# Solution to puzzle 29: xx

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

This question is amenable to a reductio ad absurdum proof.

Assume x is rational but not an integer; that is, x can be written as a/b, irreducible, with b > 1.  Assume (a/b)a/b = c/d is irreducible.

Raising each side to the power b we get (a/b)a = (c/d)b.
Hence aa db = cb ba.

Now we use the Fundamental Theorem of Arithmetic, which states that every integer greater than 1 can be written uniquely as a product of finitely many prime numbers.

Since b > 1, it will have at least one prime factor, p > 1.  Consider the number of times p occurs in the prime factorization of each of the above terms.  Letting b = pru, where p and u are relatively prime, then:

• ba: ra times
• aa: 0 times, since a and b are relatively prime
• db: sb times, where s is the number of times p occurs in d  (since p occurs on the right-hand side, it must also occur on the left-hand side, so it must be a factor of d)
• cb: 0 times, since c and d are relatively prime

By the Fundamental Theorem of Arithmetic, ra = sb.  But since a and b are relatively prime, b must divide r.
That is, r b pr, which is absurd for p > 1.

This completes the reductio ad absurdum proof.
Hence (a/b)a/b is irrational.

Of course, if b = 1, then x is an integer for any integer a, and xx is rational.

Therefore, if x is a positive rational number, xx is irrational unless x is an integer.

## Remark: xx = 2

The above result, in conjunction with the Gelfond-Schneider Theorem, can be used to show that the positive real root of xx = 2 is a transcendental number.

Clearly xx = 2 does not have an integer solution.  Hence, by the above result, x cannot be rational.

Now, using the Gelfond-Schneider Theorem, if x is algebraic and irrational, xx is transcendental, and so cannot be equal to 2.

Therefore the positive real root of xx = 2 is transcendental.  The root is approximately equal to 1.559610469462369349970388768765; see Sloane's A030798.