Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
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. – Daniel Fischer Jul 22 '13 at 13:04
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. – LamonteCristo Jul 22 '13 at 16:38
@makerofthings7: comments to other questions are not the way to ask new questions. Use the "ask question" link. – Chris Eagle Jul 22 '13 at 17:17
@makerofthings7 Might be because $\exists$ is a prefix operator, and then the habit carries over into one's language. – Jack M Jul 22 '13 at 17:18
@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. – Billy Jul 22 '13 at 18:55
up vote 59 down vote accepted

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

share|cite|improve this answer
Ah, of course! So obvious now that you've said it! – Armen Tsirunyan Jul 22 '13 at 12:46
I suppose the obvious followup question is how to prove that $\log_2 3$ is not rational. – Ilmari Karonen Jul 22 '13 at 17:00
@IlmariKaronen : if $2^{p/q}=3$, then $2^p=3^q$. – user10676 Jul 22 '13 at 17:24
Gotta be one of the shortest best answers. – LarsH Jul 22 '13 at 18:50
And by the Fundamental theorem of arithmetic... – Malvolio Feb 15 '14 at 0:50

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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.