Gelfond-Schneider transcendental Kuzmin

Question

The so-called Gelfond-Schneider constant $2^\sqrt{2}$ is transcendental, as shown by Kuzmin (1930).

What can you say about its square root? Is it also transcendental?

square roots of transcendentals are very likely to be transcendentals too. — Angina Seng, Dec 20 '17 at 14:29

score 1 · Answer 1 · answered Dec 20 '17 at 14:32

Its square root $2^{\sqrt2/2}$ satisfies the same criteria as stated in Gelfond–Schneider theorem, so it is transcendental.

Or as someone pointed out in a comment, square roots of transcendental numbers are transcendental:

Let $p(\sqrt x) =0 $ for some polynomial $p \in \Bbb Q[X]$.

Then, $p(\sqrt x) p(-\sqrt x)$ is a polynomial in $x$. Denote it by $q \in \Bbb Q[X]$.

Then, $q(x) = 0$.

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