abstract algebra - Showing numbers are irrational.

I was playing around and thought of the following question:

If given $a \not=1 \in \mathbb{R}^{+}$, prove that there are infinitely many integers $n$, such that $\sqrt[n]a$ is irrational.

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I have proven a very simple case:

If $1 < m \in\mathbb{N}$, then there are infinitely many positive integer $n$ such that $\sqrt[n]m$ is irrational. To show this, I argue by noting the polynomial, $$ p(x) = x^n -m$$ is irreducible in $\mathbb{Z}$[x] for $n > m$, hence it is irreducible in $\mathbb{Q}[x]$ as it is also primitive. So there are no rational solutions to $x^n = m$. And $\sqrt[n]m$ is irrational.

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How do I prove the general case? As it does seem quite intuitive.

Answer

The “simple” case is all you need.

Suppose $a>0$ is irrational; then $\sqrt[n]{a}$ is irrational as well, otherwise $a=\bigl(\sqrt[n]{a}\bigr)^n$ would be rational.

For a general rational $a=p/q$, consider that
$$
\sqrt[n]{a}=\frac{\sqrt[n]{pq^{n-1}}}{q}
$$
Therefore $\sqrt[n]{a}$ is rational if and only if $\sqrt[n]{pq^{n-1}}$ is rational.