Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $[-13]$?

share|improve this question

3 Answers 3

up vote 2 down vote accepted

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

share|improve this answer
    
thanks. so the equivalence class would be the set of all odd integers. –  Krysten Jan 27 '11 at 22:54
1  
yes it would be. –  PEV Jan 27 '11 at 22:58
1  
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.

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.