-
Instructor’s Manual
to accompany
Elements of Modern Algebra, Eighth Edition
Linda Gilbert and the late Jimmie Gilbert
University of South Carolina Upstate
Spartanburg, South Carolina
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Contents
Preface ix
Chapter 1 Fundamentals 1Section 1.1: True/False . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . 1Exercises 1.1: 1, 2,
3, 4, 5, 6, 7, 8, 9, 10, 11, 36, 37, 38, 40, 41, 42, 43 . . . . .
1Section 1.2: True/False . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . 4Exercises 1.2: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 28 . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . 4Section 1.3: True/False . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . 9Exercises 1.3: 1, 2, 3, 4, 5, 6,
7, 9, 12 . . . . . . . . . . . . . . . . . . . . . . . 9Section
1.4: True/False . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . 12Exercises 1.4: 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12 . . . .
. . . . . . . . . . . . . . . 12Section 1.5: True/False . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . 13Exercises 1.5: 1,
2, 3, 4, 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. 13Section 1.6: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 15Exercises 1.6: 1, 2, 3, 4, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 22(b), 25, 26, 27, 30 15Section 1.7: True/False
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17Exercises 1.7: 1, 2, 3, 4(b), 5(b), 6, 7, 8, 9, 10, 11, 12, 13,
14, 15, 16, 17, 18,
19, 20, 21, 22, 23, 25, 26, 28 . . . . . . . . . . . . . . . . .
. . . . . . . . 17
Chapter 2 The Integers 23Section 2.1: True/False . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . 23Exercises 2.1: 21,
30, 31, 32, 35 . . . . . . . . . . . . . . . . . . . . . . . . . .
23Exercises 2.2: 33, 37, 39, 40 . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 24Section 2.3: True/False . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . 26Exercises 2.3: 1, 2, 3, 4,
5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 25, 29, 30 . . .
26Section 2.4: True/False . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . 27Exercises 2.4: 1, 2, 3, 4, 6, 21, 30(a), 31 . .
. . . . . . . . . . . . . . . . . . . 27Section 2.5: True/False . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Exercises
2.5: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18,
19, 20,
21, 22, 23, 24, 29, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50,
51, 52, 53, 56 28Section 2.6: True/False . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . 29Exercises 2.6: 1, 2, 3, 4, 5,
6, 7, 8, 10, 11, 12, 13, 14, 19, 20(b), 21 . . . . . . 29
v
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vi Contents
Section 2.7: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 33
Exercises 2.7: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 14, 17, 18, 19,
20, 22, 23, 24, 25, 26 . 33
Section 2.8: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 34
Exercises 2.8: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 16,
17, 18, 19, 20, 21, 22,23, 25, 26 . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . 34
Chapter 3 Groups 36
Section 3.1: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 36
Exercises 3.1: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20,21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31,
32, 33, 34, 35, 36, 40, 41, 42(b),43, 44, 45, 46, 47, 48, 49, 50 .
. . . . . . . . . . . . . . . . . . . . . . . . 36
Section 3.2: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 40
Exercises 3.2: 5, 6, 7, 8, 9, 11(b), 13, 14, 21, 23, 27, 28 . .
. . . . . . . . . . . 40
Section 3.3: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 42
Exercises 3.3: 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 20, 21, 22, 23,
32, 35, 38, 40, 42, 45 . 42
Section 3.4: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 46
Exercises 3.4: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
15(b,c), 18, 19, 20, 21, 22,23, 24, 25, 26, 27, 35, 36, 37 . . . .
. . . . . . . . . . . . . . . . . . . . . 46
Section 3.5: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 52
Exercises 3.5: 2(b), 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15,
18, 25, 26, 27, 32, 36 52
Section 3.6: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 56
Exercises 3.6: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 17, 22
. . . . . . . . . . . 56
Chapter 4 More on Groups 57
Section 4.1: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 57
Exercises 4.1: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20,21, 22, 24, 26, 27, 28, 30(b,c,d) . . . . .
. . . . . . . . . . . . . . . . . . 57
Section 4.2: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 60
Exercises 4.2: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10(c), 11(c), 12,
13(c) . . . . . . . . . . 60
Section 4.3: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 65
Exercises 4.3: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20,22, 23, 24, 25, 26, 27, 28, 29 . . . . . . .
. . . . . . . . . . . . . . . . . . 65
Section 4.4: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 71
Exercises 4.4: 1, 2, 3, 4, 5, 6, 7, 8, 11, 12, 19, 20, 21, 22,
23, 24 . . . . . . . . 71
Section 4.5: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 73
Exercises 4.5: 1, 9, 10, 11, 12, 13, 14, 15, 25, 26, 29, 30, 32,
37, 40 . . . . . . 73
Section 4.6: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 74
Exercises 4.6: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 25, 26, 27, 30 74
Section 4.7: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 82
Exercises 4.7: 1, 2, 7, 8, 17, 18, 19 . . . . . . . . . . . . .
. . . . . . . . . . . 82
Section 4.8: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 84
Exercises 4.8: 1, 2, 3, 4, 5, 6, 9, 10, 12, 14(b), 15(b) . . . .
. . . . . . . . . . 84
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Contents vii
Chapter 5 Rings, Integral Domains, and Fields 85Section 5.1:
True/False . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . 85Exercises 5.1: 2, 3, 4, 5, 6, 7, 8, 18, 19, 20, 21(b,c), 22,
25, 26, 32, 33, 34, 35,
36, 38, 41, 42(b,c), 43(b), 51(d), 52, 53, 54, 55 . . . . . . .
. . . . . . . . 85Section 5.2: True/False . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . 91Exercises 5.2: 1, 2, 3, 4, 5,
6(b,c,d,e), 7, 8, 9, 10, 11, 12, 13, 15, 19, 20 . . . . 91Section
5.3: True/False . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . 93Exercises 5.3: 9, 10, 11, 15, 18 . . . . . . . . . . . .
. . . . . . . . . . . . . . 93Section 5.4: True/False . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . 96
Chapter 6 More on Rings 96Section 6.1: True/False . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . 96Exercises 6.1: 3,
6, 9, 11, 18, 23, 27, 28(b,c,d), 29(b), 30(b) . . . . . . . . . .
96Section 6.2: True/False . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . 98Exercises 6.2: 1, 7(b), 8(b), 9(b), 10(b), 12,
13, 17, 18, 25, 26, 27, 30(b) . . . 98Section 6.3: True/False . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . 101Exercises
6.3: 1, 2, 4, 9(b), 11, 12 . . . . . . . . . . . . . . . . . . . .
. . . . . 102Section 6.4: True/False . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . 103Exercises 6.4: 5, 6, 7, 8, 9, 10,
21, 22, 23 . . . . . . . . . . . . . . . . . . . . . 103
Chapter 7 Real and Complex Numbers 104Section 7.1: True/False .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
104Exercises 7.1: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 20, 21(a)
. . . . . . . . . . . 104Section 7.2: True/False . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . 104Exercises 7.2: 1, 2,
3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 21(b), 23, 24, 25, 26, 27,
28, 29, 30, 31, 32, 33, 34 . . . . . . . . . . . . . . . . . . .
. . . . . . . . 104Section 7.3: True/False . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . 105Exercises 7.3: 1, 2, 3, 6,
7, 8, 11, 12, 13, 14, 17 . . . . . . . . . . . . . . . . . 105
Chapter 8 Polynomials 108Section 8.1: True/False . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . 108Exercises 8.1: 1, 2,
3, 4, 5, 6, 8(b), 9(b), 11, 12, 13, 16(b,c), 17, 21, 23, 25(b)
108Section 8.2: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 110Exercises 8.2: 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15, 16, 17, 19, 20, 21,
22, 24, 35 . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . 110Section 8.3: True/False . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . 110Exercises 8.3: 1, 2, 3, 4,
7, 12, 13, 22, 27 . . . . . . . . . . . . . . . . . . . . .
110Section 8.4: True/False . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 112Exercises 8.4: 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 21,
22, 25(b), 34 . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . 112Section 8.5: True/False . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . 114Exercises 8.5: 1, 2, 3, 4,
5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25, 26, 27, 28 . . . . . . . . . . . . . . . . .
. . . . . . . . 114Section 8.6: True/False . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . 115Exercises 8.6: 1, 2, 3, 4,
5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18 . . . . . . . 116
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viii Contents
Appendix The Basics of Logic 125Exercises: 1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21,
22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37,
38, 39, 40, 41,42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54,
55, 56, 57, 58, 59, 60, 61,62, 63, 64, 65, 66, 67, 68, 69, 70, 71,
72, 73, 74 . . . . . . . . . . . . . . 125
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Preface
This manual provides answers for the computational exercises and
a few of theexercises requiring proofs in Elements of Modern
Algebra, Eighth Edition, by LindaGilbert and the late Jimmie
Gilbert. These exercises are listed in the table of contents.In
constructing proof of exercises, we have freely utilized prior
results, including thoseresults stated in preceding problems.
My sincere thanks go to Danielle Hallock and Lauren Crosby for
their careful man-agement of the production of this manual and to
Eric Howe for his excellent work onthe accuracy checking of all the
answers.
Linda Gilbert
ix
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Answers to Selected Exercises
Section 1.1
1. True 2. True 3. False 4. True 5. True 6. False 7. True
8. True 9. False 10. False
Exercises 1.1
1. a. = { | is a nonnegative even integer less than 12} b. © | 2
= 1ªc. = { | is a negative integer} d. © | = 2 for ∈ Z+ª
2. a. False b. True c. False d. False e. False f. True
3. a. True b. True c. True d. True e. True f. False
g. True h. True i. False j. False k. False l. True
4. a. False b. True c. True d. False e. True f. False
g. False h. True i. False j. False k. False l. False
5. a. {0 1 2 3 4 5 6 8 10} b. {2 3 5} c. {0 2 4 6 7 8 9 10} d.
{2}e. ∅ f. g. {0 2 3 4 5} h. {6 8 10} i. {1 3 5}j. {6 8 10} k. {1 2
3 5} l. m. {3 5} n. {1}
6. a. b. c. ∅ d. e. f. ∅ g. h. i. j. k. l. ∅ m. n. ∅
7. a. {∅ } b. {∅ {0} {1} }c. {∅ {} {} {} { } { } { } }d. {∅ {1}
{2} {3} {4} {1 2} {1 3} {1 4} {2 3} {2 4} {3 4} {1 2 3} {1 2 4} {1
3 4} {2 3 4} }
e. {∅ {1} {{1}} } f. {∅ } g. {∅ } h. {∅ {∅} {{∅}} }8. a. Two
possible partitions are:
1 = { | is a negative integer} and2 = { | is a nonnegative
integer} or
1 = { | is a negative integer} 2 = { | is a positive integer} 3
={0}
1
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2 Answers to Selected Exercises
b. One possible partition is 1 = { } and 2 = { } Another
possiblepartition is 1 = {} 2 = { } 3 = {}
c. One partition is 1 = {1 5 9} and 2 = {11 15} Another
partition is1 = {1 15} 2 = {11} and 3 = {5 9}
d. One possible partition is 1 = { | = + where is a positive
realnumber, is a real number} and 2 = { | = + where is anonpositive
real number, is a real number}. Another possible partition is1 = {
| = where is a real number}2 = { | = where isa nonzero real number}
and 3 = { | = + where and are bothnonzero real numbers}
9. a. 1 = {1} 2 = {2} 3 = {3} ;1 = {1} 2 = {2 3} ;1 = {2} 2 = {1
3} ;1 = {3} 2 = {1 2}
b. 1 = {1} 2 = {2} 3 = {3} 4 = {4} ;1 = {1} 2 = {2} 3 = {3 4} ;
1 = {1} 2 = {3} 3 = {2 4} ;1 = {1} 2 = {4} 3 = {2 3} ; 1 = {2} 2 =
{3} 3 = {1 4} ;1 = {2} 2 = {4} 3 = {1 3} ; 1 = {3} 2 = {4} 3 = {1
2} ;1 = {1 2} 2 = {3 4} ; 1 = {1 3} 2 = {2 4} ;1 = {1 4} 2 = {2 3}
; 1 = {1} 2 = {2 3 4} ;1 = {2} 2 = {1 3 4} ; 1 = {3} 2 = {1 2 4} ;1
= {4} 2 = {1 2 3}
10. a. 2 b.!
! (− )!11. a. ⊆ b. 0 ⊆ or ∪ = c. ⊆
d. ∩ = ∅ or ⊆ 0 e. = = f. 0 ⊆ or ∪ = g. = h. =
36. Let = { } = {} and = {} Then ∪ = = ∪ but 6=
37. Let = {} = { } and = { } Then ∩ = {} = ∩ but 6=
38. Let = { } and = { } Then ∪ = { } and { } ∈ P ( ∪)but { } ∈ P
() ∪ P ()
40. Let = { } and = {} Then − = {} and ∅ ∈ P (−) but ∅ ∈P ()− P
()
41. ( ∩0) ∪ (0 ∩) = ( ∪) ∩ (0 ∪0)
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Answers to Selected Exercises 3
42. a.
∪ : Regions 1,2,3 − : Region 1 ∩ : Region 2 − : Region 3
( ∪)− ( ∩) : Regions 1,3 (−) ∪ ( −) : Regions 1,3+ : Regions
1,3
Each of + and (−) ∪ ( −) consists of Regions 1,3.
b.
: Regions 1,4,5,7 + : Regions 1,2,4,6
+ : Regions 2,3,4,5 : Regions 3,4,6,7
+ ( + ) : Regions 1,2,3,7 (+) + : Regions 1,2,3,7
Each of + ( + ) and (+) + consists of Regions 1,2,3,7.
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4 Answers to Selected Exercises
c.
: Regions 1,4,5,7 ∩ : Regions 5,7 + : Regions 2,3,4,5 ∩ :
Regions 4,7
∩ ( + ) : Regions 4,5 ( ∩) + ( ∩ ) : Regions 4,5Each of ∩ ( + )
and ( ∩) + ( ∩) consists of Regions 4,5.
43. a. + = ( ∪)− ( ∩) = − = ∩0 = ∅b. +∅ = ( ∪∅)− ( ∩∅) = −∅ =
∩∅0 =
Section 1.2
1. False 2. False 3. False 4. False 5. False 6. True 7. True
8. False 9. True
Exercises 1.2
1. a. {( 0) ( 1) ( 0) ( 1)} b. {(0 ) (0 ) (1 ) (1 )}c. {(2 2) (4
2) (6 2) (8 2)}d. {(−1 1) (−1 5) (−1 9) (1 1) (1 5) (1 9)}e. {(1 1)
(1 2) (1 3) (2 1) (2 2) (2 3) (3 1) (3 2) (3 3)}
2. a. Domain = E Codomain = Z Range = Z
b. Domain = E Codomain = Z Range = E
c. Domain = E Codomain = Z
Range = { | is a nonnegative even integer} = (Z+ ∩E) ∪ {0}d.
Domain = E Codomain = Z Range = Z−E
3. a. () = {1 3 5 } = Z+ −E −1 ( ) = {−4−3−1 1 3 4}b. () = {1 5
9} −1 ( ) = Z c. () = {0 1 4} −1 ( ) = ∅
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Answers to Selected Exercises 5
d. () = {0 2 14} −1 ( ) = Z+ ∪ {0−1−2}4. a. The mapping is not
onto, since there is no ∈ Z such that () = 1 It is
one-to-one.
b. The mapping is not onto, since there is no ∈ Z such that () =
1 It isone-to-one.
c. The mapping is onto and one-to-one.
d. The mapping is one-to-one. It is not onto, since there is no
∈ Z suchthat () = 2
e. The mapping is not onto, since there is no ∈ Z such that () =
−1. Itis not one-to-one, since (1) = (−1) and 1 6= −1.
f. We have (3) = (2) = 0 so is not one-to-one. Since () is
always even,there is no ∈ Z such that () = 1 and is not onto.
g. The mapping is not onto, since there is no ∈ Z such that () =
3 It isone-to-one.
h. The mapping is not onto, since there is no ∈ Z such that () =
1Neither is one-to-one since (0) = (1) and 0 6= 1
i. The mapping is onto. It is not one-to-one, since (9) = (4)
and 9 6= 4j. The mapping is not onto, since there is no ∈ Z such
that () = 4 It isone-to-one.
5. a. The mapping is onto and one-to-one.
b. The mapping is onto and one-to-one.
c. The mapping is onto and one-to-one.
d. The mapping is onto and one-to-one.
e. The mapping is not onto, since there is no ∈ R such that () =
−1 It isnot one-to-one, since (1) = (−1) and 1 6= −1
f. The mapping is not onto, since there is no ∈ R such that () =
1 It isnot one-to-one, since (0) = (1) = 0 and 0 6= 1
6. a. The mapping is onto and one-to-one.
b. The mapping is one-to-one. Since there is no ∈ E such that ()
= 2the mapping is not onto.
7. a. The mapping is onto. The mapping is not one-to-one, since
(1) = (−1)and 1 6= −1
b. The mapping is not onto, since there is no ∈ Z+ such that ()
= −1The mapping is one-to-one.
c. The mapping is onto and one-to-one.
d. The mapping is onto. The mapping is not one-to-one, since (1)
= (−1)and 1 6= −1
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6 Answers to Selected Exercises
8. a. The mapping is not onto, since there is no ∈ Z such that
|+ 4| = −1The mapping is not one-to-one, since (1) = (−9) = 5 but 1
6= −9
b. The mapping is not onto, since there is no ∈ Z+ such that |+
4| = 1The mapping is one-to-one.
9. a. The mapping is not onto, since there is no ∈ Z+ such that
2 = 3 Themapping is one-to-one.
b. The mapping is not onto, since there is no ∈ Z+ ∩ E such that
2 = 6The mapping is one-to-one.
10. a. Let : E→ E where () = b. Let : E→ E where () = 2
c. Let : E→ E where () =⎧⎨⎩ 2 if is a multiple of 4 if is not a
multiple of 4.
d. Let : E→ E where () = 211. a. For arbitrary ∈ Z 2 is even and
(2) = 22 = Thus is onto. But
is not one-to-one, since (1) = (−1) = 0.b. The mapping is not
onto, since there is no in Z such that () = 1 Themapping is not
one-to-one, since (0) = (2) = 0
c. For arbitrary in Z 2− 1 is odd, and therefore
(2− 1) = (2− 1) + 12
=
Thus is onto. But is not one-to-one, since (2) = 5 and also (9)
= 5
d. For arbitrary in Z, 2 is even and (2) = 22 = Thus is onto.
But is not one-to-one, since (4) = 2 and (7) = 2
e. The mapping is not onto, because there is no in Z such that
() = 4Since (2) = 6 and (3) = 6 then is not one-to-one.
f. The mapping is not onto, since there is no in Z such that ()
= 1Suppose that (1) = (2) It can be seen from the definition of
that theimage of an even integer is always an odd integer, and also
that the image ofan odd integer is always an even integer.
Therefore, (1) = (2) requiresthat either both 1 and 2 are even, or
both 1 and 2 are odd. If both 1and 2 are even,
(1) = (2)⇒ 21 − 1 = 22 − 1⇒ 21 = 22 ⇒ 1 = 2
If both 1 and 2 are odd,
(1) = (2)⇒ 21 = 22 ⇒ 1 = 2
Hence, (1) = (2) always implies 1 = 2 and is one-to-one.
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Answers to Selected Exercises 7
12. a. The mapping is not onto, because there is no ∈ R − {0}
such that () = 1 If 1 2 ∈ R− {0}
(1) = (2) ⇒ 1 − 11
=2 − 12
⇒ 2 (1 − 1) = 1 (2 − 1)⇒ 21 − 2 = 12 − 1⇒ −2 = −1⇒ 2 = 1
Thus is one-to-one.
b. The mapping is not onto, because there is no ∈ R − {0} such
that () = 2 If 1 2 ∈ R− {0}
(1) = (2) ⇒ 21 − 11
=22 − 1
2
⇒ 2− 11= 2− 1
2
⇒ − 11= − 1
2
⇒ 1 = 2Thus is one-to-one.
c. The mapping is not onto, since there is no ∈ R−{0} such that
() = 0It is not one-to-one, since (2) = 25 and
¡12
¢= 25
d. The mapping is not onto, since there is no ∈ R−{0} such that
() = 1Since (1) = (3) = 12 then is not one-to-one.
13. a. The mapping is onto, since for every ( ) ∈ = Z × Z there
exists an( ) ∈ = Z×Z such that ( ) = ( ) To show that is
one-to-one,we assume ( ) ∈ = Z× Z and ( ) ∈ = Z× Z and
( ) = ( )
or( ) = ( )
This means = and = and
( ) = ( )
b. For any ∈ Z ( 0) ∈ and ( 0) = Thus is onto. Since (2 3) =(4
1) = 5 is not one-to-one.
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8 Answers to Selected Exercises
c. Since for every ∈ = Z there exists an ( ) ∈ = Z × Z such that
( ) = the mapping is onto. However, is not one-to-one, since (1 0)
= (1 1) and (1 0) 6= (1 1)
d. The mapping is one-to-one since (1) = (2) ⇒ (1 1) = (2 1) ⇒1
= 2 Since there is no ∈ Z such that () = (0 0) then is not
onto.
e. The mapping is not onto, since there is no ( ) ∈ Z×Z such
that ( ) =2 The mapping is not one-to-one, since (2 0) = (2 1) = 4
and (2 0) 6=(2 1)
f. The mapping is not onto, since there is no ( ) ∈ Z×Z such
that ( ) =3 The mapping is not one-to-one, since (1 0) = (−1 0) = 1
and (1 0) 6=(−1 0)
g. The mapping is not onto, since there is no ( ) in Z+ × Z+
such that ( ) = = 0 The mapping is not one-to-one, since (2 1) = (4
2) =2
h. The mapping is not onto, since there is no ( ) in R×R such
that ( ) = 2+ = 0 The mapping is not one-to-one, since (1 0) = (0
1) = 21
14. a. The mapping is obviously onto.
b. The mapping is not one-to-one, since (0) = (2) = 1
c. Let both 1 and 2 be even. Then 1 + 2 is even and (1 + 2) = 1
=1 · 1 = (1) (2) Let both 1 and 2 be odd. Then 1 + 2 is even and (1
+ 2) = 1 = (−1) (−1) = (1) (2) Finally, if one of 1 2 is evenand
the other is odd, then 1+2 is odd and (1 + 2) = −1 = (1) (−1) = (1)
(2) Thus it is true that (1 + 2) = (1) (2)
d. Let both 1 and 2 be odd. Then 12 is odd and (12) = −1 6=(−1)
(−1) = (1) (2)
15. a. The mapping is not onto, since there is no ∈ such that ()
= 9 ∈ It is not one-to-one, since (−2) = (2) and −2 6= 2
b. −1 ( ()) = −1 ({1 4}) = {−2 1 2} 6= c. With = {4 9} −1 ( ) =
{−2 2} and ¡−1 ( )¢ = ({−2 2}) = {4} 6=
16. a. () = {2 4} −1 ( ()) = {2 3 4 7}b. −1 ( ) = {9 6 11} ¡−1 (
)¢ =
17. a. () = {−1 2 3} −1 ( ()) = b. −1 ( ) = {0} ¡−1 ( )¢ =
{−1}
18. a. ( ◦ ) () =⎧⎨⎩ 2 if is even2 (2− 1) if is odd b. ( ◦ ) ()
= 23
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Answers to Selected Exercises 9
c. ( ◦ ) () =⎧⎨⎩
+ ||2
if is even
||− if is oddd. ( ◦ ) () =
e. ( ◦ ) () = (− ||)2
19. a. ( ◦ ) () = 2 b. ( ◦ ) () = 83 c. ( ◦ ) () = + ||2
d. ( ◦ ) () =⎧⎨⎩ 2 − 1 if = 4 for an integer otherwise e. ( ◦ )
() = 0
20. 21. ! 22. (− 1) (− 2) · · · (−+ 1) = !(−)!
28. Let : → where and are nonempty.Assume first that
¡−1 ( )
¢= for every subset of For an arbitrary ele-
ment of let = {} The equality ¡−1 ({})¢ = {} implies that −1
({})is not empty. For any ∈ −1 ({}) we have () = . Thus is
onto.Assume now that is onto. For an arbitrary ∈ ¡−1 ( )¢ we
have
∈ ¡−1 ( )¢ ⇒ = () for some ∈ −1 ( )⇒ = () for some () ∈ ⇒ ∈
Thus ¡−1 ( )
¢ ⊆ For an arbitrary ∈ there exists ∈ such that() = since is
onto. Now
() = ∈ ⇒ ∈ −1 ( )⇒ () ∈ ¡−1 ( )¢⇒ ∈ ¡−1 ( )¢
Thus ⊆ ¡−1 ( )¢ and we have proved that ¡−1 ( )¢ = for an
arbitrarysubset of
Section 1.3
1. False 2. True 3. False 4. False 5. False 6. False
Exercises 1.3
1. a. The mapping ◦ is not onto, since there is no ∈ Z such that
( ◦ ) () =1 It is not one-to-one, since ( ◦ ) (1) = ( ◦ ) (−1) and
1 6= −1
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10 Answers to Selected Exercises
b. The mapping ◦ is not onto, since there is no ∈ Z such that (
◦ ) () =0 The mapping ◦ is one-to-one.
c. The mapping ◦ is not onto, since there is no ∈ Z such that (
◦ ) () =1 The mapping ◦ is one-to-one.
d. The mapping ◦ is not onto, since there is no ∈ Z such that (
◦ ) () =1 The mapping ◦ is one-to-one.
e. The mapping ◦ is not onto, since there is no ∈ Z such that (
◦ ) () =1 It is not one-to-one, since ( ◦ ) (−2) = ( ◦ ) (0) and −2
6= 0
f. The mapping ◦ is both onto and one-to-one.g. The mapping ◦ is
not onto, since there is no ∈ Z such that ( ◦ ) () =−1 It is not
one-to-one, since ( ◦ ) (1) = ( ◦ ) (2) and 1 6= 2
2. a. The mapping ◦ is not onto, since there is no ∈ Z such that
( ◦ ) () =−1 It is not one-to-one since ( ◦ ) (0) = ( ◦ ) (2) and 0
6= 2
b. The mapping ◦ is not onto, since there is no ∈ Z such that (
◦ ) () =1 The mapping ◦ is one-to-one.
c. The mapping ◦ is not onto, since there is no ∈ Z such that (
◦ ) () =1 The mapping ◦ is one-to-one.
d. The mapping ◦ is not onto, since there is no ∈ Z such that (
◦ ) () =1 The mapping ◦ is one-to-one.
e. The mapping ◦ is not onto, since there is no ∈ Z such that (
◦ ) () =−1 It is not one-to-one, since ( ◦ ) (−1) = ( ◦ ) (−2) and
−1 6= −2.
f. The mapping ◦ is not onto, since there is no ∈ Z such that (
◦ ) () =0 The mapping ◦ is not one-to-one, since ( ◦ ) (1) = ( ◦ )
(4) and1 6= 4
g. The mapping ◦ is not onto, since there is no ∈ Z such that (
◦ ) () =1 It is not one-to-one, since ( ◦ ) (0) = ( ◦ ) (1) and 0
6= 1.
3. () = 2 () = −4. Let = {0 1} = {−2 1 2} = {1 4} Let : → be
defined by () =
+1 and : → be defined by () = 2 Then is not onto, since−2 ∈ ()
The mapping is onto. Also ◦ is onto, since ( ◦ ) (0) = (1) = 1 and(
◦ ) (1) = (2) = 4
5. Let and be defined as in Problem 1f. Then is not one-to-one,
is one-to-one,and ◦ is one-to-one.
6. a. Let : Z→ Z and : Z→ Z be defined by
() = () =
⎧⎨⎩ 2 if is even if is odd.The mapping is one-to-one and the
mapping is onto, but the composition ◦ = is not one-to-one, since (
◦ ) (1) = ( ◦ ) (2) and 1 6= 2
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Answers to Selected Exercises 11
b. Let : Z → Z and : Z → Z be defined by () = 3 and () = The
mapping is one-to-one, the mapping is onto, but the mapping ◦ given
by ( ◦ ) () = 3 is not onto, since there is no ∈ Z such that( ◦ )
() = 2
7. a. Let : Z→ Z and : Z→ Z be defined by
() =
⎧⎨⎩ 2 if is even if is odd () = The mapping is onto and the
mapping is one-to-one, but the composition ◦ = is not one-to-one,
since ( ◦ ) (1) = ( ◦ ) (2) and 1 6= 2
b. Let : Z → Z and : Z → Z be defined by () = and () = 3The
mapping is onto, the mapping is one-to-one, but the mapping ◦ given
by ( ◦ ) () = 3 is not onto, since there is no ∈ Z such that( ◦ )
() = 2
9. a. Let () = () = 2 and () = || for all ∈ Zb. Let () = 2 () =
and () = − for all ∈ Z
12. To prove that is one-to-one, suppose (1) = (2) for 1 and 2
in Since ◦ is onto, there exist 1 and 2 in such that
1 = ( ◦ ) (1) and 2 = ( ◦ ) (2)
Then (( ◦ ) (1)) = (( ◦ ) (2)) since (1) = (2) or
( ◦ ) ( (1)) = ( ◦ ) ( (2))
This implies that (1) = (2)
since ◦ is one-to-one. Since is a mapping, then
( (1)) = ( (2))
Thus( ◦ ) (1) = ( ◦ ) (2)
and1 = 2
Therefore is one-to-one.
To show that is onto, let ∈ Then () ∈ and therefore () = ( ◦ )
()for some ∈ since ◦ is onto. It follows then that
( ◦ ) () = ( ◦ ) ( ())
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12 Answers to Selected Exercises
Since ◦ is one-to-one, we have = ()
and is onto.
Section 1.4
1. False 2. True 3. True 4. False 5. True 6. True 7. True
8. True 9. True
Exercises 1.4
1. a. The set is not closed, since −1 ∈ and −1 ∗ −1 = 1 ∈ b. The
set is not closed, since 1 ∈ and 2 ∈ but 1 ∗ 2 = 1− 2 = −1 ∈ c. The
set is closed.
d. The set is closed.
e. The set is not closed, since 1 ∈ and 1 ∗ 1 = 0 ∈ f. The set
is closed.
g. The set is closed.
h. The set is closed.
2. a. Not commutative, Not associative, No identity element
b. Not commutative, Associative, No identity element
c. Not commutative, Not associative, No identity element
d. Commutative, Not associative, No identity element
e. Commutative, Associative, No identity element
f. Not commutative, Not associative, No identity element
g. Commutative, Associative, 0 is an identity element. 0 is the
only invertibleelement and its inverse is 0
h. Commutative, Associative, −3 is an identity element. −− 6 is
the inverseof
i. Not commutative, Not associative, No identity element
j. Commutative, Not associative, No identity element
k. Not commutative, Not associative, No identity element
l. Commutative, Not associative, No identity element
m. Not commutative, Not associative, No identity element
n. Commutative, Not associative, No identity element
3. a. The binary operation ∗ is not commutative, since ∗ 6=
∗
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Answers to Selected Exercises 13
b. There is no identity element.
4. a. The operation ∗ is commutative, since ∗ = ∗ for all in b.
is an identity element.
c. The elements and are inverses of each other and is its own
inverse.
5. a. The binary operation ∗ is not commutative, since ∗ 6= ∗b.
is an identity element.
c. The elements and are inverses of each other and is its own
inverse.
6. a. The binary operation ∗ is commutative.b. is an identity
element.
c. is the only invertible element and its inverse is
7. The set of nonzero integers is not closed with respect to
division, since 1 and 2are nonzero integers but 1÷ 2 is not a
nonzero integer.
8. The set of odd integers is not closed with respect to
addition, since 1 is an oddinteger but 1 + 1 is not an odd
integer.
10. a. The set of nonzero integers is not closed with respect to
addition defined onZ, since 1 and −1 are nonzero integers but 1+
(−1) is not a nonzero integer.
b. The set of nonzero integers is closed with respect to
multiplication definedon Z.
11. a. The set is not closed with respect to addition defined on
Z, since 1 ∈ 8 ∈ but 1 + 8 = 9 ∈
b. The set is closed with respect to multiplication defined on
Z.
12. a. The set Q− {0} is closed with respect to multiplication
defined on Rb. The set Q− {0} is closed with respect to division
defined on R− {0}
Section 1.5
1. True 2. False 3. False
Exercises 1.5
1. a. A right inverse does not exist, since is not onto.
b. A right inverse does not exist, since is not onto.
c. A right inverse : Z→ Z is defined by () = − 2d. A right
inverse : Z→ Z is defined by () = 1− e. A right inverse does not
exist, since is not onto.
f. A right inverse does not exist, since is not onto.
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14 Answers to Selected Exercises
g. A right inverse does not exist, since is not onto.
h. A right inverse does not exist, since is not onto.
i. A right inverse does not exist, since is not onto.
j. A right inverse does not exist, since is not onto.
k. A right inverse : Z→ Z is defined by () =⎧⎨⎩ if is even2+ 1
if is odd.
l. A right inverse does not exist, since is not onto.
m. A right inverse : Z→ Z is defined by () =⎧⎨⎩ 2 if is even− 2
if is odd.
n. A right inverse : Z→ Z is defined by () =⎧⎨⎩ 2− 1 if is even−
1 if is odd.
2. a. A left inverse : Z→ Z is defined by () =⎧⎨⎩ 2 if is even1
if is odd.
b. A left inverse : Z→ Z is defined by () =⎧⎨⎩ 3 if is a
multiple of 30 if is not a multiple of 3.
c. A left inverse : Z→ Z is defined by () = − 2d. A left inverse
: Z→ Z is defined by () = 1−
e. A left inverse : Z→ Z is defined by () =⎧⎨⎩ if = 3 for some ∈
Z0 if 6= 3 for some ∈ Z
f. A left inverse does not exist, since is not one-to-one.
g. A left inverse : Z→ Z is defined by () =⎧⎨⎩ if is even+ 1
2if is odd.
h. A left inverse does not exist, since is not one-to-one.
i. A left inverse does not exist, since is not one-to-one.
j. A left inverse does not exist, since is not one-to-one.
k. A left inverse does not exist, since is not one-to-one.
l. A left inverse : Z→ Z is defined by: () =⎧⎨⎩ + 1 if is
odd
2 if is even.
m. A left inverse does not exist, since is not one-to-one.
n. A left inverse does not exist, since is not one-to-one.
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Answers to Selected Exercises 15
3. !
4. Let : → where is nonempty.
has a left inverse ⇔ is one-to-one, by Lemma 1.24⇔ −1 ( ()) =
for every subset of by
Exercise 27 of Section 1.2.
5. Let : → where is nonempty.
has a right inverse ⇔ is onto, by Lemma 1.25⇔ ¡−1 ( )¢ = for
every subset of by
Exercise 28 of Section 1.2.
Section 1.6
1. True 2. False 3. False 4. False 5. False 6. False 7. True
8. False 9. False 10. False 11. True 12. True
Exercises 1.6
1. a. =
⎡⎢⎢⎢⎣1 0
3 2
5 4
⎤⎥⎥⎥⎦ b. =⎡⎢⎢⎢⎢⎢⎢⎣−1 −21 2
−1 −21 2
⎤⎥⎥⎥⎥⎥⎥⎦ c. =⎡⎣ 1 −1 1 −1−1 1 −1 1
⎤⎦
d. =
⎡⎢⎢⎢⎣0 1 1 1
0 0 1 1
0 0 0 1
⎤⎥⎥⎥⎦ e. =⎡⎢⎢⎢⎢⎢⎢⎣2 0 0
3 4 0
4 5 6
5 6 7
⎤⎥⎥⎥⎥⎥⎥⎦ f. =⎡⎢⎢⎢⎢⎢⎢⎣1 0 0
0 1 0
0 0 1
0 0 0
⎤⎥⎥⎥⎥⎥⎥⎦
2. a.
⎡⎣ 3 0 −48 −8 6
⎤⎦ b.⎡⎣ 1 9−3 2
⎤⎦ c. Not possible d. Not possible
3. a.
⎡⎣ −5 78 −1
⎤⎦ b.⎡⎢⎢⎢⎣−10 2 1−14 6 −216 −1 −2
⎤⎥⎥⎥⎦ c. Not possible d.⎡⎢⎢⎢⎣
7 −1112 6
−2 20
⎤⎥⎥⎥⎦
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16 Answers to Selected Exercises
e.
⎡⎣ 4 23 7
⎤⎦ f.⎡⎣ 1 3−4 10
⎤⎦ g. Not possible h. Not possible
i. [4] j.
⎡⎢⎢⎢⎣−12 8 −4−15 10 −518 −12 6
⎤⎥⎥⎥⎦4. =
3X=1
(+ ) (2 − )
= (+ 1) (2− ) + (+ 2) (4− ) + (+ 3) (6− )= 12− 6 − 3 + 28
6.
⎡⎣ 1 6 −3 24 −7 1 5
⎤⎦⎡⎢⎢⎢⎢⎢⎢⎣
⎤⎥⎥⎥⎥⎥⎥⎦ =⎡⎣ 90
⎤⎦
7. a. b. (− 1) c. 0d. if 1 ≤ ≤ 1 ≤ ≤ ; 0 if or
8.
·
9. (Answer not unique) =
⎡⎣ 1 23 4
⎤⎦ =⎡⎣ 1 11 1
⎤⎦10. A trivial example is with = 2 and an arbitrary 2 × 2
matrix. Another
example is provided by =
⎡⎣ 1 11 1
⎤⎦ and =⎡⎣ 2 33 2
⎤⎦ 11. (Answer not unique) =
⎡⎣ 1 21 2
⎤⎦ =⎡⎣ −6 −6
3 3
⎤⎦
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Answers to Selected Exercises 17
12. (−) (+) =⎡⎣ 10 12 1
⎤⎦ and 2 − 2 =⎡⎣ 2 6−4 9
⎤⎦ (−) (+) 6=2 −2
13. (+)2=
⎡⎣ 22 530 7
⎤⎦ 2+2+2 =⎡⎣ 30 036 −1
⎤⎦ (+)2 6= 2+2+214. = −1 15. = −1−1
22. b. For each in of the form
⎡⎣ 0 0
⎤⎦ then =⎡⎣ 1 10 0
⎤⎦ For each in of the form
⎡⎣ 0 0
⎤⎦ then =⎡⎣ 0 01 1
⎤⎦ 25. Let =
⎡⎣ 1 11 1
⎤⎦ and =⎡⎣ 2 00 7
⎤⎦ Then the product =⎡⎣ 2 72 7
⎤⎦ is notdiagonal even though is diagonal.
26. Let =
⎡⎣ 0 01 1
⎤⎦ and =⎡⎣ 0 10 1
⎤⎦ Then the product =⎡⎣ 0 00 2
⎤⎦ isdiagonal but neither nor is diagonal .
27. c. Let =
⎡⎣ 1 11 1
⎤⎦ and =⎡⎣ 1 −1−1 1
⎤⎦ Then the product =⎡⎣ 0 00 0
⎤⎦is upper triangular but neither nor is upper triangular.
30. (Answer not unique) =
⎡⎣ 1 00 0
⎤⎦ =⎡⎣ 2 34 5
⎤⎦ =⎡⎣ 2 36 7
⎤⎦ Section 1.7
1. True 2. False 3. True 4. False 5. True 6. False
Exercises 1.7
1. a. This is a mapping, since for every ∈ there is a unique ∈
such that( ) is an element of the relation.
b. This is a mapping, since for every ∈ there is 1 ∈ such that (
1) is anelement of the relation.
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18 Answers to Selected Exercises
c. This is not a mapping, since the element 1 is related to
three different values;11 13 and 15
d. This is a mapping, since for every ∈ there is a unique ∈ such
that( ) is an element of the relation.
e. This is a mapping, since for every ∈ there is a unique ∈ such
that( ) is an element of the relation.
f. This is not a mapping, since the element 5 is related to
three different values:51 53 and 55
2. a. The relation is not reflexive, since 2 /2 It is not
symmetric, since 4R2 but2 /4 It is not transitive, since 4R2 and
2R1 but 4 /1
b. The relation is not reflexive, since 2 /2 It is symmetric,
since = − ⇒ = − It is not transitive, since 2R(−2) and (−2)R2, but
2 /2
c. The relation is reflexive and transitive, but not symmetric,
since for arbi-trary and in Z we have:
(1) = · 1 with 1 ∈ Z(2) 6 = 3 (2) with 2 ∈ Z but 3 6= 6 where ∈
Z(3) = 1 for some 1 ∈ Z and = 2 for some 2 ∈ Z imply = 2 =
(12) with 12 ∈ Zd. The relation is not reflexive, since 1 /1 It
is not symmetric, since 1R2 but2 /1 It is transitive, since and ⇒
for all and ∈ Z
e. The relation is reflexive, since ≥ for all ∈ Z It is not
symmetric,since 53 but 3 /5 It is transitive, since ≥ and ≥ imply ≥
for all in Z
f. The relation is not reflexive, since (−1) /(−1) It is not
symmetric, since1R (−1) but (−1) /1 It is transitive, since = ||
and = || implies = || = |||| = || for all and ∈ Z
g. The relation is not reflexive, since (−6) /(−6) It is not
symmetric, since3R5 but 5 /3 It is not transitive, since 4R3 and
3R2, but 4 /2
h. The relation is reflexive, since 2 ≥ 0 for all in Z It is
also symmetric,since ≥ 0 implies that ≥ 0 It is not transitive,
since (−2)0 and 04but (−2) /4
i. The relation is not reflexive, since 2 /2 It is symmetric,
since ≤ 0implies ≤ 0 for all ∈ Z It is not transitive, since −12
and 2 (−3)but (−1) /(−3)
j. The relation is not reflexive, since |− | = 0 6= 1 It is
symmetric, since|− | = 1⇒ | − | = 1 It is not transitive, since |2−
1| = 1 and |1− 2| =1 but |2− 2| = 0 6= 1
k. The relation is reflexive, symmetric and transitive, since
for arbitrary and in Z we have:
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Answers to Selected Exercises 19
(1) |− | = |0| 1(2) |− | 1⇒ | − | 1(3) |− | 1 and | − | 1⇒ = and
= ⇒ |− | 1
3. a. {−3 3} b. {−5−1 3 7 11} ⊆ [3]4. b. [0] = { −10−5 0 5 10 }
[1] = { −9−4 1 6 11 }
[2] = { −8−3 2 7 12 } [8] = [3] = { −7−2 3 8 13 }[−4] = [1] = {
−9−4 1 6 11 }
5. b. [0] = { −14−7 0 7 14 } [1] = { −13−6 1 8 15 }[3] = { −11−4
3 10 17 } [9] = [2] = { −12−5 2 9 16 }[−2] = [5] = { −9−2 5 12 19
}
6. [0] = { −2 0 2 4 } [1] = { −3−1 1 3 }7. [0] = {0±5±10 }
{±1±4±6±9} ⊆ [1] {±2±3±7±8} ⊆ [2]8. [0] = { −4 0 4 8 } [1] = { −7−3
1 5 } [2] = { −6−2 2 6 } [3] = { −5−1 3 7 }
9. [0] = { −7 0 7 14 } [1] = { −13−6 1 8 } [2] = { −12−5 2 9 }
[3] = { −11−4 3 10 } [4] = { −10−3 4 11 } [5] = { −9−2 5 12 } [6] =
{ −8−1 6 13 }
10. [−1] = { −3−1 1 3 } [0] = { −2 0 2 4 }11. The relation is
symmetric but not reflexive or transitive, since for arbitrary
integers and , we have the following:
(1) + = 2 is not odd;
(2) + is odd implies + is odd;
(3) + is odd and + is odd does not imply that + is odd. For
example,take = 1 = 2 and = 3
Thus is not an equivalence relation on Z
12. a. The relation is symmetric but not reflexive or
transitive, since for arbitrarylines 1 2 and 3 in a plane, we have
the following:
(1) 1 is not parallel to 1 since parallel lines have no points
in common;
(2) 1 is parallel to 2 implies that 2 is parallel to 1;
(3) 1 is parallel to 2 and 2 is parallel to 3 does not imply
that 1 is parallelto 3 For example, take 3 = 1 with 1 parallel to
2
Thus is not an equivalence relation on Z
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20 Answers to Selected Exercises
b. The relation is symmetric but not reflexive or transitive,
since for arbitrarylines 1 2 and 3 in a plane, we have the
following:
(1) 1 is not perpendicular to 1;
(2) 1 is perpendicular to 2 implies that 2 is perpendicular to
1;
(3) 1 is perpendicular to 2 and 2 is perpendicular to 3 does not
imply that1 is perpendicular to 3
Thus is not an equivalence relation.
13. a. The relation is reflexive and transitive but not
symmetric, since for arbi-trary nonempty subsets and of we
have:
(1) is a subset of ;
(2) is a subset of does not imply that is a subset of ;
(3) is a subset of and is a subset of imply that is a subset
of
b. The relation is not reflexive and not symmetric, but it is
transitive, sincefor arbitrary nonempty subsets and of we have:
(1) is not a proper subset of ;
(2) is a proper subset of implies that is not a proper subset of
;
(3) is a proper subset of and is a proper subset of imply that
is aproper subset of
c. The relation is reflexive, symmetric and transitive, since
for arbitrary non-empty subsets and of we have:
(1) and have the same number of elements;
(2) If and have the same number of elements, then and have
thesame number of elements;
(3) If and have the same number of elements and and have the
samenumber of elements, then and have the same number of
elements.
14. a. The relation is reflexive and symmetric but not
transitive, since if and are human beings, we have:
(1) lives within 400 miles of ;
(2) lives within 400 miles of implies that lives within 400
miles of ;
(3) lives within 400 miles of and lives within 400 miles of do
notimply that lives within 400 miles of
b. The relation is not reflexive, not symmetric, and not
transitive, since if and are human beings we have:
(1) is not the father of ;
(2) is the father of implies that is not the father of ;
(3) is the father of and is the father of imply that is not the
fatherof
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Answers to Selected Exercises 21
c. The relation is symmetric but not reflexive and not
transitive. Let and be human beings, and we have:
(1) is a first cousin of is not a true statement;
(2) is a first cousin of implies that is a first cousin of ;
(3) is a first cousin of and is a first cousin of do not imply
that isa first cousin of
d. The relation is reflexive, symmetric, and transitive, since
if and arehuman beings we have:
(1) and were born in the same year;
(2) if and were born in the same year, then and were born in
thesame year;
(3) if and were born in the same year and if and were born in
thesame year, then and were born in the same year.
e. The relation is reflexive, symmetric, and transitive, since
if and arehuman beings, we have:
(1) and have the same mother;
(2) and have the same mother implies and have the same
mother;
(3) and have the same mother and and have the same mother
implythat and have the same mother.
f. The relation is reflexive, symmetric and transitive, since if
and arehuman beings we have:
(1) and have the same hair color;
(2) and have the same hair color implies that and have the
samehair color;
(3) and have the same hair color and and have the same hair
colorimply that and have the same hair color.
15. a. The relation is an equivalence relation on × Let and
bearbitrary elements of
(1) ( ) ( ) since =
(2) ( ) ( )⇒ = ⇒ ( ) ( ) (3) ( ) ( ) and ( ) ( ) ⇒ = and =
⇒ = ⇒ = since 6= 0 and 6= 0⇒ ( ) ( )
b. The relation is an equivalence relation on × Let ( ) ( ) (
)be arbitrary elements of × .(1) ( ) ( ) since =
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22 Answers to Selected Exercises
(2) ( ) ( )⇒ = ⇒ = ⇒ ( ) ( ) (3) ( ) ( ) and ( ) ( ) ⇒ = and = ⇒
= ⇒
( ) ( )
c. The relation is an equivalence relation on × Let and
bearbitrary elements of
(1) ( ) ( ) since 2 + 2 = 2 + 2
(2) ( ) ( )⇒ 2 + 2 = 2 + 2 ⇒ 2 + 2 = 2 + 2 ⇒ ( ) ( ) (3) ( ) ( )
and ( ) ( ) ⇒ 2 + 2 = 2 + 2 and
2 + 2 = 2 + 2
⇒ 2 + 2 = 2 + 2⇒ ( ) ( )
d. The relation is an equivalence relation on × Let ( ) ( ) and(
) be arbitrary elements of ×(1) ( ) ( ) since − = − (2) ( ) ( )⇒ −
= − ⇒ − = − ⇒ ( ) ( ) (3) ( ) ( ) and ( ) ( ) ⇒ − = − and − = −
⇒
− = − ⇒ ( ) ( ) 16. The relation is reflexive and symmetric but
not transitive.
17. a. The relation is symmetric but not reflexive and not
transitive. Let and be arbitrary elements of the power set P () of
the nonempty set (1) ∩ 6= ∅ is not true if = ∅(2) ∩ 6= ∅ implies
that ∩ 6= ∅(3) ∩ 6= ∅ and ∩ 6= ∅ do not imply that ∩ 6= ∅ For
example, let
= { } = { } = { } and = { } Then ∩ ={} 6= ∅ ∩ = {} 6= ∅ but ∩ =
∅
b. The relation is reflexive and transitive but not symmetric,
since for arbi-trary subsets of we have:
(1) ⊆ ;(2) ∅ ⊆ but * ∅;(3) ⊆ and ⊆ imply ⊆
18. The relation is reflexive, symmetric, and transitive. Let
and be arbitraryelements of the power set P () and a fixed subset
of
(1) since ∩ = ∩ (2) ⇒ ∩ = ∩ ⇒ ∩ = ∩ ⇒
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