Is $(\sqrt 2)^\sqrt3+(\sqrt 5)^\sqrt{7}$ an irrational number? is there any result about rationality\irrationality of numbers of the form $a^b+c^d$ where $a,b,c$ and $d$ are well-known irrational numbers (like $\sqrt{2}, \pi$ and...)
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1It is transcendental. See https://math.stackexchange.com/questions/446647/irrationality-of-sqrt2-sqrt2 – Landuros Dec 27 '17 at 10:02
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Also see Ramanujan's constant: http://mathworld.wolfram.com/RamanujanConstant.html – For the love of maths Dec 27 '17 at 10:15
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While the transcendentality (and even the irrationality) of $\sqrt 2 ^\sqrt3 + \sqrt 5 ^\sqrt7$ is still unknown, we can prove that $\sqrt 2 ^\sqrt3$ and $\sqrt 5 ^\sqrt7$ are transcendental using the Gelfond-Schneider theorem. The theorem states that:
If $a$ and $b$ are algebraic numbers with $a ≠ 0,1$ and $b$ non-rational, then any value of $a^b$ is a transcendental number.
This not only gives rise to trivial results such that $2^{\sqrt 2}$ is irrational, but to some more interesting ones as well. For instance, by taking $a=-1$ and $b=-i$, where $i=\sqrt {-1}$, we can obtain the following result:
$$ (-1)^{-i} = \left( e^{i \pi} \right)^{-i} = e^\pi, $$
showing that $e^\pi$ is transcendental.
Klangen
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1While $x:=\sqrt{2}^\sqrt{3},,y:=\sqrt{5}^\sqrt{7}$ are both transcendental, how do you know $x+y$ is? – J.G. Jun 10 '19 at 11:55
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1Oh sorry I misread the question. Indeed the transcendentality (and even the irrationality) of $x+y$ is still unknown. I will edit the response. – Klangen Jun 10 '19 at 11:59