Math challenge: Prove there exist two irrational numbers such that?

Question

Prove that there exist two (not necessarily distinct) irrational numbers a and b such that a^b is rational. Please only use facts about irrationality know to the ancient Greeks. For example, the ancient greeks would not have known about logs.

Answer 1

We know √2 is irrational. The Greeks knew this as well so it's a valid starting point.

First let
a = √2
b = √2

Now (√2)^(√2) is either rational or irrational (it's irrational, but was not proven so till later I believe, so we must cover both cases). So proof by cases

case1: (√2)^(√2) is rational. Well if so we found a and b
q.e.d

case2: (√2)^(√2) is irrational.
well then
a = (√2)^(√2)
b = (√2)

Therefore
a^b = [(√2)^(√2)]^(√2)

Using properties of exponents
a^b = [(√2)^[(√2)(√2)]]
a^b = [(√2)^2]
a^b = 2
2 is rational
q.e.d

Therefore such an a and b exist. though without knowing whether (√2)^(√2) is irrational we don't know which two, just that they exist, which is all the problem asks for.

Answer 2

Let b > 0 be irrational and put f(x) = x^b for x >= 0. Then f is continuous, increasing and assumes every value - so every irrational - in [0, oo). These irrationals form an uncountable set. Since the set {x^b | x is rational} is countable, it doesn't contain all irrationals of [0, oo). Hence, for some irrational a, a^b must be irrational. I think the ancient Greeks knew cardinality.

Answer 3

You could let a=root[sqrt(2)](2) [The sqrt(2) root of 2] and b=sqrt(2). Then a^b=2. Though I am not sure if the ancient Greeks had any concept of irrational powers or roots.