Talk:Proof that π is irrational

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A shorter proof than all of the ones on the page

http://mathforum.org/kb/plaintext.jspa?messageID=1614449 — Preceding unsigned comment added by 98.155.236.135 (talk) 20:46, 14 March 2015 (UTC)[reply]

This link doesn't work. Does anyone have more info on that proof? George Albert Lee (talk) 19:14, 29 October 2020 (UTC)[reply]

Jones' proof

Jones's proof has been cited in two books: More Calculus of a Single Variable (Undergraduate Texts in Mathematics) 2014th Edition by Mercer and The Heart of Calculus: Explorations and Applications (Classroom Resource Materials) by Philip M. Anselone (Author), John W. Lee (Author). — Preceding unsigned comment added by 98.208.142.38 (talk) 00:59, 30 June 2016 (UTC)[reply]

It has been several years since all of these comments were made. In that time the Mathematical Monthly has not retracted, corrected, or commented on Jones' article. Zhou's article "On Discovering and Proving the Irrationality of Pi" has not been published.

Meantime Jones' second proof has been published by the Journal of College Mathematics. Does this mean anything? — Preceding unsigned comment added by 98.208.142.242 (talk) 19:02, 27 November 2014 (UTC)[reply]

It can be argued that Jones' proof is not sufficiently new or noteworthy to be included in the list of proofs. However, the simplicity of his proof and its geometric character will make the subject matter more approachable and enjoyable to a greater audience. It can also be argued that the proof (the first one) is really too much like Niven's proof. This, however, can be countered by saying that Jones' proof is much easier to understand than Niven's. It is shorter and it is arranged much differently; it argues in a different way without defining functions. I belief it is just a matter of time before those who have a vested interest in Niven and other proofs will have to give way to this new proof. The question to me is not if they will but when. I say now is a good time. I use it in my calculus course. He gave me his second proof and to me it makes for a very good application of complex calculus and motivates the transcendence of pi. Currently there is no good motivation for the technique.

Here is the proof again -- parking it here for purposes of discussion.

Assuming $\pi =p/q$ we note that $\sin {\text{ x}}$ and the quadratic $x(p-qx)$ will have the same shape, roots, and symmetry in the interval $[0,\pi ]$ . The function $f(x)\sin {\text{ x}}$ with $f(x)=x^{n}(p-qx)^{n}$ will share these properties. The maximum of this function occurs at $p/2q$ . We have

{\begin{aligned}0<\int _{0}^{\frac {p}{q}}f(x)\sin x{\text{ dx }}\leq {\frac {p}{q}}\left({\frac {p^{2}}{4q}}\right)^{n},\end{aligned}}

where the upper and lower bounds follow from the symmetry of the curve. Evaluating this integral using integration by parts (or tabular integration) gives

{\begin{aligned}\int _{0}^{\frac {p}{q}}f(x)\sin x{\text{ dx }}=\sum _{k=0}^{n}(-1)^{k}f^{(2k)}(\pi ,0),\end{aligned}}

where $f^{(2k)}(\pi ,0)=f^{(2k)}(\pi )+f^{(2k)}(0)$ . Using the symmetry of $f(x)$ we have $f^{(2k)}(0)=f^{(2k)}(\pi )$ . As $f(x)$ is a polynomial whose least power of $x$ is $n$ , the first $n$ derivatives evaluate to $0$ at $0$ . After this all derivatives have an $n!$ in all their coefficients. We conclude that $n!$ divides the sum in the last equation. As factorial growth exceeds polynomial growth we have a contradiction.

— Preceding unsigned comment added by 76.101.178.158 (talk • contribs) 14:22, 11 January 2011

Ok, you justified it last time, and nobody said anything. If nobody argues this time, I won't remove it again, as long as you make it clear that it's not related to the Jones in the infobox. --SarekOfVulcan (talk) 19:17, 15 January 2011 (UTC)[reply]

I've reverted the last couple of edits, because 999ers inserted his comments in the middle of the IP's. This is against Wikipedia guidelines, so please just leave comments at the end of the thread, or elsewhere as described by WP:TALK. Thanks.--SarekOfVulcan (talk) 14:38, 25 January 2011 (UTC)[reply]

Rebuttal of Jones' fallacious claims

Thanks to SarekOfVulcan for removing T. W. Jones' self-claimed proofs from the list. It's a most appropriate decision.

If only mathematicians read these pages, then there is no need for me to waste time to debunk Jones at all, because they can see easily the faulty logic in Jones' claims. However, Jones is trying to peddle his own misunderstandings in math and his faulty logic to mathematically unsophisticated 'greater audience' and 'calculus students' (more to satisfy his ego than to enlighten the students), even after I have explained these mistakes to him in great depths and details.

Jones' self-claimed 1st proof

1) It's not simpler, it's not more approachable, it's not more enjoyable, and it's not easier to understand than Niven's. It's the same as Niven's. It's deliberately made to look shorter by suppressing the details Niven took care to present. The only trivial difference is whether to present the proof with integration-by-parts or the equivalent product-rule. Even if this trivial difference really counts for anything, then Jones does not have any right to claim for credit, because partial integration has been used by many mathematicians ever since Hermite. For example, see Miklós Laczkovich, Conjecture and proof, MAA, 2001, pp. 6--7 , or Frits Beukers, A rational approach to pi, Nieuw Archief voor Wiskunde, Dec(2000). These other mathematicians have the good sense and math competency not to claim anything new, which shows that the size of one's ego is inversely proportional to one's math talent.

2) If Jones had just retyped Niven's proof, then it would have been much less harmful and offensive. He didn't understand what makes Niven's proof work (because Niven polished Hermite's work and removed the clue) and introduced, in the retyping, serious red herrings and logical fallacies. His claimed 'geometric character' is nothing but faulty logic. The shared symmetry between $f(x)=x^{n}(p-qx)^{n}$ and $\sin x$ has nothing to do with the proof. The lower and upper bounds of the integral are easy consequences of $0<f(x),0<\sin x\leq 1,x^{n}<(p/q)^{n},(p-qx)^{n}<p^{n}$ on the interval $(0,p/q)$ . It is very important to notice the irrelevance of the graphical similarity, because the students need to see how to generalize later to harder proofs (transcendence of $e$ and $\pi$ , irrationality of $\cos r$ for rational nonzero $r$ , etc.), where $\sin x$ can be replaced by non-symmetrical $e^{x},\cos x,e^{x}\cos x$ , etc., and $x,p-qx$ can have different powers. See Zhou's article for a detailed discussion of this and other red herrings in Jones' claims.

Jones' self-claimed second proof

There is a well-known proof of the transcendence of $\pi$ after Lindemann, Hurwitz, and Hilbert. Jones went to this standard proof and retyped it for the simplest case of linear equation (instead of a degree-n equation satisfied by $\pi$ ) and claimed it as 'his own' again. It doesn't motivate the transcendence of $\pi$ at all. Let me give the reader an analogy. Imagine a school kid who is taught the quadratic formula. He then plugs in three concrete numbers a=2, b=5, c=-4 into the formula and gets an answer. Can he claim that he came up with a 'new' and 'his very own' way of solving $2x^{2}+5x-4=0$ ? Can he claim that his plugging-in has motivated the formula? Of course not! To motivate the formula, he has to see how the formula could be sensibly discovered. Therefore, to motivate the proof of the transcendence of $\pi$ , we have to figure out how the ingenious integral could be sensibly discovered, instead of cheating on the masters' integral and taking it as a starting point for granted. There are of course good motivations for Hermite, Lindemann, and Hurwitz's techniques, but it takes hard work and math competency to understand. It's a typical trait of mathematical cranks (see Underwood Dudley's book for definition) to believe that if a problem is easy to understand, then its solution is easy to find (think of all the trisectors and circle-squarers); and that if a proof looks easy (like Niven's), then its motivation is also easy. This is what led Jones to use his unexplainable graphical sixth sense about the similarity between $x(p-qx)$ and $\sin x$ as the motivation for Niven's proof. He has neither the competency nor the patience to understand continued-fractions, recurrences, and Hermite's works to discover the real motivation (see Zhou's article).

It's of course forgivable for a mathematician to make honest mistakes and has logical lapses, so long as he acknowledges them once pointed out. But it's very much unacceptable and unethical to knowingly perpetuate falsehood. So 'I say now is a good time' to stop his nonsense. 999ers (talk) 18:56, 28 January 2011 (UTC)[reply]

The AMS does not agree that Jones' proof is bad

I've just checked the review of Jones' proof as given by the American Mathematical Society via its MathSciNet service. This review, just completed in the last week or so, does not indicate any errors in Jones' proof. Not all proofs are reviewed and sometimes those that are are given unfavorable reviews and sometimes errors are indicated. The fact that this article was reviewed and the fact that this review is not unfavorable and the fact that it does not indicate any errors to me means that two highly reputable mathematical organizations, the MAA and the AMS, do not agree with the assessments given above.

It is also apparent that the simplicity of Jones' proof will allow more people to appreciate the irrationality of pi. Presently the Wikipedia page is in my mathematical opinion appealing exclusively to professional mathematicians. Niven's original proof is half a page; his proof as given in the Wikipedia article is much longer and much more complicated; it delves into the esoterica. Wikipedia will be the main entry point for many amateur and student mathematicians and I believe Wikipedia by its very nature can and should serve both audiences. —Preceding unsigned comment added by 76.101.178.158 (talk) 13:09, 30 January 2011 (UTC)[reply]

Jones' new logical fallacy: Appeal_to_authority

Jones not only lacks basic logic and math competency, but also lacks simple knowledge about the math profession. All articles and notes from Amer. Math. MONTHLY are automatically included in the MR (Math Review at MathSciNet) database. Most of these reviews are just the author's abstract and publication data. It doesn't even mean anyone has read Jones' paper yet, much less that AMS is lending support to Jones' claims. If you think I'm wrong about this, then please contact MR and find out who reviewed your MONTHLY paper, and I'll be very glad and ready to discuss your fallacies with the reviewer. After all, we appeal to reason, not to authority in math. By the way, in our personal communications before, you also appealed to authority (some 'math PhD highly respected in the field'). Would you care to disclose this authority as well, so that I can have a similar discussion with him/her?

Now let me inform you the situation with MAA MONTHLY. The Editor of MONTHLY has always agreed that your paper is an exposition of Niven's proof, not a new proof. After my discussions with him, he further realized that your symmetry claim is irrelevant and a red herring. The only reason that MONTHLY hasn't published a correction yet is format and space. The Editor wants to limit my criticisms to one page for his Editor's Endnotes. I believe a paper so flawed (a rare blunder in the history of MONTHLY) requires a more detailed and careful refutation (see Zhou's article). There are backlogs with many math journals, and a criticism of your paper may need to wait for some months before getting published. So be patient. Meanwhile, it only makes you look more like a charlatan to claim that you have AMS and MAA's supports to rob Niven (ultimately Hermite).

I do agree that Niven's proof is presented too long at Wikipedia (although some parts are connections and motivations rather than proof), but it does not mean that any trivial editing of it should be claimed as someone's brand new proof. In fact, the integral proof should only belong to Hermite (see Zhou's article), and no one else has much left to claim (Cartwright never claimed anything for herself). The simplest way to present Hermite's proof is:

u_{n}(x)\sin x+v_{n}(x)\cos x={\frac {x^{2n+1}}{2^{n}n!}}\int _{0}^{1}(1-z^{2})^{n}\cos(xz)\,dz,

where $u_{n}(x),v_{n}(x)$ are polynomials in $x$ of integer coefficients and degrees at most $n$ . Letting $x=\pi /2$ completes the proof. The rest are all trivialities.

Jones doesn't have much right to criticize others' presentation of Niven's proof either. Niven's paper is half-page long. Jones stretches it to three pages in his MAA MONTHLY paper, with two and half of those three pages filled with confusions, superstitions (about the magic power of graphical similarity), red herrings, and logical fallacies. 'Many amateur and student mathematicians' will be seriously misled and deceived to believe that the nonsensical 'visual concept that the product of two symmetric functions is symmetric' is what makes Niven's proof work! 999ers (talk) 03:49, 31 January 2011 (UTC)[reply]

The AMS reviews are not as described

I have found a proof of the irrationaity of pi that the reviewers indicated had an error. They further said that the error could not be fixed. Not all articles are reviewed. To me if the article was as bad as indicated above -- in other people's opinion -- it would have been indicated in the review. What is the purpose of the review other than to inform the mathematical community about the article's value. However I will leave it to some other of the math community to post the proof, if they should see fit. —Preceding unsigned comment added by 169.139.115.67 (talk) 22:46, 1 February 2011 (UTC)[reply]

MathSciNet reply

I've contacted the MR Editor and got a reply on the MR review of Jones' paper (MR2662709): 'In actual fact, the paper was not sent out for review - the author's abstract was taken; when a correction to MONTHLY is published, it will be indexed in MathSciNet and linked to the listing of the paper.' I believe now that it's also appropriate to remove Jones' MONTHLY paper from 'Further reading', considering the potential deception done to a casual browser of the article page. 999ers (talk) 20:52, 4 February 2011 (UTC)[reply]

Done--SarekOfVulcan (talk) 21:13, 4 February 2011 (UTC)[reply]

Zhou's article referring to Jones' paper should also be removed. I don't really see why it was ever put in in the first place. It is just referring to an obviously bad paper. — Preceding unsigned comment added by Theonanda (talk • contribs) 18:06, 5 February 2011 (UTC)[reply]

Zhou's article is not just a criticism of Jones's article, it also contains a very good historical discussion of the origin and motivation of Hermite's work (which Niven polished but not referenced in 1947). Read it and you may find very interesting things and it may deepen your understandings on the topic. In fact, it was first posted in the 'Article' page on Dec. 3-5, 2010. You can see that before then, the page did not have the origin of Cartwright's exam problem, and had no mention of Hermite. After Zhou's reference was posted on Dec. 3-5, 2010, many insights, connections, and motivations between Lambert's, Hermite's, Cartwright's, and Niven's proofs came along. You may see good reasons for this from a careful reading of Zhou's article. 999ers (talk) 03:47, 6 February 2011 (UTC)[reply]

Also, Jones' paper is not 'obviously bad'. The logical fallacies had deceived the author himself and the MAA MONTHLY referees and Editor. They become 'obvious' only after Zhou's insightful analysis. It's like a magic trick which can be quite deceptive, but obvious after pointed out. Zhou's criticisms really poke Niven's proof more in depth (than ever before, with new variations of integrals and examples) to reveal what's really crucial and what's irrelevant. The reader can benefit much from these examples and comparisons. 999ers (talk) 15:32, 6 February 2011 (UTC)[reply]

Jones' second proof

This proof uses Euler's identity and the ideas of Hermite, Lindemann, Hurwitz, and Niven. Assume $\pi =m/n$ . Define

${\begin{aligned}f(z)=z^{p-1}(nz-mi)^{p}\end{aligned}}$

where $p$ is a prime; the function is a complex polynomial. Define $F(z)$ as the sum of the derivatives of $f(z)$ . Using Leibniz' formula we find that if $p>m$ , then $(p-1)!$ does divide the complex part of $F(0)$ , but $p$ does not divide the complex part of $F(0)$ . We also find that $p!$ divides the complex part of $F(\pi i)$ . We have then that the complex part of $F(0)+F(\pi i)$ is not zero. If the sum were zero then $p$ would have to divide $F(0)$ .

We use the function $e^{-z}F(z)$ next. We have

${\begin{aligned}{\frac {d}{dz}}\left(e^{-z}F(z)\right)=-e^{-z}F(z)+e^{-z}F'(z)=-e^{-z}f(z),\end{aligned}}$

as $F(z)-F'(z)=f(z)$ . Integrating this, we have

${\begin{aligned}\int _{0}^{\pi i}{\frac {d}{dz}}\left(e^{-z}F(z)\right)={\Bigr |}_{0}^{\pi i}\left(e^{-z}F(z)\right)=\int _{0}^{\pi i}e^{-z}f(z){\text{ dz}}.\end{aligned}}$

With some algebraic manipulation this gives ${\begin{aligned}e^{\pi i}F(0)=F(\pi i)+\int _{0}^{\pi i}e^{-z}f(z){\text{ dz}}.\end{aligned}}$

Adding $F(0)$ to both sides, we have

${\begin{aligned}0=F(0)(e^{\pi i}+1)=F(0)+F(\pi i)+\int _{0}^{\pi i}e^{-z}f(z){\text{ dz}},\end{aligned}}$

where we have used $e^{\pi i}+1=0$ , Euler's identity. The complex part of $F(0)+F(\pi i)$ remains a non zero upon division by $(p-1)!$ . As the integral goes to $0$ with this division, we have a contradiction.

This proof generalizes to all powers of $\pi$ and also yields the transcendence of $\pi$ . —Preceding unsigned comment added by 76.101.178.158 (talk) 13:23, 1 March 2011 (UTC)[reply]

Pi is not irrational

Pi is not irrational. $π$ is commonly defined as the ratio of a circle's circumference $x$ to its diameter $y$ :^[1] The ratio $x / y$ is constant, regardless of the circle's size. For example, if a circle has twice the diameter of another circle it will also have twice the circumference, preserving the ratio $x / y$ . This proves that pi=x/y.

\pi ={\frac {x}{y}}

Base pi: 3.14₁₀=10₃.₁₄₁₆ — Preceding unsigned comment added by 83.28.62.68 (talk • contribs)

A rational number is one that can be expressed as a ratio of two integers. Pi cannot be thus expressed i.e. your x and y cannot both be integers. This is a mathematical fallacy.BethNaught (talk) 11:42, 18 May 2014 (UTC)[reply]

But ((-1/2)!)^2 is rational, which is pi. — Preceding unsigned comment added by 83.24.97.120 (talk) 08:51, 15 June 2014 (UTC)[reply]

References

^ Arndt & Haenel 2006, p. 8

You failed to prove that x and y are integers (any number with no decimal, ie 1, 42, -9, 0, etc.). A rational numbers is a number that can be expressed as x/y **where x and y are integers**, not just any x and y since if any x and y were allowed every number would be a rational number. You have only shown that pi is a number

Because x and y come from geometric lengths we can know that they are non-negative real numbers but we do not know if they are able to be expressed as integers or not JGHFunRun (talk) 22:24, 5 December 2022 (UTC)[reply]

And no, ((-1/2)!)² is not rational either. Please tell me where you got that information JGHFunRun (talk) 22:25, 5 December 2022 (UTC)[reply]

Notation

Does $n!!$ mean the double factorial or $(n!)!$ ? 95.49.72.119 (talk) 17:33, 5 March 2020 (UTC)[reply]

[Arndt-1] Arndt & Haenel 2006, p. 8

[1]

A shorter proof than all of the ones on the page

Category

Jones' proof

Pi is not irrational

Notation