Prove the relation 'x divides y' on the natural numbers is antisymmetric but not on the integers.

Question

Here is my proof so far:

Let x,y be in the Natural Numbers. Then, if x divides y and y divides x, there exist integers k, q such that y = kx and x = qy. So, y = kqy.

Here is where I'm stuck. I know I am trying to get x=y in order to show the relation is antisymmetric but I don't think that's possible with what I've written so I must have messed up somewhere.

And then to show that this relation does not hold for integers, can I let x=-2 and y =2. So, x divides y means there exists an integer k such that y=kx. Let k = -1 so 2 = -2 * -1. And then y divides x means there exists an integer q such that x =qy. Let q = -1 so -2 = 2 * -1. Since -2 does not equal 2, x does not equal y and thus the relation is not anti-symmetric.

Answer 1

Since $y=kqy$, $1=kq$. But the only two natural numbers $k$ and $q$ whose product is $1$ are $k=q=1$. So, $x=y$.

What you did for the integers is fine, of course.

Answer 2

What is the definition of antisymmetric? If aRb and bRa, then a=b. Suppose there are any two natural numbers x and y, where x divides y to give another integer z. That is, y=xz. That implies y>=x.

Let us take a contradictory assumption that R is not antisymmetric, and that y divides x, while x is not equal to y. That means x=my. That also implies x>y.

Now, since we have taken the same x and y in both cases, by logic, both of these statements can be true simultaneously, we can infer that our assumption was wrong, and that x=y, which in turn leads us to our conclusion that R is in fact antisymmetric. Hope that helps.