   Chapter 8.3, Problem 43ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# In Example 8.3.12, define operations of addition (+) and multiplication (·) as follows: For every ( a ,   b ) ,   ( c ,   d ) ∈ A , [ ( a ,   b ) ] + [ ( c ,   d ) ] = [ ( a d + b c ,   ​ b d ) ] [ ( a ,   b ) ] ⋅ [ ( c ,   d ) ] = [ ( a c ,   b d ) ] . a. Prove that this addition is well defined. That is, 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 ' ) ] . b. Prove that this multiplication is well defined. That is, show that if [ ( a ,   b ) ] = [ ( a ' ,   b ' ) ] and [ ( c ,   d ) ] = [ ( c ' ,   d ' ) ] . then [ ( a c + b d ) ] = [ ( a ' c ' ,   b ' d ' ) ] . c. 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. Find an identity element for multiplication. That is, find (i, j) in A so that for every (a, b) in A ,   [ ( a , ​   b ) ] ⋅ [ ( i ,   j ) ] = [ ( i ,   j ) ] ⋅ [ ( a , ​   b ) ] = [ ( a , ​   b ) ] . e. For any ( a ,   b ) ∈ A , show that [(-a, b)] is an inverse for [ ( a , ​   b ) ] for addition. That is, show that [ ( − a ,   b ) ] + [ ( a ,   b ) ] = [ ( a ,   b ) ] + [ ( − a ,   b ) ] = [ ( 0 ,   1 ) ] . f. Given any ( a ,   b ) ∈ A with a ≠ 0 , find an inverse for [ ( a , ​   b ) ] for multiplication. That is, 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).

To determine

(a)

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

Explanation

Given information:

The operations of addition (+)and multiplication (·) as follows: For all (a,b),(c,d)A,

Proof:

Let us assume that [(a,b)]=[(a,b)] and

[(c,d)]=[(c,d)].

In general: a[a] for all aA:

(a,b)[(a,b)]=[(a',b')](c,d)[(c,d)]=[(c',d')]

By the definition of the equivalence class:

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 a0. 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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