# prove a number is irrational [duplicate]

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?

• $\sqrt{2}^{\sqrt{2}}$ 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? Apr 22, 2016 at 0:50
• @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. Apr 22, 2016 at 0:53
• @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. Apr 22, 2016 at 1:04
• @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. Apr 22, 2016 at 1:07
• Yes, it is necessary. You are right, there are nicer counterexamples than the one I've given. Apr 22, 2016 at 1:09

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.

• 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. $\qquad$ Apr 22, 2016 at 2:19
• 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.
– user325968
Apr 22, 2016 at 2:25

$e$ 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.

• 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. Apr 22, 2016 at 1:00
• @MathematicsStudent1122: Yes. Thanks. Apr 22, 2016 at 1:03
• @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. Apr 22, 2016 at 1:24
• @VincenzoOliva Ah I see, thanks. Apr 22, 2016 at 1:36
• 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.) $\qquad$ Apr 22, 2016 at 2:17