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

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Find derivatives using the product rule II

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Q. Show that there exist irrational numbers

x,y

for which

x^y

is rational

Consider $\sqrt{2}$ : Let's consider the number $\sqrt{2}$ , which is known to be irrational. We want to find another irrational number $y$ such that $(\sqrt{2})^y$ is rational.
Choose irrational number: Let's take $y$ to be the irrational number $\log_{2}(\sqrt{2})$ . The logarithm of a number to its own base is $\frac{1}{2}$ , which is rational, but the expression $\log_{2}(\sqrt{2})$ itself is irrational.
Calculate $(\sqrt{2})^y$ : Now, let's calculate $(\sqrt{2})^y$ using the value of $y$ we have chosen. We have $y = \log_2(\sqrt{2})$ , so $(\sqrt{2})^y = (\sqrt{2})^{\log_2(\sqrt{2})}$ .
Apply property of exponents: Using the property of exponents and logarithms, we know that $a^{\log_b(a)} = b$ . Applying this property to our expression, we get $(\sqrt{2})^{\log_{2}(\sqrt{2})} = \sqrt{2}^{\frac{1}{2}}$ .
Simplify expression: Simplifying $\sqrt{2}^{\frac{1}{2}}$ , we get $(\sqrt{2})^{\frac{1}{2}} = (\sqrt{2})^{\frac{2}{4}} = (\sqrt{2^2})^{\frac{1}{4}} = 2^{\frac{1}{4}}$ .
Reach contradiction: Since $2^{\frac{1}{4}}$ is simply the fourth root of $2$ , which is $\sqrt{2}$ , we have shown that our initial irrational number $\sqrt{2}$ raised to the power of another irrational number $\log_{2}(\sqrt{2})$ results in $\sqrt{2}$ , which is irrational. This is a contradiction to our goal, so we need to reconsider our approach.

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