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let $S=\Bbb R\times \Bbb R$ define a relation $R$ on $S$ by $(a,b)R(c,d)$ iff $a-b=c-d$, verify $R$ is an equivalence relation

I don't think that it is reflexive, because $a-b$ will not always $= b-a$.
For example if $a=1$ and $b= 3$, $1-3=-2$, but $3-1=2$.

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Hint: You've got the order wrong. Indeed, $R$ is reflexive. Observe that for any $(a,b)\in S$, we have $a-b=a-b$. Hence, by the definition of the relation, we have $(a,b)R(a,b)$, as desired.

To prove that $R$ is symmetric, choose any $(a,b),(c,d) \in S$, and show that if $(a,b)R(c,d)$, then $(c,d)R(a,b)$.

To prove that $R$ is transitive, choose any $(a,b),(c,d),(e,f) \in S$, and show that if $(a,b)R(c,d)$ and $(c,d)R(e,f)$, then $(a,b)R(e,f)$.

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  • $\begingroup$ Oh duh! Thanks! On the symmetric and transitive, I know I can pick specific numbers to make it work. But how do I show that it will work for ANY R? $\endgroup$ – user84849 Jul 3 '13 at 7:01
  • $\begingroup$ @user84849 Instead of starting your proof with "let $a=1$, $b=3$,...", don't specify your variables. Leave them arbitrary. I've started you off for proving that $R$ is symmetric: "Choose any $(a,b),(c,d) \in S$ such that $(a,b)R(c,d)$. Then we know that..." Your proof should end with something like: "...hence, we know that $(c,d)R(a,b)$. So $R$ is symmetric, as desired." $\endgroup$ – Adriano Jul 3 '13 at 7:04
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You should write $\mathbf x=(x_1,x_2)$ for the members of $S$, then you have the relation on $\mathbf x\mathrel R\mathbf y$. Now it's clearer what are the objects in the domain of $R$.

Another way of proving this would be to show there exists a function $f\colon S\to\Bbb R$ such that $\mathbf x\mathrel R\mathbf y\iff f(\mathbf x)=f(\mathbf y)$. First prove that such $f$ defines an equivalence relation, now it's trivial from the definition of $R$, because $f$ is obvious in it.

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Any relation $R$ defined by $aRb \iff f(a)\equiv f(b)$, where $f$ is a function and $\equiv$ is an equivalence relation, is itself an equivalence relation.

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