Define a relation $\mathbb{R}^2$ by $(a,b) \sim (c,d)$ iff $b−a = d−c$. Show that $\sim$ is an equivalence relation.

Question

In order to be an equivalence relation, it has to be reflexive, symmetric, and transitive.

Answer 1

Reflexivity: We need $\forall (a,b) \in \mathbb{R}^2, (a,b) \sim (a,b)$, and we have $(a,b) \sim (a,b) \Leftrightarrow b - a = b - a$, which is true.

Symmetry: We need $\forall (a,b), (c,d) \in \mathbb{R}^2, (a,b) \sim (c,d) \Rightarrow (c,d) \sim (a,b)$, and we have $(a,b) \sim (c,d) \Leftrightarrow b - a = d - c \Leftrightarrow d - c = b - a \Leftrightarrow (c,d) \sim (a,b)$.

Transitivity: We need $\forall (a,b), (c,d), (e,f) \in \mathbb{R}^2, \Big(\big((a,b) \sim (c,d)\big) \; \cap \big((c,d) \sim (e,f)\big)\Big) \Rightarrow \big((a,b) \sim (e,f)\big)$, and we have $(b - a = d - c) \cap (d - c = f - e) \Rightarrow b - a = f - e \Rightarrow (a,b) \sim (e,f)$

Answer 2

So do you know what reflexive, symmetric and transitive mean?

Can you write the conditions in terms of the writing $b-a=d-c$?

Example: Reflixive $\implies$ for any $(a,b)$ that $(a,b)\sim(a,b)\implies b-a = b-a$.

Can you always say that $b-a = b-a$?[1]

DO the same for the definitions of symmetric and transitive.

If you can actually write out these sentences is it literally impossible not to prove it is an equivalence relationship[2]. Literally.

.....

[1] Hint: Yes you can. Everything is equal to itself so $b-a = b-a$.

[2] It is impossible not to prove it. You'll end up having to make claims like: "$K = K;$" and "if $K = J$ then $J=K; $" or "If $K = J$ and $J = L$ then $K = L$" and, literally, you can't not have those be true.