I began thinking today about the set of all irrational numbers $x$ for which there exists a real power $p$ such that $x^p = q$ is rational. Specifically, I'm wondering if the set $$S = \{\text{irrationals } x \mid x^p \text{ is rational for some real } p\}$$ is countable or uncountable. I suspect there is a simple argument that would show this set is uncountable, but I don't have a proof and I'm thus curious to hear your thoughts.

We can certainly come up with a subset of S that is countable. For example, $\sqrt{P}$ for any prime number $P$ is irrational, while $\sqrt{P}^2 = P$ is rational. Since the primes are countable, it would follow that this subset of $S$ is countable. To that set we could form the union of all roots of primes, and this set would be countable.

Intimately related to this question is whether there exist powers of transcendental numbers like e, pi, etc. such that $x^p$ is rational. I have a feeling that the power p involved would have to be irrational since $x$ is transcendental, but I'm not quite sure of this either. If the answer is no (there do not exist such transcendental numbers) then that subset of transcendental irrationals would be excluded from the set $S$ I have in mind.

Any thoughts about the countability of $S$?


Given that exponentiation with real powers is only defined for positive bases, it's easy to see that

$$S={\mathbb R}^+\backslash {\mathbb Q},$$

that is the set of positive irrational numbers.

For any given $x \in S$ the equation

$$x^p=2$$ has the solution

$$p=\frac{\log2}{\log x},$$

which is a well defined real number.

ADDED: This means the cardinality of $S$ is uncountable, equal to the cardinality of $\mathbb R$.

In the comments the question came up what happens if $p$ is restrictued to the rational numbers. In this case, the equation becomes

$$x^p=r;\quad p,r \in \mathbb Q, x \in \mathbb R$$

which is equivalent (if we restrict ourselves to $x > 0,r > 0$) to


Since $r,p \in \mathbb Q$, this can only produce a countable number of values $x$, as $\mathbb Q \times \mathbb Q$ is still countable.

Now even if we allow negative $x$ with exponents $p=\frac{m}n$, $\gcd(m,n)=1$, $n$ odd, we only get at most the same number of negative values in addition, because we have $|x|^p=|r|$, so any negative $x$ corresponds to $-x$, a positive solution we already found before and the cardinality of which was countable.

So in the case of restricting $p$ to rationals, the answer changes: Then $S$ is infinitely countable, the infinity coming for example from the fact that all $2^\frac1n$ for $n \ge 2$ are irrational and different and satisfy $\left(2^\frac1n\right)^n=2$, so $2^\frac1n \in S$ for all $n \ge 2$.

  • $\begingroup$ (+1) You forgot to state that the set is uncountable. $\endgroup$ – Peter Foreman May 5 '19 at 21:08
  • $\begingroup$ Essentially any positive $x$ will do, and there are uncountably many irrational options for $x$. $\endgroup$ – Mark Bennet May 5 '19 at 21:15
  • $\begingroup$ Great. Thank you. In retrospect I see that allowing the power p to be a real number made this question less interesting than it possibly could be. How about if the power were restricted to be rational, p = m/n? $\endgroup$ – Jim Rockford May 5 '19 at 21:27
  • $\begingroup$ @JimRockford See my extended answer for discussion of rational p. $\endgroup$ – Ingix May 6 '19 at 8:10

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