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Are there two irrational numbers $a,b$ such that $a^{b}$ is a natural number?

Edit: After the answer of Lhf we add the following:

Are there two irrational but algebraic numbers $a,b$ with $a^{b} \in \mathbb{N}$?

Ali Taghavi
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  • Exactly. There must be by construction. For every irrational $a$ that is not itself an integer to a rational power (so everything except for roots of integers), there is an irrational $b=\log_a n$. – orion Dec 16 '16 at 11:04
  • Related: http://math.stackexchange.com/questions/1609918/what-exactly-are-those-two-irrational-numbers-x-and-y-such-that-xy-is-r – Björn Friedrich Dec 16 '16 at 11:13

2 Answers2

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$e^{\log n} = n$ for all $n \in \mathbb N$.

It is well known and easy to prove that $e$ is irrational.

If $\log n=a/b$ with $a,b \in \mathbb Z$, then $e^a=n^b$ and $e$ would be algebraic (but it is harder to prove that $e$ is not algebraic).

lhf
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There exists irrationals $a,b$ such that $a^b$ is rational.

EXAMPLE: We know that $\sqrt{2}$ is irrational and that $2$ is rational. We also have $q=\sqrt{2}^{\sqrt{2}}$ is irrational by the Gelfond-Schneider theorem.

Consider the expression $(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}$. This can be written as $(\sqrt{2})^{\sqrt{2}\times \sqrt{2}} = (\sqrt{2})^2 = 2$ which is rational.

Hence, we have $a=\sqrt{2}^{\sqrt{2}}$ and $b=\sqrt{2}$ such that $a^b$ is rational.

EDIT: You can also see other proofs here.

  • @MaxPayne Not exactly - the question asks for $a^b$ integer, not just rational. – Wojowu Dec 16 '16 at 11:28
  • Question needs a proof by construction, this is not a proof that there exists two irrational numbers = natural as you'd need to prove that $\sqrt{2}^\sqrt{2}$ is irrational – q.Then Dec 16 '16 at 15:05
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    @Ephemeral That is why I have provided a link to the Gelfond theorem. –  Dec 16 '16 at 15:06