Existence of Rational Powers of Irrational Numbers/Proof 3

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Theorem

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


Proof

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:

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

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.

$\blacksquare$


Also see


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