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!
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.
