I recently have thought of a proof but I can’t tell if it is correct or not. The proof is of $n \pi$ being irrational if n is an integer and non zero. The proof is below:
We assume that $n \pi$ is not irrational, thus meaning that $n \pi \in \mathbb Q$. Since $\sin(x)$ is an irrational number if x $\in \mathbb Q$ we can logically assume that $\sin(n \pi)$ is then irrational. But as we all know, $\sin(n \pi)$ = 0. QED
I have a lot of doubts about the proof being correct especially because of the fact that the proof seems to simple and that am usually terrible with proofs. Can someone please tell me if the proof is correct or incorrect? And if it is incorrect, (which it probably is) can someone tell me where I went wrong?
Your proof isn't wrong but it depends on the following prior result:
(You stated it without the "nonzero" condition, which makes it false, but with that correction it's true, and even more is true: if $x$ is a nonzero rational number then $\sin x$ is transcendental.) That's true, but it's just as difficult to prove as the irrationality of $\pi$. In fact, I think the only known proofs are more difficult than the easiest proofs that $\pi$ is irrational.
(There's an easier theorem -- "Niven's theorem" -- which is about when $r$ and $\sin r\pi$ are both rational. But what you need is something different.)