How often is $a^b$ rational when $a,b>0$ are irrational.

Question

This question was motivated by the proof of the existence of irrational numbers $a,b$ such that $a^b$ is rational. I am wondering how many such numbers there are.

Let $\mathbb I = \mathbb R_+ \setminus \mathbb Q$ ( I'm choosing positive numbers to avoid complex numbers). Now consider the following set $P= \{ (x,y) \in \mathbb I \times \mathbb I \, \, | \, \, x^y \in \mathbb Q \}$. Now is it true that $\overline{P} = (\mathbb R_+ \cup \{0\})\times (\mathbb R_+ \cup \{0\})$ ?

Answer 1

Let $q > 0; q \in \mathbb Q$. Let $a$ be irrational such that $a^r \ne q$ for any $r \in \mathbb Q$. (For instance let $a$ be transcendental; or if $q = 2$ and $a = \sqrt 3$) Then $b = \log_a q$ is irrational.

Then $a^b = a^{\log_a q} = q$.

There are a huge number of these and easy to find.

Second part. Let $S = \{(x,\log_x q)| x \text{ transcendental}; x > 0; q \in \mathbb Q^+\} \subset P$.

Technically $S = \cup_{x\in \mathbb T}\{x\} \times \{\log_x q| q \in Q\}$

It's obvious $\{\text {positive transcendental}\}$ is dense in $\mathbb R^{\ge 0}$. (Their complement, the algebraics, are countable for one thing). Likewise for any $x > 0; x\ne 1$ and any positive $w < z$ we can find rational $q$ so that $q$ is between $x^w$ and $x^w$ which are not equal. So $w < \log_x q < z$. And so for any positive real $x; x \ne 1$, in particular for any positive transcendental $x$, $\{\log_x q| q \in \mathbb Q+\}$ is dense in $\mathbb R^{\ge 0}$.

So $\overline S = \overline{\cup_{x\in \mathbb T}\{x\} \times \overline{\{\log_x q| q \in Q^+\}}}=\overline{\cup_{x \in \mathbb T}\{x\} \times \mathbb R^{\ge 0}} = \mathbb R^{\ge 0} \times \mathbb R^{\ge 0}$

So $(\mathbb R^{\ge 0})^2\ = \overline{S} \subset \overline{P} \subset (\mathbb R^{\ge 0})^2$

Answer 2

Take any open interval, map it forward by raising to a power $b\in\mathbb I$, then the image open set must contain some rational number $q/p$ with $p$ prime.

Answer 3

Here's a sketch of a proof.

Let $(a,b) \in (\mathbb R_+ \cup \{0\})\times (\mathbb R_+ \cup \{0\})$. If $a^b \in \mathbb{Q}$, then let $q = a^b$. Otherwise, let $\{q_i\}$ be a rational sequence approaching $a^b$. For each rational $q_i$, let $\{a_i\}$ be an irrational sequence approaching $a$. Then $(a_i, \log_{a_i} q_i)$ is a sequence in $P$ that converges to $(a,b)$.

It is well known that both the rationals and irrationals are dense in $\mathbb{R}$, so I don't explicitly define $q_i$ and $a_i$.

The remaining piece to be shown is that an irrational based log of a rational is irrational, which is true since an irrational to a rational power must be irrational.