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

Consider √2: Let's consider the irrational number √2. We know that √2 is irrational.Consider (√2)^(√2): Now let's consider another irrational number, which is (√2)^(√2). It's not immediately clear whether this number is rational or irrational.Check rationality: If (√2)^(√2) is rational, then we have found our x and y: x = √2 and y = √2, and x^y = (√2)^(√2) is rational.Set x and y: If (√2)^(√2) is irrational, then let's set x = (√2)^(√2) and y = √2. Now we will calculate x^y.Calculate x^y: Calculating x^y gives us ((√2)^(√2))^(√2) = (√2)^(√2 * √2) = (√2)^2.Simplify result: Simplifying (√2)^2 gives us 2, which is a rational number.

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

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Q. show there exist irrational numbers

x, y

for which

x^y

is rational

Consider $\sqrt{2}$ : Let's consider the irrational number $\sqrt{2}$ . We know that $\sqrt{2}$ is irrational.
Consider $(\sqrt{2})^{(\sqrt{2})}$ : Now let's consider another irrational number, which is $(\sqrt{2})^{(\sqrt{2})}$ . It's not immediately clear whether this number is rational or irrational.
Check rationality: If $(\sqrt{2})^{(\sqrt{2})}$ is rational, then we have found our $x$ and $y$ : $x = \sqrt{2}$ and $y = \sqrt{2}$ , and $x^y = (\sqrt{2})^{(\sqrt{2})}$ is rational.
Set $x$ and $y$ : If $(\sqrt{2})^{(\sqrt{2})}$ is irrational, then let's set $x = (\sqrt{2})^{(\sqrt{2})}$ and $y = \sqrt{2}$ . Now we will calculate $x^y$ .
Calculate $x^y$ : Calculating $x^y$ gives us $((\sqrt{2})^{(\sqrt{2})})^{(\sqrt{2})} = (\sqrt{2})^{(\sqrt{2} \cdot \sqrt{2})} = (\sqrt{2})^2$ .
Simplify result: Simplifying $(\sqrt{2})^2$ gives us $2$ , which is a rational number.