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show there exist irrational numbers x,yx, y for which xyx^y is rational

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Full solution

Q. show there exist irrational numbers x,yx, y for which xyx^y is rational
  1. Consider 2\sqrt{2}: Let's consider the irrational number 2\sqrt{2}. We know that 2\sqrt{2} is irrational.
  2. Consider (2)(2)(\sqrt{2})^{(\sqrt{2})}: Now let's consider another irrational number, which is (2)(2)(\sqrt{2})^{(\sqrt{2})}. It's not immediately clear whether this number is rational or irrational.
  3. Check rationality: If (2)(2)(\sqrt{2})^{(\sqrt{2})} is rational, then we have found our xx and yy: x=2x = \sqrt{2} and y=2y = \sqrt{2}, and xy=(2)(2)x^y = (\sqrt{2})^{(\sqrt{2})} is rational.
  4. Set xx and yy: If (2)(2)(\sqrt{2})^{(\sqrt{2})} is irrational, then let's set x=(2)(2)x = (\sqrt{2})^{(\sqrt{2})} and y=2y = \sqrt{2}. Now we will calculate xyx^y.
  5. Calculate xyx^y: Calculating xyx^y gives us ((2)(2))(2)=(2)(2⋅2)=(2)2((\sqrt{2})^{(\sqrt{2})})^{(\sqrt{2})} = (\sqrt{2})^{(\sqrt{2} \cdot \sqrt{2})} = (\sqrt{2})^2.
  6. Simplify result: Simplifying (2)2(\sqrt{2})^2 gives us 22, which is a rational number.

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