Quotient set of the relationship $x,y\ \in \mathbb{Q}, \ x \sim y \iff x-y \in \mathbb{Z}$

Question

$x,y\ \in \mathbb{Q} \ \\ x\sim y \iff x-y \in \mathbb{Z}$

I already know this is an equivalence relationship, in fact:

1. reflexive: $x-x =0 \in \mathbb{Z}$
2. symmetric: $x-y\in \mathbb{Z} \implies y-x \in \mathbb{Z}$
3. transitive: $x-y \in \mathbb{Z}, y-z \in \mathbb{Z} \implies x- \not y+ \not y-z \in \mathbb{Z}$

But what's the quotient set? I notice that values in this form $\{\frac{1}{2}, \frac{2}{4},...,\frac{8}{16}, ...\}$ are always in relation, for instance $\frac{1}{2}-\frac{1}{2} = 0 \in \mathbb{Z}, \frac{1}{2}-\frac{2}{4} = 0 \in \mathbb{Z}, \frac{1}{2}-\frac{8}{16} = 0 \in \mathbb{Z}, \text{...}$ however not only fractions with denominator multiple of 2 but even of 3, for instance: $\frac{2}{3} \sim\frac{8}{12}$

so what's the quotient set?

Answer 1

The quotient set is, as usual, the set of equivalence classes.

In this case, it should be straightforward to check that, for each $q \in \mathbb Q$, with $0 \leq q < 1$, there is a different equivalence class: indeed, for $q_1, q_2$ in that condition $0<q_1-q_2<1$ or $-1<q_1-q_2<0$, whence $q_1-q_2 \notin \mathbb Z$.

Conversely, for any $q \in \mathbb Q$, there is $q_0 \in \mathbb Q$ such that $0 \leq q_0<1$ and $q_0<q \in \mathbb Z$.

So the interval $$\{ q \in \mathbb Q: 0 \leq q < 1 \}$$ is a set of representatives of the equivalence classes.
For example $[0]$, the class of $0$, is the set of integers; more generally, for each $q$ in the above set, $$[q] = \{ q + k : k \in \mathbb Z \}$$ is the equivalence class of $q$.