If $x$ and $y$ are irrational numbers then $x$ to the power of $y$ is irrational I am asked to prove or disprove this statement. To do so I got an idea to use the contra-positive, for that I need to prove if $x$ to the power of $y$ is rational then $x$ and $y$ are rational. I took $\ln$ for both sides, because keeping some element in power would make the sum more difficult, now I have $y\ln(x) = \ln(p/q)$. How could I show that $x$ and $y$ are rational?
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2$\begingroup$ $\sqrt{2}^{\sqrt{2}}$ is irrational, $\sqrt{2}$ is irrational, but $\biggl(\sqrt{2}^{\sqrt{2}}\biggr)^{\sqrt{2}} = (\sqrt{2})^2=2$ is very much rational. Your take on this? $\endgroup$– Sarvesh Ravichandran IyerApr 22, 2016 at 0:50
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$\begingroup$ @астонвіллаолофмэллбэрг: It doesn’t really matter whether $\sqrt2^{\sqrt2}$ is irrational: if not, we can take $x=y=\sqrt2$, and if so, your example works. $\endgroup$– Brian M. ScottApr 22, 2016 at 0:53
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2$\begingroup$ @BrianM.Scott You are right, but I thought I would complicate things. Besides, the Gelfond-Schneider theorem says that $\sqrt{2}^\sqrt{2}$ is transcendental, forget about irrational. $\endgroup$– Sarvesh Ravichandran IyerApr 22, 2016 at 1:04
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$\begingroup$ @астонвіллаолофмэллбэрг: But someone asking this question is unlikely to be aware of the Gelfond-Schneider theorem or to have any idea whether $\sqrt2^{\sqrt2}$ is irrational, so the extra complication is probably necessary. $\endgroup$– Brian M. ScottApr 22, 2016 at 1:07
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$\begingroup$ Yes, it is necessary. You are right, there are nicer counterexamples than the one I've given. $\endgroup$– Sarvesh Ravichandran IyerApr 22, 2016 at 1:09
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2 Answers
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An elementary solution: $x = \log_2 3$ is irrational (otherwise if it equals $m/n$ for positive integers $m$ and $n$, $2^m = 3^n$ contradicting unique prime factorization). Now $\sqrt 2$ is irrational and $(\sqrt 2)^{2x} = 3$ is a counter-example.
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$\begingroup$ For this you can get by with a far weaker statement than uniqueness of prime factorizations, namely the fact that a product of two odd numbers is odd. $\qquad$ $\endgroup$ Apr 22, 2016 at 2:19
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$\begingroup$ But the way I wrote is much more general, it suggests a much greater family of similar examples. I often have such additional, hidden reasons in my writing. With that said, of course you are right. $\endgroup$– user325968Apr 22, 2016 at 2:25
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$e$ and $\log2$ are irrational but $e^{\log2}=2$ is not.
In particular, $e$ is transcendental. This theorem allows to prove the irrationality of $\log n,$ $n\in\mathbb{N\setminus\{0,1\}}$. Assume $\log n=a/b$ for some integers $a,b$. Then $e=n^{b/a}$, which makes it a solution of the algebraic equation $x^a-n^b=0$; absurd. Hence $\log n$ is irrational.
answered Apr 22, 2016 at 0:57
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3$\begingroup$ To expand on this answer, this is true because it is known that $e$ is both irrational and transcedental; the proofs are not too difficult. $\log 2$ is irrational because $\log 2$ rational would imply $e$ algebraic, contradiction. $\endgroup$ Apr 22, 2016 at 1:00
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$\begingroup$ @MathematicsStudent1122: Yes. Thanks. $\endgroup$ Apr 22, 2016 at 1:03
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1$\begingroup$ @Deepak: I don't think s/he was referring to Gelfond. Let $\log 2$ be the real number $r$ s.t. $e^r=2$ (there clearly does exist such a unique $r$). Then $e=2^{1/\log2}$. If $\log 2=a/b$ for some integers $a,b$, then $e$ would be algebraic (in particular, it would be the solution of the algebraic equation $x^a-2^b=0$), a contradiction. $\endgroup$ Apr 22, 2016 at 1:24
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$\begingroup$ How do you prove $\log_e 2$ is irrational? Should that be added to the answer? (Proofs that $e$ is irrational are found in many textbooks and are short and simple.) $\qquad$ $\endgroup$ Apr 22, 2016 at 2:17