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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.2011-01-27
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    yes it would be.2011-01-27
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    why the downvote?2011-01-27
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    @PEV: Perhaps because it doesn't make sense to talk of equivalence classes before first proving that a relation is an equivalence relation?2011-01-27
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    But that's not what I asked for2011-01-27
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    @Krysten: Whether you asked for it or not, it is a necessary part of the solution. If you proceed as in my answer you can see everything you need in one fell swoop.2011-01-28
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    I concur. I probably should have been more specific and noteed that I had already proven that R was an equivalence relation.2011-01-28
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    @Krysten: I'm curious as to how you proved it is an equivalence relation without it being obvious how to find the equivalence class of -13. Could you please elaborate.2011-01-28
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    I proved it by showing R to be reflexive, symmetric, and transitive. I just don't have much experience with equivalence classes.2011-01-29
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$a+b^2=0\ (2)$ 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.2011-01-28
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    @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.)2011-01-30
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    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.2011-01-30
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HINT $\ \ \rm b^2 \equiv -b\ \ (mod\ 2)\ $ thus $\rm\ a\ R\ b\ \iff\ a\ \equiv\ b\ \ (mod\ 2)\ $ which is an equivalence relation.