Numbers such as $e$ and $π$ are known to be transcendental, however, $e^e$ or $π^π$ are not even known to be irrational, let alone transcendental.

There are infinitely many transcendental numbers $a$ such that $a^a$ is rational, namely the solution of every $x^x = p$ where $p$ is prime.

My question is: do we know of any transcendental number $b$ such that $b^b$ is transcendental?


1 Answer 1


Let $b = 2/W(2)$, where $W$ is the Lambert W function. Note that $z = W(2)$ satisfies $z e^z = 2$. If $z$ were algebraic, then $e^z$ would also be algebraic, but this would contradict Lindemann's theorem. Therefore $z$ is trancendental, and so is $b$. Now $ W(2) + \log(W(2)) = \log(2)$, so $\log(b) = \log(2) - \log(W(2)) = W(2)$, and $$b^b = \exp(b \log(b)) = \exp(2)$$ which is transcendental.


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