Question

Theorem: there are irrational a and b such that a^b is rational. Is the following proof correct?
(note: use the fact that √(2) is irrational)

Proof: either (√(2))^√(2) is rational or irrational.
Case 1: suppose (√(2))^√(2) is rational. Let a = b = √(2), a^b = (√(2))^√(2) is rational by assumption

Case 2: suppose (√(2))^√(2) is irrational. Let a = (√(2))^√(2), and b = √(2). Then a^b = [(√(2))^√(2)]^√(2) = [√(2)]^2 = 2, which is rational.

Answer 1

I think this is like a circular reason.
In second case, (√(2))^√(2) is assumed to be rational but then the conluclusion says it's rational.

In the first case it's assumed (√(2))^√(2) is rational, but this is the conclusion of the theorem.

Answer 2

It's not circular reasoning.

Answer 3

The second case assumes (√(2))^√(2) is irrational.

Answer 4

i think i was trying to prove sqrt(2)^sqrt(2) is rational, which is not what the theorem says :D

Answer 5

The theorem says \(\exists a,b \in \mathbb R \setminus \mathbb Q \quad a^b \in \mathbb Q\).
This is much different than \(\forall a,b \in \mathbb R \setminus \mathbb Q \quad a^b \in \mathbb Q\).