Prove that $R = \{((m,n),(p,q)):m + q = n + p\}$ is an equivalence relation on $\mathbb N_0$. 
Define the relation $R$ on $\mathbb{N}_0$ by
$$R = \{((m,n),(p,q)):m + q = n + p\}.$$
(a) Prove that $R$ is an equivalence relation.

Now I need to prove it's reflexive, symmetric, and transitive!

Proving $\simeq$ is reflexive.
To Prove: $(\forall a \in S) a \simeq a$.
Form of  Proof:

*

*Let $a$ be an arbitrary (variable) element of $S$.


*Give an argument which concludes that $a \simeq q$.

Now my biggest confusion is what do I let $a$ equal? Obviously it will be an arbitrary element in $\mathbb{N}_0$.

Let $a=(x,y)$ for some $(x,y) \in \mathbb{N}_0$. Then $((x,y),(x,y)) \in R$ because $x+y=y+x$, so that $aRa$ (or $a \sim a$). And so on for symmetry and transitivity.
