# Proving equivalence relation

Let $$Q$$ be the following subset of $$\mathbb{Z}\times \mathbb{Z}$$:

$$Q=\left \{ (a,b)\in \mathbb{Z}\times \mathbb{Z}: b\neq 0 \right \}$$

Define the relation $$\sim$$ on $$Q$$ as

$$(a,b)\sim (c,d)\Leftrightarrow ad=bc$$

Proof that $$\sim$$ is an equivalence relation, and specify $$[(2,3)]$$ and more generally the equivalence class $$[(a,b)]$$. Try to give an explanation of $$Q/\sim$$

I know that an equivalence relation is reflexive, symmetric, and transitive. I am not sure on how to approach such a proof and then to specify the values.

• The fact that this relation is indeed an equivalence follows straight from the definition: you merely need to verify that it is indeed reflexive, symmetric and transitive, with the first two properties being quite obvious and the third one requiring just a little bit of work. What is important to realise is that this is the construction by which one obtains the field $\mathbb{Q}$ from the ring $\mathbb{Z}$. The class of any pair $(m, n)$ represents nothing else than the fraction $\frac{m}{n}$. (to be cont.)
– ΑΘΩ
Commented Oct 18, 2020 at 1:53
• (cont.) In general, if $k=(m; n)$ is the greatest common divisor of $m$ and $n$ and if $m=kr, n=rs$, then the class of $(m, n)$ will be given by $\{(qr, qs)\}_{q \in \mathbb{Z^{\times}}}$.
– ΑΘΩ
Commented Oct 18, 2020 at 1:55
• I see, I was having difficulties on how to formulate and write it up even though I know all the concepts. But I see it is quite simple! Thanks for the explanation! Now, to the tricky part; how would i specify the two classes?
– user831952
Commented Oct 18, 2020 at 14:04
• What exactly do you mean by "specifying" the two classes? If you are referring to describing them explicitly, then the way to go is via the description I mentioned above, which requires some notions of elementary arithmetic in order to be proved.
– ΑΘΩ
Commented Oct 19, 2020 at 4:27

• $$\sim$$ is reflexive: For all $$(a,b) \in Q$$ we have $$ab = ba$$, that is, $$(a,b) \sim (a,b)$$.
• $$\sim$$ is symmetric: For all $$(a,b),(c,d) \in Q$$ such that $$(a,b) \sim (c,d)$$ we have $$ad=bc$$ and then $$cb = da$$, that is, $$(c,d) \sim (a,b)$$.
• $$\sim$$ is transitive: For all $$(a,b),(c,d),(e,f) \in Q$$ such that $$(a,b) \sim (c,d)$$ and $$(c,d) \sim (e,f)$$ we have $$ad=bc$$ and $$cf = de$$. Thus, $$(af)\color{red}{d} = (ad)f = (bc)f = b(cf) = b(de) = (be)\color{red}{d}$$ and since $$d\neq0$$, it follows that $$af = be$$, that is, $$(a,b) \sim (e,f)$$.