(a) To show that if [ ( a , b ) ] = [ ( a ′ , b ′ ) ] and [ ( c , d ) ] = [ ( c ′ , d ′ ) ] , then [ ( a d + b c , b d ) ] = [ ( a ′ d ′ + b ′ c ′ , b ′ d ′ ) ] .

Question

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Answer 1

Question

Chapter 8.3, Problem 43ES

To determine

(a)

To show that if [(a,b)]=[(a′,b′)] and [(c,d)]=[(c′ ,d′)], then

[(ad+bc,bd)]=[(a′d′+b′c′,b′d′)].

To determine

(b)

To prove:

If [(a,b)]=[(a',b')] and [(c,d)]=[(c',d')], then [(ac,bd)]=[(a'c',b'd')].

To determine

(c)

To show that [(0,1)] is an identity element for addition. That is, show that for any (a,b)∈A,

[(a,b)]+[(0,1)]=[(0,1)]+[(a,b)]=[(a,b)].

To determine

(d)

An identity element for multiplication for the given operation. That means, find (i,f) in A so that for all (a,b) in A. [(a,b)]⋅[(i,j)]=[(i,j)]⋅[(a,b)]=[(a,b)].

To determine

(e)

To prove:

To show that [(−a,b)] is an inverse for [(a,b)] for addition if (a,b)∈A,. That means, to show that [(−a,b)]+[(a,b)]=[(a,b)]+[(−a,b)]=[(0,1)].

To determine

(f)

An inverse for [(a,b)] for multiplication for any (a,b)∈A with a≠0. That means, to find (c,d) in A so that [(a,b)]⋅[(c,d)]=[(c,d)]−[(a,b)]=[(i,j)], where [(i,j)] is the identity element you found in part (d).

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Answer 2

Question

Chapter 8.3, Problem 43ES

To determine

(a)

To show that if [(a,b)]=[(a′,b′)] and [(c,d)]=[(c′ ,d′)], then

[(ad+bc,bd)]=[(a′d′+b′c′,b′d′)].

To determine

(b)

To prove:

If [(a,b)]=[(a',b')] and [(c,d)]=[(c',d')], then [(ac,bd)]=[(a'c',b'd')].

To determine

(c)

To show that [(0,1)] is an identity element for addition. That is, show that for any (a,b)∈A,

[(a,b)]+[(0,1)]=[(0,1)]+[(a,b)]=[(a,b)].

To determine

(d)

An identity element for multiplication for the given operation. That means, find (i,f) in A so that for all (a,b) in A. [(a,b)]⋅[(i,j)]=[(i,j)]⋅[(a,b)]=[(a,b)].

To determine

(e)

To prove:

To show that [(−a,b)] is an inverse for [(a,b)] for addition if (a,b)∈A,. That means, to show that [(−a,b)]+[(a,b)]=[(a,b)]+[(−a,b)]=[(0,1)].

To determine

(f)

An inverse for [(a,b)] for multiplication for any (a,b)∈A with a≠0. That means, to find (c,d) in A so that [(a,b)]⋅[(c,d)]=[(c,d)]−[(a,b)]=[(i,j)], where [(i,j)] is the identity element you found in part (d).

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Answer 3

Question

Chapter 8.3, Problem 43ES

To determine

(a)

To show that if [(a,b)]=[(a′,b′)] and [(c,d)]=[(c′ ,d′)], then

[(ad+bc,bd)]=[(a′d′+b′c′,b′d′)].

To determine

(b)

To prove:

If [(a,b)]=[(a',b')] and [(c,d)]=[(c',d')], then [(ac,bd)]=[(a'c',b'd')].

To determine

(c)

To show that [(0,1)] is an identity element for addition. That is, show that for any (a,b)∈A,

[(a,b)]+[(0,1)]=[(0,1)]+[(a,b)]=[(a,b)].

To determine

(d)

An identity element for multiplication for the given operation. That means, find (i,f) in A so that for all (a,b) in A. [(a,b)]⋅[(i,j)]=[(i,j)]⋅[(a,b)]=[(a,b)].

To determine

(e)

To prove:

To show that [(−a,b)] is an inverse for [(a,b)] for addition if (a,b)∈A,. That means, to show that [(−a,b)]+[(a,b)]=[(a,b)]+[(−a,b)]=[(0,1)].

To determine

(f)

An inverse for [(a,b)] for multiplication for any (a,b)∈A with a≠0. That means, to find (c,d) in A so that [(a,b)]⋅[(c,d)]=[(c,d)]−[(a,b)]=[(i,j)], where [(i,j)] is the identity element you found in part (d).

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Answer 4

Question

Chapter 8.3, Problem 43ES

To determine

(a)

To show that if [(a,b)]=[(a′,b′)] and [(c,d)]=[(c′ ,d′)], then

[(ad+bc,bd)]=[(a′d′+b′c′,b′d′)].

To determine

(b)

To prove:

If [(a,b)]=[(a',b')] and [(c,d)]=[(c',d')], then [(ac,bd)]=[(a'c',b'd')].

To determine

(c)

To show that [(0,1)] is an identity element for addition. That is, show that for any (a,b)∈A,

[(a,b)]+[(0,1)]=[(0,1)]+[(a,b)]=[(a,b)].

To determine

(d)

An identity element for multiplication for the given operation. That means, find (i,f) in A so that for all (a,b) in A. [(a,b)]⋅[(i,j)]=[(i,j)]⋅[(a,b)]=[(a,b)].

To determine

(e)

To prove:

To show that [(−a,b)] is an inverse for [(a,b)] for addition if (a,b)∈A,. That means, to show that [(−a,b)]+[(a,b)]=[(a,b)]+[(−a,b)]=[(0,1)].

To determine

(f)

An inverse for [(a,b)] for multiplication for any (a,b)∈A with a≠0. That means, to find (c,d) in A so that [(a,b)]⋅[(c,d)]=[(c,d)]−[(a,b)]=[(i,j)], where [(i,j)] is the identity element you found in part (d).

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Answer 5

Chapter 8.3, Problem 43ES

To determine

(a)

To show that if [(a,b)]=[(a′,b′)] and [(c,d)]=[(c′ ,d′)], then

[(ad+bc,bd)]=[(a′d′+b′c′,b′d′)].

To determine

(b)

To prove:

If [(a,b)]=[(a',b')] and [(c,d)]=[(c',d')], then [(ac,bd)]=[(a'c',b'd')].

To determine

(c)

To show that [(0,1)] is an identity element for addition. That is, show that for any (a,b)∈A,

[(a,b)]+[(0,1)]=[(0,1)]+[(a,b)]=[(a,b)].

To determine

(d)

An identity element for multiplication for the given operation. That means, find (i,f) in A so that for all (a,b) in A. [(a,b)]⋅[(i,j)]=[(i,j)]⋅[(a,b)]=[(a,b)].

To determine

(e)

To prove:

To show that [(−a,b)] is an inverse for [(a,b)] for addition if (a,b)∈A,. That means, to show that [(−a,b)]+[(a,b)]=[(a,b)]+[(−a,b)]=[(0,1)].

To determine

(f)

An inverse for [(a,b)] for multiplication for any (a,b)∈A with a≠0. That means, to find (c,d) in A so that [(a,b)]⋅[(c,d)]=[(c,d)]−[(a,b)]=[(i,j)], where [(i,j)] is the identity element you found in part (d).

Answer 6

Chapter 8.3, Problem 43ES

To determine

(a)

To show that if [(a,b)]=[(a′,b′)] and [(c,d)]=[(c′ ,d′)], then

[(ad+bc,bd)]=[(a′d′+b′c′,b′d′)].

Answer 7

Chapter 8.3, Problem 43ES

To determine

(a)

To show that if [(a,b)]=[(a′,b′)] and [(c,d)]=[(c′ ,d′)], then

[(ad+bc,bd)]=[(a′d′+b′c′,b′d′)].

Answer 8

To determine

(b)

To prove:

If [(a,b)]=[(a',b')] and [(c,d)]=[(c',d')], then [(ac,bd)]=[(a'c',b'd')].

Answer 9

To determine

(b)

To prove:

If [(a,b)]=[(a',b')] and [(c,d)]=[(c',d')], then [(ac,bd)]=[(a'c',b'd')].

Answer 10

To determine

(c)

To show that [(0,1)] is an identity element for addition. That is, show that for any (a,b)∈A,

[(a,b)]+[(0,1)]=[(0,1)]+[(a,b)]=[(a,b)].

Answer 11

To determine

(c)

To show that [(0,1)] is an identity element for addition. That is, show that for any (a,b)∈A,

[(a,b)]+[(0,1)]=[(0,1)]+[(a,b)]=[(a,b)].

Answer 12

To determine

(d)

An identity element for multiplication for the given operation. That means, find (i,f) in A so that for all (a,b) in A. [(a,b)]⋅[(i,j)]=[(i,j)]⋅[(a,b)]=[(a,b)].

Answer 13

To determine

(d)

An identity element for multiplication for the given operation. That means, find (i,f) in A so that for all (a,b) in A. [(a,b)]⋅[(i,j)]=[(i,j)]⋅[(a,b)]=[(a,b)].

Answer 14

To determine

(e)

To prove:

To show that [(−a,b)] is an inverse for [(a,b)] for addition if (a,b)∈A,. That means, to show that [(−a,b)]+[(a,b)]=[(a,b)]+[(−a,b)]=[(0,1)].

Answer 15

To determine

(e)

To prove:

To show that [(−a,b)] is an inverse for [(a,b)] for addition if (a,b)∈A,. That means, to show that [(−a,b)]+[(a,b)]=[(a,b)]+[(−a,b)]=[(0,1)].

Answer 16

To determine

(f)

An inverse for [(a,b)] for multiplication for any (a,b)∈A with a≠0. That means, to find (c,d) in A so that [(a,b)]⋅[(c,d)]=[(c,d)]−[(a,b)]=[(i,j)], where [(i,j)] is the identity element you found in part (d).

Answer 17

To determine

(f)

An inverse for [(a,b)] for multiplication for any (a,b)∈A with a≠0. That means, to find (c,d) in A so that [(a,b)]⋅[(c,d)]=[(c,d)]−[(a,b)]=[(i,j)], where [(i,j)] is the identity element you found in part (d).

Answer 18

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Answer 19

Expert Solution & Answer

Answer 20

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Answer 21

Ch. 8.1 - If R is a relation from A to B, xA , and yB , the...Ch. 8.1 - Prob. 2TY Ch. 8.1 - Prob. 3TY Ch. 8.1 - Prob. 4TY Ch. 8.1 - If R is a relation on a set A, the directed graph...Ch. 8.1 - As in Example 8.1.2, the congruence modulo 2...Ch. 8.1 - Prove that for all integers m and n,m-n is even...Ch. 8.1 - The congruence modulo 3 relation, T, is defined...Ch. 8.1 - Define a relation P on Z as follows: For every...Ch. 8.1 - Prob. 5ES

(a) To show that if [ ( a , b ) ] = [ ( a ′ , b ′ ) ] and [ ( c , d ) ] = [ ( c ′ , d ′ ) ] , then [ ( a d + b c , b d ) ] = [ ( a ′ d ′ + b ′ c ′ , b ′ d ′ ) ] .

Solution Summary: The author explains the operations of addition (+)and multiplication () as follows: for all (a,b)in A,

Videos

(a)

To show that if [(a,b)]=[(a′,b′)] and [(c,d)]=[(c′ ,d′)], then

[(ad+bc,bd)]=[(a′d′+b′c′,b′d′)].

(b)

To prove:

If [(a,b)]=[(a',b')] and [(c,d)]=[(c',d')], then [(ac,bd)]=[(a'c',b'd')].

(c)

To show that [(0,1)] is an identity element for addition. That is, show that for any (a,b)∈A,

[(a,b)]+[(0,1)]=[(0,1)]+[(a,b)]=[(a,b)].

(d)

An identity element for multiplication for the given operation. That means, find (i,f) in A so that for all (a,b) in A. [(a,b)]⋅[(i,j)]=[(i,j)]⋅[(a,b)]=[(a,b)].

(e)

To prove:

To show that [(−a,b)] is an inverse for [(a,b)] for addition if (a,b)∈A,. That means, to show that [(−a,b)]+[(a,b)]=[(a,b)]+[(−a,b)]=[(0,1)].

(f)

An inverse for [(a,b)] for multiplication for any (a,b)∈A with a≠0. That means, to find (c,d) in A so that [(a,b)]⋅[(c,d)]=[(c,d)]−[(a,b)]=[(i,j)], where [(i,j)] is the identity element you found in part (d).

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Chapter 8 Solutions

(a) To show that if [ ( a , b ) ] = [ ( a ′ , b ′ ) ] and [ ( c , d ) ] = [ ( c ′ , d ′ ) ] , then [ ( a d + b c , b d ) ] = [ ( a ′ d ′ + b ′ c ′ , b ′ d ′ ) ] .

Solution Summary: The author explains the operations of addition (+)and multiplication () as follows: for all (a,b)in A,

Videos

(a) To show that if [(a,b)]=[(a′,b′)] and [(c,d)]=[(c′ ,d′)], then [(ad+bc,bd)]=[(a′d′+b′c′,b′d′)].

(b) To prove: If [(a,b)]=[(a',b')] and [(c,d)]=[(c',d')], then [(ac,bd)]=[(a'c',b'd')].

(c) To show that [(0,1)] is an identity element for addition. That is, show that for any (a,b)∈A, [(a,b)]+[(0,1)]=[(0,1)]+[(a,b)]=[(a,b)].

(d) An identity element for multiplication for the given operation. That means, find (i,f) in A so that for all (a,b) in A. [(a,b)]⋅[(i,j)]=[(i,j)]⋅[(a,b)]=[(a,b)].

(e) To prove: To show that [(−a,b)] is an inverse for [(a,b)] for addition if (a,b)∈A,. That means, to show that [(−a,b)]+[(a,b)]=[(a,b)]+[(−a,b)]=[(0,1)].

(f) An inverse for [(a,b)] for multiplication for any (a,b)∈A with a≠0. That means, to find (c,d) in A so that [(a,b)]⋅[(c,d)]=[(c,d)]−[(a,b)]=[(i,j)], where [(i,j)] is the identity element you found in part (d).

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Chapter 8 Solutions

(a)

To show that if [(a,b)]=[(a′,b′)] and [(c,d)]=[(c′ ,d′)], then

[(ad+bc,bd)]=[(a′d′+b′c′,b′d′)].

(b)

To prove:

If [(a,b)]=[(a',b')] and [(c,d)]=[(c',d')], then [(ac,bd)]=[(a'c',b'd')].

(c)

To show that [(0,1)] is an identity element for addition. That is, show that for any (a,b)∈A,

[(a,b)]+[(0,1)]=[(0,1)]+[(a,b)]=[(a,b)].

(d)

An identity element for multiplication for the given operation. That means, find (i,f) in A so that for all (a,b) in A. [(a,b)]⋅[(i,j)]=[(i,j)]⋅[(a,b)]=[(a,b)].

(e)

To prove:

To show that [(−a,b)] is an inverse for [(a,b)] for addition if (a,b)∈A,. That means, to show that [(−a,b)]+[(a,b)]=[(a,b)]+[(−a,b)]=[(0,1)].

(f)

An inverse for [(a,b)] for multiplication for any (a,b)∈A with a≠0. That means, to find (c,d) in A so that [(a,b)]⋅[(c,d)]=[(c,d)]−[(a,b)]=[(i,j)], where [(i,j)] is the identity element you found in part (d).