n'th roots of 2

We should immediately feel that the answer should be $1$ ; $sqrt{2}$ is irrational and there seems to be no obvious natural $n$ such that $\sqrt [ n ]{ 2 }$ is rational.

We wish to prove that $n=1$ is the greatest natural number such that $\sqrt [ n ]{ 2 }$ is rational. We will do this by proving an equivalent statement- that $\sqrt [ n ]{ 2 }$ is irrational for all $n \geq 2$ .

Firstly, recall that $\sqrt{2}$ is irrational. The proof is elementary, so I won't repeat it, though a selection of proofs can be found here .

So, we now only need to show that $\sqrt [ n ]{ 2 }$ is irrational for all $n \geq 3$ . Take $n \geq 3$ and suppose that $\sqrt [ n ]{ 2 }$ is rational. Then, there exist $p, q \in \mathbb{Z}$ , with nonzero $q$ , such that $\sqrt [ n ]{ 2 } = \frac{p}{q}$ . Now, raising both sides to the power of $n$ preserves equality: $\big( \frac{p}{q} \big) ^n = 2$ i.e. $p^n = 2q^n$ . Note that if $q$ is nonzero then so is $p$ .

So, for $n \geq 3$ , there exist $p, q \in \mathbb{Z}$ , with nonzero $p, q$ , such that $p^n = q^n + q^n$

This contradicts Fermat's Last Theorem , which asserts that no $n \geq 3$ satisfy $p^n = q^n + q^n$ for nonzero $p,q \in \mathbb{Z}$ .

$\blacksquare$