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

Question

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 $[-13]$?

score 2 · Accepted Answer · edited Jan 27 '11 at 22:46

2

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

edited Jan 27 '11 at 22:46

Arturo Magidin

answered Jan 27 '11 at 22:42

NebulousReveal

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. – NebulousReveal Jan 27 '11 at 22:58
1

why the downvote? – NebulousReveal 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
@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. – Bill Dubuque Jan 28 '11 at 02:20
I concur. I probably should have been more specific and noteed that I had already proven that R was an equivalence relation. – Krysten Jan 28 '11 at 05:57
@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. – Bill Dubuque Jan 28 '11 at 06:41
I proved it by showing R to be reflexive, symmetric, and transitive. I just don't have much experience with equivalence classes. – Krysten Jan 29 '11 at 01:23

score 0 · Answer 2 · answered Jan 27 '11 at 22:51

0

$a+b^2=0\ (2)$ is the same as "$a$ and $b$ have the same parity", so the set of odds is $-13$'s class.

answered Jan 27 '11 at 22:51

msh210

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 07:57
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

score 0 · Answer 3 · edited Oct 01 '17 at 20:37

0

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

edited Oct 01 '17 at 20:37

davidlowryduda

answered Jan 27 '11 at 22:59

Bill Dubuque

3 Answers3