# Equivalence Classes of this Relation on the integers : $a + b^2 \equiv 0\pmod{2}$.

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? –  Bill Dubuque 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.

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That's saying what the equivalence class is, but not answering the question which is how to go about finding that class. –  Mitch Jan 28 '11 at 21:11
@Mitch, hm? I found it by first finding that a+b^2=0 (2) is the same as "a and b have the same parity"! (Not many details there, I admit, but the method was clearly outlined.) –  msh210 Jan 30 '11 at 7:57
Just to explain what I'm getting at, Krysten asked for 'how' and if she's doing that then she'd probably also need a step or two of 'how' in order to get to your parity statement. That is, -how- do you know that the strange looking sum leads to the parity statement. And that's what Bill's answer helps with. Sorry making so much of not much, I think those missing details are what the OP hoped for. –  Mitch Jan 30 '11 at 23:09

HINT $\ \ \rm b^2 \equiv -b\ \ (mod\ 2)\$ thus $\rm\ a\ R\ b\ \iff\ a\ \equiv\ b\ \ (mod\ 2)\$ which is an equivalence relation.

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