Can anyone give me a counterexample to this statement?

Question

Statement: Let $n$ and $m$ be two irrational numbers. Then $n^m$ is always irrational.

I think this statement is correct, otherwise can someone give me a counterexample?

Thanks!

@DannyCheuk I modified your edit because it uses very nonstandard notation. — Potato, Jul 25 '13 at 18:01
Cool, I was actually struggling to decide what symbol to use for irrational numbers — , Jul 25 '13 at 18:02

score 11 · Answer 1 · answered Jul 25 '13 at 17:49

11

A simple counterexample: $$e^{\ln 2}$$

answered Jul 25 '13 at 17:49

user87690 · Answer 2 · 2013-07-25T18:54:54.357

7

Exactly one of these is a counterexample: $√3^{√2}$, $(√3^{√2})^{√2} = 3$.

Hint: What happens if $√3^{√2}$ is rational? What happens if it's irrational?

edited Jul 25 '13 at 18:54

answered Jul 25 '13 at 18:08

user87690

score 2 · Answer 3 · answered Jul 25 '13 at 17:49

2

Counterexample:

Let $a$ be a number such that $\log a\notin\mathbb{N}:e^{\log a}\in\mathbb{Q} $

answered Jul 25 '13 at 17:49

score 2 · Answer 4 · answered Jul 25 '13 at 17:57

2

$x = 2^\sqrt2 $, $y=1/\sqrt2$ , $x^y=2$

answered Jul 25 '13 at 17:57

newzad

4 Answers4