The relation R defined on Z by aRb if 3 | (2a + 7b) is an equivalence relation, I have the equivalence classes but I have no idea how they get them, please explain step by step in detail, I have seen many examples but I really dont get it
The relation R defined on Z by aRb if 3 | (2a + 7b) is an equivalence relation, I have the equivalence classes but I have no idea how they get them, please explain step by step in detail, I have seen many examples but I really dont get it
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter1: Fundamentals
Section1.7: Relations
Problem 11E: Let be a relation defined on the set of all integers by if and only if sum of and is odd. Decide...
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The relation R defined on Z by aRb if 3 | (2a + 7b) is an equivalence relation, I have the equivalence classes but I have no idea how they get them, please explain step by step in detail, I have seen many examples but I really dont get it
![Let's see what the equivalence classes look like for the equivalence relation defined
We begin with the integer 0. Here,
.
In addition,
[0] = {x € Z: x R 0} = {x € Z: 3|(2x+7.0)}
= {x € Z : 3 | 2x} = {x € Z : 3 | x} = {0, ±3, ±6, ...}.
For the integer 1,
[1] = {x € Z: x R 1} = {x € Z : 3 | (2x+7-1)}
= {x € Z: 3 | (2x + 7)} = {1, −2, 4, −5, 7, 8, 10, ...}.
[2] = {x € Z: x R 2} = {x € Z : 3 | (2x+7-2)}
= {x € Z: 3 | (2x + 14)} = {2, —1, 5, —4, 8, —7, 11, ...}.
It turns out that these are the only distinct equivalence classes.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F42044228-cff6-4391-9596-005bad4f5c7f%2F6d94c5f6-332c-4742-9b7d-9f20f922ba2d%2Fmvzqu4q_processed.png&w=3840&q=75)
Transcribed Image Text:Let's see what the equivalence classes look like for the equivalence relation defined
We begin with the integer 0. Here,
.
In addition,
[0] = {x € Z: x R 0} = {x € Z: 3|(2x+7.0)}
= {x € Z : 3 | 2x} = {x € Z : 3 | x} = {0, ±3, ±6, ...}.
For the integer 1,
[1] = {x € Z: x R 1} = {x € Z : 3 | (2x+7-1)}
= {x € Z: 3 | (2x + 7)} = {1, −2, 4, −5, 7, 8, 10, ...}.
[2] = {x € Z: x R 2} = {x € Z : 3 | (2x+7-2)}
= {x € Z: 3 | (2x + 14)} = {2, —1, 5, —4, 8, —7, 11, ...}.
It turns out that these are the only distinct equivalence classes.
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