Another tricky exercise about equivalence relations.

Question

We have the following relation $R \subset \mathbb{Z} \times \mathbb{Z}$ defined as follows: $aRb$ $\iff$ $|a+b|$ is an even number.

No problem showing the relation is reflexive and symetrical but I cannot still figure out why the relation is transitive. I mean, if $|a+b|=2k$ and $|b+c|=2m$ for $k,m \in \mathbb{Z}$ how it happens that $|a+c|=2n$ for some $n \in \mathbb{Z}$? Thanks.

Answer 1

Hint: $|a+b|$ is even if and only if $a$ and $b$ are of the same parity.

Answer 2

Got it! If $|a+b=x|=2k$ and $|b+c=y|=2m$ with $k,m \in \mathbb{Z}$ then we have that $a$ and $b$ have the same parity, the same argument with $b$ and $c$. Then $a$ and $b$ have the same parity, doesnt matter if both are odd or even we obtain that $|a+c|$ is even. Thanks mates. :)