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?
$\require{cancel}$
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 $$(x, y), (y,x) \in R \implies x=y$$