Prove whether $ x \sim y \iff x + y \in 2\mathbb{Z}$ is an equivalence relation

Question

Prove whether this is an equivalence relation:

• For $ x, y \in\mathbb{Z}$
  $$ x \sim y \iff x + y \in 2\mathbb{Z}.$$

I think it is not an equivalence relation because the transitive property doesn't work but I'm not sure.

Because $1+5 \in \mathbb{2Z}$ and $2 +2 \in \mathbb{2Z}$ but $2+5 \notin \mathbb{2Z}$

Is my counterexample correct ?

Answer 1

You showed that $1 \equiv 5$ (since $1 + 5 \in 2 \mathbb{Z}$) and $2 \equiv 2$ (since $2 + 2 \in 2\mathbb{Z}$). But $2 \not \equiv 5$, since $2 + 5 \notin 2\mathbb{Z}$.

This is correct, but it is not a counterexample to transitivity. To find a counterexample to transitivity, you would have to give integers $a,b,$ and $c$ such that $a \equiv b$, $b \equiv c$, and $a \not\equiv c$.

In fact, transitivity holds, and this is an equivalence relation. See if you can check all the properties.

Answer 2

The relation is in fact transitive. Suppose $x\sim y$ and $y\sim z$. Then, $x+y=2k$ for some $k\in\mathbb{Z}$ and $y+z=2m$ for some $m\in\mathbb{Z}$. So, $$ x+z=(2k-y)+(2m-y)=2(k+m)-2y=2(k+m-y)\in2\mathbb{Z}. $$