Is this relation anti-symmetric? I was wondering if the following relation is anti-symmetric. I have done some work, but not sure if this is correct.

Given:$\;\;  R$ is a relation on $\mathbb Z^+$ such that $(x, y) \in R$ if and only if $y$ is divisible by $x$.
Hint:  An integer $y$ is divisible by an integer $x$ (with $x\neq 0$) if and only if there exists an integer $k$ such that $y= kx$.

Let $(x,y)$ be in the relation $R$.  Then $y = kx$.
Let $(y, x)$ be in the relation $R$. Then $x=ky$
$y=kx$ and $x=ky$. If you substitute $x$ in $y = kx,$ then $ y = k^2y$ and you can solve $k$ for 1 which would conclude $x =y.$ Is this the right way to prove this?
 A: Your proof doesn't quite work. You need to start by assuming that x is some multiple of y and y is some multiple of x, with no assumptions on what those multipliers might be: i.e. $x=ky$ and $y=lx$, for some constants $k$ and $l$ that might be different.
A: $\require{cancel}$
You've certainly got some intuition heading in the right direction; the only   mistake you made is having used   $k$ to represent the positive integer multiple of both $x,$ and $y$.
Once we assign $y=kx$ (y being a multiple of x), we can't assume the same $k$ to represent x being a multiple of y.  So, we pick a different integer multiple.  
Otherwise, the proof follows along the lines of your proof.

So suppose $(x, y), (y, x) \in R$.
(1) Then there exists an integer we call $m$ such that $y = mx$.  
(2) And there exists an integer, not necessarily $m$, which we'll call $n$ such that $x = ny$.  
Since we know $y= mx$ for some integer $m$ (1), then with (2)  $$x= ny \iff x = nmx \iff 1=nm$$ $$\implies n = m = 1$$  Hence from (1), we have $y=x$, and from (2) we have $x = y$.
The relation is anti-symmetric because $$(x, y), (y,x) \in R \implies x=y$$
