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

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

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

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

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