms, if
XllX'l
0, then
Xl
=
XIX2
=
51 (Since any isomorphism of finite Boolean algehras must correspond atoms.)
(b)
13.
Xl,X',l
= f(x) +
(a)
(b)
fey; =
I:
S
i ES
11, (Xl
+ X2 + . , . +
o
that
+ Xz
.."AA".......Ai"'" 10) we
one-to-one and onto.
for all x, y E HI. [Follows
Supplementary Exercises
1.
denotes the Boolean sum of Xl and X2' For n ~ 2) we U<:JJ.J.U<:J
Xl + X2 + .. .
Xn
X n +!
by (Xl + X2
Xn)
[A
definition
can be given for the Boolean product.]
For n = 2, Xl + X2 = Xl X2 is true, for this is one of the
Laws. Assume the
result for n = k
and
the case of 11, = k
+ X2 + ... + Xk + Xk+l =
X2 + ... +
X2
- ' " XkXk+!. Consequently,
the result follows for all n 2:: 2 by the Principle of Mathematical Induction.
(b) Follows from part (a) by duality.
(a)
VVl1en n = 2, Xl
y=
3.
z = 7;
x =
X2
or 25
Let v, w, x, y, z indicate that Eileeen invites Margaret, Joan, Kathleen, Nettie, and Cathy,
respectively. The conditons
(a) - (e) can then be expressed as
(a)
(v ~ w) {;;:::9 (v + w)
(b)
(d)
(c) wz+wz
(x ~ vy) {;;:::9 (x + vy)
yz+yz
(e)
x+y+xy ¢=} x
(v + w)(x + vy)(wz wz)(yz + Ii z)(x y) ¢=} (v + w)(xy + vy)(wz wz)(yz
~ (v + w)(xy + vy)( wy z
wy z) ¢=} (v + w)( w xy z + wvy z) ¢=;> v W xy z
y
y z)
Consequently, the only way Eileen can have her party and satisfy conditions (a) - (e) is to
invite only Nettie and Cathy out of this group of five of her friends.
4.
h = E m(2, 4, 6, 8) + d(O, 10, 12, 14)
5.
Proof: If x < z and 11 S z then from Exercise 6(b) of Section 15.4 we have x + 11 S z z.
And by the idempotent law we have z + z = z.
Conversely, suppose that x + 11 S z. We find that x S x + y, because x( x y) = x xy
(by the idempotent law) ::::;: x (by the absorption law). Since x
x y and x y z we
have x
a partial order is traul'ilitive. [The proof
z follows in a similar
way.}
S. If
Proof:
XSy=>x zSy+z=>1:511 x=>y X=X y=1.
x(x 'II) = x . 1 ::::;:}
0) xy = x ::::;:} xy = x ::::;:} x y.
:J:
jj ::::;:}
= x ::::;:}
-
-
=.: x .
0 ::; O.
y, z E B~
z
x(y + V) = xy + xy = xv, and x = xy :::} x :::; y.
Proof: If x = y then xy + xy = xx
xx = 0 + 0 = O.
Conversely, suppose that xfj + xy == O. Then
x = x +0 x + (xy xy)
(x
xfj) + xy, by the Associative Law of
x + xfj, by the Absorption Law (Theorem 15.3 (b)')
- (x + x)(x + y), by the Distributive Law of over·
l{x +y)
- x+y
(x + y)1
- (x+y)(y+y)
- xy + y, by the Distributive Law of + over·
(and the Commutative Law of +)
xfi + (xy + y), by the Absorption Law (Theorem 15.3 (by)
(xfj + xy) + y 1 by the Associa.tive La.w of +
- O+y = y.
9.
(a)
wx
yz
00
01
11
10
f(w,x,y,z) = w x + xy
00
01
11
10
(b)
wx
yz
00
01
11
10
wx
yz
00
01
11
(v = 1)
(v = 0)
xz
(a) g(a,b,
e) = abee (1)
(2)
a:bee
00 01
11
10
(c)
g( a, b, c, e) = L mel, 2,4,8) + d(lO, 11, 12,13, 14, 15)
ab
ce
00
01
11
10
00
01
11
10
g( a, b, c, e) : alice
9
11.
(a)
12.
1
13.
(a) 60 = 22 .3.5 so there are 12 divisors of 60. Since 12 is not a power of 2 these divisors
cannot
a Boolean
(b) 120 = 23 .3.5 and there are 16 divisors of 60. Let x = 4. Then x = 120/4 = 30
and x' x = gcd of x
"if = gcd(4, 30) = 2,
Hence although
= 24
divisors of 120 do not yield a Boolean algebra.
14.
ac = c, so
- H"...... 'U· = a( b+c). Couversely, if
= a(b+c) = ab+ac,
then ac = ac o = ac ( ab +
= (ab + ac) + ab = (all c)
=c (ab ab) = c, and
c:5 a,
ac = c:::::;::;:} c:5 a.
CHAPTER 16
GROUPS, CODING THEORY,
POLYAlS METHOD OF ENUMERATION
Section 16.1
(a) Yes. The identity is 1 and each element is its ovm .n""""...."",
(b) No. The set is
dosed under addition and there is no identity.
(c) No. The set is not dosed under addition.
(d) Yes. The identity is OJ the inverse of lOn is lO(-n) or -lOn.
(e) Yes. The identity is lA and the inverse of 9: A ~ A is 9- 1 : A ~ A.
(f) Yes. The identity is OJ the inverse of a/(2 n ) is (-a)!(2 n ).
2.
(c)
ab = ae = } a- 1 (ab) = a- 1 (ac) ==} (a- 1 a)b = (a- 1 a)e ==} eb = ec = } b = e
(d)
ba = ca = } (ba)a- 1 = (ea)a- 1 = } b(aa- 1 ) = c(aa- 1 ) ==> be = ce ==> b = c
3.
Subtraction is not an associative (dosed) binary operation - e.g.) (3 -- 2) - 4 = -3 =1= 5 =
3 - (2 - 4).
4.
(i) For all a, b, c E G l
(a 0 b) 0 c = (a + b + ab) 0 c = a b + ab c + (a + b + ab)c = a + b ab + c + ac + be + abc
a 0 (b 0 c) :::::: a 0 (b + e + be) = a + b c + be + a( b + c + be) = a b e + be + ab ac + abc.
Sinc.e
0 b) 0 e = a 0 (b 0 c) for all a, 0, c E G it follows that the (dosed) binary operation
is associative.
(ii) If x,y E G; then x 0 y = x y xy = y x yx = 11 0 x, so the (dosed) binary
operation also 'VU'~ .u",'u.
(iii) Can we find a E G 80 that x:::::: X 0 a for
x E G?
x = x 0 a -;:::} x = x + a + xu ==} 0 = a(1 +
:} a :::::: 0, because x
so 0 is
the
binary operation.
(iv) For
yf
xoy=
O=xoy=x Y
:::::::=}
-x = y(l
x)-\
of x IS
+
It '-"'Mi.'''''' ""
•
each x E Z, we have -x·~-2 E Z and xo( -x-2) = x+( -x80 -x - 2
the
x
o.
(Z,
1=-1
(-x-2)
xl,
an abelian group.
6.
(i) Forall (a,b),(u,v),
y) E S we have
(a, b) 0 [(u, 0 (x, V)] = (a, b) 0 (ux, vx y) = (aux, bux vx y)
[Ca, b) 0 (u, v)] 0 (x, = (au, bu v) 0 (x, y) = (aux, (ou + v)x + y) = (aux, bux vx + y),
so the given (dosed) binary operation is associative.
(ii) To
the identity element we need (a, b) E S such that (a, b) 0 (u, v) = (u , v) =
(u, v) 0 (a, b) for all (u, v) E
(u, v) = (u, v) 0 ( a, b) = (ua, va + b) ===> u = ua and v = va b => a = 1 and b = 0.
In addition, (1,0) 0 (u, v) = (1 . u,O . u v) = (u, v), so (1,0) is the identity for this
(closed) binary operation.
(iii) Given (a, b) E 8 can we find (c,d) E S so that (a, b)o(c,d) = (c,d)o(a, b) = (1,0)?
(1,0) = (a, b) 0 (e, d) = (ae, be + d) ===> 1 = ac, 0= bc + d:=:::} c = a-I, d = _00- 1 •
Since (a-1,-ba- 1 )o(a,b) = (a- 1 a,(-OO- 1 )a b) = (1,0), (a-l,-ba- 1 ) is the inverse of
(a, b) for this (closed) binary operation.
From (i)-(iii) it follows that (8,0) is a group. Since (1,2), (2,3) E S and (1,2) 0 (2, 3) =
(2, 7), while (2,3) 0 (1,2) = (2,5), this group is nonabelian.
7.
lJ20 = {1,3, 7,9,11,13,17,19}
lJ24 = {1,5,7,11,13,17,19,23}
8.
9.
Proof: Suppose t.hat G is abelian and that a, bEG. Then (o,b) '}; = (ab)( ab) = a( ba)b =
a(ab)b:::;:: (aa)(bb) = o, 2 b2 , by using the associative property for a group and the fact that
this group is abelian.
Conversely, suppose that G is a group where (ab):! = a2 b2 for all G, bEG. If x, y E G,
then (xy)2 = x 2 y2 =} (xy)(xy) = x 2y2 => x(yx)y = x(xy2) =} (yx)y = xy2 (by Theorem
16.1 (c» => (yx)y = (xy)y ==> yx = xy (by Theorem 16.1 (d». Therefore, the group G is
abelian.
(a) The
follows from. Theorem 16.1(b) since hoth
a-I.
-
=;;::.}
1t.
{U}; {O,6};
{l}i {l,lO};
12.
are
-
=bo,=::::>
Z12'
a
11'1,
1fi
counterclockwise rotation through i(900), 0 ~ i ~ 3; 1'1 is the refiectin
the vertical;
horizontal; r'3
diagonal from lower
to
reflection in
reflection
1"2
refiection in the diagonal from upper
to lower right.
right; and 1'4
(b)
0
1fo
'iri
1f2
'irs
1'1
1'2
1'3
1"4
11"0
1ro
1r1
'ira
1f3
'1'1
1'2
1'3
r4
1r1
1fl
1f2
1r3
1ro
7'3
7"'4
1"2
r1
1f2
11"2
11"3
1ro
1r1
1'2
1'1
1'4
1'3
1r2
1'4
1"3
1'}
7'2
1r3
11"3
1ro
1rl
1'1
. '1'1
'1"4
1'2
1'3
1ro
1r2
11"3
11"1
'1"2
1"2
1'3
1'1
1'4
1r2
11"0
11"1
11"3
1'3
1"3
1'1
1'4
"2
11"1
11"3
11"0
11'"2
1'4
1'4
1'2
1'3
1'1
11'"3
1rl
1r2
1ro
11'"0
is the group identity.
The inverse of each reflection is the same reflection. The inverse
the rotation 11'"1 is
rotation 11"3, and conversely. The inverse of the rotation 11"2 is itself. Also, the inverse of
1ro IS 1ro.
13.
(a) There are 10: five rotatiollS through i(72°),O:::; i :::; 4, and five refiections about lines
containing a vertex and the midpoint of the opposite side.
(b) For a regular n-gon (n 2: 3) there are 2'11 rigid motions. There are the n rotations
through i(360° I'll), 05 i ~ n - 1. There are n reflections. F01: n odd each reflection is
about a line through a vertex and the midpoint of the opposite side. For n even, there
are n/2 reflections about lines through opposite vertices and nj2 reflections about lines
through the midpoints of opposite sides.
14.
a{J
f14
e
2345
)
11)234'
{Ja
= p1345)
12034·!
a- 1
-
~.
15.
e
e
2345
32514 ) '
a3
-
= C2345)
3124.5 ,
p-l
-- (! ),
:::::;
-
-
2345
12345) '
16.
(a)
w == (1/ J2)(1
i)
=t
4
3
w = (1/V2)(-1
5
= (1/.;2)( -1 - i)
wi
= (1/.;2)(1 - i)
w
w =-1
w6 =
w8 = 1
(b) Let S={w"11:::;n:::;8}. Then for all 1:::;i,k:::;8,wi.w k =wm ,where m=i k
(mod 8)
1 m 8. So S is
under the binary operation of multiplication,
which is commutative and associative for
complex numbers - so, ill particular, the
complex numbers is S.
-element w8l = 1 is the identity element and, for alII:::; n :::; 7, we have (w n
- w8- n ,
so every element of S has a multiplicative inverse in S.
Consequently, S is an abelian group under multiplication.
17.
Let (!il,h 1 ),(9""h 2 ) E G x H. Then (9l,h 1 ) · (921hz) = (91 0 92,hl *' h2 ), where
91092 E G, hl * h'J E H, since (G,o) and (H, *) are closed. Hence G x H is closed.
For (9], h 1 ),(92, h2)' (g3, h s) E GxH, (91,hd'(92,h 2 )]'(93,h3 ) = (91 0 92, h 1 *h 2 )'(93,h3 ) =
«91 092) 0931 (h l * h 2) * ha) == (91 0 (92 093), hI * (h2 * hs) = (91, hd . (g2 093, h2 * h3) ==
(111, h l ) . [(92, h 2 ) • (g3) h3 )], since the operations in G and H are associative. Hence,
G x H is a.'38ociative under '.
Let eG, eH denote the identities for G, H, respectively. Then (cG,eH) is the identity
in G x H.
Finally, let (g, h) E G x H. If g-1 is the inverse of 9 in G and h- 1 is the inverse of
h in H, then (g-1,h- 1 ) is the inverse of (g,h) in G x H.
(b) (i) 216
(ii) H t = Hz, 0, O)lz E Z6} is a subgroup of order 6; H2 == {(x, 'II, O)lx, 'II E Zfh Y = 0,3}
is a subgroup of order 12; H3 = {(x, y, O)lx, y E Z6} has order 36.
(iii) -(2,3,4) = (4,3,2); -(4,0,2) = (2,0,4); -(5,1,2) = (1,5,4).
18.
wehave e E HnK and HnK 0. Now let x, 'II E H n K.
nK=}x,yEH and x, 'liE==>
xy E K
K are
E H
( a) Since e E H and e E
~xyEHn
nK = } x E
are
ERn
x
= 1, x = 4:
x ::;;:
(mod p) ;::;}
X=
::::;..
=°
x=
-1 _
°
p) "'* x E 1
_
{}
1
(mod
=
result is true
p = 2,
(2 - 1)1 = 1!
(mod 2).
p
elements 1,2, ... ,p -1 in (Z;, .), The
2,3, ... ,p - 2 yield (p - 3)/2 pairs of
the
x,
when p = 11 we
that 3,4, ... 9 yield
four pairs
2,6; 3,4; 5,9; 7,8.] Consequently, (p - I)! (1)(1)(p-3)/2(p -1) p - 1 --1 (mod p).
(d)
=
20.
=
.) we
32 =
so 3 = 3-\ and 52 = 1, so 5 =
2
(b) In (UH",) we have 7 = 1,807 = 7- 1 , and 92 = 1, 809 = 9- 1 ,
(c) Let x = (2 k - 1 - 1) in (U2k, .). One finds that
= (2 k - 1 2·21.:-1 + 1 = (2k)(2k-2) - 2k 1 = 0(2k-2) - 0 + 1 = 1, so x =
x = (211:-1 + 1).
j
=
(a)
-
1) =
. This is also true
Section 16.2
1.
(c)
If n = 0, the result follows from part (a) of Theorem 16.5. So consider n E Z+.
For n = 1, J(a n ) = f(a 1 ) = f(a) = (f(a)F = [J(a)J'\ so t.he result follows for n = 1.
Now assume the result true for n = Ii: (;?: 1) and consider n = k + 1. Then J( an) =
l(a H1 ) = J(a k -a) = /(ak).J(a) = (/(a»)I.:. I(a) = (f(a)]kH = [J(a)Jn. So by the Principle
of Mathematical Induction, the result is true for all n;::: 1.
we have a- n = (a-1Y, - as defined in the material following Theorem 16.1. So
I(a-") = fl(a- I )",] = [/(a-lW~ by our previous work. Then [j(O-l)JI'l, = [(J(a»-I]n =
[J(a)J-n -- by part (b) of The,orem 16.1. Hence /(0-'11,) = [f(a)J-n.
For n;?:
Consequently, f(a n ) = [/(a)]\ for aU a E G and all n E Z.
2.
(a)
A' = [
n-
A = ~ -~ 1, and A4 = [~
-~ 1'n 4,Am.
=
= AT, where 1:5 :5 4
3
0
(b) For
Hence the
1
{A , A2 , A 1 A4}
3
=
m
r (mod 4).
Clolse<1 under the binary
of matrix multiplication.
operation
2x 2
four matrices.
A'n+n
ffi 1
l'
-- A2 1
I:
A---+i
I:
or
=i
AS ---+
= i3
A4 ---+ 1 = i4
2
---+
A ---+ -i
A2 ---+ -1 =
AS ---+ i = (_i)3
---+ 1 = ( _i)4
In either case I is an isomorphism for the two given cyclic groups of order 4.
3.
1(0) = (0,0)
J(3) = (0,1)
J(l) = (1,
J(4):::
0)
J(2) = (2,0)
1(5) = (2,
4.
Let x, y E H. Since J
onto, there exist a, bEG with f(a) = x, feb) = y. Then
xy = f(a)f(b) = f(ab) = f(ba) (since G is abelian) = f(b)f(a) = yx, so H is abelian.
5.
We need to express the element (4,6) of Z x Z in terms of the elements (1,3) and (3,7),
so let us write
(4,6) = a(l, 3) e b(3, 7), where a, bE Z.
Then f(4, 6) = f(a(l, 3) EB b(3, 7» = f(a(l, 3» f(h(3, 7» = af(l, 3) bf(3, 7).
With (4,6) = a(l, 3) e b(3, 7) we have 4 = a + 3b and 6 = 3a + 7b, from which it follows
that a = -5 and b = 3.
Consequently, f( 4,6) = -591 392.
6.
(a) For each k E Z, we find that (k,O) E Z x Z and. f(k,O) = k - 0= k, so the function f
is onto Z. Furthermore, if (a, b), (c,d) E Z x Z, then f«a,b) e{c,d» = f(a+c,b+d) =
(a c) - (0 + d) = (a - b) (c - d) = f(a, b) + f(c,d). Consequently, the function f is a
homomorphism onto Z.
(b) If f(a, b) = 0, then since f(a, b) = a - 0, it follows that a = b. Also, a = b => a - b =
0,* tea, b) = O. Hence f(a, b) = 0 if and only if a = b, or 1-1(0) = {(a, a)la E Z}.
(c) Since /-1(7) = Ha, b)lJ(a, b) = a - b = 7}, here we may also write /-1(7) =
{(b + 7, b)lb E Z} = Ha, a - 7)la E Z}.
(d)
f(a~ b) = a -
(a, b) E Z x Z. We find that (a, b) E f-l(E) if
1.
nE
(b)
--
16,6)
)=
U
)=
) = Q( 1is) :=
(.I
= 2.
n=2:(~ 21 33 4:4 ~)
{( ~
2 3 4
1 3 4:
2
~ ), ( ~ 2 3 4 ~) }
2 3 4
1,
;:.;: o(r'l) =
n E
b is an
n=3:
3 4 5)
(~ 32 145
order 3
{( 1 2 3 4 5) (~ 2 3 4 5 )
231 4 5 ' ~ 1 2 4 5 '
n=4:
(1
5)}
s.o.
234
1 234 5
(~ 32 43 41 ~ ) has order 4 and generates the cyclic subgroup
5) (1
{( 1 2 3 4
234 1 5
'
2 345)(12345) (1234
3 4 125'
41235'
1234
n=15:(~ 32 43 45 ~ ) has order 5 and generates the cyclic subgroup
{(~ ~ ! : ~),(~ ! ~ ~ ~),(! ! i ~ ~),
(~ i ~ ~ !), (~ ~ ~ : ~)} of S5'
9.
(a) The elements of order 10 are 4, 12, 28, and 36.
(b) The elements of order 10 are a\ a 12, a'll!" and a 36 •
10.
(a) U14 = {1,3,5,9, 11, 13} = {a E Z+ lIsa 5 13 and gcd(a, 14) = l}.
(b) Since
31 = 3
32 = 9
33 = 13
34 = 11
35 = 5
36 = 1,
we
f,hat
18
We also
that
=5
54 =9
80 UH
12.
= (5).
are no
=3
~ )} of S5'
the given groups. For any x,v E G, (gof)(x
y) =
= (g(1(x» . (g(l(y)) =
0 I)(x» .
0 f)(y»), since f, 9
are homomorphisHlS. Hence, 9 0 I : -+
18 a
(H,*), (K,·)
g(/(x + y» = g(1(x) *
«9
From Exercise 16
that
G = (w ) = {w ) = (w"f).
3
«9
G = (w). It is
true
5
= [n), 1 :::; n :::;
1 < k, m
8, then w k =
wm ~ k = m ~ [kJ = [mJ {:::::::::> I(wi» = I(w m), so I is a one-to-one function. Since
IGI = IZs!, it follows
5.11 that I is also onto. Finally, for 1 k, m :::; 8,
m
f(w"· w ) = f(wk+m) = (k m] = [k] [m] = I(w k ) f(w m), so I is an isomorphism.
I : G -+ Zs by
Note: Three other isomorphisms are also possible here. They are determined, in each case,
by the imag-e of w. We find these to be:
fl : G -+ Zg, where flew) = [3];
!2: G -+ Zg, where 12(W) = [5];
h : G -+ Zg, where h(w) = [7].
15.
(Z12l +) = (I} = (7) = {ll}
(Z161 +) = (I) = (3) = (5) == {7} == (9) = (11) = (13) = (15)
(Z24! +) = {I) = (5) = (7) = {HI = {13} = {17} = (19) == (23)
(a)
(h) Let G = (ale). Since G = (a), a = (ak)S for some s E Z. Then a1- k$ = e, so
1- ks = tn since Q( a) = n. 1- ks = in ==:} 1 = ks + tn ==:} gcd( k, n) = 1. Conversely, let
G = {a} where an E G and gcd(k, n) = 1. Then (a k ) ~ G. gcd(k, n) = 1 ==? 1 = ks+tn,
for some 5, t E Z ==:} a = at = ak.s+ nt = (ak)"(any = (ak)ll(e)t = (a k )3 E {a k}. Hence
G (aN). So G = {ale), or ak generates G.
(c)
16.
¢(n).
II k J n, let n = qk T', 0 < r < k. Then I(an) = fCe-G) = eH and J(a n ) = (/(a»n =
(f(a»9HT = (f(a)k)Q(1(aY) = (f(a)Y. But (I(a»" = en with 0 < l' < k contradicts
o(/(a» = k. Consequently,
{( ~ 4 14) ' (1342314) (1 2 3 4) (1 2 3 :) }
:), (; 4) (1 4 ~ ),( ~
~) {(!
(~
2 :3
3
2 3
(!
2
j
1 4
=
2 3
2 1
2 3
2 3
1) = {( ~
2 3
3 4
4 1 2 3
)
2 3
3 2 1
1 2 3
$
1
2 :3
3
2 3
1 4:
:)}
~ ), ( ~ 4 1 ~ ), ( ~ 1 2 ! ), (! 2 3:3 i )}
2 3
2 3
2
2.
(~
2 3
2 4:
(~
(i
(~
2 3
3 2
!)
2 3
4: 2
~)H={(~ 1 3
2 3
2 3
: )H=H
~)H={(~ 3 1
2 3
= {( ~ 4 3 ~
2 3
2 3
Let K = (P} = {(
:), ( ~
), (~
! ), ( ~
2 3
4 2
2 3
1 4
2 3
2 4
~
!
~), (!
1), (!
)
1 (
2 3
1 3
2 3
2 1
2 3
3 1
~ ), ( ~ 2 4 ~) }
2 :3
! ), ( ~
~ ), ( ~
2 3
3 4 ) (1 2 3
~ 2
!
)
,(
~
2 3 !) }
3 1 4 ' 3 1 2
(~
2 3
2 3
! )K=K
(~
2 3
4 2
~)K={(~ 4 3 i ), (! 4 1 ~
(~
(!
2 3
2 4:
~)K={(~ 3 4 ~ ),( ! 1 4 ~ ), ( ~ 2 4 !) }
2 3
1 2
!)K={(!
(~
2 3
1 3
:)K={(~ 2 1
(;
2 3
3 4
~)K={(~ 1 4: ~ ) (! 2 4 ~ ) ( ~ 3 4:
(;
2 3
3 2
:)K={(~ 1 3 :), (~ 2 1 : )) ( i 3 2
(!
2 3
1 3
~)
2 3
2 3
2 3
1 (
2 3
2 3
2 3
2 3
~ 4 2 :) }
2 3
)
2 3
~ ),( ! 3 31 ~
2
:), ( ~
2 3
:3 2
!
4) (1
4 ' 2
)
j
2 3
2 3
(
2 3
1 2
2 3
1 3
2 3
)
2 3
2 3
2 3
4 3 2
~ ), (! 1 3 ~) }
2 3
4) (1
= {(! 2 1 3
2 3
'
!) }
:) }
~) }
:) }
2 3
2 3
3 2
2 3
4 2
:) }
!) }
= {4} = {O, 4, 8,
16,20}
1 +K == {1,5,9,13,17,21}
2+1( = {2,6, 10,14, 18,22}
3
7,1 15, 19,23}
Lagrange's Theorem we know that IKI = 66(= 2·3·11)
IHI and that
divides IGI = 660(::::; 22 ·3·5·1
Consequently, since K
Hand H i= G, it follows that
IHI is 2(2·3 ·11) = 132 or 5(2·3·11) = 330.
6.
Let G
the set of units in R. u E G ::::;;:::} G i= 0. Also, the elements
G are
associative under multiplication (inherited from
multiplication
R). If x,y E G
then X-t,y-l E R (and in G), and (xy)(y- 1 X- 1 ) = u = (y-l x -l)(xy), so xy E G.
Consequently, G
a multiplicativ'e group.
7.
(a)
(1)(2)(3)(4)
(12)(34)
(13)(24)
(14)(23)
(1)(2)(3)(4)
(1 )(2)(3)( 4)
(12)(34)
(13)(24)
(14)(23)
(12)(34)
(12)(34)
(1 )(2)(3)( 4)
(14)(23)
(13)(24)
(13)(24)
(13)(24)
(14)(23)
(1)(2)(3)(4)
(12)(34)
(14)(23)
(14)(23)
(13)(24)
(12)(34)
(1)(2)(3)( 4)
It follows from Theorem 16.3 that H is a subgroup of G. And since the entries in the
above table are symmetric about the dia,gonal from the upper left to the lower right, we
have H an abelian subgroup of G.
(b) Since IG! = 4! = 24 and IHI = 4, there are 24/4 = 6 left cosets of H in G.
( c) Consider the function f: H.....-j. Z2 X Z2 defined by
f: (12)(34) - f
f: (1)(2)(3)(4).....-j. (0,0),
0),
l: (13)(24) - t (0,
f: (14){23) - f (1,1).
This function J is one-to-one and onto, and for all x, y E H we find that
f(x· y) = f(x) ED fey).
are
8.
I=
.k
so
I=p·
of
p.
10.
Corollary 16.1. o(a) = I(a)l. By
Theorem I(a)l divides IGI, so o(a)l!GI.
bea
with IGI = p, a
G is cydic.
Corollary 16.1, o{x) = P, so G = (x)
x E G, x
e. By
11.
(a) Let x E H n K. x E H => o(x)I1O => o(x) = 1,2,
Q(x) = 1,3,7, or
Hence o(x) = 1
x = e.
12.
(a) For all a E G 1
a = e E H, so aRa and R is reflexive. If a, bEG and aRb,
1
then aRb => a- b E H => (a- 1 bt 1 E H (because H
a subgroup) => b-1a E
H => liRa, so R
symmetric. Finally, let
h, c E
with aRb and bR.c. Then we
have a- 1 b, b-1c E H and since H is closed under the group operation, (a-1b)(b-1c) =
a--1(bb-1)c = a-1(e)c = a-Ie E
so aRc and 'R. is transitive. Hence 'R. is an equivaleuce
relation.
or 10. x E
=> o(x)121
?
(b) a'Rb => a-1b E H => a- 1 b = h, where h E H => bH = (ah)H = a(hH) = aH.
Conversely, all = bH => a E bH => a = bh for some h E H => h- 1 = a-1b, where
h- 1 E H =>
bE H and aRb.
(c) Let x E [a]. Then xRa so x-1a E H. Since H is a subgroup, (x- 1 a)-1 = a-Ix E H.
So a,-l x = h E H aud x = ah E aH. Hence [a] ~ aH. Conversely, if yEaH then
y = aht, for some hI E H. Y = ah 1 => a- I y = hI E H => any. With n symmetric we
also have yRa, and so y E raj. So here we find aH ~ fa}. With both inclusions established
it now follows that aB = [aJ.
(d) Define f : aH ~ H by f(ak) = h, for hE H. ah 1 = ah2 ¢:=:;;> hi = h2 { = } f(ah 1 ) =
l(ah 2 ), so I is a one-to-one function. Also, for h E H, f-l(h) ;2 {ah}, so 1-1 (h) 1= 0,
and I is onto. Hence f is bijective and laBI = !H!.
(e) Since 'R is an equivalence relation on G, R induces a partition of G as
.li.JiC,t!l'!Aj
{ail ::::: (jiB for aU 1 S i ::; t, and
IGI = tlHI, and IHI divides IGI·
.)
= IH I = m for all 1
i S t. Consequently,
[xl E
a E Z, if p I a
1
b
1
Section 16.4
1.
Here n = 2573 and e = 7.
The a.<3signment for the
IN
VE
IN
0813 2104 1819 0813
plaintext
ST
OC
1819 1402
Since
{08Ia? mod 2573 = 0462
(2104)1 mod 2573 = 0170
(1819yt mod 2573 = 1809
(0813r mod 2573 = 0462
1018
(1819)7 mod 2573 = 1809
2573 = 1981
(1402)7
(1018? mod 2573 0305,
=
the ciphertext
0462 0170 1809 0462 1809 1981 0305
2.
Here n = 1459 and e = 5.
The assignment for the given plaintext is:
OR
DE
1417 0304
RA
PI
ZZ
AX
1700 1508 2525
0023
Since
(1417)5 mod 1459 = 0152
(0304)5 mod 1459 = 0466
(1700)5 mod 1459 = 1318
(150sl' mod 1459 = 1177
(2525/' mod 1459 = 0055
(0023)5 mod 1459 = 0694
the ciphertext is
0152 0466 1318 1177 0055 0694.
3.
Here n = 2501 = (41)(61), so r = .p(n) = (40)(60) = 2400. Further, e = 11 is a unit in
1
Z2400 and d = e- = 1091.
the encrypted ", .. ,-,"''''. .
14181436
IS
=0411
4.
= (43)(71), so r = .p(n) = (42}(70) =
e=
1105 1232 2281 2967 0272 1818 2398 1153,
0986 3029
we calculate the following:
(0986)173 mod 3053 = 1907
(3029)173 mod 3053 = 0417
(1134 )173 mod 3053 = 0408
(1105)113 mod 3053 = 1818
(1232)113 mod 3053 = 0005
(2281)173 mod 3053 = 0419
(2967)113
3053 = 2408
(0272)113 mod 3053 = 1313
(1818)173 mod 3053 = 2012
(2398)113 mod 3053 = 0104
(1153)173 mod 3053 = 1718
Consequently, the assignment for the original message is
1907 0417 0408 1818 0005 0419 2408 1313 2012 0104 1718
and this reveals the message as
THERE IS SAFETY IN NUMBERS ~
5.
Here n = pq = 121,361 and r = tjJ(n) = 120,432.
Since p + q = n - " + 1 = 930 and p - q =
-/379,456 = 616, it follows that
p = 157 and q = 773.
6.
= -/864,900 - 485,444 =
Here n = pq = 5,446,367 and r = <p(n) = 5,441,640.
Sincep+q = n-r+l = 4728 andp-q = jCn - r + 1)2 - 411, = -/22,353,984 - 21,785,468 =
-/568516 = 754, it follows that
p = 1987 and q = 2741.
Section 16.5
e = 0001001
2.
(b) r = 1111011
(b)
(0.95)1(0.05)2
(d)
(~)(O.95)'1(0.05)2
(f)
s.
(c)
(c) c = 0101000
D(r) = 01
(e)
(d) 0000000000, 1000000000, 0000000001
(e)
Sections 16.6 and
= {101010,001010,111010,100010,l01110,101000,101011}
5(11
2.
11,
-
1,101111,110111,1
111101,111110}
5(000000,1)= {OOOOOO,lOOOOO,OlOOOO,001000,OOOlOO,
OOOOOl}
5(010101,1)= {010lOl,ll0101,OOOlOl,011101,01000l,010lll,Ol0100}
3.
(a)
D(110l01) = 01
(b)
D(1010ll) = 10
ee)
D(OOl
(d)
D(nOOOO) = 00
(a)
IS(x, 1)1 = 11;
IS(x, 2)1 = 56;
IS(x, 3)1 = 176
(b)
IS(x, k)1 = 1
(~)
(~) = £::::0 (~)
1) = 00
(;) + ...
4.
k = 8;
5.
(a) The minimum distance between code words is 3. The code can detect all errors of
weight :5 2 or correct all single errors.
(b) The minimum distance between code words is 5. The code can detect all errors of
weight :5 4 or correct all errors of weight :::; 2.
( c) The minimum distance between code words is 2. The code detects all single errors
but has no correction capability.
(d) The minimum distance between code words is 3. The code can detect all errors of
weight
2 or
errors.
n=4
(i)
(ii)
,so c=1
= (OOO)t1·, so C 110101
= (OlO)h', so C =
I so C =
=1
=
= (lOO)tl",
(1) if
if
c=
c=
if 1
c=
H·
H·
-
SOl
C
D(c) = 110
=1
=
-
Assuming a double error,
if
1 = 110 001, then c = 110101
D(c) =
(2) if 111 = 011 + 100, then c = 000000 and D(c) = OOOi and
(3) if 111 = 101 010,
c = 011
D(c) =
(b)
7.
No. The results
(vi) and (viii) are not unique.
(a) C = {OOOOO, 10110,01011, 11101}. The minimum distance between
so the code can detect all errors of weight S; 2 or correct all single errors.
(b)
H=
words is
10100]
1 1 0 1 0
[
o 1 001
(c) (i) 01
(ii) 11
(v) 11
(vi) 10
For (iii) and (iv) the syndrome is (111)tll' which is not a column of H. Assuming a double
error, if
l)tr = (Hoyr + (OOl)tr, then the decoded received word is 01 (for (iii»
10
(for (iv)). If (lllyr = (OUlr (100)t", we get 10 (for (iii») and 01 (for (iv».
8.
( a)
G= [
°100° 11 ~l]
1
0 1
00110
c= {OOOOOO, 100111,010010,001101, 110101,101010,011111, 1110OO}
(b)
No. Tbe second and fift,h columns of H are the sam.e.
9.
G = [IslA] where Ig is the 8 x 8 multiplicative identity matrix and A is a column of
eight 1'8. H == [Atrll} =:: [1111111111J.
10.
(a) For each x E {O,l}, xG = xxxxxxxxx.
(b) H = [Ails] where
eight l's.
is the 8 x 8 multiplicative identity and A is a column of
11.
error
l'
If r E Z2'
as an incorrect
c as well as those rp"',Alv~,..t ~1'"",,·.rU;t
o
> !M(n,k)HE?!o (7)1,
d" E Zi where d(c*,c) >
aU code
c. So we can
set of code
and
a
larger code where the minimal distance between code words is still 2k 1. This, however,
IA1(n, 1.:)1 so
:5 pv/(n,
(:1,(:2'
there is an
(7)].
Sections 16.8 and 16.9
1.
e~6) calculations are needed to find the minimum distance between
words. (A
calculation here
the distance between a pair of code words.) If
is a group
homomorphism we
to calculate the weights
255 nonzero code words.
2.
(a)
H =
[1 01 1 0 0]
1 1 001 0
1 100 1
o
Received word r
000011
100011
111110
100001
001100
011110
001111
111100
H . r tr(Oll)tr
(101 )tr
(110)tr
(1 l1)tr
(OOlt'
(oooyr
(OlOyr
(lOO)tr
e
010011
101011
011110
D(e}
110101
110
e=".
001101
011110
001101
111000
010
101
all
001
all
001
(b) If 100001 is
(in the last row of Table 16.8) as the coset leader instead of 010100,
r = 100001,
r = 100001
x=
c == 000000 (not 11(101) and D(c):::;;: 000 (not 110).
3.
(a)
000
1
011
100
Coset
00000
10110
01000
00100
00010
00001
11000
01100
11110
10010
10100
101
01110
11010
001
101
111
01011
11011
00011
01111
01001
01010
10011
00111
11101
10101
11001
11111
11100
00101
10001
[The last two rows are not unique.]
(b)
Received Word
11110
11101
11011
10100
10011
10101
11111
01100
Code Word
10110
11101
01011
10110
01011
11101
11101
00000
Decoded Message
·10
11
01
10
01
11
11
00
4.
G-
[~
0 0 0 1
1 0 0 1
0 0 1 0 0
0 0 0 1 1
1
0
1
1
II
(a)
(1011) G = 1011010
-
=1
= 1110000
11
=1
c
1110111
001'0001
0011100
1
1111111
1
0011100
1
11
0010
0011
(c)
100
010
001
Coset Leader
0000000
1000000
0100000
0010000
0001000
0000100
0000010
0000001
Same results as
part (b).
000
110
101
011
111
(d)
5.
(a) G is 57 x 63; H is 6 x 63
(b) The rate is 57/63.
6.
The rate of the (S,l) triple repetition code is l/S. The rate for
4/7. Since (4/7) > (1/3) the Hamming code is more efficieut.
7.
(a) The Hamming (7,4) code corrects all single errors in transmission, so the probability
of the correct decoding of 1011 is (0.99)7 + (;)(0.99)6(0.01)
(b)
[(0.99)7 + G)(0.99)6(0.01)]5
Section 16.10
1.
(a)
_
-
(C10;'C3C4CSCeC7CaCsClOCl 1C12 C13C14 C15 CHI)
C1C"O),CSC4C"C6CrCgCHCU1CU,C12C13CHCIG
=
0:=
f3 = (135)(2674)
"/=
6=
=3;
a=
0(6) =
Hamming (7,4) code is
4.
Here G is the group of Example 16.7.
(a)
W(r.~)
= 23
W(ri) = 22
~(1rn = 2
w(r;) = 22
w(r.;) = 2
W(T;) = 22
(b)
~(r.o)
= 33
W(ri) = 3
2
w(?}) = 3
W(Ti) = 32
3 +3
3(32 )] = 10.
For 0:; i :; 4, let 7Ti denote a clockwise rotation through i(72°). Also, there are five
reflections 'f'i, 1 :; i :; 5, each about a line through a vertex and the midpoint of the
opposite side. Here IGi = 10.
(a) W(7T~) = 25
'lJ(7Ti) = 2, 2:; i:; 4
3
w(r;} = 2 ,
1 :; i :; 5.
The number of distinct configurations
6.
3(22)] = 4.
W(r. 2) = 3
w(r;) = 32
The number of distinct colorings is (1/6)[33
5.
2
of distinct colorings is (1/6)(23 + 2
The
(b)
39
(a)
(i)
(1/10)[25
Free to move in t'\lllO dimensions:
Here
4(2) + 5(23 )] = 8.
- {7To, 7Th 7T2, 7T3} where the 1ri,
o :; i ~ 3, are as in Example 16.28.
\lJ(1r;) = 34, W(1rr) ::::;: 'Ii(1r;) ::::;: 3, \lI(1ri) = 32 ,
The number of distinct configurations is (1/4)[3 4
(ii)
2(3)
Free to move in three dimensions: Here G is the group of Example 16.28.
'[1(11'0) ::::;: 3\ W(1rI) = W{1r;) = 3, W(1I';) ::::;: 32 ,
W(ri) = 33 ::::;: 'li('f';), w(r;) = 32 ::::;: W(r:).
The .lAW,UVIC£. of distinct configurations is (1/8)(34 + 2(3)
(b)
32] = 24.
(i)
51
1.
4] = 70.
431
= 18;
(
(b)
G = {1f0 =
(1/2)[34 + 32] = 45;
9.
(1/2)(43
42 ] =
(~ ~ ~ !), 1f = (! ~ ; ~)}
1
(1/2)[44
42 ] = 136
(c)
n odd: (lj2){3 n + 3(n+1)/2); (1/2)[41'
n even: (1/2)[3'1't 3n/2]; (1/2}[4n +
(d)
(a) (1/2)[3-2-2+3- =9;
(b) (1/2)[3·2·2 - 2 + OJ = 12;
4 +1)/21
10
(1/2)[4·3-3 4·3] =24
(1/2)[4 - 3 - 3 - 3 + O} = 54.
Triangular Figure:
(a) G={1fO,1I"b1f2}
(1/3)[2 4 +2 2 +22 ]=8
(b) G = {1fo) 1f}, 11"2, t'b 1"2, r3} (1/6)[24 + 22 + 22
3(23 )] = 8
Square Figure:
(a) G = {1I"0, 1ft, 1i'21i'3}
(1/4)[25 + 2(22) + 23 J = 12
(b) G = {1I"0, 1fh 1i'2, 11"3,1"1,1"2,1"3, r ,d (1/8)[2 5 + 2(22)
(1/4)[4I"+2(4 2)
4
(1/4)[4(3 ) + 2(4)(3) + 4(3 2 )] = 96
23 + 2(23 ) + 2(24)] = 12
43]=280
10.
G={1i"O,1fl,1I":h1l"S}
11.
(a)
12.
(a) G = {'lio 1 1r}, 'liz, 1I"s}
(1/4)[216 + 2(24) + 28 J = 16456
(b) G = {'liO,1i'l,1I"2,1r3,rl,r2,r3,r,d
(1/8)[216 + 2(24) 28 + 2(28 ) + 2(210)} = 8548
13.
G = {1riIO i:5 6}, where 'Iii is the (clockwise) rotatioll through i· (360°/7).
(1/7){3 7 6(3)J:= 315
(b)
140
14.
102
theu
= x = 1ri(x),
(x) = x =
+5 2 +
(b)
2(5)
2(5 2 )
=X=
E
= 120
(a) (1/5)[5 5 4(5») = 629
(b) (1/lO){5 5 + 4(5) + 5(53 )] = 377
3.
(Triangular Figure):
(a) G = {'irO,1il, 11"2}
(1/3)[44 2( 42 )] = 96
" 1f:h 1'17 1
'}
(1/'6)[4 4 + 2(42 )
(b) G = {1ro, "'/'1,
2,1'3
3(43 )J -- 80
(Hexagonal Figure):
rotatiollthrough i·180",i=0,1.
(a) G={1fo,lfd where 1ri is
{1/2)[49 + 45} = 131,584
(b)
= {1rO,1rl,rl,r2} where 1'1(1'2) is the vertical (horizolltal) reflection.
(1/4)[49 + 45 + 45 + 47] = 70,144
4.
(a) (1/12)[36 + 2(3) + 2(3 2 ) + 4(3 3 ) + 3(3 4 ») = 92
(b) (1/12)[m6 + 2m + 2m 2 4m3 + 3m 4 ] is the munber of ways to m-color the vertices
of a regular hexagon that is free to move in space.
5.
(a) (1/6)[56 + 2(5) +
(b) (1/12)[56 2(5)
-
(c)
SOy vOa
R
R
G
W
G
W
R
R
6.
(1/3)[36
-
f'3}
+
rA, 1'2"
-
Figure):
-
-
I'll
=954
=n.
~ certain
must
~
For example,
WW
aD
W
W
R
0
even
W
R
w
I
R
W
w
R
0
is equivalent to
.I
W
W
a
W
13
W
to
W
R
W
Section 16.12
1.
(a)
(i) (1/4)[(1' W)4 2(1'4 w4 ) + (1'2 + w2 ?J = 1'4 + w4 + 1'3 w 21' 2 w 2 + rw 3
(ii) (1/8)[(1' W)4 + 2(1. 4 + w4 ) + 3(1'2 + W 2 )2 + 2(1' + w)2(1'2 + w 2 ») = 1'4 + w4 +
r3 w + 2y,2W 2 + rw 3
(i) (1/4)[(r + b W)4 2(r4 + b4 + w 4 ) + (1'2 + b2 w 2 )2J
(ii) (1/8)[(1' b W)4 2(r 4 +b4 +w 4 ) 3(1,2 b2+w2r~ 2(1'
2.
b+w)2(r 2
"".....,,,.... elements are as follows:
we COf,lSld.er
b2 +W 2 )J
answer is (1/10)[30
5(2)] = 4
(See Example 16.35)
3.
(1)
(2)
(3)
(4 )
Rigid Motion
Identity
Rotation through 90°
Rotation
1800
Rotation through 2700
Rotations of 1800
Rotations of 1200
There are then (1/24)[26 + 6(23 ) + 3(24) + 6(23) + 8(22)1 = 10 distinct 2-colorings of the
faces of the cube.
(b) (1/24)[(r+ wY'
6(r +Wy;J(T·4+W4) +3(r
w)2(r2+w2)2
6(r2+w 2)3
8(r3 +w3)2]
(c) For three red and three white faces we consider the coefficients of the sm:nmands that
involve r3 w 3:
(1' + W)6 :
3(1' + wY"(r 2
8(1'3 + W 3 Yl :
(:) = 20
2
W )2:
12
16
161 = 2
The answer is (1!24){20 + 12
4.
(36) - (1/12)[34 + 8(3 2 ) 3(32 )J = 36 - (1/12)[180} = 21 compounds have at least one
hromine atom.
For the compounds with exactly three hydrogen atoms we need the coefficients of w 3 x
and w 3 y in the pattern inventory.
(w x y z}':
4 )+(3,0,1,0" )-8
( 3,1,0,0
8( w
x
11 + z)( w 3 + x3
answer IS
/12)[8 +
y3
Z3):
8(1
1) = 16
= 2
+
+
6.
Here G = {1fiiO
Uenol;e the
i . (360° /7).
1'+
, so
of
(b)
answer is (1/7)(3,~.2) = 30.
(1/7)[7 (for w'l') + C'i,~,l) (for wl'ibr) + (3,~,2) (for w3 b2 r2) + (1,~,3) (for wb3 r 3 )] + 42 210 + 140] =
is
number of ways to .-......~~v.. the seven horses on
(n '7
all integf~t;~ 7 divides
7.
180° rotation.
(b)
+ 24} = 136 distinct ways to 2-color the squares of the chessboard.
(1/2)[(1' w? + (1'2 + w2)4]
(c)
Four red and four white faces: (1/2)[(:) + (~)] = 38
(1/2)[~
Six red and two white faces: (1/2)[(:) + (~)] = 16
8.
Here G = {1I"i!0 SiS 3} where 1ri is a (clockwise) rotation through i(900).
2(22) + 24] = 70
(a)
(1/4)[~
(b)
(1/4)[38 + 2(3 2 ) + 34J = 1665
( c) For the pattern inventory denote the colors as follows: b: black; g: gold; and
u: blue. Then the pattern inventory is given by (1/4)[(b + 9 + U)8 + 2(b4 + g4 + U4)2+
(If
9 2 + U2)4J.
two gold, and two
re~ions we
pattern inventory. This
(1/4)[(4'~'2) + b,~,l)] 108.
}1or four
the
=
n1,
colors.
x E 1{,
- en. For all
)=
9 E
for all x E K, 9 E G, we
2.
Let
that
EK.
+ denote the operation in G, H, and K.
°
Let S =
O)lh E
Here is
in G. S is a nonempty subset of G.
for H (and K)
18
The function f: G -4 G
by
k) = (h,O) is a homomorphism with f(G) =
S is a subgroup of G. The function g: S -4 H defined
by g( h, 0) = h provides an isomorphism between Sand H.
80 by part (d) of Theorem 16.5
In like manner, {(O, k)
E K} is a subgroup of G that is isomorphic to
3.
Let a, bEG. Then a 2 b2 = ee = e = (00)2 = abab. But a 2 b2 = abab ==> aabb = abab ==>
ab = 00, so G is abelian.
4.
Since G has even order, G - {e} is odd. For each 9 E G, 9
e, if 9 =1= g-1, remove
{g, g-l} from c.onsideration. As we continue
process we must get to at least one
element a E G where a = a-I.
5.
Let G =< 9 > and let h = f(g). If hI E
then hI = f(gn) for some nEZ, since f
is onto. Therefore, hI = f(gn) = [f(g»)n = hn, and H = {h}.
6.
(a) Since (1, 0) ffi (0, 1) = (1,1), it follows that (1, O)ffi(O, 1)ffi(l, 1) = (1,1) ffi(l, 1) = (0,0).
(b) Here we have «1, O,O)ffi(O, 1, l»ffi«O, 1, O)tB(l, 0, l»ffi«O, 0,1)63(1,1, O»tB(l, 1, 1) =
(1,1,1) EEl (1, 1, 1) EEl (1, 1, 1) EEl (I, 1,1) = (0,0,0).
(c) Let n E Z+, n > 1. Consider the group (Z2 x Z2 x ... X Z:h ffi), where we have n copies
of Z;'h and the group operation ffi is oomponentwise addition modulo 2. The sum of aU the
nonzero (or non-identity)
this group is (0,0, ... ,0), the identity element of the
group.
Proof:
group there are
- 2
each. such
at
one 0 and at least one 1. These .- 2 elements can be considered in (1/2 )(2n - 2) = 2,,-1_1
pairs x, y
:t
y = (1, ... , 1))
all n COIn J)()ne~l1tS
Therefore tbe sum of
2"' -'2 elements res\uts in
nmnber of
1, ... ,I).
1O,t.\,:;'U,L<;;J,-,," (1 'I 1, ... 'I 1) we
all
nonzero elemems
a,b E
) 0
ao
8.
that n + 1
For i = 0 we
permutations.
we have Q(n, k)
a cycle (of length 1) by
Now let i = 1. Here 11, + 1 is in a cycle of length 2. The
:in n = (~) ways, and we have (~)Q(n - 1, k) permutations.
element can be selected
When i = 2, then n 1 is in a cycle of length 3.
other two elements can be selected
in (;) ways, and these three elements can
arranged in a
of length three in 21
ways. This gives us
(;) 2!Q( n - 2, k) permutations of 1 2, ... , n 1 represented a.'3
a product of disjoint cycles of length
most k, where n 1 is in a cycle of length 3.
j
In general, for i = t - 1) where 1:S; t :s; k, we find n 1 in a cycle of length t. The
other t - 1
can be selected
ways, and
these t elements can be
(t:l)
(t:l)(t -
arranged in a cycle of length t in (t - 1)! ways. Then
l)!Q(n - (t - 1), k)
counts the permutations of
2,3, ... , n + 1 represented as a product of disjoint cycles of
length at most k, where n 1 is in a cycle of length t.
vVe have counted the same set of permutations in two ways, 80 it follows that
Q(n
1, k) = I: ~ (i!)Q(n -- i, k).
k-l (
i=O
9.
)
~
(a) Consider a permutation IY that is counted in P(n+l,k). If (n+l) is a cycle (oflength
1) in q, then q (restricted to {I, 2, ... ,n}) is counted
pen, k -1). Otherwise, consider
any permutation 1" that is counted in Pen, k). FOl' each cycle of 1", say (a1(12'" (1".), there
are r locations in which to place tt 1 - (1) Between at and a2i (2) Between (12 and
a3; ... ; (1' -1) Between a"'-1 and a".j aud (r) Between a". and aI- Hence there are n
locations,intotal,tolocate n+l in 1". Consequently, P(n+l,k) = P(n, k-l)+nP(n, k).
Lk=l P( n, k) counts all of the permutations in
whi('h has nI
Sn
1 ~ i::; 'n,
-~ r(i)j;;::: 0, so
r) = 0 ¢=> maxIO'(i) - rei)! = 0,1::; i:S; n ~
(a)
(1,1" E
(ii)
... )
(ill
l::5:i
(iv)
l:S;i:Sn~(J'=T'.
,. <
1_
:: t(p(i)-q(i»+
=n
or
(c)
If 1i(n) = n
we may regard
<&'.-1
with d( 1r,
of
an-z, n 2:: 2, al = 1, <12 = 2
an = a n --l
number.
1
S; 1 (f ill Sn-2), there are a n-2 such permutations. Therefore,
= 1),
an
=
Fibonacci
(a)
that 11, is composite.
consider two cases.
n = m· 1', where 1 < '1n < r:< n: Here (n - 1)1 = 1· 2··· (m -1)· m· (m
- 1) . r . (:r + 1) ... (n - 1) 0
n).
(n ¢.
(mod n).
(2) n =
where q is a prime: If (n - I)! -1 (mod n) then 0 == q(n - I)!
n - q 0 (mod n). So in th.is case we also have (n - 1)1 -1 (mod n).
=
j
=
1)···
=
1) =
(b) From Wilson's Theorem, when p is an odd prime, we find that
= (p - 1)! == (p - 3)!(p - 2)(p - 1) = (p _.- 3)!(p2 12.
2)
= 2(p-
(mod p).
G = {'lIO, 'lit, 'liZ, 'lI3}
(a) (1/4)[58 +5 2 +5 4
52J=97,825
(b) Here four colors are a.ctually used. Nicole can select four colors in (!) = 5 ways.
For one selection of four colors let cj,l'::; i .::; 4, denote that the i-th color is not used.
Then using the principle of inclusion and exclusion we have
N = (1/4)[48 + 2(42 ) + 44] = 16,456
N(e;) = (1/4)f3 8 + 2(3 2 ) + 34 ] = 1665, 1 i < 4
N(CiCj) = (1/4)[28 2(22) + 24 J = 70, 1 S; i < j
4
8
N(CiCjCk) == (1/4)[1
2(12) + 1"] = 1, 1 .::; i < j < k .::; 4
N(CICaC3C4) := 0
N(CICZC3C4) = N - 8 1 + S2 - 8 3 s,$. = 16,456 (1665) + (~)(70) - (:)(1) + 0 = 10,212.
G)
The answer
(5)(10,
CHAPTER
FINITE FIELDS AND
COMBINATORIAL DESIGNS
Section
1.
l(x)+g(x)=2x 4 +
2X4 + 5x 3 + x 2 + 5
+3)x3+(3+5)x2+(1+6)x+(4+1)=2x4
I(x) - g(x) = 2X4 (2 - 3)x 3 + (3 - 5)x 2
2x4 + (-1)x 3 + (-2)x 2 (-5)x + 3 = 2X4
5x 3 +8x 2 +7x
5=
(1- 6)x (4 -1) =
6x 3 + 5x 2 + 2x 3
f(x)g(x) = (2)(3)X7 [(2)(5) + (2)(3)Jx 6 + [(2)(6) + (2)(5) + (3)(3)]X5 [(2)(1) + (2)(6)
[(3)(1) (1)(6) + (4)(5)]x 2 +
(3)(5) + (1)(3)]x4 + [(2)(1) (3)(6) + (1)(5) (4)(3)Jx 3
6
5
4
3
[(1)(1) (4)(6)Jx + 4 = 6x i' 16x + 31x
32x + 37x
29x 2 + 25x + 4 =
6 x i' 2x6 3x 5 4X4 + 2x 3 x 2 + 4x + 4.
2.
There are four such polynomials:
(3) x 2
3.
(10)(11)2;
4.
(a)
.5.
(10)(11)3;
(10)(11)4;
f(x) = 4x
8, g(x) = 3x 2
hex) = 4x 5
X,
1
(4) x 2 + x + 1
(lO)(ll)n
k(x) = 3x 2
(Theorem 17.1) We shall prove one the distributive laws. Let I(x) = E~o aixi, g(x) =
ie
EJ!=o bjx j , hex) =
Ck;X , where Tn?:: p. For 0
t
of xt in
f(x)[g(x) hex)] is
ai(bj
where the sum
'n, () :; j =::; m
i j = t. But this is
0 =::; j =::; m,
i + j = t, because
+17.1)
()
R
f(;l!)g(x) = g(x)J(:c)
is oomulUtative.
Let 1 QeD.Olie
1 or
is
unity
Let R be an integral domain and let f(x) = Ei:::::o{LiXi, 9(X) =
an
0, Om
f(X)9(X) = then (Lnbm = 0 contradicting
as an Hlli,ellrw
Conversely, if R[x] is an integral domain and a, b E R with a 0
= (aJ}»(bxO) i: 0 and R is an integral domain.
(c)
6.
7.
8.
(a) q( x) = x 5
(b) q(x) = x 2 x
(c) q( x) = x 2 +
(a) and (b) I(x) = (x 2
U."'U...'CUL....
roC x) =
- 9X2 - 30x - 3
rex) = 1
r( x) = x + 2
2
4)(x - 2)(x
2);
roots are
(c)
f(x) = (x + 2i)(x - 2i)(x - 2)(x + 2);
roots are
(d)
(a) f(x) = (x 2 - 5)(x 2 5); no rational roots
(b) J(x) = (x - ..;5)(x + .;5)(;;;2 + 5); the roots are ±y5
(c) f(x) = (x - V5)(x + .;5)(x - y'5i)(x + y'5i)i the roots are
(a) 0,2,6,8
No -- Z12 is not a field.
(b)
x(x
±2i
±vI5, ±iv5
4) = (x - O)(x - 8) = J(x) = (x - 2)(x - 6)
(c)
f(3) = 8060
(b)
f(1) = 1
f(-9) = f(2) = 6
9.
(a)
10.
(a) J(x) = x 3 + 5x 3 + 2x + 6 = (x - l)(x - 3)(x - 5)
(b) f(x) = x'l - X = x(x - l)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)
11.
4; 6; p-1
12.
(a) H x-I is a factor of J(x), then 1 is a
J(I) = an + a n -1 ... + aa + a1 + ao.
(c)
of the polynomial. Consequently, 0 ::::::
Conversely, an + a n-1 + ... + aa + al + ao = 0 ::::}
o = an(1)n + {Ln-l (1 )71,-1 ••• + (La (1 )2 + 0,1 (1)1 + ao(l)O = f(l) ::::}
1 is a root
I( x) ::::} x-I is a T"' ......
r....
I(x), then -1 is a root of f(x). Therefore, 0 = an( _l)n +
~-1
+ ...
+
+11.0.
n is even it follows
an al~_a + ...
0::::: an -,. a'n-l an -2 - an -3
o = an(-l)n +
+
13.
Let l(x) =
aix' and h( x) = Ef""o bix i , where aoj E R
hi E R for 0 ~ i :::;
m :$
Then f(x) + hex) =
bi)Xi , "fh""....
am+! = a m +2 == ••• = ak = Z,
zero of R, so G(f(x) hex)~ = GO:~:::::o(ai bi)X i ) =
E7=og(ai+bi)xi = E~o[g(ai)+g(bi)]xi =
g(ai)x i D""og(bi)xi = G(f(x» G(h(x».
m+k
m+k
i:::;:O
i""O
G(f(x)h(x» = G(l: Cixi) = I: g(Ci)X i •
m+k
m
k
L: g(Ci)X i = (I:g(ai)xi)(L9(b )xi) = G(f(x»· G(h(x».
i
i::::;O
i=O
Consequently, G: R[x] - - t S[x} is a ring homomorphism.
14.
If f(x) is a unit in R[x] then there exists $l(x) in R[x] where f(x)g(x) == 1 (the
unity of R[xD. But f(x)g(x) == 1 and R an. integral domain imply that deg 1 == 0 =
degf(x)g(x) = degf(x) + degg(x). So degf(x) = 0 = degg(x), and each of f(x),y(x)
are constants, and consequeutly units in R.
15.
In Z4[xl, (2x 1)(2x 1) = 1, so (2x + 1) is a unit. This does not contradict Exercise
14 because (Z41 +,.) :is not an integral domain.
16.
If a
b (mod n), then by mathematical induction it follows that aft
bk (mod n)
for all k E Z+, Also, c(a k )
c(bk ) (mod n) for each c E Z, by the definition of
multiplication in Zn. Finally, again by mathematical induction (on the degree of l(x))
an.d the definition of addition
Zn, it follows that f(a) feb) (mod n).
=
=
=
=
17.
First note that
f(x) = anx?'b a n _lx n - 1 +... a"x 2 +atx+ao! we have an an-l
(1,2
0,1
ao = 0 and
if f(l) = O. Since the zero polynomial is
S,
set S is
empty, With f(x) as given here, let g(x) == bmx'ff~+bm_1Xm-l+ .. ·+~x2+blX+bo E
'm n, and for m < n we
= bm +2 = '" = bn = 0.)
f(l)-g(l} = 0-0 = 0
so
-g(X)E
hex) =
E
E
so h(x)f(x) E
aiEl
0
= an =
=am +2 = ...
EI
~
rl(lt-l! 1'2 at-::I)"" 'fe-lab TtaO E I
it
that Ct E I and h(x)f(x) E
In a similar way it follows that f(x)h(x) E I{x]. Consequently, I[x] is an ideal in R[xJ.
1.
(8.) x 2 +
x2
3.
Over R,
- 1 is irreducible over
- (( -3 + Ji3)/2)][x - « -3 - M)/2)].
3x - 1 =
x4 - 2
(b)
irreducible over Q.
Over R, X4 - 2 = (x - {I'2)(x {I2)(x 2 + J2);
x4 - 2 = (x - {12)( x
~)( x - {l2i)( x
V'2i) over
(c)
+ x 1 = (x
2
x + x + 1 :.:::: (x
(d)
x4 + x 3 + 1 is irreducible over
(e)
x 3 + 3x2 -
Degree 1:
Degree 2:
Degree 3:
X
2)(x + 2) over Za. Over Z5, x 2
5)( x 3) over Z7.
x + 1 is irreducible;
Z2'
+ 1 is irreducible over Zs.
x; x + 1
x2 x + 1
x 3 + x 2 + , x3
X
+ 1
4.
f(x) = (2x2 + 1)(5x3 - 5x 3)(4x - 3) = 2(x 2 + 4)(5)(x3 5(x 2 + 4)(x3 - X + 2)(x - 6), since 40 = 5 in Z.,.
5.
75
6.
(Theorem 17.7)
X
+ 2)(4)(x - 6) =
f( x) ::; 1.
were ...",.-111'1"
(a.)
E
with degg(x), hex) ?: 1. Then 1 degf(x) = degg(x) +degh(x) 2:: 2.
(b)
'T.
(x-r)
of f(x).
f(x} = (x-1')g(x)
f(x) is
Conversely,
x) ?: 1.
= 2 or one
hex) = ax + b, bE F, a O. Then
(a)
=* f(x) =
= a(x )f(x) +
BO
m(x) A f(x),
fex) = q(x)m(x)
where rex) 0 and 0:::; degr(x) <
degm(x). m(x) = s(x)f(x) + t(x)g(x) 80 rex) = f(x) - q(x)[s(x)f(x) + t(x)g(x)] =
(1- q(x»s(x)f(x) - q(x)t(x)g(x), so rex) E S. With
rex) < degm(x) we contradict
choice
m(x). Hence rex) = 0 and m(x)!f(x).
8.
(Theorem 17.9)
From the last equation rk(x) divides rlc-l(x). The next to last equation yields 'ik(X)
divides rk-Z(x). Continuing backwards we
1'k(X) divides 1'z(x) and 'il(X), so 'i,lc(x)
divides 7'(.X). From the second equation i'k(X) divides f( x); 'ik(X) then divides g( x)
from the
equation.
establish condition (b) of Definition 17.6, let k(x) E F[x]
where k( x) divides r( x). From
second equation, k( x) divides 'il (x) since it divides
f( x) and r( x). Continuing down the list of equations we get to where k( x) divides
l'k-2(X) and rk-l( x), and, consequently, 'ik( x).
(Theorem 17.10)
For aU fex) E F[x}, f(x) - f(x) = 0 = o· sex), so 'R is reflexive. To show that 'R is
symmetric,let f(x),g(x) E F[x] with f(x)1?.g(x). f(x)R.g(x) ===} f(x)-g(x) =t(x)s(x),
for some t(x) E F[x] = } g(x) - I(x) = [-t(x)Js(x), -t(x) E F[xJ = } g(x)1?.f(x) , so
'R is symmetric. Finally, let f(x),g(x),h(x) E F[xJ with f(x)'Rg(x) and g(x)'Rh(x).
Then f(x) - g(x) = t(x)s(x), g(x) - hex) = u(.-r)s(x), so [I(x) - g(x») [g(x) - hex)] =
f(x) - hex) = [t(x) u(x)]s(x) and 'R is transitive.
9.
(a)
By the long division of polynomials we have
+ x 3 + x 2 - x -1 = (x 3 - 2x 2 5x - 8)(x 2 + X - 2) (17x -17)
x 2 x - 2 = (17x - 17)[(1/17)x + (2/17)xJ,
so the gcd of f(x),g(x) is
(x -1) = (1/17)(x 5 - X4 + + x 2 - x-I) - (1/17)(x 2 x - 2)(x 3 - 2x2 + 5x - 8).
Xli -
X4
(b)
The gcd is 1 = (x + 1)(x4 + x 3
(c)
10.
gcd
1) + (x 3 + x 2 + x)(x 2 + X + 1)
x 2 +2x+l=(x4 +2x 2
2x
2)+(x
2)(2x3 +2x2
x+1)
If
then x "'- a would be a factor
hoth f(x) and g(x), so x - a would
divide the gcd of
g(x). This contradicts f(x),
being
pnme.
= 0, or a = 0
444
- b=
a = 0, b = 0;
12.
(a)
a=
x 2 =x+l
X4
sex)~ = }
+ x3 + X + 1 =x
(Also note that
13.
b = 1.
1
sex»~ ==>
+x=l
x +1
= 0 (mod
», so
X4
=x
x + 1 E [0].
x + 1 = (x 2 + 1)(x2 + xl).)
X4
x3
+ 1]
(b)
x 3 + x'2
1E
(c)
x'+x 3
x 2 +1=x2 (x 2 +x
1)+1,80 x4 +x 3 +x 2 +1E[1].
=
f(x) flex) (mod sex»~ => f(x) = hex) h(x)s(x);
g(x) = gl(X) (mod s(x» = } g(x) = gl(X) k(x)s(x)
Hence f(x) + g(x) = flex) + gl(X) + (h(x) k(x»s(x), so f(x) g(x) = flex) gl(X)
(mod s(x», aud f(x)g(x) = ft(X)gl(X) + (h(x)k(x) gl(x)h(x) + h(x)k(x)s(x»s(x), so
f(x)g(x) = f1(x)gl(x) (mod sex)).
(a)
(b) These properties follow from the corresponding properties for F[x]. For example, for
the distributive law,
[j(x)] ([g(x)J
= [f(x)][g(x) + hex)] = [j(x)(g(x) + hex»~)
= [J(x)g(x) + f(x)h(x») = iJ(x)g(x)] + iJ(x)h(x)J
[hex)])
= [J(x)J[g(x)]
[J(x)J[h(x)J
If not, there exists g(x) E F[x] where degg(x) > 0 and g(x)lf(x), sex). But then
sex) would be reducible.
(c)
(d) A nonzero element of F[xl/(s(x» has the form [I(x)} where f(x) =/: 0 and
degf(x) < degs(x). Witb I(x), sex) relatively prime, there exist rex), t(x) with
1 = f(x}r(x) + s(x)t(x), so 1 = f(x)r(x) (mod sex)~ or [1] = [J(x)J[r(x)l. Hence
[r{x)J = (f(x )]-1.
(e)
(l~
(a.)
Xl
1 = (x +
1]
(e)
11j --
1]
(c)
a
16.
= 2 (mod 3) so
= {2x + 2J-l = [2xJ.
I
09(0) = 1 = s(l), so sex) has no root in Z2 or
factor in Zz[xJ. But
we can
as f(x)g(x) where deg
= degg(x) = 2. If so we
2
2
3
sex) =
+x
1 = f(x)g(x) = (x
ax b)(x
cx+
where a,b,c,dE Zz. Then
(x 2 +ax+b)(x z +cx+d) =
(a c)x3 (b ac+d)x 2 +(bc ad)x bd=X 4 +X3 I=}
a c = 1, b ac d = 0 =
+ bd = l.
(a)
bd=l=}b=d=l.
b+ac d=O:::::.:::}ac=O:=.:}a=c=O
a = c = 0 =:;.- a + c = 0, contradicting a + c = 1. Consequently, s( x) is irreducible.
(b)
Uax
Z2[X]J(S(x)) is 2" = 16 since Zz[xl/(s(x»
bx + ex + dJla, b, e, dE Z2}'
The order
3
=
2
(c) [x 2 x l][ax 3 +bx 2 cx+dJ = [1) => [ax 5 +(a+b)x 4 (a b+c)x3 +(b+c+d)x 2 +
(c d)x+dJ = [a(x 3 x l)+(a b)(x3 l)+(a b+c)x 3 +(b+c d)x 2+(c+d)x+d) =
[(a c)x3 (b+c d)x 2 +(a e+d)x+(b+d)]=[l]=:;.-a+c=:::O=-=b c+d=a+c+d
(mod 2), b + d = 1 (mod 2). b + d == 1 (mod 2)
"> c = 1 (mod 2) =:;.- a = 1
(mod 2) = } d = 0 (mod 2) => b = 1 (mod 2). Hence [x 2 + x + 1]-1 = [x 3 + x2 + xl.
(d) [x 3 +x 1Hx2 1]=[x 5 +x 2 +x 1]= [(x 3 x+l)+x 2 +x+l]=[x 3 +x2 ]
11.
(a) Zp[xJ!(s(x)) = {aO+alx+a2x2+ ... +an_lXn-llao,at,a2, ... ,an_l EZp} whichhas
order prJ,.
(b) The multiplicative group of nonzero elements of this field is a cyclic group of order
pn -1, so it has <p(p'll, - 1) generators.
18.
(a)
11
19.
(a) 6
(b)
(b)
12
11
(c)
(c)
12
20.
446
0
(d)
(d) 1cm( m, 16 )
0
(e)
0
23.
For sex) = J;3 +
+ X + 2 E Z3[J;] one finds that s(O) = 2, s(l) = 2, and ..,(2) = 1. It
follows from
parts (b)
3
Z3[X]J(S(x»
a finite field with 3 =
elements.
24.
dxD::::} a + hi = c di::::} a = c and b = d::::} a bx = c dx::::}
[a + bJ;] [c + dx]' so h is one-to-one,
For all a + bi E C, where a, bE R, we find that [0:
E R[x}/(x 2 + 1) and k([o: + bxD =
a + hi. Consequently, the function h is also onto,
Finally, if [a+bxl, [c+dx] E R[x]l(x 2 +1),
h(ia+bxl+[c+dxJ) = h([(a+bx)+(c+dx)]) =
h([(a c) (b d)x]) = (a c) (b d)i =
bi) + (c di) = hGa bx)) h([c dxD,
so h preserves the operation of addition.
(a) kGa + bx)) = h([e
=
(b) Let Up, UK denote the unity elements of fields F and K, respectively. Then 9(UF) =
g(UF' UF) = g(UF) 0 g(UF), so g(UF) 0 UK = g(up) = g(up) 0 g(up). If ZF, ZK denote
the zero elements of F and K, respectively, then g(ZF) = ZK (from part (a) of Theorem
14.15). Since 9 is one-to-·one, g(UF) f:. ZK, so by cancellation
K we have g(UF) 0 UK =
g( UF) 0 g( UF) ::::} UK = g( UF).
Now if a E F and a f:. ZF, then a- 1 E F and g(UF) = g(a. a- 1) = g(a) 0 g(a- 1). But
9(UF) = 'UK = g(a) 0 [g(a)]-l because g(a) E K. Siuce g(a) f:. ZK, by cancellation in K we
have g( a-I) = [g( a )]-1.
25.
(a) Since 0 = 0+0../2 E Q[J2], the set Q[2] is nonempty. For a bV2, c+dJ2 E Q[J2],
we have
(a + bV'2) - (c + dJ2) = «(.\ - c) (b - d)V2, with (a - c),(b - d) E Q; and
(a + by'2)(c + dv'2) = (ae + 2bd) + (ad + bc)Ji, with ac 2bd, ad + be E Q.
Consequently, it follows from part (a) of Theorem 14.10 that Q[-/2] is a 8ubring of R.
(b) In o:rde:r to show that Q[V2] is a subfield ofR we ne.ed to find in Q[v'21 a multiplicative
each nonzero
in Q [-12].
b = 0, then a f:. 0 and a-I E Q Let a + bV2 E Q[v'2] with a bv'2 f. o.
o· E Q[v'2j. For b
we
to
e
E Q[v'2} 80
(a + bV2)(e
dV2) = 1
= 1 ::;;.} ae
21:i~d = b ::::} d
- 2) -jo Q[2] by
=a
= 1 and
= b/(21J2 -
an argument similar to
one given in Example 17.10
follows that J is an isomorphism.
26.
Exerc.ise24 it
part
(a) Here we want to write x 2 bx + c as the product
- rx)(x - 1'2) where 1"1,1'2 E Zp.
Since
- 1'2) =
- 1'1)'
we may have 7"1 = 1'2,
we
the nmnber of selections of size 2 from the set Zp = {O,1,2, ... ,p - I}, with repetitions
Consequently, the number
these monic quadratic polynomials is (P+~-l) =
(V;l) = (!)(p + l)(p) = (!)(p2 + p).
(b) Since ax 2 + bx
c is a quadratic polynomial we have a =J O. Then with Zp a field
it
that ax:.! bx + c = a(x 2 + a-1bx
c), so we want to
able to factor the
monic quadratic polynomial x 2 a- 1 ba:
x 2 + b1 x +(1) into linear factors. We have
returned to part (a) of the problem where we found the arlswer to be (l)(p2 p). So here
the answel' is (p - 1)( ~ )(p2 + p) = )p(p2 - 1), because there are p - 1 nonzero choices for
a.
(!
(c) Since there are (l)(p)(p) = p2 mornc quadratic polynomials over ZPl by using the
result from part (a) it follows that there are p2 - (~)(p2 p) = t(p2 - p) irreducible monic
quadratic polynomials over Zp.
(d) Here we use the result from part (b), and find that there are (p-l)(p)(p)-(~)p(p2-1) =
)p(p - 1)2 irreducible quadratic polynomials over Zp.
Section 17.3
1.
(a)
2.
llot,
the
that
1
2
2 1
4: 3
3 4:
3
4:
2
1
4
3
1
2
18 an
(b)
1
3
2
4:
2 3 4
4 1 2
143
3 2 1
(c)
1
4:
3
2
3
2
1
4
4: 2
1 3
2 4:
3 1
(j, k), 1 j, k :5 n, that appears more than once when
are superimposed. Let
2.
3.
to one
5.
L3:
4: 5 1
2 3 4
5 1 2
3 4 5
1 2 3
'2
L4:
3
5 1
3 4
1 '2
4 5
In standard form
£II
l'
3 4 5
4 5 1
5 1 2
1 '2 3
1 '2 3 4
£I2'
L'3'. 1 2 3 4 5
£I.
4'
4:
2
5
3
6.
Lil
2
3
4:
5
2
3
4:
5
5
3
1
4
'2
5
1
2
5
3
1 2
4:
5 1 '2 3
4 5 1 2
3 4 5 1
2 3 4 5
1 '2 3 4
1 ::; i ::;
'2
3 4
5 1
'2 3
4 5
I
1
5
4
3
2
3
1
4
4
3
'2
1
5
2
1
5
4
3
3 4
'2 3
4 5
1 2
5
5
4
1
3
3 4
2 3
1 2
5 1
4: 5
5
4
3
2
1
5
1
Z1::::: {I, 2, 3, ... , 7}, so
f1::::: 1, f7 ::::: 7, as in the pmof of Theorell1 17.16. Here there are
a total of six 7 x 7 Latin squares Lk = (a~~)), 1 k ::; 6, where a~J)::::: Jkfi + h.
For k = 1, aU) =:: fIli. + Ii = Ii + Ii. This results in the Latin square L1 (and Li is L1
in standard form).
L1 : 2 3 4 5 6 7 1
V"l ' 1 2 3 4 5 6 7
3 4: 5 6 7 1 2
2 3 4: 5 6 7 1
4 5 6 7 1 2 3
3 4: 5 6 7 1 2
()
7
1
2
3
4:
4: 5 6 7 1 2 3
5
6 7 1 2 3 4: 5
5 6 7 1 2 3 4:
7 1 2 3 4 5 6
6 7 1 2 3 4 5
1 2 3 4 5 6 7
7 1 2 3 4 5 6
aW = /2Ji
For k='2 we
3
4
5 6
7 1
'2 3
4 5
6 7
1 2
k;
5 6
7 1
'2 3
4 5
6 7
1 2
3 4:
-
7 1
2 3
4 5
(I 7
1 '2
3 4
5 (I
2
4
6
1
3
5
.-
h, and this gives
1
:3
5
7
2
7
4
6
-
we
2
4
4
6
1
3
()
3
5
7
'2
4:
5 6
7 1
1
3
5
7
2
5 6
"'l'
f
1
2 3
4 5
6 7
1 2
3 4:
7
'2
4:
()
1
:3
5
4
1
3
6
2
5
5
6 7
1 2 3
4 5 6
7 1 2
3 4 5
6 7 1
1 2 3 4:
a third
(a)
Neither of
L"'3-
1 2 3
4 5 6
7 1 2
3 4: 5
6 7 1
2 3 4
5 6 7
1
4
7
3
6
2
5
as four types
2
3
5 6
1 2
4 5
7 1
3 4
6 7
5
7
1 2 3
4 5 6
7 1 2
3 4 5
6 7 1
2 3 4:
6
transmission fluid or four types of tires.
3x3 Latin squares in Example
(b) The 4 x 4 Latin square
4
7
3
6
2
5
1
Example 17.15( c)
.15(b) is self-orthogonal.
self-orthogonal.
H not, let 6kk = 6 mm for some 1 S k < m S n. Then when L
are superimposed we get the ordered pair (akA;,akk) = (amm,a mm ) and
orthogonal.
(e)
and V r
are not
lines.
3.
lines
(i)
into four
Slope
O.
y = 0; y = ,y = 2
(ii)
Infinite slope
x = OJ x = Ii x = 2
(iii) Slope 1
y = x; y = x
1; Y = x
2
(iv) Slope 2 (as shown in the figure).
(1)
y
= 2x
(2) y' = 2x
(3)
The Latin square corresponding to the
1
2x + 2
parallel class is
1
3
2
3
2
y
1
2
1
3
4.
(0,4)
(1,4)
(2,4)
(3,4)
(4,4)
Here there are 25 points and
30 lines. These lines faU
into six parallel classes.
(O,3)
(1,3)
(2,3)
(3,3)
(4,3)
(i) Slope 0:
(0,2)
(1,2)
(2,2)
(3,2)
(4,2)
y = 0; 11 = 1; y = 2;
11 = 3; y = 4
(0,1)
(1,1)
(2,1)
(4,1)
Y = Xi Y = x
y=x
(4,0)
y=
y=
y=
y = 4x; y =
Infinite Slope: x =
y=
2; y =
1; y = 3x
11;::::;
1; 11 =
x 2; x = 3; x ::;:;; 4.
=
y=
y=
Y
=
5
4
3
2
1
5.
6.
4
3
2
3
2
2
1
1
1
5
1
5
4
5
4
3
5
4
3
2
(a) y = 4x + 1
(c) y = lOx or
(b)
Y = 3x + 10 or 2x
3y
3=0
= 11x
(a) (AI) fails because the distinct points (2,4) and (5,4) are 011 both y = 2x and
y = 4x
(A2) fails because the point (2,4) is not on the line '!I = 3, yet this point is on both
= 4x + 2. However, neither y = 2x nor y = 4x + 2 has a point in common
with the line '!I = 3.
'!I = 2x and y
(A3), however, still holds. The points (0,0), (1,0), (0,1), and (1,1) are such that no
three of these points are on the same line.
(b) In this "geometry", each of the 42 lines contains six points, and each of the 36 points
is on seven lines.
7.
(a) Vertical line: x = c. The line y = mx + b intersects this vertical line at the unique
point (c,mc + b). As b takes on the values of F, there are no two column entries (on
the line x = c) that are the same.
=
Horizontal line: y = e. The line y mx b intersects this horizontal line at the unique
point (m-l(e - b), c). As b takes on the values of F, no two row entries (on tIle line
y c) are the same.
=
Let Li be the Latin square for the parallel class of slope mi, i = 1, 2, mi t 0, mi
If an ordered pair (j, k) appears more than once when L 1 , L2 are superimposed,
there are two pairs lines: (1) y = mIX lJ., 11 = m2X ba;
(2) y = mIX
'JJ = maX + b~ which both intersect at (j, . But then hi
k - r1'ilj
b~ and
= k -_. m,;
(b)
=
=
Section 17.5
v=
,-
b = 12t 1'=
I 2 3
A=2
1 234:
1 256
1 2 4:
1- -
1 3 4:
1 3 5
",.
i
1 4 6 '1
),=1.
234
-
2 3 6 7
24:5 7
3 4: 5 6
=
4.
b
4
12
v
4
9
13
r
3
4
9
4
13
30
21
10
k
3
3
3
4
7
A
2
1
2
1
3
These results follow from the information given in
and (2) A(v -1) = r(k - 1).
table and the equations (1) vr = ble;
3.
(a) vr = bk ==> 4v = 28(3) ==> v =
A(V -1) = r(1e -1) ==> 20A = 4(2) = } A ~ Z+,
so no such design can
(b) vr = bk ==> (17)(8) = 5b ==> b rt Z+, so no such design can exist in this case either.
6.
With v = b and vr = bk, we have r = k. Then A(V - 1) = r(1e - 1) = k(k - 1), where
one of k and k - 1 must be even. Hence A( v-I) is even. With v even, it follows that
v-I is odd, so ),.( v-I) even ==>),. even.
7.
(a) )"Cv -1) = r(k -1) = 21" ==> A(v b(3)(2) => 6/AV( v-I).
IS even.
AV(v -1) = vr(k -1) = bk(k -1) =
(b) Here ),.=1. By part (a) 6Iv(v-l)=>3Iv(v-1)==>3Iv or 31(v-l),since3is
prime. Also, by part (a) A(v - 1) = (v -1) is even, so v is odd.
(i) 31v ==> v = 3t, todd => v = 3(209 1) = 6s 3 and v 3 (mod 6).
(ii) 31(v-l)==>v-l=3t,t even =}v-l=6x=>v=6x+l and v=l (mod 6).
=
8.
=
Here v = 9, k = 3, b 12, l' = 4 and A(v -1) = r(k -1) = 4(2) => 8A = 8 ==> ). = 1,
so the design is a Steiner triple system.
k = 3, ).
10.
(a)
v=
=
b=
=> v = 9, r = 4.
P2)
P3)
b= 21;
r=7
14.
Here
n, b = p, k = m.
r = bk/v =
fJ =
(a)
(b) >.(v -·-1) = r(k 15.
::::;. >.(n -1) = (pm/n)(m --1) ::::;. >. = [pm(m -l)J/[n(n -1»).
1 = 6 => n = 5, so
(b) n + n + 1 = 57 ::::;. n =
(a) n
are
2
+n
1 = 31 points in this projective plane.
are n + 1 = 8 points on each line of this plane.
16.
The lines y = x and y = x z, for example, would intersect at the two distinct points
(0,0,0) and (1,1,0). This contradicts conditions (PI) and (P2) of Definition 17.
17.
(a) v = b = 31; r = k = 6; >. = 1
18.
(b)
v = b = 57; r = k = 8; >. = 1
(c)
v = b = 73; 'f' = k = 9; >. = 1
(a) There are rune points:
(0,0), (1,0), (2,0)
(0,1), (1,1), (2,1)
(0,2), (1,2), (2,2)
and 12 lines:
y=o
°
x=l
y=l
x =
x=2
y=2
y=x
y=x 1
y=x+2
y = 2x
= 2x 1
y = 2x +2.
y
Here there are four parallel classes, and the parameters for the associated balau.ced incomplete block design are v = 9, b = 12, r = 4, k = 3, ). = 1.
(b)
From the :nine points in part (a) we get
(0,0,1), (1,0,1), (2,0,1)
(0,1,1), (1,1,1), (2,1,1)
(0,2,1), (1,2,1), (2,2,1).
To these nine we adjoin
the line z 0. Co:nsequently,
=
and the
additional poi:nts (1,0,0) and
,0), (1,1,0), (2,1,0) for
"''''+'nT'''' plane has 9
3 1=
+ 3 1 = 13 points
x = 0:
y=
x = z:
y = z;
x = 2z:
y = 2z:
y=x:
y = x +z:
y = x 2z:
y= 2x:
y = 2x + z:
y = 2x +2z:
z = (Roo):
°
{(O,O,1 ),(O,l,l ),(0,2,1 ),(0,1,0)}
{(0,0,1 ),(1,0,1 ),(2,0,1 ),(1,0,0)}
{(I,O,l ),(1,1,1 ),(1,2,1 ),(O,l,On
{(0,1, 1),(1,1,1 ),(2,1,1 ),( 1 ,O,O)}
{(2,0,1 ),(2,1,1 ),(2,2,1 ),(O,l,O)}
{(O,2,1 ),(1,2,1 ),(2,2,1 ),(1,0,0)}
{(O,O,l ),(1,1,1 ),(2,2,1 ),(l,l,O)}
{(1,2,1 ),(2,0,1 ),(0,1,1 ),(1,1,0)}
{(1,O,1 ),(0,2,1 ),(2,1,1 ),(1,1,0)}
{(0,0,1 ),(1,2,1 ),(2,1,1 ),(2,1,0)}
{(O,l,l ),(1,0,1 ),(2,2,1 ),(2,1,0)}
{(0,2,1 ),(1,1,1 ),(2,0,1 ),(2,1,0)}
{(I ,0,0 ),(0,1,0),(1,1,0 ),(2,l,0)}
Since there a.re four points on £00 there are four parallel classes. Finally, the parameters
for the associated balanced incomplete block design are v:::;.: b :::;.: 13, " = k = 4, A = 1.
Supplementary Exercises
1.
n=9
2.
(a) 0 = fer /13) = an(r l.s)1/, a n-l (1' /s )71-1 + ... + al(,-js) + ao =? 0 = anrn + an_lrn-ls +
'" + a]rs n - 1 + atJ!ll. Since /3 divides 0 aud s is a factor of all summands except the
first, it follows that s divides a n 7,n.With gcd(r,.9) = 1, sian. In similar fashion, rlao.
(b) (i) f(x) = 2x 3 3x 2 - 2x - 3.
From part (a) the possible rational roots are ±1, ±3, ±1/2, +3/2.
f(l) = 2(1 3 )
3(12) - 2(1) - 3 = 0, so 1 is a root of f(x) and x - I is a factor.
By long division of polynomials (or synthetic division) f(x) = (x - 1)(2x2 ox
(x - 1)(2x 3)(x 1), so the other roots of f(x) are -3/2
(ii) f (x) == X4 + x 3 - x 2 - 2x - 2.
The possible rat:iOU,W
-2=?a
so
c=l,
= -1,
x-n~
are 31
(b)
b)=x 2 +2x-n:=.:::?b-a=2
2 we find that n = ab S; 960, so
are 30
= n. When 1 a S; 30
values of n
this case.
-a)(x
b=a.
(c) In this case there are 29 ,..alues of n. Each n has
form a(a
5) for 1 S; a <
If (x - a)(x b) = g(x),
b- a = k
= n. When k = 1000, b = a 1000 and
2
ab = a
1000a > 1000. For k = 999, with a = 1 and b = 1000 we have n = ab = 1000 and
2
x +999x-lOOO =
1000)(x-l). In fact, for each 1 S; k :s; 999,
a = 1.
b = k+l
andn=ab=k 1, and it follows thatx 2 kx-n=x 2 kx-(k l)=[x (k 1)](x-1) .
......"',..."''"' the smallest positive integer k for whicll g( x) cannot be so factored is k = 1000.
If F = Z:'h then j(l) = 1 1 + 1 + 1 = 0, so 1 a root of 1 and (x = (x + 1)
a factor. If Ff:.Z2 then -lEF and -1 1. Here 1(-1)=1-1-1+1=0,80
is a root
( x l ) is a factor.
18
5.
For all a E ZPl aP = a (See part (a)
Exercise 13 at the end
Section 16.3), 80 a is a
P
root of x - x and x - a is a factor of xl' - x. Since (Zp, +, .) is a field, the polynomial
xl' - x can have at most p roots. Therefore XV - x = nl.EZp(x - a).
6.
lex) = Xn + a'll._lXn-1 +...
alX
(a) The coefficient of xn - 1
comparing coefficients we have
+ aO = (x - r'l)(X - r2)'" (x - r'n)'
(x - f'l)(X - 1':d" . (x - '1'n) is
an-l = -1'1 - 1'2 - ••• - 1'n , or
(b) The constant term in (x - rd(x - 1'2)'" (x - Tn)
comparison of coefficients we find that
ao
= (-1)101"17'2'" r'H
IS
-1'1 -
,,'), -
(-1)"1'11'2'"
••• -- Tn, SO by
Tn'
Again by
or
e-l)"ao = (_1)2n r1r2 ·•• T'r., = 1'17'2'" 1'10'
{1,2,4}, {2,3,5}, {4,5,1}
8.
Ii: =
..\ =
n
v=
==> r = 31 , b =
1=
==> n = 8 ==> n
1 = 10 :=.:::? n
= 9 :=.:::?
»2
1 = 9,
+ n + 1 ::;;::: 91,
to
k
r
(b) A . Jt.
a vX b
whose (i,j) entry is k, since there are k 1'8
1. Hence J'I}' A =
each column of A and every entry in J11
(c) The (i, j) entry in A·
is obtained from the componentwise multiplication of rows
j
i = j this results
the
1'8 in row i, which is r.
i
j, the number of 1'8 is the number
times Xi and Xj appear in the same block by).. Hellce A· Atr = (r - >')111 )'Jv'
this is
't
(d)
).
;\
>.
>.
r
).
).
).
r
)..-'1' A-r A-r
I Ar r-A
0
0
>..
>..
>..
r
).
).
A
).
).
)..
'f'
>.
>..
..\
).
)..
)..
r
r
r-A
0
0
7' -)..
o
o
o
o
o
r-..\
0
o
o
r -)..
0
o
r -
o
0
0
).
'1'->..
..\
..\
>..
fr + (v -1);\](r - ..\)V-l = (r _ >..)11-1
o
0
0
o
o
o
(v - 1)>"
I
). - r
.0
0
o
o
o
>..
r-A
r(k - 1)1 = rk(r _ >"),,-1
(1) Multiply column 1 by -1 and add it to the other v - 1 columns.
(2) Add rows 2 through v to row L
12.
(a) Here V = {I, 2, ... ,9} and the 12 blocks are
3
2
2
2
5 7 8 9
6 7 8 {)
3 5 {) 8 9
3 4: 5 (} 7
4:
4:
r,t
= k' ::::;
1
1
1
1
-
we
;;;;;;:b-
; ; ; : v-k,
5
3
3
2
6 7 8 9
4: () 7 9
4: 5 6 8
4. 5 7 8
124: 5 6 9
1 236 7 8
123 5 7 8
1
234:
8
9
THE
APPENDICES
1
LOGARITHMIC FUNCTIONS
(b)
2.
(a.) 125- 4/ 3 = 1/(125)4/3 = 1/[(125)1/3)4 = 1/54 = 1/625
(b) 0.027 2 / 3 = [(0.027)1/3P~ = (0.3)2 = 0.09
(c) (4/3)(1/8)-2/3 = (4/3)[1/(1/8)2/3] = (4/3)[1/[(1/8)1/3]2] = (4/3)[1/(1/2)2) =
(4/3)[1/(1/4)J = (4/3)(4) = 16/3
3.
(a.) (53/4)(513/4) = 5[{3/4)+(13/4)} = 516/ 4 = 54 = 625
(b) (73/5)/(718/5) = 7[(3/5)-(18/5)J = 7{3-18)/5 = 7- 15/ 5 = 7-3 = 1/73 = 1/343
(c) (51 / 2 )(201 /2) = (51 / 2 )(4.5)1/2 = (51/2)(41/2)(51/2) = 2(51 /2)2 = 2(5) = 10
4.
(a) 5&:2 = 553)+2 ::} 3x 2 = 5x
2=} 3x 2 - 5x - 2 = O::} (3x
l)(x - 2)
= 0 ::} x = -1/3
or x = 2.
(b)
= (1/2)'~X-l ::} 2 2{x-l} = 2-(4;:;-1) ::} 2(x fix = 3 ::} x = 1/2
= -(4x -1) ::} 2x - 2 = -4x
5.
2
11
o
x=
1 ::}
8.
Proof: Let x = 10gb rand 'tI - 10gb s.
y = logbs ~ bY = s, we have
Then, because x - 10gb r
~
bX
-
rand
= b:& /bY = b:&-'!J,
part (2) of Theorem ALL
= ba:- y ~ 10&,('1'/8) = X - 'til
it follows that
'1'-
8.
(a) Proof (by Mathematical Induction):
For n = 1 the statement is 10gb '1'1 = 1 . 10gb r, so the result is true for this first case.
Assuming the result for n = k C;;::: 1) we have: 10gb rk = k 10gb r. Now for the case where
n = k + 1 we find that 10gb rk+ 1 = log",(r . rio)
lo&, '1' + 1ogb'1'k (by part (a) of Theorem
A 1.2) = 10gb r + k 10gb r (by the induction hypothesis) = (1 + k) 10gb r = (k + 1) 10gb r.
Therefore the result follows for all n E Z+ by the Principle of Mathematical Induction.
=
(b) For all n E Z+, lo&, r- n = logb(l/'1' n ) = 10gb 1-1o~ '1'n (by part (b) of Theorem A1.2)
= 0 - nlogb r (by part (a) above) = (-n) lo~ r.
10.
(a) log21O = log2(2· 5) = log22 + loga 5 = 1 + 2.3219 = 3.3219
(b) log2100 = log2102 = 21og2 10 = 2(3.3219) = 6.6438
(c) logz(7/5) = log27 -loga5 = 2.8074 - 2.3219 = 0.4855
(d) log2175 = log2 (7 ·25) = logz 7
11.
21og2 5 = 2.8074 + 2(2.3219) = 7.4512
(a) Let x = loga 3. Then 2 X = 3 and x(ln 2) =
1.0986/0.6931 :... U5851.
(b) logs 2:=
5=
(&)
2/1n5 = 0.6931/1.6094 ...:.. 0.4307
3=
:r, ;: loglO(2 . 5) =:;} x = 10
:... 1.4650
2:1: = In 3, so loga 3 = x =
3/ln2 =
(c) 2 = log3(x2 4x + - 10g3(2x - 5) = loga[(x 2 4x
2
(X
4x 4)/(2x ..-- 5) ::=> 9(2x - 5) = x 2 4x 4::::.=;}
x 2 -14x 49 = O::=> (x - 7)2 = O::::.=;} x =
14.
4)/(2x - 5)J ==> 32 = 9 =
- 45 = x 2
4 ::=>
log2 x = (1/3){10g2 3 - log:;! 5] + (2/3) log2 6 + 10g2 17 = log2(3/5)1/3
10g2[17(108/5)1!3} ::=> x = 17(108/5)1/3
log'}, 6'2/3
Proof: Let x = aiogb C and y =
b. Then
x = alor;b C==> 10gb X = 10gb [alD~ C] = {10gb C )(logb a) and
y=
b = } lo&y = 10gb [dog" a] = (lo&a)(logb c),
Consequently, we find that 10& x = 10gb Y from which it follows that x = y.
j
(a)
(c)
2.
0 =
[~ ~ :]
(B
C) =
[~ ~ !]
(b)
+ 0 = [~ ~ ~]
(d)
(f)
2A
(h)
5C = [2
(j)
A + 2B - 3C = [
(e)
2A =
[4-2 02 8]
6
(g)
20
30 =
(i)
2B - 4C = [ _
(k)
2(3B) =
3a
B)
3251
=[0 2 7 J
A
4 = 2,
- 12 =
[2~ 20 ~i51
5
~8 :122 ;~]
[~ 162 2~]
a = -2/3;
c=
- 8 = 0,
3d - 8 =
[ :]
-11 2
<1] [ 1 2
[
o3
1 3
3B =
[71 56 11]
18
5 ~i5]
0
(1)
, 0 1"
:=
(A
. 3)B =
;0
_4 -80 200]
[~ 1~ 2~]
b = 8/3i
d=
[9 1
[-
P! f1
[: ;]
5] [ ~ ~][ 0
1
:
3
3 10 24]
[ 3 9 15
[
A(B
1
386
C) = [
0
=
5.
-12 33
4]([12-4]
[ 0 -7 -! ])
~
+
3
1 3
5
-2
[~
CA =
16
=[
-12 33
; -:] [ ~ ] + [ -~ -7 -! 1[ ~]
3
1
0
1
0
-4] + [ 3 -~] = [ 4-4]
25
-5
21
-1 -4
11
(a) (-1/5) [_~
-~]
-3
21
(b)
[~ ~]
( c) The inverse does :not exist.
[~ ~ 1
01
=[22r[1 0]=
2] [i J = [~ - ]
=[; :f([ ~] [~ ~]) = [ 2] [~ ~ J -~~ ]
=
~ -~ ]
AB= [
~]
[! 1
[
231
3
1.
1
[ 1 5-6] [-1~ 4]~ = [4 ]
(d)
6.
16
-2~ 18 =
-6
J
(b) BA
= [~
--21 50
n[ 5 11] =! 50]
1
0
= [
-31 26] [
A-I
5
(1/2) [
-
3
= [
--1
[~
]
-
[~ ]
(a)
-50
[! l[:l=[~l
[!] ) [ ~][~]=[~]
[:]=[!
[~ 3][:]=[325]
9.
r
[:l=[~ r [
=
(b)
_~
325] '" (-1/19) [ -2
51[21=[~]
(c) 21
(a) 21
(b) 21
775
11- det(2A) =22(31) = 124, det(5A) = 5 (31) =
10.
2
12.
1 0
.3 1 -1
4 1
2
(a)
= 3( _1)2+ 1 10 -21 + I( _1)2+21 1 -22
1
2
4
1! ~ = -3(2) + (10) 1 5
1 0 -2
!~
+( -1)( -1)2+·3
=
I
(-1)( _1)2+3 1
3 1 -1 = (-2)( _1)1+3
4 1
2
.31 ()1
I 1 31
1=
1
I~ 5
11
2
I
"",I
I
31_
o rq1 1
2 3
j
o5
2(1) = 5
2
-
j
I~
2 ==
11
1
2
21
1=
!~I
I
1
1'
(d)
63
0 2
6 -2 1
4:
3 2
1
13.
(a)
2(18
(-4 - 3)
= (1)(8) =
4: 7 0
4 2 0 = (2)(3 6 2
(c)
1 2
0 1
3 3
+
I
3
31 =
i
=45
I! ~
o 1=(1)(_1)1+1 \13
21
~ 1+ 2( -1 )1+ 1 ~
3
1= (2)(8 - 28) = -40
12 = 14
21 = 2
1
14.
(a) (i) 0
(ii) 0
(iii) 0
(iv) 0
(b) Let A be a 3 x 3 matrix. If two rows of A are identical, or if two columns of A are
identical, th.en det(A) = O. In fact, for each n E Z+ where n > if A is an n X n matrix
with two identical rows or two identical columns, then det(A) = O.
15.
(a) (i)
1
2
1
!
=2(_1)3+ 1 1
0 -1 -1
2
3
0
11+ 3(_1)3+2 01
!
2
-1 -1
1
i
= 2( - 2 - ( -1 » - 3( -1) = 2( -1)
16.
(ii) 5
(iii) 25
(b) (i)
(U) 306
3 = 1.
(iii) 510
There are n 2 entries in the matrix product AB. For each entry we perform n multiplications
and n - 1 additions. Therefore,
we perform n 3 multiplications and n 2(n - 1) =
- f!i additions.
APPENDIX 3
COUNTABLE
UNCOUNTABLE
1.
(Ii},) True
(b) False
(c) True
(d) True
(e) True
(f) True
(g) False: Let
= Z+ U (0,1] and B = (0,1]. Then A, B are both uncountable, but
A - B = {2, 3, 4, ... } is countable.
2.
(a) The function I: z+ ~ A defined by I(n) = n 2 is a one-to-one correspondence.
(b) Let y: z+ ~ {2,6, 10, 14, ... } be defined
yen) = (n-l)4+2. Then 9 is a one-to-one
correspondence.
3.
If B were countable, then by Theorem A3.3 it would follow that A is countable. This leads
us to a contradiction since we are given that A is uncountable.
4.
The set [ of irrational numbers is uncountable. If not, then R = Q U [ would be countable
- by virtue of Theorems A3.8 and A3.9 (or, A3.7).
5.
Since S, T are countably infinite, we know from Theorem A3.2 that we can write S =
{ShSa,S31"'} and T = {tb t",
... } - two (jnfinit-e) sequences of distinct terms. Define
the function
f: S X T ~ z+
=
by f(Si,tj)
2i ai, for all i,j E Z+. If i,j,k,l E Z+ with !(Si,t;) = f(Sk,tl), then
j'(Si' tj) = j'(81:;, tt) ::::}
= 2k3£ ::::} i = k, j = I. (By the Fundamental Theorem of
:::¢-- 8;, = 13k
tj = t.e ::::} (Si' tj) = (Sk'
f is a
functioll and S x
"" f(8 x T) C Z+.
from Theorem A3.3 we know that S X T is
x
x
= f(a21 b2 , C2),
-
::::}
= ~,Cl ::.'::: C2
- f(Z+ X Z+ X
f(o" b, c) =
18 nnf"-T.n,-or~p
(Z - {O}) X Z X Z is COUIU,RD.l.e.
468
for all (a, b, c) E (Z - {O}) x Z x Z there are at most two (distinct) real solutions for the
quadratic
ax 2 ox + c =
it
set
bx + c =
where a, OJ c E Z
a
all real solutions of the quadratic
is countable.
(a) f(x)=3x,
O<x<l
= 5x + 2, 0 < x < 1
(e) hex) = (b - a)x a, 0 < x < 1