a. Recall that the set Z of integers is the set of equivalence classes {[(a, 6)] | a, 8 e N}, where the equivalence relation v on the set N x Nis defined as (a, b) ~ (c, d) A a+d = 6+ e for any (a, 5), (c, d) eN x N so that the equivalence class of (a, b) is gives as [(a, 6)] = {(e, d) | (a, 8) ~ (c, d)} . For instance, 1 = [(1,0)] according to this notation of integers. Let * = [(28, 34)] , y = [(8, 44)] € Z. Find a + y and a - y. * + y =
a. Recall that the set Z of integers is the set of equivalence classes {[(a, 6)] | a, 8 e N}, where the equivalence relation v on the set N x Nis defined as (a, b) ~ (c, d) A a+d = 6+ e for any (a, 5), (c, d) eN x N so that the equivalence class of (a, b) is gives as [(a, 6)] = {(e, d) | (a, 8) ~ (c, d)} . For instance, 1 = [(1,0)] according to this notation of integers. Let * = [(28, 34)] , y = [(8, 44)] € Z. Find a + y and a - y. * + y =
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter1: Fundamentals
Section1.7: Relations
Problem 8E: In Exercises 610, a relation R is defined on the set Z of all integers. In each case, prove that R...
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![a. Recall that the set Z of integers is the set of equivalence classes
{[(, 6)] | a, 5 e N},
where the equivalence relation v on the set N x Nis defined as
(a, 6) ~ (e, d) = a+d = 6+e
for any (a, b) , (c, d) eN x N so that the equivalence class of (a, b) is gives as
[(a, 5)] = {(c, d) | (a, 5) ~ (6, d)} .
For instance, 1 = [(1,0)] according to this notation of integers.
Let * =
[(28, 34)] , y = [(8, 44)] € Z. Find * + y and a y.
*+ y =
b. Recall that the set Q of integers is the set of equivalence classes
{ [(a, b)] | a, b e Z, b # 0} ,
where the equivalence relation = on the set Z x (Z- {0}) is defined as
(a, b) = (c, d) e a • d = 6.c
for any (a, 6) , (e, d) e Z x (Z - {0}) so that the equivalence class of (a, b) is gives as
[(a, 6)] = {(e, d) | (a, 6) = (e, d)}.
For instance, 2 = [(2, 1)] according to this notation of rational numbers.
Let g =
[(36, 49)] ,r = [(-16, –15)] € Q. Find g+r and g - r.
q+r =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcc482d4e-e84c-42ce-b2ed-ac27cfa2fcb3%2F6baddfde-f58a-4e91-a02e-8f2741eefe53%2Fr4b01l_processed.png&w=3840&q=75)
Transcribed Image Text:a. Recall that the set Z of integers is the set of equivalence classes
{[(, 6)] | a, 5 e N},
where the equivalence relation v on the set N x Nis defined as
(a, 6) ~ (e, d) = a+d = 6+e
for any (a, b) , (c, d) eN x N so that the equivalence class of (a, b) is gives as
[(a, 5)] = {(c, d) | (a, 5) ~ (6, d)} .
For instance, 1 = [(1,0)] according to this notation of integers.
Let * =
[(28, 34)] , y = [(8, 44)] € Z. Find * + y and a y.
*+ y =
b. Recall that the set Q of integers is the set of equivalence classes
{ [(a, b)] | a, b e Z, b # 0} ,
where the equivalence relation = on the set Z x (Z- {0}) is defined as
(a, b) = (c, d) e a • d = 6.c
for any (a, 6) , (e, d) e Z x (Z - {0}) so that the equivalence class of (a, b) is gives as
[(a, 6)] = {(e, d) | (a, 6) = (e, d)}.
For instance, 2 = [(2, 1)] according to this notation of rational numbers.
Let g =
[(36, 49)] ,r = [(-16, –15)] € Q. Find g+r and g - r.
q+r =
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ISBN:
9781285463230
Author:
Gilbert, Linda, Jimmie
Publisher:
Cengage Learning,