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I suggested the following problem to my friend: prove that there exist irrational numbers $a$ and $b$ such that $a^b$ is rational. The problem seems to have been discussed in this question.

Now, his inital solution was like this: let's take a rational number $r$ and an irrational number $i$. Let's assume

$$a = r^i$$ $$b = \frac{1}{i}$$

So we have

$$a^b = (r^i)^\frac{1}{i} = r$$

which is rational per initial supposition. $b$ is obviously irrational if $i$ is. My friend says that it is also obvious that if $r$ is rational and $i$ is irrational, then $r^i$ is irrational. I quickly objected saying that $r = 1$ is an easy counterexample. To which my friend said, OK, for any positive rational number $r$, other than 1 and for any irrational number $i$ $r^i$ is irrational. Is this true? If so, is it easily proved? If not, can someone come up with a counterexample?

Let's stick to real numbers only (i.e. let's forget about complex numbers for now).

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    $\begingroup$ But of course, a cardinality argument shows that for $0 < r \neq 1$, $r^i$ is irrational for most (all but countably many) irrational $i$. So the example isn't irrepairably wrong. $\endgroup$
    – Daniel Fischer
    Jul 22, 2013 at 13:04
  • $\begingroup$ Can anyone explain why math people say "prove that there exist irrational numbers a and b..." instead of the more intuitive "prove that irrational numbers a and b exist...". The second approach makes more sense to me. $\endgroup$
    – makerofthings7
    Jul 22, 2013 at 16:38
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    $\begingroup$ @makerofthings7: comments to other questions are not the way to ask new questions. Use the "ask question" link. $\endgroup$
    – Chris Eagle
    Jul 22, 2013 at 17:17
  • $\begingroup$ @makerofthings7 Might be because $\exists$ is a prefix operator, and then the habit carries over into one's language. $\endgroup$
    – Jack M
    Jul 22, 2013 at 17:18
  • $\begingroup$ @makerofthings7 It's probably a combination of many things: (a) you speak the way people around you speak, in mathematics or anywhere else - this is just a kind of jargon, (b) there are lots of non-native English speakers reading and writing English papers, so clarity and consistency is very important, (c) "prove that x exists such that..." sounds very sloppy to me, because "such" modifies "x" ("prove that such an x exists that..."??), and in any case it might mislead you into thinking you were given a formula or algorithm for x earlier on, and are being asked to check it makes sense. $\endgroup$
    – Billy
    Jul 22, 2013 at 18:55

2 Answers 2

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Consider $2^{\log_2 3}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

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    $\begingroup$ I suppose the obvious followup question is how to prove that $\log_2 3$ is not rational. $\endgroup$
    – Ilmari Karonen
    Jul 22, 2013 at 17:00
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    $\begingroup$ @IlmariKaronen : if $2^{p/q}=3$, then $2^p=3^q$. $\endgroup$
    – user10676
    Jul 22, 2013 at 17:24
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    $\begingroup$ Gotta be one of the shortest best answers. $\endgroup$
    – LarsH
    Jul 22, 2013 at 18:50
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    $\begingroup$ And by the Fundamental theorem of arithmetic... $\endgroup$
    – Michael Lorton
    Feb 15, 2014 at 0:50
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    $\begingroup$ Fundamental Theorem of Arithmetic is way overkill - just note one side is odd and the other even for a contradiction $\endgroup$
    – Aaron
    Dec 4, 2019 at 23:59
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A supplement to the answer by Chris above:

Let $r$ be a positive rational number and $i$ a positive irrational number. If $r^{i}$ is rational, then $r^i=\frac{a}{b}$ for $a,b\in \mathbb{Z}$ such that $b\neq 0$. In particular, $i=\log_{r}\left(\frac{a}{b}\right)$. Therefore, Chris Eagle's answer is, in fact, prototypical. (Note also, that if $r=1$, then we get a contradiction as you observed.)

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    $\begingroup$ Do you know an elementary proof of this, please? $\endgroup$
    – Alex M.
    Sep 23, 2016 at 21:26

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