There exist irrational a and b such that a^b is rational

Theorem

There exist irrational $a$ and $b$ such that $a^b$ is rational.

Proof

• $\sqrt 2$ is irrational
• $2$ is trivially rational, as $2 = \dfrac 2 1$.

Consider the number $q = \sqrt 2^{\sqrt 2}$, which is irrational by the Gelfond-Schneider Theorem.

Thus $q^{\sqrt 2} = \left({\sqrt 2^{\sqrt 2}}\right)^{\sqrt 2} = \sqrt 2 ^{\left({\sqrt 2}\right) \left({\sqrt 2}\right)} = \sqrt 2^2 = 2$ is rational, so $a = q = \sqrt 2^{\sqrt 2}$ and $b = \sqrt 2$ are the desired irrationals.

$\blacksquare$

Alternative Proof

Given that $2$ is rational and $\sqrt 2$ is irrational, consider the number $q = \sqrt 2^{\sqrt 2}$.

We can prove the proposition using a similar argument to the first proof without the Gelfond-Schneider Theorem by considering two cases.

1. If $q$ is rational then $a = \sqrt 2$ and $b = \sqrt 2$ are the desired irrational numbers.
2. If $q$ is irrational then $q^{\sqrt 2} = \left({\sqrt 2 ^{\sqrt 2}}\right)^{\sqrt 2} = \sqrt 2 ^{\left({\sqrt 2}\right) \left({\sqrt 2}\right)} = \sqrt 2^2 = 2$ is rational, so $a = q = \sqrt2 ^{\sqrt 2}$ and $b = \sqrt 2$ are the desired irrationals.

$\blacksquare$

Sources

• Michael R.A. Huth and Mark D. Ryan: Logic in Computer Science: Modelling and reasoning about systems (2000): $\S 1.2.5$: Theorem $1.26$