Question

123

Answer 1

Have you tried anything yet? I think a proof by contradiction might work best here, but that's my impression based on the proof for the irrationality of \(\sqrt2\).

Answer 2

Personally I wouldn't call this a simple proof... the only proofs I've seen are probably well beyond the grasp of most first year undergrad maths students (at least in the country I'm studying in). The usual approach seems to be to prove that it is transcendental, and since transcendental implies irrational, the number is irrational. The result follows immediately from the Gelfond–Schneider theorem, which is not at all easy to prove and probably requires complex analysis.

You mention case by case analysis. I wonder if you've actually been asked to prove something like, "Prove that there are irrational numbers x, y such that x^y is rational"? :-)

Answer 3

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Answer 4

There are other numbers you can work with to show that result.

One would be \(\sqrt{e}^{~\ln6}=(e)^{\frac{1}{2}\ln6}=e^{\ln3}=3\).

Answer 5

Certainly an irrational to an irrational power can be irrational. It is pretty easy to come up with an example, as @SithsAndGiggles has given above, although the numbers should be
\[\sqrt{e}^{\ \ln 6} = e^{\frac{1}{2}\ln 6} = e^{\ln \sqrt{6}} = \sqrt{6}
\]which is indeed irrational.

The more interesting question is whether an irrational to an irrational power can be rational. It's a classic argument and features in many logic and analysis courses, and would fit perfectly into a discrete maths course. This is probably the sort of argument they're looking for.

Consider \(\sqrt{2}^{\sqrt{2}}\). There are two cases: either it is rational, or it is irrational. (In a formal logic course, you may have to justify this by saying, "by the law of excluded middle".) If it is rational, then we've found the example we're looking for. If it is irrational, then consider
\[(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = \sqrt{2}^{\sqrt{2}\sqrt{2}}
= \sqrt{2}^{2} = 2
\]which is rational.

It may be possible to modify the argument to show that an irrational to an irrational power can be irrational. The beauty in this argument is that nowhere do you need to know if \(\sqrt{2}^{\sqrt{2}}\) is actually irrational, and because you never have to find specific numbers that fit the conditions!

As for whether \(\sqrt{2}^{\sqrt{2}}\) is actually irrational, the answer is yes. The proof of this is clearly not what the question is looking for, and unless you're familiar with complex analysis, there's no real way of proving it as far as I know. If you're interested, you can read about the Gelfond–Schneider theorem, a proof of which is given here:

http://www.proofwiki.org/wiki/Gelfond-Schneider_Theorem

The proof is not simple, and was in fact a Hilbert problem for many years. I do not claim to understand the proof myself, since I am not particularly well-versed in complex analysis...