My question is:
Let $H$ and $J$ be the Equivalence relations in $\mathbb R \times \mathbb R$. that is:
$$H = \bigg{\{}[(a,b),(c,d)]: b-a = d-c\bigg{\}}$$ $$J = \bigg{\{}[(a,b),(c,d)]: a+b = c+d\bigg{\}}$$ $(a,b,c,d \in \mathbb R)$
Prove that $H \circ J$ = $J \circ H$ and conclude that $H \circ J$ is an equivalence relation. And describe the equivalent class modulo $H \circ J$.
I cannot even figure out how to start this. Thanks in advance for all your help!
I suppose the usual definition of composition of equivalence relations (or binary relations, in general) applies. So $$H\circ J = \left\{ ((a,b),(c,d)) \in (\mathbb R^2)^2 : \exists (u,v) \in \mathbb R^2 \bigl( ((a,b),(u,v)) \in H \,\&\, ((u,v),(c,d)) \in J \bigr) \right\}.$$ So let $(a,b),(c,d)$ be arbitrary elements of $\mathbb R^2$ and define $$u=\frac{a-b+c+d}{2}\quad\text{and}\quad v=\frac{-a+b+c+d}{2}.$$ It follows that $$v-u=\frac{2b-2a}{2}= b-a$$ and $$u+v=\frac{2c+2d}{2}=c+d,$$ whence $$((a,b),(u,v)) \in H \quad\text{and}\quad ((u,v),(c,d))\in J,$$ and therefore $((a,b),(c,d))\in H\circ J$. It follows that $H\circ J = (\mathbb R^2)^2$.
Similarly, $J\circ H = (\mathbb R^2)^2$.
So $H\circ J = J \circ H$ is an equivalence relation with only one equivalence class, that is $\mathbb R^2$.