Irrational Powers of Irrational Numbers

Given two irrational numbers $a$ and $b$ we might assume that $a^b$ is also an irrational number. This is not true however! There exists irrational numbers $a$ and $b$ for which $a^b$ is a rational number!

Lemma 1: $\sqrt{2}$ is an irrational number.

• Proof: Suppose not. Then there exists integers $a$ and $b$ with $b \neq 0$ such that $\sqrt{2} = \frac{a}{b}$. Let $a$ and $b$ be chosen such that the greatest common divisor of $a$ and $b$ is $1$.

• We square both sides of the equation above to get $2b^2 = a^2$. This shows that $a^2$ is an even number. But then $a$ is an even number. So $a$ is of the form $a = 2m$ for some integer $m$. Hence:

(1)

\begin{align} \quad 2b^2 &= (2m)^2 \\ &= 4m^2 \end{align}

• So $b^2 = 2m^2$. So $b^2$ is an even number. But then $b$ is an even number.

• Since $a$ and $b$ are both even numbers, their greatest common divisor must be at least $2$ which is a contradiction. Therefore the assumption that $\sqrt{2}$ is not irrational is false. So $\sqrt{2}$ is irrational. $\blacksquare$

Theorem 2: There exists irrational numbers $a$ and $b$ such that $a^b$ is a rational number.

• Proof: Let $a = \sqrt{2}^{\sqrt{2}}$.

• First suppose that $a$ is rational. Let $A = \sqrt{2}$ and $B = \sqrt{2}$. From Lemma 1, $A$ and $B$ are irrational. So there exists irrational numbers $A$ and $B$ such that $A^B$ is rational.

• Now suppose that $a$ is irrational. Let $b = \sqrt{2}$. From Lemma 1, $b$ is irrational. Consider $a^b$. We have that:

(2)

\begin{align} \quad a^b = (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = (\sqrt{2})^{\sqrt{2} \cdot \sqrt{2}} = (\sqrt{2})^{2} = \sqrt{2} \cdot \sqrt{2} = 2 \end{align}

• So there exists irrational numbers $a$ and $b$ such that $a^b$ is rational. $\blacksquare$

There exists many pairs of irrational numbers $a$ and $b$ such that $a^b$ is rational. For example, let $a = \sqrt{e}$ and let $b = \ln (4)$. We first show that $a$ and $b$ are irrational.

If $\sqrt{e}$ is rational then set $\sqrt{e} = \frac{a}{b}$ (where $a$, $b$ are integers and $b \neq 0$, and their greatest common divisor is $1$) so that $e = \frac{a^2}{b^2}$. Then $eb^2 = a^2$. Since $a$ and $b$ are rational, $a^2$ and $b^2$ are rational. But since $e$ is irrational, $eb^2$ is irrational. So the equation $eb^2 = a^2$ says that an irrational number is equal to a rational number - a contradiction. So $\sqrt{e}$ is irrational.

If $\ln(4)$ is rational then set $\ln(4) = \frac{a}{b}$ (where $a$, $b$ are integers and $b \neq 0$, and their greatest common divisor is $1$). Then:

(3)

\begin{align} \quad e^{\ln(4)} = e^{\frac{a}{b}} \end{align}

Then $4 = e^{\frac{a}{b}}$. Since $a$ is a nonzero integer, $e^a$ is irrational. From a similar proof as the one above, this implies that $e^{\frac{a}{b}}$ is irrational. So the equation $4 = e^{\frac{a}{b}}$ says that an irrational number is equal to a rational number - a contradiction. So $\ln (4)$ is irrational.

Lastly we show that $a^b$ is rational. We have by the properties of logarithms that:

(4)

\begin{align} \quad a^b = \sqrt{e}^{\ln(4)} = e^{\frac{1}{2} \ln (4)} = e^{\ln \sqrt{4}} = e^{\ln 2} = 2 \end{align}