Existence of Rational Powers of Irrational Numbers

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Theorem

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


Proof 1

We have that:

$\sqrt 2$ 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.

$\blacksquare$


Proof 2

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

We consider the two cases.

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

$\blacksquare$


Proof 3

Consider the equation:

$\left({\sqrt 2}\right)^{\log_{\sqrt 2} 3} = 3$


We have that $\sqrt 2$ is irrational and $3$ is (trivially) rational.


It remains to be proved $\log_{\sqrt 2} 3$ is irrational.

We have:

\(\ds \log_{\sqrt 2} 3\) \(=\) \(\ds \frac{\log_2 3}{\log_2 \sqrt 2}\) Change of Base of Logarithm
\(\ds \) \(=\) \(\ds \frac{\log_2 3}{1/2}\) Logarithms of Powers: $\sqrt 2 = 2^{1/2}$ and $\log_2 2 = 1$
\(\ds \) \(=\) \(\ds 2 \, \log_2 3\)

From Irrationality of Logarithm, $\log_2 3$ is irrational.

Therefore $2 \, \log_2 3$ is irrational.

Thus there exist two irrational numbers $a = \sqrt 2$ and $b = \log_{\sqrt 2} 3$ such that $a^b$ is rational.

$\blacksquare$