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I have already proven that the relation $R=\{(x,y) \in \mathbb Z \times \mathbb Z \mid x+y\text{ is even}\}$ is an equivalence relation by showing reflexive, symmetric, and transitive properties of the relation. But when I try to find the equivalence classes of it I'm stuck.

Since $x$ and $y$ are either both even or both odd, does this simply come down to the equivalence classes of "congruence modulo $2$"?

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  • $\begingroup$ What are all the numbers equivalent to $0?$ $\endgroup$ – Will Jagy Jul 19 '15 at 21:40
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Your intuition is correct. To describe the equivalence classes explicitly, it often helps to find the equivalence classes of numbers that are easy to work with.

Let's find, for example, the equivalence classes $[0]$ and $[1]$.

$0+a$ is even for which integers $a$? All these will form $[0]$.

$1+b$ is even for which integers $b$? All these will form $[1]$.

Are there any other equivalence classes? (You can be sure you've found them all when your equivalence classes form a partition of $\Bbb Z$.)

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Suppose $x+y$ is even. What does that tell you about $x$ and $y$? HINT: Write down some examples . . .

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  • $\begingroup$ it would either tell that both x and y are even, or both x and y are odd. ex. 2+4 = 6. 3+5 = 8. etc... $\endgroup$ – user3467433 Jul 19 '15 at 21:42
  • $\begingroup$ So from this, can you tell me what the equivalence classes are? (Hint: there's two of them . . .) $\endgroup$ – Noah Schweber Jul 20 '15 at 5:59
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$x+y$ is even if and only if $x-y$ is even because $x-y=(x+y)-2y$. So the even integers form one equivalence class, and the odd integers form the other.

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