Need help proving some equivalance classes.

Question

For the equivalence relation mod m on the integers prove that if x=y(mod m) then [x]=[y].

The farthest I can get is:

Assume that x=y(mod m).
This means that x-y=mq for some integer q.
=> x = mq+y.

And then I stuck. Anybody able to help me?

Answer 1

Suppose $R$ is an equivalence relation on some set $X$ and $x R y$ for $x, y \in X$. Let $z \in [x]$. Then $z R x$. Then we have $z R x$ and $x R y$, so by transitivity, $z R y$. That is, $z \in [y]$. Hence $[x] \subset [y]$.

The reverse inclusion is similar.