My friend Aaron Brown asked me a question a while back:

Suppose $x > 0$ is some rational number, and that $2^x$ is also rational.

Does this mean that both of them must actually be integers?

Let’s see what happens if we proceed in a way similar to

showing that $\sqrt{2}$ is irrational.

Let

$ \displaystyle x = \frac{m}{n}$ and $ \displaystyle 2^x = \frac{k}{l}$

where $m, n, k, l$ are all positive integers.

Make sure to choose $k$ and $l$ so that $ \displaystyle \frac{k}{l}$ is in
lowest terms,
meaning that $k$ and $l$ do not have any common positive integer divisors (other than the number $1$ if you want to count that).

Then

$ \displaystyle 2^{\frac{m}{n}} = \frac{k}{l} $

and so by taking $n$-th powers,

$ \displaystyle \left(2^{\frac{m}{n}}\right)^n = \left(\frac{k}{l}\right)^n $

This gives

$ \displaystyle 2^m = \frac{k^n}{l^n} $

and so

$ \displaystyle 2^m l^n = k^n $

Now using unique factorization into primes, write $k$ and $l$ as products of powers of distinct prime numbers:

$ \displaystyle k = p_1^{a_1} p_2^{a_2} \cdots p_K^{a_K} $ and $ \displaystyle l = q_1^{b_1} q_2^{b_2} \cdots q_L^{b_L} $

Substitute into $ \displaystyle 2^m l^n = k^n $ and you get

$ \displaystyle 2^m \left( q_1^{b_1} q_2^{b_2} \cdots q_L^{b_L} \right)^n = \left( p_1^{a_1} p_2^{a_2} \cdots p_K^{a_K} \right)^n $

and so

=== BEGIN THE GREEN EQUATION ======================

$ \displaystyle 2^m q_1^{b_1 n} q_2^{b_2 n} \cdots q_L^{b_L n} = p_1^{a_1 n} p_2^{a_2 n} \cdots p_K^{a_K n} $

=== END THE GREEN EQUATION ========================

Crucial observations from the green equation:

- The right side $ \displaystyle p_1^{a_1 n} p_2^{a_2 n} \cdots p_K^{a_K n} $ is a prime factorization of $k^n$.
- No prime $p_i$ is the same as a prime $q_j$, because the fraction $ \displaystyle \frac{k}{l}$ is in lowest terms. Otherwise there would be a common prime factor of $k$ and $l$ …
- The equality of the left and right sides shows that the prime $2$ divides (is a factor of) the right side, and so is one of the primes appearing in the prime factorization of the right side.
- (From #3 and unqiue factorization into primes)

**Exactly one of**the $p_i$ equals $2$. - (From #2 and #4)

**None of**the $q_j$ equals $2$.

Thus the left side is also a unique prime factorization (of the same number as the right side). - (From #4 and unqiue factorization into primes)

The exponent of $2 = p_i$ in the prime factorization of the right side $ \displaystyle p_1^{a_1 n} p_2^{a_2 n} \cdots p_K^{a_K n} $ equals $a_i n$. - (From #5 and unqiue factorization into primes)

The exponent of $2$ in the prime factorization of the left side $ \displaystyle 2^m q_1^{b_1 n} q_2^{b_2 n} \cdots q_L^{b_L n} $ equals $m$. - Again by unique prime factorization, the exponent of the prime $2$ must be the same on the left and the right sides.
- (From #6, #7, and #8)

**Conclusion:**if $p_i = 2$, then $a_i n = m$.

We’re done.

From $a_i n = m$ we deduce $ \displaystyle a_i = \frac{m}{n} = x$ is an integer, and $ \displaystyle 2^x$ must also be.