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 From https://testbankgo.eu/p/Solution-Manual-for-Elements-of-Modern-Algebra-8th-Edition-by-Linda-Gilbert
30
Embed
Instructor s Manual to accompany Elements of Modern ... · Instructor s Manual to accompany Elements of Modern Algebra, Eighth Edition Linda Gilbert and the late Jimmie Gilbert University
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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
From https://testbankgo.eu/p/Solution-Manual-for-Elements-of-Modern-Algebra-8th-Edition-by-Linda-Gilbert
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}
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 () = 1
Neither is one-to-one since (0) = (1) and 0 6= 1i. 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 is
not one-to-one, since (0) = (1) = 0 and 0 6= 16. a. The mapping is onto and one-to-one.
b. The mapping is one-to-one. Since there is no ∈ E such that () = 2
the 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
From https://testbankgo.eu/p/Solution-Manual-for-Elements-of-Modern-Algebra-8th-Edition-by-Linda-Gilbert
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− 1
1= 2− 1
2
⇒ − 11= − 1
2
⇒ 1 = 2
Thus 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) = 2
5and
¡12
¢= 2
5
d. The mapping is not onto, since there is no ∈ R−{0} such that () = 1Since (1) = (3) = 1
2 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.
From https://testbankgo.eu/p/Solution-Manual-for-Elements-of-Modern-Algebra-8th-Edition-by-Linda-Gilbert
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
() = () =
⎧⎨⎩ 2if 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
From https://testbankgo.eu/p/Solution-Manual-for-Elements-of-Modern-Algebra-8th-Edition-by-Linda-Gilbert
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 2
are 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 odd
integer but 1 + 1 is not an odd integer.
10. a. The set of nonzero integers is not closed with respect to addition defined on
Z, 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 defined
on 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.
From https://testbankgo.eu/p/Solution-Manual-for-Elements-of-Modern-Algebra-8th-Edition-by-Linda-Gilbert
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 but
2 /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 but
2 /1 It is transitive, since and ⇒ for all and ∈ Ze. 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 ≤ 0
implies ≤ 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:
From https://testbankgo.eu/p/Solution-Manual-for-Elements-of-Modern-Algebra-8th-Edition-by-Linda-Gilbert
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 arbitrary
elements of the power set P () and a fixed subset of
(1) since ∩ = ∩ (2) ⇒ ∩ = ∩ ⇒ ∩ = ∩ ⇒
From https://testbankgo.eu/p/Solution-Manual-for-Elements-of-Modern-Algebra-8th-Edition-by-Linda-Gilbert