Question

Is there a rational number `x` such that `sqrt(x)` is irrational, but `sqrt(x)^sqrt(x)` is rational?

The motivation for this question is to consider: `sqrt(2)^sqrt(2)` and `(sqrt(2)^sqrt(2))^sqrt(2)`.

Note that `sqrt(2)` is irrational, and `(sqrt(2)^sqrt(2))^sqrt(2) = sqrt(2)^2 = 2`.

So either `sqrt(2)^sqrt(2)` is rational and we find that this rational number raised to an irrational power `sqrt(2)` is also rational, or `sqrt(2)^sqrt(2)` is irrational and we find that this irrational number raised to an irrational power is rational.

Going back to the question asked, I think there is no such rational number `x`, but I also suspect that it is difficult to prove.

Answer 1

No. If `sqrt(x)` is irrational, then `sqrt(x)^sqrt(x)` is transcendental by the Gelfond-Schneider theorem.

Explanation:

The Gelfond-Schneider theorem says that if `a, b` are algebraic numbers with `a != 0`, `a != 1` and `b` irrational, then `a^b` is transcendental (which implies that it is irrational).

A proof of the Gelfond-Schneider theorem can be found at http://people.math.sc.edu/filaseta/gradcourses/Math785/Math785Notes8.pdf

In our example, `sqrt(x)` is irrational, so it will not be equal to `0` or `1` and is algebraic, since it's a root of `t^2-x = 0` with `x` rational.

Hence with `a=b=sqrt(x)` we can deduce that `sqrt(x)^sqrt(x) = a^b` is transcendental, so irrational.