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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?

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    $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Apr 14 '10 at 18:27
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    $\begingroup$ 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. $\endgroup$ – Felipe Voloch Apr 14 '10 at 20:59
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    $\begingroup$ @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. $\endgroup$ – Barry Apr 14 '10 at 22:51
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    $\begingroup$ 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 $\endgroup$ – Will Jagy Apr 15 '10 at 1:56
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    $\begingroup$ I think it's awesome that a sentence can start with, "Second, $\pi ^2/6 = \prod _p (1 - 1/p^2)^{-1}$..." $\endgroup$ – Amit Kumar Gupta Dec 4 '10 at 10:16
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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.

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    $\begingroup$ This may be the most interesting and surprising answer I have read on MO so far! :-D $\endgroup$ – Andrea Ferretti Apr 15 '10 at 16:12
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    $\begingroup$ "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." It sure is, in algebraic geometry. Fantastic answer! $\endgroup$ – Allen Knutson Aug 7 '10 at 1:49
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    $\begingroup$ 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. $\endgroup$ – Gerald Edgar Dec 4 '10 at 1:01
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    $\begingroup$ There is no a priori reason to assume that IOpen can define $\pi$, just like there is no reason to assume it can define, say, Lebesgue measure. The theory is simply too weak for the question to be meaningful. $\endgroup$ – Emil Jeřábek Feb 22 '11 at 14:05
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    $\begingroup$ (That is precisely the conclusion of my post.) $\endgroup$ – François G. Dorais Feb 22 '11 at 17:19
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Barry, this post is coming a "little bit" after your class ended, but I want to direct you to a paper on the Lindemann-Weierstrass theorem by Beukers, Bezivin, and Robba in the Amer. Math. Monthly in 1990: https://www.jstor.org/stable/pdf/2324683.pdf. They give a proof of LW theorem that does not explicitly involve a choice of an auxiliary prime number, and it would be a good exercise to work out what the proof is saying in the special case of the number $\pi$ to see more directly how it proves $\pi$ is transcendental, or more generally how it shows $e^\alpha$ is transcendental when $\alpha$ is a nonzero algebraic number (the Lindemann part of LW) without dealing with the more general setting of a finite set of algebraic $\alpha_i$ (the Weierstrass part of LW). As indicated at the start of the paper by Beukers et al., the proof was inspired by ideas from $p$-adic analysis, but their paper is a simplified version that does not have any explicit dependence on $p$-adic notions.

I don't agree with the post claiming infinitude of primes "might be necessary" to prove transcendence of $\pi$. Just because you are working within an axiomatic framework that would allow you to prove infinitude of primes doesn't mean you ought to be required to do that for what you want, so that whole matter seems irrelevant to answering your question. If I am proving the Eisenstein irreducibility criterion in $\mathbf Z[x]$ you can't seriously tell me that I must go on a detour in the middle of the proof and prove $\mathbf Z[i]$ is a Euclidean domain just because it is logically possible with the axioms I am allowing. In any case, I think the paper I linked to should settle your question in an affirmative way that you were seeking.

UPDATE: The impression that you need infinitely many primes to prove transcendence results is incorrect, not because there are proofs that avoid it like the comparatively recent proof I mention above or older proofs I did not mention, but because even the proofs that mention it don't really need it. For example, look at the two proofs of transcendence of $e$ and $\pi$ here (pp. 3-5) and here. (In case those links change in the future, the proofs are essentially the same as those on pp. 4-6 of Baker's book "Transcendental Number Theory".) The way the proofs work is that an expression $J_p$ is introduced that depends on a prime $p$ (and some fixed data like a hypothetical algebraic relation for the number to be proved transcendental) and $J_p$ is shown to satisfy two properties: (i) a growth estimate $|J_p| \leq c^p$, where $c$ is independent of $p$ and (ii) $J_p$ is a sum of two integers, one being a multiple of $p!$ and the other being a multiple of $(p-1)!$ that for all large prime $p$ is not a multiple of $p!$, so $J_p$ is an integer multiple of $(p-1)$! that is not zero. Therefore $|J_p| \geq (p-1)!$, which contradicts the growth estimate in (i) since we can't have $(m-1)! \leq c^m$ for large $m$. How is the primality of $p$ really used? Its only purpose is to be an arbitrarily large number that is relatively prime to a couple of specific nonzero integers. But there is no need to rely on prime numbers to achieve that: if you want a huge integer $m$ relatively prime to two nonzero integers $a$ and $b$, just take $m = |ab|M + 1$ for a huge integer $M$. So you can work with numbers in such an arithmetic progression in place of a large prime number $p$ in the proof. (Personally I'd also pass to the integer $J_m/(m-1)!$ so the bounds $0 < |J_m/(m-1)!| \leq c^m/(m-1)!$ lead to the "no integers between $0$ and $1$" contradiction once $c^m/(m-1)!< 1$, but that's a matter of taste.)

Of course the idea of creating numbers relatively prime to particular numbers by taking a multiple of their product and adding $1$ is the main idea in Euclid's proof of the infinitude of the primes, so we aren't bypassing a method that can lead to infinitely many primes but we are bypassing any need for prime numbers in the proof that uses them.

The idea of using a large prime in proofs of the transcendence of $e$ and $\pi$ is due to Hurwitz, who added this detail to a proof by Hilbert that relies on divisibility properties but not on primes. Mahler discusses this in his Springer Lecture Notes in Math "Lectures on Transcendental Numbers" (1976). The volume is available online here, and while most chapters are behind a paywall the last part called Back Matter is not, and that's the part relevant to us. Mahler presents Hilbert's proof starting on p. 237. On p. 243 he mentions Hurwitz's idea of using a prime instead of a number in a simply constructed arithmetic progression and explains why he thinks this "seemingly simpler" approach of Hurwitz is actually not a good idea.

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    $\begingroup$ While I agree with what you say in your second paragraph, you may be misunderstanding Francois' claim. The "might be necessary" claim is in terms of reverse mathematics, not in terms of the words used in any given proof. In other words, any proof of the transcendence of $\pi$ might require the proof system to also contain as a true statement the infinitude of primes. [From what Francois says in the comments to his post, he may believe this is true only because $\pi$ is difficult to even define without having enough power to prove the infinitude of primes.] $\endgroup$ – Pace Nielsen Nov 14 '16 at 19:37
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    $\begingroup$ @PaceNielsen I understood that was his intent, but I feel that does not address the purpose of the original question. $\endgroup$ – KConrad Nov 14 '16 at 20:34
  • $\begingroup$ Another way to put your point is this: Let $P$ = "There are infinitely many primes". Then $P$ has the character of a theorem rather than an axiom. So even if (1) there is some weak base theory $T$ that doesn't prove $P$ and (2) there is some axiom $A$ such that $T+A$ does prove $P$ and (3) the transcendence of $\pi$ implies $A$ (and therefore $P$) over $T$, it doesn't follow that any proof of the transcendence of $\pi$ in $T+A$ must take a detour through $P$. $\endgroup$ – Timothy Chow Nov 14 '16 at 21:12
  • $\begingroup$ @TimothyChow But doesn't every statement $P$ have the property that it can be thought of as a theorem, rather than an axiom, in some theory $T$? For example, if we are proving the Jordan curve theorem using the Brouwer fixed point theorem (over the theory ${\bf RCA}_0$), we don't need to make a detour through the fact that countable commutative rings have prime ideals. Nevertheless, that fact is equivalent (over ${\bf RCA}_0$) to both the Jordan curve theorem and the Brouwer fixed point theorem. $\endgroup$ – Pace Nielsen Nov 14 '16 at 21:31
  • $\begingroup$ @KConrad As I said above to Timothy, any specific statement (or even axiom) can be avoided in an alternate proof by using different axioms. If Barry's question was simply "Can I prove it without ever referencing primes?" the answer is going to be (trivially) yes. The real question in my opinion is "Are the primes hiding somewhere?" But I suppose the original question might also be talking about "naturality" which is harder to quantify. $\endgroup$ – Pace Nielsen Nov 14 '16 at 21:39
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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.

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  • $\begingroup$ 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. $\endgroup$ – Barry Apr 14 '10 at 19:20
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    $\begingroup$ 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)! $\endgroup$ – Victor Miller Apr 15 '10 at 2:39
  • $\begingroup$ That's a nice paper, Victor, and not an area I knew about. $\endgroup$ – Will Jagy Apr 15 '10 at 3:56
  • $\begingroup$ Yes, an entertaining paper. $\endgroup$ – Barry Apr 15 '10 at 15:23
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    $\begingroup$ 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... $\endgroup$ – Wadim Zudilin Apr 16 '10 at 10:45
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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).

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  • $\begingroup$ 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. $\endgroup$ – Li Zhou Dec 4 '10 at 16:20
  • $\begingroup$ Your recursive irrationality proofs are nice. I have one question about something that surprised me in your arXiv post (arxiv.org/abs/0911.1933) and published paper (jstor.org/stable/10.4169/…): the author names are not in alphabetical order. That is rather unusual for a math paper. Why were the author names not alphabetized? $\endgroup$ – KConrad Nov 13 '16 at 18:44

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