The relation R defined on Z by aRb if 3 | (2a + 7b) is an equivalence relation. I know the equivalence classes but I have no clue how they find those numbers, really show step-by-step how they get the numbers, I am absolutely clueless, explain what you do each step
The relation R defined on Z by aRb if 3 | (2a + 7b) is an equivalence relation. I know the equivalence classes but I have no clue how they find those numbers, really show step-by-step how they get the numbers, I am absolutely clueless, explain what you do each step
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The relation R defined on Z by aRb if 3 | (2a + 7b) is an equivalence relation.
I know the equivalence classes but I have no clue how they find those numbers, really show step-by-step how they get the numbers, I am absolutely clueless, explain what you do each step
![[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,
In addition,
[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%2F0dd01737-50ee-4fc8-804a-e0c24ff93aa7%2Fksgmaqt_processed.png&w=3840&q=75)
Transcribed Image Text:[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,
In addition,
[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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