Is an irrational to an irrational rational? [duplicate]

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I am working on this logical proof for class and trying to either prove or disprove that an irrational to an irrational power is also irrational. Please don't answer the problem for me but I'm completely stuck on how to begin. Any hints would be appreciated.

marked as duplicate by JimmyK4542, Tunk-Fey, tomasz, Did, Davide GiraudoSep 4 '14 at 8:22

• math.stackexchange.com/questions/104119/… – Heisenberg Sep 4 '14 at 5:52
• But the problem is that an irrational to an irrational power can be both rational and irrational. You are not going to get much far with logic alone here! – Mariano Suárez-Álvarez Sep 4 '14 at 5:52
• Consider the equation $x^x=2$. – alex.jordan Sep 4 '14 at 5:54
• I once read a great answer to this question on MSE that is actually "constructive" in the sense that it is both constructive and elementary to show that a, b are irrational and that $a^b$ is rational. I'll see if I can find it back or reconstruct it. – Myself Sep 4 '14 at 6:40
• @MarianoSuárez-Alvarez: Well, to be more precise, it can be either rational or irrational. It can't really be both. – tomasz Sep 4 '14 at 7:09

If an irrational positive real raised to an irrational power were always irrational, then whenever $$\text{irrational}^{x}=\text{rational}\text{,}$$ $x$ would have to be rational. This is equivalent to saying that $\log_{\text{irrational}}(\text{rational})$ would have to be rational. So then for any fixed rational $r$, the map $$\text{irrational}\mapsto\log_{\text{irrational}}({r})=\frac{\ln(r)}{\ln(\text{irrational})}$$ would be a decreasing map from irrationals greater than $1$ to rationals greater than $0$. Can you have a monotonic (and hence one-to-one) map from an uncountable subset of the reals to a countable one?

• Better yet, if you have a function whose domain contains a real interval, continuous into reals and which is not constant, it must have both rational and irrational values. You don't need cardinality arguments: Darboux property and non-constantness of the function and density are all that's needed. – tomasz Sep 4 '14 at 7:12

I think I learned this first from MSE but I could not retrieve in what question. I googled for it and found it here, on the blog of Andrej Bauer, 2009. It is probaby a well-known folklore fact, albeit a lot less known than the standard $\sqrt{2}^{\sqrt{2}}$ based proof.

Here it goes: take $a = \sqrt{2}$ and $b = \log_2{9}= 2\, \log_23$. Then $$a^b = \sqrt{2}^{2 \log_2 3} = 2^{\log_2 3} = 3$$ is rational, but it is well known and easy to show that both $\sqrt{2}$ and $\log_2 9$ are irrational. (Both follow from unique factorisation into primes in some sense.)

You can disprove the statement by simply finding a situation where an irrational number to the power of an irrational number is a rational number.

For example, consider

$$\left(\sqrt2^\sqrt2\right)^\sqrt2$$

• Without knowing whether $\sqrt2^{\sqrt2}$ is rational or irrational, this is essentially a non-constructive proof. – Henry Sep 4 '14 at 6:28
• @Henry The goal is to conclude that either $\sqrt2^\sqrt2$ or $\left(\sqrt2^\sqrt2\right)^\sqrt2$ is in the form of irrational^irrational=rational. We do not need to know which one in particular fits that bill, but that one of them does. – user137794 Sep 4 '14 at 6:32