I got to wondering if a non-zero integral multiple of any irrational number is guaranteed to be irrational? This seems intuitive but I can't prove it to myself.
There is an answer on this site with regard to factors, but not multiples of irrationals.
If someone could sketch a quick proof, my day would be golden. Thanks!
Yes, it is guaranteed. Suppose for the sake of contradiction that $\alpha$ is irrational, and that $n\alpha$ is rational for some non-zero integer $n$. In other words, suppose $$n\alpha=\frac{a}{b}$$ for some non-zero integer $n$ and rational number $\frac{a}{b}$. Then this implies $$\alpha=\frac{a}{bn}$$ contradicting the definition of $\alpha$ being irrational.