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 May 16th, 2013, 10:06 PM #1 Senior Member   Joined: Jan 2012 Posts: 118 Thanks: 1 Equivalence Relations! Hey Everyone! In my abstract algebra class we are starting to learn how to determine if a relation is an equivalence relation. I know to test for Reflexivity, Symmetry, and Transitivity, but I am a little confused with this one. 1. Let $A= \mathbb{R}$. We define a ~ b if and only if $3 | (a + 2b)$. Okay, so when checking for reflexivity you need to check is a~a right? So I went $3|(a+2a)$ = $3|(3a)$ therefore obviously 3 can divide 3a so a~a. Next when checking for symmetry is where I get confused. So you assume a ~ b. And then you want to show that b~a. So if $3|(a+2b)$ does $3|(b+2a)$? I feel like this is true. I couldn't find a counterexample that proved it false. But is there an addition or product law or something that I can use to prove that those two statements are equal? Any help would be great! Thanks After I posted this I started working on another and I am confused again. 2. Let $A= \mathbb{R}$. We define a~b if and only if $a + b \in \mathbb{Z}$. Just need some clarification here. The first part where it says that A = R. Does that mean any a's and b's that we decide to work with need to be real numbers? If that is the case, then right away wouldn't this relation fail to be an equivalence relation due to not having reflexivity? Because what if a = 1.25 (which is a real number), then a + a = 2.5 which is not an integer? Am I correct in saying that, or am I interpreting the initial A = R statement wrong?
May 17th, 2013, 05:01 AM   #2
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Re: Equivalence Relations!

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 Originally Posted by Jet1045 Next when checking for symmetry is where I get confused. So you assume a ~ b. And then you want to show that b~a. So if $3|(a+2b)$ does $3|(b+2a)$? I feel like this is true. I couldn't find a counterexample that proved it false. But is there an addition or product law or something that I can use to prove that those two statements are equal?
You say you're working in $\mathbb{R},$ in which case 3 divides everything. If instead you're working in $\mathbb{Z}$ then there are 9 cases: a = b = 0 mod 3, a = 0 mod 3, b = 1 mod 3, ..., a = b = 2 mod 3. See if any of these have 3 | (a+2b) but not 3 | (2a+b).

(You can reduce this to three cases by taking each value of b and then choosing the value of a such that 3 | (a+2b).)

Well, I guess maybe you are working over the reals but the | is over Z, so you need to consider things like a = 1/3, b = 4/3. But then it's not even reflexive.

Quote:
 Originally Posted by Jet1045 2. Let $A= \mathbb{R}$. We define a~b if and only if $a + b \in \mathbb{Z}$. Just need some clarification here. The first part where it says that A = R. Does that mean any a's and b's that we decide to work with need to be real numbers? If that is the case, then right away wouldn't this relation fail to be an equivalence relation due to not having reflexivity? Because what if a = 1.25 (which is a real number), then a + a = 2.5 which is not an integer? Am I correct in saying that, or am I interpreting the initial A = R statement wrong?
I agree that this is not reflexive over $\mathbb{R}.$

 May 18th, 2013, 09:43 PM #3 Senior Member   Joined: Jan 2012 Posts: 118 Thanks: 1 Re: Equivalence Relations! Perfect, thanks for the help! I believe I solved the first one correctly now! I have another question I am stumped on now.... Suppose $p>0$ is prime. Show that if $a^{2} \equiv b^{2} (mod p)$ then $a \equiv b (modp)$ or $a \equiv -b (modp)$ this one I actually don't know where to start, so any help would be great.
May 19th, 2013, 01:58 AM   #4
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Re: Equivalence Relations!

Quote:
 Originally Posted by Jet1045 Perfect, thanks for the help! I believe I solved the first one correctly now! I have another question I am stumped on now.... Suppose $p>0$ is prime. Show that if $a^{2} \equiv b^{2} (mod p)$ then $a \equiv b (modp)$ or $a \equiv -b (modp)$
Transform to

$(a+b)(a-b) \equiv 0 (mod \hspace{4} p)$

In general if p | nm, what can we conclude?

May 19th, 2013, 02:46 AM   #5
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Re: Equivalence Relations!

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 Originally Posted by Jet1045 I actually don't know where to start
Just think that you are working over Z instead of Z/pZ, which is essentially the same except the fact that Z/pZ is finitely generated.

 May 19th, 2013, 01:03 PM #6 Senior Member   Joined: Jan 2012 Posts: 118 Thanks: 1 Re: Equivalence Relations! thanks for the help, I think I got it. $a^{2}\equiv b^{2}$ (mod p) p| (a^2 -b^2) p | (a-b)(a+b) p | (a-b) OR p | (a+b) therefore $a \equiv b$ (mod p) OR $a \equiv -b$ (mod p)

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