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

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!

• 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. – Aaron Rotenberg Feb 14 at 14:27

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.