This question is a classic and is on Stack Exchange several times, but I am looking for some atypical answers. The basic question, as you all already know is, "Find two irrational numbers $a$ and $b$ such that $a^b$ is rational."

There are two very common answers. The first being the classic $(\sqrt{2}^{\sqrt{2}})^\sqrt{2} = 2$ argument (in which the irrationality of $\sqrt{2}^{\sqrt{2}}$ happens to be irrelevant) and the second being the $\sqrt{2}^{2\log_2(3)} = 2$ example. These are both trivial and traditional proofs, but are there any other examples not usually given? A bunch more examples would be nice. It would also be helpful to show that the two numbers $a$ and $b$ are irrational, as some of these proofs, like $\pi$ and $e$ are not elementary.


  • $\begingroup$ The proof of $\pi$'s irrationality is not trivial, but that of $e$ is quite accessible, as long as you know the Taylor series expansion. For example: math.stackexchange.com/a/2520799/151732 $\endgroup$ – Deepak Nov 1 '20 at 9:46
  • $\begingroup$ I have never seen this proof before! So you could do this question with $e$ and $\ln2$. Are there any other ways to do it besides these that you can think of? $\endgroup$ – user817934 Nov 1 '20 at 9:59
  • $\begingroup$ We can use $$ a^{\log_2 b} = b^{\log_2 a} $$ to get a lot of other examples of the same type as your $\sqrt{2}^{2\log_2(3)}$ example. $\endgroup$ – GEdgar Nov 1 '20 at 10:05
  • $\begingroup$ That is interesting, but they're really part of the same gimmick as that example, as you mention. I'm trying to think up some other interesting examples. There is also an interesting proof that the solution to $x^x = 2$ is irrational, but what about two distinct irrational numbers? $\endgroup$ – user817934 Nov 1 '20 at 10:07
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    $\begingroup$ I realized that you have already posted a comment regarding $x^x$, but perhaps you may find this interesting. It is fairly easy to show that $x^x$ is an irrational number when $x>0$ is a non-integer rational number. (See, for instance, ocf.berkeley.edu/~wwu/cgi-bin/yabb/…); granted this, then the solution to the equation $$x^{x+n}=q\,$$ where $n\ge 0$ is an integer and $q>0$ is a non-integer rational number, is an irrational number. $\endgroup$ – Jack LeGrüß Nov 1 '20 at 10:37

It is easy to establish that $x^x$ is irrational when $x$ is a positive non-integer rational number (see for instance, https://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1031393807). Consider the equation $$x^{x+n}=q\,,$$ where $n\ge 0$ is an integer and $q>0$ is a non-integer rational number (Note: A solution to this exists by the Intermediate Value Theorem).

Claim: $x$ is an irrational number

Proof: Suppose $x$ is a rational number, then we have $$x^x=x^{-n}q\,.$$ Thus, $x^x$ is a rational number and thus $x$ must be a positive integer; but $x^{-n}q$ cannot be an integer, which is absurd.

  • $\begingroup$ This is helpful. An interesting solution to the question that I have not seen before. (Also how dumb of me not to think of multiplying both sides by $x^{-n}$, haha) $\endgroup$ – user817934 Nov 1 '20 at 12:28

Let $x=e^{i\pi/\sqrt2}=\cos(\pi/\sqrt2)+i\sin(\pi/\sqrt2)$ and $y=\sqrt2$. We have that $x$ is irrational since $\sin(\pi/\sqrt2)\not=0$ (so $x$ isn't even real, much less rational), and $y$ is irrational by for the usual elementary reasons. And $x^y=e^{i\pi}=-1$ is rational.

Added later: I am tacitly using a definition of irrational number as a number that is not rational. I see however that the set of irrational numbers is often explicitly defined to be the subset of real numbers that are not rational. So this may or may not be a suitable example, depending on what you mean by irrational.

  • $\begingroup$ I have never seen complex numbers being shown to be irrational or rational. It seems in this argument you say all complex numbers with nonzero imaginary part are irrational. This makes sense, but why is this the case? How can you even say a number like $1+i$ is rational or irrational? $\endgroup$ – user817934 Nov 1 '20 at 12:16
  • $\begingroup$ Any complex number that is not real and rational is considered to be irrational. It makes the theorems easier to state. $\endgroup$ – Gerry Myerson Nov 1 '20 at 12:28

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