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:
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.
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.