Let $R$ be a relation on $\mathbb{N}$ defined by $(m, n) \in R$ if $m \mid (2n)$ or $n \mid (2m)$. Is $R$ is an equivalence relation on $\mathbb{N}$.

Question

So far, I have that R is reflexive, by saying,

Let $m,n \in \mathbb{N}$. Since $m|2m$ means $\frac{2m}{m}=2$, and $n|2n$ means $\frac{2n}{n}=2$, then $m|2m$ and $n|2n$ are true. So $mRm$ and $nRn$, and thus $R$ is reflexive.

For showing R is symmetric, I have: Let $m,n \in \mathbb{N}$ and assume $mRn$. Therefore, $m|2n$ or $n|2m$. This means $m$ is divisible by $2n$ or $n$ is divisible by $2m$.

From here I know I now have to show the opposite, but to do this would I try and show $n|2m$ or $m|2n$, which is symmetric to the original problem, or would I try and show $2n|m$ or $2m|n$?

For transitive I need help.

Answer 1

It is in fact not transitive. Indeed, we have $(3,2) \in R$ and $(2,5) \in R$, but $(3,5)\not\in R$.

Answer 2

By the way, for reflexive, you really needed to consider only one representative $m\in \mathbb N$.

For symmetric, you assume $mRn$ (i.e., $m|2n$ or $n|2m$), and then you want to show $nRm$, which means $n|2m$ or $m|2n$. Because logical "or" is commutative, that is easy.

$R$ is not transitive, because, for example, $3R12$ and $12R4$ but not $3R4$.