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Thu, 26 Jan 2006
More irrational numbers
As long as I am on the subject, undergraduates are sometimes asked whether there are irrational numbers a and b such that a^{b} is rational. It's easy to prove that there are. First, consider a = b = √2. If √2^{√2} is rational, then we are done. Otherwise, take a = √2^{√2} and b = √2. Both are irrational, but a^{b} = 2. This is also a standard example of a non-constructive proof: it demonstrates conclusively that the numbers in question exist, but it does not tell you which of the two constructed pairs is actually the one that is wanted. Pinning down the real answer is tricky. The Gelfond-Schneider theorem establishes that it is in fact the second pair, as one would expect.
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