irrational to the power of irrational can be rational

Prove that there exists two irrational numbers \( a , b \) such that \( a^{b} \) is rational

## show

Consider \( a = \sqrt{2} \) and \( b = \sqrt{2} \), then there are two cases, either \( \sqrt{2}^{\sqrt{2}} \) is rational, and the proof is over, or \( \sqrt{2}^{\sqrt{2}} \) is irrational and in that case consider \( a = \sqrt{2}^{\sqrt{2}} \) and \( b = \sqrt{2} \), then \( a^{b} = \left ( \sqrt{2}^{\sqrt{2}} \right )^{\sqrt{2}} = \sqrt{2}^{\sqrt{2} \cdot \sqrt{2}} = \sqrt{2}^{2} = 2 \), so in any case we've found an \( a , b \) that are irrational such that \( a^{b} \) is rational.