If x is a positive rational number, show that x^x is irrational unless x is an integer.

We prove the following more general proposition:

(1) When (p/q)^(m/n) is rational, with reduced fractions p/q and m/n, then p and q are n-th powers.

From this, the problem statement follows immediately: Say (p/q)^(p/q) is rational, then q is a power of q, i.e q=a^q for some integer a. This can only work for q=1, which is when p/q is an integer.

Now to the proof of (1): Say we have

(2) (p/q)^(m/n) = s/t

with reduced fractions p/q, m/n and s/t. We take the n-th power on both sides and multiply by q^m*t^n and get

(3) p^m*t^n = q^m*s^n

Let d be any prime divisor of p with p=d^i*p', such that not d|p' (i.e. i is maximal). To show that p is an n-th power, we need to show that i is a multiple of n. First we see from (3) that d|s (p and q being coprime). Therefore we can write s as

with not d|s' (j is maximal). Now we can insert the representation for p and s into (3) and get
(4) d^(im)*p'^m*t^n = d^(jn)*s'^n*q^m

In (4), no factor, except the explicit d-powers, is divisible by d. Therefore

im=jn <=> m/n = j/i

Since m/n is a reduced fraction, i is a multiple of n. This completes the proof that p is an n-th power.

In the exact same way, thanks to the symmetry of (3), we see that q is an n-th power, This proves the puzzle's statement.

Edited on February 2, 2020, 3:49 pm

Posted by JLo on 2020-02-02 15:40:37