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What is the set $\{x\in\Bbb R\mid \forall q\in\Bbb Q: q^x\in\Bbb Q\}$?

Of course $\Bbb Z$ is a subset of this set.

Are there any other? if not what is the proof? is there a good reference for it?

Asaf Karagila
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rohit
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    I think the OP wants to find $\left{x\in \Bbb R : (\forall q \in \Bbb Q )(\exists p \in \Bbb Q)(x^q=p)\right}$ – Git Gud May 01 '13 at 11:56
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    @Asaf, I think the OP wants it to be the other way round: $(\frac{1}{2})^2$ should be rational. So the set required is ${x \in \mathbb{R} : q^x \in \mathbb{Q} \forall q \in \mathbb{Q}}$. This should be true for all integers, as for integer $n$,$p$,$q$, $(p/q)^n = p^n/q^n \in \mathbb{Q}$. – user73445 May 01 '13 at 12:04
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    The question should probably restrict to the positive rationals, since exponentiation of a negative number to a non-integer (real) power is not well or uniquely defined, and definitions based one a particular branch cut of the complex logarithm will not even give a real result (see the answer by GEdgar for the confusion this causes). – Marc van Leeuwen May 01 '13 at 14:23

3 Answers3

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It is known that if $2^c$, $3^c$, and $5^c$ are all rational then $c$ is a nonnegative integer. It follows from the Six Exponentials Theorem, q.v.

On the 1971 Putnam, it was asked to show that if $2^c,3^c,4^c,\dots$ are all integers then $c$ is a nonnegative integer.

MJD
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Gerry Myerson
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    I know it's the intense heat which is baking my brain, but how can $2^c$ not be an integer when $c$ is a nonnegative integer itself? – Asaf Karagila May 01 '13 at 13:26
  • @AsafKaragila yes this implication is straithforward if $c$ is a non negative integer then $2^c,3^c,5^c$ are rationals but this is the other direction. – clark May 01 '13 at 13:45
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    but even if 2^c 3^c and 5^c are rationals c can be negative integer also! c= -1 works , as far as six exponential thm is concerned , i am confused about how to apply it, we are taking same c each time . – rohit May 01 '13 at 13:51
  • @clark: I knew that something was off. I didn't figure it was my reading. (And this is the third mis-read I had on the site in the past few hours...) – Asaf Karagila May 01 '13 at 13:52
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    @AsafKaragila hehe, yes my first reading left me with the same impression but as for me it is a more common occurrence the first thing I questioned was my reading ability... – clark May 01 '13 at 13:57
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Expansion of Gerry Myerson/MJD answer.

If $x$ is rational but not integer, then $2^x$ is irrational by unique factorization. Now suppose $x$ is irrational.

Let's use the "Six Exponentials Theorem", which states:
Let $x_1, x_2, x_3, y_1, y_2$ be complex numbers, assume $x_1, x_2, x_3$ are linearly independent over $\mathbb Q$ and $y_1, y_2$ are linearly independent over $\mathbb Q$. Then at least one of $\exp(x_iy_j)$ is transcendental.

Take $x_1 = \log 2, x_2 = \log 3, x_3 = \log 5, y_1 = 1, y_2 = x$. Then $\log 2, \log 3, \log 5$ are linearly independent by unique factorization. And $1,x$ are linearly independent since $x$ is irrational. Conclusion: at least one of $$ 2, 3, 5, 2^x, 3^x, 5^x $$ is transcendental. So, in particular, they are not all rational.

user642796
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GEdgar
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Note that $-1$ is rational. If $x$ satisfies the condition, in particular $(-1)^x$ is rational. Taking that to mean $e^{i\pi x}$ is rational, and thus real, I conclude that $x$ is an integer.

GEdgar
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  • But I would say $(-1)^{1/3}=-1$ is real (and rational) even though $e^{i\pi/3}$ is not. @Marc van Leeuwen has made this point in the comments on the question. – Gerry Myerson Feb 19 '18 at 00:09