# Equivalence Classes

Let $R$ be the relation defined on $\mathbb{Z}$ where $a\; R\; b$ means that $a + b^2 \equiv 0\pmod{2}$.

How would I go about finding the equivalence class $[-13]$?

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You would find all $a \in \mathbb{Z}$ such that $a+(-13)^2 = 0 \pmod{2}$ since $[-13] = \{a \in \mathbb{Z}: aR -13 \}$.

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 thanks. so the equivalence class would be the set of all odd integers. – Krysten Jan 27 '11 at 22:54 yes it would be. – PEV Jan 27 '11 at 22:58 why the downvote? – PEV Jan 27 '11 at 23:05 @PEV: Perhaps because it doesn't make sense to talk of equivalence classes before first proving that a relation is an equivalence relation? – Gone Jan 27 '11 at 23:10 But that's not what I asked for – Krysten Jan 27 '11 at 23:34
$a+b^2=0\ (2)$ is the same as "$a$ and $b$ have the same parity", so the set of odds is $-13$'s class.
HINT $\ \ \rm b^2 \equiv -b\ \ (mod\ 2)\$ thus $\rm\ a\ R\ b\ \iff\ a\ \equiv\ b\ \ (mod\ 2)\$ which is an equivalence relation.