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I need someone to check my proof.

Question: On the set $\mathbb{R}^2$ of ordered pairs define the 2-plane relation $\sim$ as follows $(a,b)\sim(c,d)$ if and only if $a^2+b^2=c^2+d^2$. Prove that $\sim$ is an equivalence relation on R2

My Answer: We must show reflexivity, symmetry, and transitivity.

Symmetry: $(a,b)\sim(b,a) \Rightarrow a^2+b^2=b^2+a^2$

Transitivity: Suppose $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$ Then $a^2+b^2=c^2+d^2$ and $c^2+d^2=e^2+f^2$ Therefore $a^2+b^2=e^2+f^2$

Reflexivity: I am not sure - might need a little help with this.

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For syymetry, you have to show $(a,b)\sim(c,d)$ implies $(c,d)\sim(a,b)$ for all ordered pairs $(a,b)$ and $(c,d)$. –  Michael Greinecker Jan 28 '13 at 12:34
    
So what about this for symmetry: –  Gamecocks99 Jan 28 '13 at 13:18
    
(a,b)~(c,d) than f(a,b)=g(c,d). Thus, f(g(c,d))=f(c,d)=c^2+d^2 and g(f(a,b))=g(a,b)=a^2+b^2. Therefore f(c,d)=g(a,b). f(a,b)=g(c,d) implies f(c,d)=g(a,b) which equals c^2+d^2=a^2+b^2. Thus, (a,b)~(c,d) ==> (c,d)~(a,b). Or is that wrong? –  Gamecocks99 Jan 28 '13 at 13:21
    
I don't understand where the g and f is coming from. But it is not hard: $(a,b)\sim(c,d)\iff a^2+b^2=c^2+d^2\iff c^2+d^2=a^2+b^2\iff (c,d)\sim(a,b).$ –  Michael Greinecker Jan 28 '13 at 13:24
    
Oh well thanks! I thought there was more to it than that but I guess I just over complicated something so trivial. I appreciate it. –  Gamecocks99 Jan 28 '13 at 13:27

4 Answers 4

For reflexivity, it is enough to write that for all $(a,b) \in \mathbb{R}^2$, $a^2+b^2=a^2+b^2$...

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In this cant we say that (a,b)~(c,d) if they are on the same circle about the center because the radius square is equal. So this must be a equivalence relation with the classes being the circles about the origin.

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It is also useful to note that if $A$ is a set, and we have a relation $\equiv$ such that $a\equiv b$ iff $f(a)=f(b)$ (Where $f$ is a function whose domain is $A\times A$), then $\equiv$ is an equivalence relation. In this case $f$ is the function that sends $(a,b)$ to $a^2+b^2$

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This is called the (equivalence) kernel of $f,$ and the classes are called fibers or level-sets (here circles). –  Math Gems Jan 28 '13 at 15:45
    
@MathGems I didint know these terms before –  Amr Jan 28 '13 at 16:31

Your transitivity reasoning is correct, but for symmetry, as was commented, you rather need $(a,b)\sim (c,d) \implies (c,d)\sim (a,b)$.

For reflexivity, you need $(a,b)\sim (a,b)$, which is even more trivial than the others.

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