I read this:
At first, I thought each class was an integer, that is: There were infinite classes of equivalence. Now I am a bit confused and it looks like there are two classes of equivalence. But I guess having two classes of equivalence would make this group finite.
The set of equivalence classes is in bijection with $A = [0, 1)\cap \mathbb Q$.
To see this, note that any distinct $a,b$ in $A$ are not equivalent, and that for any rational $x$, we have $x\sim x - \lfloor x \rfloor \in A$.
It might help to think about this in terms of decimals. Let $a,b\in\Bbb{Q},$ and write $a = \sum_{i\ll\infty}a_i 10^i$ and $b = \sum_{i\ll\infty}b_i 10^i,$ where $0\leq a_i,b_i\leq 9$ are digits. That is, $a$ has decimal expansion $a_n a_{n-1}\cdots a_1 a_0.a_{-1}a_{-2}\dots,$ and similarly for $b$. Then $a$ and $b$ live in the same equivalence class if and only if when you subtract them, their decimal parts cancel out. Another way to think about this is that $a$ and $b$ are in the same equivalence class if and only if their decimal parts agree after the decimal point; i.e., if and only if $a_i = b_i$ for all $i < 0.$ (Actually, this isn't quite true -- you need to make sure that you avoid the issue of $0.\overline{9}= 1,$ but you can decide to avoid decimal representations ending in repeating $9$'s and only use the representations ending in repeating $0$'s.)
So, each equivalence class in this group has a unique representative of the form $0.a_{-1}a_{-2}\dots < 1$ such that this decimal is a rational number. Another way to phrase this is that each equivalence class has a unique representative $\alpha\in\Bbb{Q}\cap [0,1)$.
For any rational number $r$ there is a unique rational number $s\in[0,1)$ such that $r-s\in\mathbb{Z}$. This implies that $\mathbb{Q}/\mathbb{Z}=\{s+\mathbb{Z}\mid s\in [0,1)\cap\mathbb{Q}\}$.
