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Given the Equivalence relation R = { x, y $\in$ $\Bbb{Z}$ : (x+y) mod 2 = 0}, what are equivalence classes of 1 and 2?

I can't really see the equivalence classes of infinite sets. Only by having a drawing of all elements can I distinguish the answers, wich is not the case in the above mentioned example.

What would be the best way to tackle such problems?

Thanks!

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    $\begingroup$ The question as written doesn't type-check to me. The set you describe is a set of pairs of integers, but then you ask what are the equivalence classes containing single integers. $\endgroup$ – Aaron Rotenberg Feb 14 at 14:27
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There is no general way to answer such a question other than to repeat the definition. So one might answer for example for $1$:

The equivalence class that $1$ belongs to is the set of all integers $n$ such that $1 + n \equiv 0 \mod 2$.

You might simplify this by noting that $1 + n \equiv 0$ implies $n \equiv 2-1$ or $n \equiv 1$ modulo $2$, giving:

The equivalence class that $1$ belongs to is the set of all integers $n$ such that $n \equiv 1 \mod 2$.


Now the answer that the asker is looking for will probably be "the odd integers". If they disagree that the above answer is correct I would argue that the question is too vague however, as in my view the above is perfectly correct.

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