Relations $R$ and $S$ on $\mathbb{Z}$,both $R,S$ are both symmetric and transitive. Show that $|\mathcal{R}-\mathcal{S}|=|\mathcal{S}-\mathcal{R}|=1$

Question

Give an example of two relations $R$ and $S$ on $\mathbb{Z}$ with the following properties: (i) both $R$ and $S$ are both symmetric and transitive. (ii) if $\mathcal{R} \subseteq \mathbb{Z} \times \mathbb{Z}$ and $\mathcal{S} \subseteq \mathbb{Z} \times \mathbb{Z}$ represent $R$, respectively $S$, then $$ |\mathcal{R}-\mathcal{S}|=|\mathcal{S}-\mathcal{R}|=1 $$

How does this $1$ come? Don't even have the intuition of $ |\mathcal{R}-\mathcal{S}|=|\mathcal{S}-\mathcal{R}|=1 $.

Really hope someone could help\hint me a little bit!

Answer 1

$\mathcal R=\{(0,0)\},\mathcal S=\{(1,1)\}$.