The Universe of Disco

More irrational numbers
Gaal Yahas has written in with a delightfully simple proof that a particular number is irrational. Let x = log₂ 3; that is, such that 2^x = 3. If x is rational, then we have 2^a/b = 3 and 2^a = 3^b, where a and b are integers. But the left side is even and the right side is odd, so there are no such integers, and x must be irrational.

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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