Prove that $2n\mid m$ is asymmetric

Question

Let $R$ be a relation on $\Bbb{N}$ given by $$nRm\iff2n\mid m.$$ Prove that $R$ is asymmetric.

---

I know that a symmetric relation is $\forall a,b\in A(aRb\to b\not Ra)$ i.e. $\forall a,b\in A\;\neg(aRb\wedge bRa)$.

From this, suppose that $2n\mid m$ i.e. there exists a $k\in\Bbb{Z}$ such that $m=2kn$. We need to prove that $n\neq2pm$, where $p\in\Bbb{Z}$ i.e. $2m\not\mid n$, right?

If it is right then I am not able to prove it. If $m=2kn$ then $(2p)m=(2p)2kn$. What is next?

EDIT Since $2pm=4pkn$ we conclude that $2pm\neq n$ i.e. $2m\not\mid n$ but $2pm=4pkn$, right?

Answer 1

Try with contradiction. Say exists $a,b$ such that $aRb$ and $bRa$ so $2a\mid b$ and $2b\mid a$.

Then $2a\leq b$ and $2b\leq a$ so $4a\leq a$ and hence $3a\leq 0$ a contradiction.

Answer 2

My first thought at seeing this is that if $2n \mid m$ and $2m \mid n$ (if it were symmetric) then $4m \mid 2n \mid m$. So, because $\mid$ is transitive, we have $4m \mid m$. The point of this is to reduce from two variables ($m$ and $n$) to just one variable.

Now we can continue the argument: $4mk = m$ and if $m \ne 0$ then $4k = 1$, which is impossible.

Answer 3

Since we are working in $\mathbb{N}$, $2n\mid m\Rightarrow m=2nk, k\in\mathbb{N}\Rightarrow n<m.$

Now, if $2m\mid n$, then $n=2mr,r\in\mathbb{N}\Rightarrow m<n,$ so we found a contradiction.