I have this relation: $S = \{ (x, y) \mid ((x - y) / 3) \in\Bbb Z \}$
I already proved that it's reflexive. I don't know how to prove symmetry.
If I'm not mistaken it should be something like this:
H: $∀ x, y ∈ \Bbb Z: x \mathop{\rm S} y \iff ((x - y)/3) ∈ \Bbb Z$
T: $∀ x, y ∈ \Bbb Z: y \mathop{\rm S} x \iff ((y - x)/3) ∈ \Bbb Z$
but I don't know how to prove it.
If $\frac{x-y}{3} \in \mathbb{Z}$,
It means $\exists k \in \mathbb{Z}, \frac{x-y}{3}=k$
and hence $\frac{y-x}{3}=-k$
Since $-k \in \mathbb{Z}$ as well.
Hence $\frac{y-x}{3} \in \mathbb{Z}$