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!
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3$\begingroup$ See this. $\endgroup$– David MitraJul 25, 2013 at 17:49
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$\begingroup$ @DannyCheuk I modified your edit because it uses very nonstandard notation. $\endgroup$– PotatoJul 25, 2013 at 18:01
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$\begingroup$ Cool, I was actually struggling to decide what symbol to use for irrational numbers $\endgroup$– user67258Jul 25, 2013 at 18:02
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4 Answers
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A simple counterexample: $$e^{\ln 2}$$
answered Jul 25, 2013 at 17:49
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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?
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$\begingroup$ Very nice example! I'd add "Hint" to this. +1 $\endgroup$ Jul 25, 2013 at 18:15
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Counterexample:
Let $a$ be a number such that $\log a\notin\mathbb{N}:e^{\log a}\in\mathbb{Q} $
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$x = 2^\sqrt2 $, $y=1/\sqrt2$ , $x^y=2$
