If $a$ is irrational, does there exist a natural number $n$ such that $na$ is rational?

Question

For some irrational $a$, does there exist an $na$ which is contained within the rational numbers for some natural $n$?

Answer 1

Only $n=0$. Any $n > 0$ would contradict the assumption that $a$ is irrational, as if $na = \frac{p}{q}\in\mathbb{Q}$ then $a = \frac{p}{nq}\in\mathbb{Q}$.

Answer 2

No, there isn't, other than $n = 0$. If there is an $n > 0$ such that $na$ is rational, then that means $a$ is rational as well.

Consider for example $a = \frac{22}{7}$. Then set $n = 7$, and you get $na = 22$. But obviously $a$ was rational to begin with.

Now consider $a = \pi$. Now we have a number that is not only irrational, it is non-algebraic (transcendental). There is no value of $n$ that will give you $na$ rational. For that matter, no value of $n$ can make $a^n$ rational either.