Existence of Rational Powers of Irrational Numbers/Proof 3

Theorem

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

Proof

Consider the equation:

$\paren {\sqrt 2}^{\log_{\sqrt 2} 3} = 3$

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.

$\blacksquare$

Also see

• Positive Rational Number as Power of Number with Power of Itself for an interesting related result.

Sources

• November 2008: Nick Lord: Maths bite: irrational powers of irrational numbers can be rational (The Mathematical Gazette Vol. 92: p. 534)  www.jstor.org/stable/27821850