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Exponentiation break closures. For example with the integers, exponentiation of an integer (other than $0$) to a negative integer can be a rational number that is not equal to any integer. With the rational numbers exponentiation of a rational number(other than $0/1$) to a rational number is a root which is an algebraic number that is not necessarily a rational number.

  1. Can exponentiation of an algebraic number (other than $0$) by a non rational algebraic number (example: $2^\sqrt{2}$) also result in a non-algebraic number?

  2. Are the reals > $0$ closed under exponentiation? if Yes, what's the proof?

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Zuhair
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    For point 2), the positive reals (or really the non-negative reals if you accept $0^0=1$) are closed under exponentiation. $(-1)^{1/2}$ is not a real number, and complex exponentiation takes a lot of care to get working nicely. – Arthur Apr 28 '19 at 10:31

1 Answers1

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  1. The Gelfond–Schneider theorem says that if $a$ and $b$ are algebraic numbers with $a \neq 0$, $a \neq 1$, and $b$ irrational, then any value of $a^b$ is a transcendental number.
  2. Yes: is $a,b>0$, then $a^b=\exp(b\log a)\in\mathbb R$.
José Carlos Santos
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