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This is a really nice proof, and I love this kind of "you could have discovered X if you'd just had this core idea!" kind of exposition. Perhaps a hard ask, bu

by mjw 14y ago

This is a really nice proof, and I love this kind of "you could have discovered X if you'd just had this core idea!" kind of exposition. Perhaps a hard ask, but do you know of any similarly explicable proofs for transcendental-ness of "well known" numbers?

ColinWright 14y ago
Results for specific known numbers such as e and pi are known to be hard to come by. http://en.wikipedia.org/wiki/Transcendental_number#Numbers_proved_to_be_transcendental http://en.wikipedia.org/wiki/Transcendental_number#Numbers_p... The main tools are the Gelfond-Schneider theorem and the Lindeman-Weierstrass theorem: http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theor... http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_t... gjm11 will know much more about these than I, and is better placed to offer insight into them.
e3pi 14y ago
Felix Klein's lecture made Lindemann's pi and Hermite's e transcendental proof accessible and interesting to German `HS' students, and was my own first persuasion. It was a Dover soft cover: Felix Klein's Famous Problems of Elementary Geometry, Dover, 1956.
tzs 14y ago
e can be done reasonably with material from a good first year college calculus course. Such a proof is given around page 437 of the 3rd edition of Spivak's "Calculus".
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 ago
  Another 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 ago
    But 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?)

by mjw 14y ago

ColinWright 14y ago

Results for specific known numbers such as e and pi are known to be hard to come by. http://en.wikipedia.org/wiki/Transcendental_number#Numbers_proved_to_be_transcendental http://en.wikipedia.org/wiki/Transcendental_number#Numbers_p... The main tools are the Gelfond-Schneider theorem and the Lindeman-Weierstrass theorem: http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theor... http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_t... gjm11 will know much more about these than I, and is better placed to offer insight into them.

e3pi 14y ago

Felix Klein's lecture made Lindemann's pi and Hermite's e transcendental proof accessible and interesting to German `HS' students, and was my own first persuasion. It was a Dover soft cover: Felix Klein's Famous Problems of Elementary Geometry, Dover, 1956.

tzs 14y ago

e can be done reasonably with material from a good first year college calculus course. Such a proof is given around page 437 of the 3rd edition of Spivak's "Calculus".

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 ago
Another 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 ago
  But 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?)