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Show that there exist a rational number a and an irrational number b such that a^b is irrational.

User Ramanr
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Answer:

In explanation below.

Explanation:

Presumably, the proof you have in mind is to use a=b=2–√a=b=2 if 2–√2√22 is rational, and otherwise use a=2–√2√a=22 and b=2–√b=2. The non-constructivity here is that, unless you know some deeper number theory than just irrationality of 2–√2, you won't know which of the two cases in the proof actually occurs, so you won't be able to give aa explicitly, say by writing a decimal approximation.

User Genero
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