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As Colin says, transcendence proofs tend to be hard except for numbers carefully constructed to make transcendence proofs easy :-). His mention of the (beautif
by gjm11 14y ago
As Colin says, transcendence proofs tend to be hard except for numbers carefully constructed to make transcendence proofs easy :-).
His mention of the (beautiful but difficult) Gelfond-Schneider theorem reminds me of the following much easier, much less deep but rather pretty observation.
Question: Is it possible to take an irrational number, raise it to an irrational power, and get a rational number?
Answer: Yes. Write s=sqrt(2) -- which is of course irrational -- and consider s^s. If it's rational then we're done. Otherwise, (s^s)^s is an irrational number to an irrational power; but it equals s^(s^2) = s^2 = 2, which is rational!
Thanks to Gelfond and Schneider, we know that actually s^s is not only irrational but transcendental.
(In 1919, the great mathematician David Hilbert mentioned three big unsolved problems in number theory: the Riemann Hypothesis, Fermat's "last theorem", and the transcendence of 2^sqrt(2). He said RH would probably be proved within a few years, FLT maybe in his lifetime, and the transcendence of 2^sqrt(2) probably not in the lifetime of anyone in his audience. In fact he got the order exactly wrong: the transcendence of 2^sqrt(2) was proved in 1930, FLT in 1995, and RH still isn't done despite its great importance.)
- tzs 14y agoAnother answer I've seen to the irrational^irrational question is sqrt(10)^log(4)=2, where log is the common logarithm. This avoids upsetting the constructivists.
- gjm11 14y agoBut upsetting the constructivists is the whole point! (That is: what's so cute about that proof is that it tells you that there's a solution to irrational^irrational=rational, and that it's one of two things, but it doesn't tell you which. Also, I think it's a little shorter and more elegant than using sqrt(10)^log10(4). Don't you?)