Prove that $n^{1/n}$ is irrational for all $n \in \Bbb Z$ and $n > 1$

Question

Short question:

For all integers $n>1$, is $\sqrt[n]{n}$ always irrational? If so, how can you prove it?

Or in other words: is the following true? (And is the typography right?)

$\forall n \ni n \in \mathbb{N},n>1:\sqrt[n]{n} \notin \mathbb{Q}$

(For all values $n$, such that $n$ is a natural number and larger than 1, the $n^{th}$ root of $n$ is not rational.)

The format could also be $$\forall n\in \mathbb{Z}_{>1}, \ \sqrt [n] {n} \notin \mathbb{Q}$$ or for the end part, $$\nexists \frac{p}{q} : \sqrt [n] {n} = \frac{p}{q}, \ (p, q)\in \mathbb{Z} \ \land \ q \neq 0$$. — Mr Pie, Sep 08 '17 at 00:46
Interesting. But for $1= 1^n < n < 2^n$ so $\sqrt[n]{n}$ is not an integer. It's easy to prove that $\sqrt[n]{k}$ is either an integer or irrational for integer $k$ . So for $n > 1$ $\sqrt[n]{n}$ is irrational — fleablood, Sep 08 '17 at 00:54
side question. Under what circumstance is $\sqrt[q]q$ rational for rational $q$. Or $\sqrt[x]x$ rational for real $x$? — fleablood, Sep 08 '17 at 01:03
Dupe of Prove that if $n \geq 2$, then $\sqrt[n]{n}$ is irrational. Hint, show that if $n \geq 2$, then $2^{n} > n$. — Bill Dubuque, Aug 21 '23 at 15:52

lhf · Accepted Answer · 2017-09-08T00:57:10.390

14

Hint:

If $m$ is an integer, then $\sqrt[n]{m}$ is rational iff it is an integer, that is, iff $m$ is an $n$-th power.
$n$ is never an $n$-th power because $n = t^n > 2^n$ is always false.

Or use that $1 < \sqrt[n]{n} < 2$ implies that $ \sqrt[n]{n} $ is never an integer.

edited Sep 08 '17 at 00:57

answered Sep 08 '17 at 00:40

lhf

1 Answers1