verify R is an equivalence relation 
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 $3-1=2$.
 A: 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 $(a,b)R(e,f)$.
A: 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.
A: 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.
