Definition of an equivalence relation:
It has to be:
Reflexive: $(x \sim x) \forall x$
Symmetric: $(x\sim y)\iff (y \sim x)$
Transitive: $(x \sim y) \land (y \sim z) \Rightarrow(x \sim z)$
Prove the equivalence relation on the set $\mathbb{Z}$:
$x\sim y :\iff x-y$ is an even number
Proof of symmetry:
There exists a $k\in\mathbb{Z}$ such that $x-y =2k$ which is the relation $x\sim y$.
The changed relation $y\sim x$ of our given equation gives us $[y-x=(-2)k]$ $\lor$ [$y-x=2(-k)]$
Thus it is symmetric. Proved.
My idea was: If $x-y=2k$ then $2k$ is positive because of $x>y$. And when we change the equation to: $y-x$ then I have to the choice between saying: $(-2)k$ or $2*(-k)$
Am I also allowed to define my variables like that? $x=2$ and $-y=k$
So I would get $-x=-2$ and $y=k$
What you want to verify is $x-x=2\times0$, $x-y=2k\implies y-x=2(-k)$ and $x-y=2k,\,y-z=2l\implies x-z=2(k+l)$.