radicals - How would you prove that $sqrt[n]{2}$ is irrational?

How would you prove that $\sqrt[n]{2}$ is irrational?, where $n \in \{2, 3, 4, \ldots\}$.

Answer

You have probably already seen the proof for $n=2$. So, let's assume that $n \geq 3$.

Seeking a contradiction, suppose that $\sqrt[n]{2}$ is rational. This, together with the positivity of $\sqrt[n]{2}$ imply that there exist $p,q \in \mathbb{N}$ such that $\sqrt[n]{2}=p/q$. Raising both sides to the $n^{th}$ power, we see that

$$2=\frac{p^n}{q^n}.$$
Multiplying through by $q^n$ and using $2q^n = q^n + q^n$, we have
$$q^n + q^n = p^n.$$
This violates Fermat's Last Theorem, giving a contradiction. Thus $\sqrt[n]{2}$ is irrational.