Prove that there can exist irrational numbers a & b such that a^b is rational?

Question

Proof of this can be written in a small space, which I will do at end of question period, if none are offered.

Okay, good answer guys, I should have said, "REAL irrational numbers a & b". Sorry about that.

Answer 1

let b = √2, an irrational number
then let a^√2 = 2
therefore, a=2^(1/√2) = √2^√2
If √2^√2 is irrational, then we are done.
If √2^√2 is rational, then let a=√2 and b=√2 and we are done, since √2^√2 is rational.
(It just so happens that √2^√2 is irrational, but we dont need to prove that to prove the conjecture.)

Answer 2

enable b > 0 be irrational and placed f(x) = x^b for x >= 0. Then f is non-end, increasing and assumes each and every fee - so each and every irrational - in [0, oo). those irrationals type an uncountable set. because of the fact the set {x^b | x is rational} is countable, it would not incorporate all irrationals of [0, oo). for this reason, for some irrational a, a^b might desire to be irrational. i think of the classic Greeks knew cardinality.

Answer 3

Euler showed that e^(i x pi) = -1

Answer 4

A= i = sqrt(-1)
B= i = sqrt(-1)
i ^ i = e ^ (-pi/2) = .20788