proof by contradiction

Question

I have the following theorem:

and I want to prove it by contradiction. I have started by negating the consequence, so I will have that

so if I rearrange these terms I will have that:

which will be against the closure property of integer numbers, because the substraction of both integers (which are integers because their division by the gcd will not affect that condition) is giving me a number which is not an integer, so there is a contradiction.

Is my proof ok?

Answer 1

Your proof is correct, but there is no reason to use contradiction here: it just makes the argument unnecessarily complicated. You’ve basically taken the natural direct proof and turned it into a proof by contradiction.

We know that $g\mid x$, so there is an integer $m$ such that $x=mg$. Similarly, there is an integer $n$ such that $y=ng$. But then $x-h=mg-ng=(m-n)g$, and $m-n$ is an integer.