Transcendence of PI

Question

Can anyone suggest me an ingenious proof of the transcendence of $\pi$. I have seen Lindemann's proof but it appears intricate.

Answer 1

There is a very nice book, "Irrational Numbers" by Ivan Niven. Available in paperback from the M.A.A. Evidently he gives a proof in the M.A.A.'s American Mathematical Monthly, volume 46 (1939) pages 469-471. His comment in the notes for chapter 9 of the book has "Proofs of the transcendence of $e$ and $\pi$ are not so difficult as the proof of the more general Theorem 9.1" And his 9.1 is indeed Hermite-Lindemann-Weierstrass. $$ $$ See also Proof that pi is transcendental that doesn't use the infinitude of primes which had a specific emphasis.

Answer 2

I have never seen a proof of the transcendence of $\pi$ which is anything but ingenious: that's rather the problem. I gather you want a proof which is as "transparent" as possible.

So let me mention a highly recommended book with that as its stated goal: Making Transcendence Transparent by Burger and Tubbs. For instance, this book received a rave review from (sometime MOer and full-time arithmetic geometer) Álvaro Lozano-Robledo on the MAA website:

http://www.maa.org/reviews/brief_feb06.html

I had a little trouble electronically searching for "transcendence of $\pi$" in this book. Rather, they view that result as a special case of the Lindemann-Weierstrass theorem, which seems like a good way to go.

Let me disclose that I have not read the book myself, but it is where I would start if I wanted to gain a better understanding of these issues.

Answer 3

An unfortunate thing is that proving the transcendence of $\pi$ has the same complexity as the Lindemann--Weierstrass theorem. (Proving the transcendence of $e$ is much easier.) I am quite surprised to see that the book [A.I. Galochkin, Yu.N. Nesterenko, and A.B. Shidlovskiĭ, Введение в теорию чисел [Introduction to number theory], 2nd edition, Moscow State Univ., 1995. 160 pp. ISBN: 5-211-03075-3. MR1367734 (96i:11001)] is not translated into English (I remember that there was an attempt to negotiate the translation about 10 years ago). I put here an extract from Yu. Nesterenko's lectures (in Russian!) on the Lindemann--Weierstrass theorem which are "cleaned", so I believe this is the simplest version of proof.

There are more exotic ways to state the transcendence of $\pi$, of course, these proofs are more involved. One example in this direction is [V.N. Sorokin, On the measure of transcendency of the number $\pi^2$, Sb. Math. 187 (1996), no. 12, 1819--1852] where the author shows the transcendence of $\pi^2$ (rather than $\pi$ itself) by constructing linear forms involving the numbers $$ \sum_{n_1\ge n_2\ge\dots\ge n_r\ge1}\frac1{n_1^2n_2^2\cdots n_r^2} $$ which are (now) known to be simple rational multiple of $\pi^{2r}$, $r=1,2,\dots$.