Prove $\forall x,y \in \mathbb{Z} (x \equiv _m y$ $\rightarrow$ $\bar{x} = \bar{y}$)

Question

pf. Assume x $\equiv _m y$. Then $\bar{x}$ = y (mod m). By equivalence y = x (mod m). Then y $\in $ [0,m-1]. Thus, $\bar{y}$ = y (mod m). Therefore, $\bar{x}$ = $\bar{y}$.

Is this proof correctly written? I just did a direct proof. I feel like this statement is obvious, but when that is the case the proofs seem difficult.

Let R be an equivalence relation on a set A. For x $\in$ A, the equivalence class of x determine by R is the set. x/R = {y $\in$ A : xRy}. x/R is equivalent to [x] and $\bar{x}$

Edit: Assume x $\equiv$ $_m$ y. Then x=y+mk for some k $\in$ $\mathbb{Z}$. There exists some z such that z $\equiv _m x$, then $$z=x+ml$$ $$z=y+mk+ml$$ $$z=y+m(k+l)$$ $$z\equiv _m y$$ Therefore $\bar{x} = \bar{y}$.

Answer 1

No, this proof doesn't really make sense.

I'm not entirely clear on the definitions you're using, but let's take some usual ones:

• We say that $x \equiv_m y$ if there is some integer $k$ for which $x = y + m k$.
• We write $\overline{x}$ for the equivalence class of $x$ under the relation $\equiv_m$, i.e. $\overline{x} = \{ x' \; | \; x' \equiv_m x\}$.

You need to assume that $x \equiv_m y$, i.e. that $x = y + m k$ for some $k$, and show that $\overline{x} = \overline{y}$, i.e. that if $z \equiv_m x$ then $z \equiv_m y$. (This amounts to showing that $\equiv_m$ is a transitive relation.)