A relation $R$ is defined on $\mathbb{Z}$ by $aRb$ if and only if $2a + 2b\equiv 0\pmod 4$. Prove that $R$ is an equivalence relation.
My method:
Let $a \in \mathbb{Z}$ be given. So, for any $a \in \mathbb{Z}$, we have $2a +2a = 4a$. Since $4a$ is divisible by $4$, it is congruent to $0\pmod 4$. It follows that $aRa$. So, $R$ is reflexive. Then, for any $a, b \in \mathbb{Z}$, we have $2a + 2b = 2b + 2a$. It follows that if $aRb$ then $bRa$, then $R$ is symmetric. Finally, we assume that $aRb$ and $bRc$, so we have $(2a + 2b) + (2b + 2c) = (2a + 2c) + 4b\equiv 2a + 2c\pmod 4$, since $4b\equiv 0\pmod 4$. Then if $2a + 2c\equiv 0\pmod 4$ and $2b + 2c\equiv 0\pmod 4$, it follows that $2a + 2c\equiv 0\pmod 4$. Then $R$ is transitive.
I think this needs some work, but I am not sure what else needs to be done. Would I need a different way to find if it's reflexive, symmetric, or transitive?
Best Answer
Your proof is right.
I would clarify the steps in the transitivity part by saying something like this:
Assume $aRb$ and $bRc$. Now $2a+2b\equiv 0 \mod 4$ and $2b+2c\equiv 0 \mod 4$, which means $2a+2b=4m$ and $2b+2c=4n$ for some integers $m$ and $n$. Now by adding the two equations together we have $2a+2b+2b+2c=4m+4n$, i.e. $2a+2c=4(m+n-b)$, and thus $2a+2c\equiv 0 \mod 4$. Hence $aRc$.
But those steps should be fairly clear if not stated.