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

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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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.
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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