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This is a nice proof showcasing an important idea, but for this particular theorem I prefer the beautiful constructive proof found by Liouville. It can be sketc

by gjm11 14y ago

This is a nice proof showcasing an important idea, but for this particular theorem I prefer the beautiful constructive proof found by Liouville. It can be sketched in a few paragraphs, so let me do that. 0. The basic idea is to show that an algebraic number -- i.e., one that's a root of a polynomial with integer coefficients -- can't be very well approximated by rational numbers ... and then to construct a number that can be very well approximated by rational numbers. 1. Suppose you've got a polynomial p, of degree d, with integer coefficients. Let x be one of its roots, and let a/b be any rational number. How close can it be to x? Well, let's ask a slightly different question: how close can p(a/b) be to 0? Answer: it's always (some integer)/b^d, and unless x actually equals a/b that integer isn't zero, so is >=1 or <=-1; so abs(poly(rational)) >= 1/denominator^degree. 2. OK, so how close can a/b be to x? Well, roughly speaking the ratio between p(a/b)-0 (which we know isn't too small) and a/b-x (which is what we care about) is close to p'(x), provided a/b is close to x. So (after filling in the details) we find that: if x is a root of an integer polynomial of degree d then there's a constant C such that every rational number has abs(rational-x) >= C denominator^d. 3. Now let's construct a number that is better-approximated by rationals than that allows. Specifically, take a decimal that looks like 0.110001000... where the spacing between the 1s grows very fast. The traditional thing is to take 10^-1! + 10^-2! + 10^-3! + ... so let's do that. Then truncating after 10^-k! leaves an error that's approximately 10^-(k+1)!; in other words we have a rational number with denominator N=10^k! and an error on the order of N^(k+1), so for any integer polynomial of degree d, once k gets somewhat larger than d this approximation is "too good", so this number can't be a root of that polynomial. So it can't be the root of any integer polynomial; in other words, it's transcendental.

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.)
ColinWright 14y ago
That is beautiful and elegant. It's also powerful. The article submitted here easily shows that in any interval on the real line there is at least one transcendental. Your proof here shows that you can create a transcendental "close" to any given rational. Both have a power beyond just "transcendentals exist". The reason I like the one submitted is that it's more visual and less algebraic, hence easier to explain to those less comfortable with polynomials and the like, but certainly both have their merits. Thanks for showing us this one.
- Evbn 14y ago
  What is the difference between "any interval" and "close to a given rational"? Rationals are dense, after all.
  - ColinWright 14y ago
    The difference is in the way it's being described, and how that description feeds in to the subsequent definition and proof. The submitted item says: take an interval, here's how to construct a transcendental inside your chosen interval. The proof that gjm11 gives says: Here's a proof that if something is unreasonably close to a rational then it's not algebraic, and here's how to construct something like that. The concept of "unreasonably close" then has to be expressed in terms of sizes of denominators, and hence is a relative term. That forms a key part of the proof.

by gjm11 14y ago

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.)

ColinWright 14y ago

That is beautiful and elegant. It's also powerful. The article submitted here easily shows that in any interval on the real line there is at least one transcendental. Your proof here shows that you can create a transcendental "close" to any given rational. Both have a power beyond just "transcendentals exist". The reason I like the one submitted is that it's more visual and less algebraic, hence easier to explain to those less comfortable with polynomials and the like, but certainly both have their merits. Thanks for showing us this one.

Evbn 14y ago
What is the difference between "any interval" and "close to a given rational"? Rationals are dense, after all.
- ColinWright 14y ago
  The difference is in the way it's being described, and how that description feeds in to the subsequent definition and proof. The submitted item says: take an interval, here's how to construct a transcendental inside your chosen interval. The proof that gjm11 gives says: Here's a proof that if something is unreasonably close to a rational then it's not algebraic, and here's how to construct something like that. The concept of "unreasonably close" then has to be expressed in terms of sizes of denominators, and hence is a relative term. That forms a key part of the proof.