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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.

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    $\begingroup$ what about x = 1, y = pi, z = 0, h = pi-3 $\endgroup$ – shai horowitz Oct 1 '16 at 10:42
  • $\begingroup$ If you know that your relation is an equivalence relation, you do not have to verfiy it because it is implied. What about $e$ or $\sqrt{2}$: are they also in the same equivalence class as $\frac{1}{3}$? $\endgroup$ – Moritz Oct 1 '16 at 10:43
  • $\begingroup$ @Moritz Okay these are irrationals but I can't yet see what you mean by this. $\endgroup$ – JohnK Oct 1 '16 at 10:45
  • $\begingroup$ The symbol "$=$" is not a drop-in replacement for the English word "is". It means is equal to, and there are no $x$ and $y$ such that $x-y$ is equal to the word "rational". If you want to write symbolically that $y-x$ is a rational number, write "$y-x\in\mathbb Q$". $\endgroup$ – Henning Makholm Oct 1 '16 at 10:52
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    $\begingroup$ @shai et al. Read $x, y \in (0, 1]$ and see that $0$ is not included. $\endgroup$ – Namaste Oct 1 '16 at 11:02
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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.

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