Is $x \sim y$ iff $y - x \in \mathbb Q$ an equivalence relation?

Question

I can easily show that the relation $x\sim y$ iff $y - x \in \mathbb Q$ and $x,y \in (0,1]$ is an equivalence relation since it is reflexive (the number zero is rational), symmetric (the negative of a rational number is rational) and transitive as the sum/difference of two rational numbers is rational but I get stuck in what follows. Specifically, I know that equivalence relations induce mutually disjoint equivalence classes but I can't verify that here. My silly counterexample is that if $x=\frac{1}{3}$ then clearly $y=\frac{3}{3}$ is in the equivalence class but $y=\frac{3}{3}$ is also in the equivalence class of $x=\frac{2}{3}$ as that difference too is rational. My question therefore is, what have I misunderstood here? Thank you.

Answer 1

The numbers $\frac{1}{3}, \frac{2}{3}, \frac{3}{3}$ are all in the same equivalence class, they are all equivalent to each other.