Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I just taught the classical impossible constructions for the first time, and in finding my class a reference for the transcendence of pi, I found a dearth of distinct proofs. In particular, those that I read all require the existence of infinitely many primes, which strikes me as extraneous. Is there a known proof that requires only knowledge that I would "expect", namely, integral calculus to get your hands on the actual constant and algebraic properties of polynomials in connection with the assumption that the constant is algebraic?

share|improve this question
3  
The proof that there are infinitely many primes is both beautiful and easy---why avoid it? It seems especially appropriate if your students haven't seen it yet. –  Joel David Hamkins Apr 14 '10 at 18:27
46  
Two comments. First, transcendence theory as is commonly understood (incl. the transcendence of $\pi$) is number theory. Second, $\pi^2/6 = \prod_p (1-1/p^2)^{-1}$ so the irrationality of $\pi^2$ implies the infinitude of primes. –  Felipe Voloch Apr 14 '10 at 20:59
4  
@Vladimir and Felipe: We all lump algebraic/transcendental numbers under the number theory umbrella, but I do not accept that every pair of results under that umbrella must be seen as entwined. Example from algebra: Elliptic curves are related to the j-function, which are related to the monster group. If I said I was shocked that there was a connection between elliptic curves and the monster group, someone could say that they both fall under "algebra". I do not believe your comments are quite this extreme, but I still don't believe that saying "both are number theory" is a strong point. –  Barry Apr 14 '10 at 22:51
4  
Bit of a coincidence, the April 2010 M.A.A. Monthly has short proofs that $\zeta(2) = \pi^2 / 6,$ pages 352-353, and that $\pi^2$ is irrational, pages 360-362 –  Will Jagy Apr 15 '10 at 1:56
8  
I think it's awesome that a sentence can start with, "Second, $\pi ^2/6 = \prod _p (1 - 1/p^2)^{-1}$..." –  Amit Kumar Gupta Dec 4 '10 at 10:16
show 12 more comments

3 Answers 3

up vote 78 down vote accepted

The infinitude of primes (more precisely, the existence of arbitrarily large primes) might actually be necessary to prove the transcendence of $\pi$. As I explained in an earlier answer, there are structures which satisfy many axioms of arithmetic but fail to prove the unboundedness of primes or the existence of irrational numbers. Shepherdson presented a simple method for constructing such models, I will present such a model where $\pi$ is rational!

The Shepherdson integers $S$ consist of all Puiseux polynomials of the form $$a = a_0 + a_1T^{q_1} + \cdots + a_kT^{q_k}$$ where $0 < q_1 < \cdots < q_k$ are rationals, $a_0 \in \mathbb{Z}$, and $a_1,\dots,a_k \in \mathbb{R}$. This is a discrete ordered domain, where $a < b$ iff the most significant term of $b-a$ is positive; this corresponds to making $T$ infinitely large. This ring $S$ satisfies open induction axioms $$\phi(0) \land \forall x(\phi(x) \to \phi(x+1)) \to \forall x(x \geq 0 \to \phi(x))$$ where $\phi(x)$ is a quantifier free formula (possibly with parameters). So the ring $S$ satisfies the same basic axioms as $\mathbb{Z}$, but only a very limited amount of induction. In the field of fractions of $S$, $\pi$ is equal to the ratio $\pi T/T$. In other words, $\pi$ is a rational number!

Is $\pi T/T$ really $\pi$? The integers form a subring of $S$, and if $p,q \in \mathbb{Z}$ then $p/q < \pi T/T$ in $S$ if and only if $p/q < \pi$ in $\mathbb{R}$. So $\pi T/T$ defines the same Dedekind cut as $\pi$ does, which is a very accurate description of $\pi$. Indeed, any proof of the transcendence of $\pi$ must ultimately be based on the comparison of $\pi$ and its powers with certain rational numbers, which $\pi T/T$ will accomplish just as well as the real number $\pi$. However, the usual definitions of $\pi$ are not easily formalizable in this basic theory, so there is much room for debate here and I wouldn't claim that $\pi T/T$ satisfies all reasonable definitions of $\pi$. Shepherdson only presented this argument for real algebraic numbers like $\sqrt{2}$, which have a finitary description in this theory and leave little room for debate. In any case, the conclusion to draw from this is that basic arithmetic with open induction does not suffice to prove that $\pi$, or any other real number, is irrational (never mind transcendental).

What about primes? In the ring $S$, the only primes are the ones from $\mathbb{Z}$. Although there are infinitely many primes in $S$, it is not true that there are arbitrarily large primes. For example, there are no primes larger than $T$. Thus $S$ is a model where the unboundedness of primes fails and so does the irrationality of $\pi$. This only shows that basic arithmetic with open induction does not suffice to prove either result. A possible line of attack to show that the unboundedness of primes is necessary to prove the transcendence of $\pi$ would be to show that the minimum amount of induction necessary to prove that $\pi$ is transcendental also suffices to prove the unboundedness of primes. Unfortunately, I do not know how much induction is necessary to prove the transcendence of $\pi$. (And the minimum amount of induction necessary to prove the unboundedness of primes is still an open problem.)


Well, here is a partial answer, which is a bit of a bummer. There is another Shepherdson domain $S_0$ similar to the above where $\pi$ is transcendental over $S_0$ and $S_0$ does not have arbitrarily large primes. This shows that the transcendence of $\pi$ does not imply the unboundedness of primes over basic arithmetic with open induction. The ring $S_0$ is the subring of $S$ where the coefficients of the Puiseux polynomial are restricted to be algebraic numbers. The unboundedness of primes fails in $S_0$ because the real algebraic numbers form a real closed field just like $\mathbb{R}$. The number $\pi$ is transcendental over $S_0$ because it is transcendental over the field of real algebraic numbers.

This is not entirely surprising since open induction is a very weak base theory and the Shepherdson type rings are very pathological. To constrain such pathologies Van Den Dries suggested requiring that the domain is integrally closed in its field of fractions; he called such domains normal but I don't know if this is standard terminology. Neither $S$ nor $S_0$ are normal. More convincing examples would be normal discrete ordered domains. The methods of Macintyre and Marker (Primes and their residue rings in models of open induction, MR1001418) suggest that normal analogues of $S$ and $S_0$ might exist.

The conclusion that I draw from this is that open induction is probably too weak a base theory to study this question. Stronger base theories run into the difficulty that it is still not known just how little induction is necessary to prove the unboundedness of primes. The next reasonable candidate is bounded-quantifier induction (IΔ0), which is not known to imply the unboundedness of primes. Using the Euler product $\pi^2/6 = \prod_p (1-p^{-2})^{-1}$ looks promising, but so far I can only make sense of this product in IΔ0 + Exp which is known to prove the unboundedness of primes.

share|improve this answer
1  
As Felipe pointed out, the fact that $\pi^2 = 6\prod_p(1-p^{-2})^{-1}$ suggests that the transcendence of $\pi$ does imply the infinitude of primes over a relatively weak base theory. –  François G. Dorais Apr 14 '10 at 21:21
1  
Thank you François. I can only report that I got the gist of your entire response, but I take from it that you believe it possible that the infinitude of primes is necessary in the proof and have a potential explanation for this. If this is indeed the case, then I would find it fascinating. This has been bugging me the way needing analysis to prove the fundamental theorem of algebra bugs some people, but at least there, I understand why getting your hands on $\mathbb{C}$ without analysis is difficult. –  Barry Apr 14 '10 at 23:06
4  
This may be the most interesting and surprising answer I have read on MO so far! :-D –  Andrea Ferretti Apr 15 '10 at 16:12
5  
Dedekind cut, eh? Since this model is nonarchimedean, a cut corresponds to many different elements; some rational and some irrational. So you need a more convincing reason that this particular element is pi. –  Gerald Edgar Dec 4 '10 at 1:01
2  
(That is precisely the conclusion of my post.) –  François G. Dorais Feb 22 '11 at 17:19
show 11 more comments

I recommend the book Irrational Numbers by Ivan Niven, one of the M.A.A. Carus Monographs and available as a paperback. He proves irrationality of $\pi$ and $\pi^2$ much earlier, then shows that $\pi$ is also transcendental with the Lindemann theorem, chapter 9. I really like this book.

As you know from teaching your class, impossibility of the compass and straightedge constructions does not need anywhere near the full weight of transcendence, merely that the associated constant not lie in a tower of fields that expresses the idea of taking square roots, see

On using field extensions to prove the impossiblity of a straightedge and compass construction

or Appendix C in Galois Theory by Joseph Rotman, where he uses "only elementary field theory; no Galois theory is required." However, in line with your complaint, I should admit that I do not personally know of any proof that shows $\pi$ is not among the "constructible numbers" except for proofs of transcendence.

share|improve this answer
    
Thank you. I'll check out Niven's proof. Do you know if he uses the infinitude of primes? I'm guessing his proof is similar to that in his Monthly article, in which case he does. –  Barry Apr 14 '10 at 19:20
2  
I recommend the aritcle "A rational approach to Pi" by Frits Beukers staff.science.uu.nl/~beuke106/Pi-artikel.ps . Although this is concerning irrationality measures it occurred to me that it might be possible to prove transcendence of pi by showing that its irrational measure is $> 2$, and then using Roth's theorem (only half kidding)! –  Victor Miller Apr 15 '10 at 2:39
    
That's a nice paper, Victor, and not an area I knew about. –  Will Jagy Apr 15 '10 at 3:56
    
Yes, an entertaining paper. –  Barry Apr 15 '10 at 15:23
1  
Frits needs the estimate for the least common multiple of $1,2,\dots,n$. It's hard without primes. A proof of the irrationality of $\pi$ which I consider "most elementary" and which does not touch the primes in an obvious way is in Robert Breusch's note in the Amer. Math. Monthly 61 (1954) 631-632. The transcendence proofs for $\pi$ all require primes... –  Wadim Zudilin Apr 16 '10 at 10:45
add comment

My feeling is similar to Barry's: the infinitude of primes may not be necessary. For example, in Chapter 2 of Niven's Irrational Numbers, he also used the infinitude of primes to prove that cos(r) is irrational for nonzero rational r. But our recent proof (Monthly, April/2010, 360-362, mentioned by Will Jagy earlier) does not need this at all. By the way, our proof (half-page long) can replace more or less the entire Chapter 2 of Niven's book (except the transcendence of e).

share|improve this answer
    
I'm now more convinced that the infinitude of primes is not essential. It's used in the proof of the transcendence of pi for very similar purposes as in the proof of the irrationality of cos(r) by Niven, thus may be replaced by recurrences as in our Monthly paper (also available at arxiv.org/abs/0911.1933). The recurrences, however, will be very very messy. –  Li Zhou Dec 4 '10 at 16:20
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.