Talk:Square root of 2/Archive 2

Archive 1

Archive 2

Constructive proof

The proof says 'Errett Bishop (1985)' I would like to see a proper citation including page number please. Does this proof use 'valuations' as used here? The reason I'd like to see that there is a specific proof there is that there are other simpler proofs here that can be easily changed to start with p/q and end up with that it can't be the square root of 2 which cuts out the business of the exclude middle. So I'd like to see that this specific method was used. Dmcq (talk) 07:15, 12 July 2010 (UTC)

OK, done. Which modification of which proof are you thinking about? Tkuvho (talk) 12:42, 12 July 2010 (UTC)

Any of them. Just take the infinite descent one to start with

Take any rational number a/b

reduce it to lowest terms.

compare a² and 2b², both are integers

show that a b can't be in lowest terms as in the proof if these integers are equal as then a would be even

therefore the two expressions are different integers

therefore they differ by at least 1

therefore a²/b² differs from 2 by at least 1/b²

therefore one can show a/b differs from sqrt(2) by at least 1/3b²

As to citations it is best to stick <ref>...</ref> around them at where they are referred to and then the citation gets stuck automatically in the references with a link. Please see WO:CITE or look through the source for some examples to copy. I looked up Bishop initially and found a different book by him in 1985 rather than what yoiu put in for 1985. I can't see the text but does it actually use valuations of p-adic numbers to prove the square root of 2 is irrational as shown? Dmcq (talk) 11:01, 13 July 2010 (UTC)

Your proof is fine. Just one remark: there is no need to reduce to lowest terms. The infinite descent proof as stated in the article differs from what you presented in two essential ways, which make it constructively suspect: (1) it relies on an argument by contradiction, hence on LEM; (2) it does not contain any explicit lower bound for the difference with a/b, hence not constructively convincing. Other than that, your proof is a good paraphrase of the constructive proof. Bishop does not use the term "valuation"; I used it because the page I linked it to discusses parity and defines the valuation in the context of p=2. Certainly there is no need to speak of p-adics here. Tkuvho (talk) 12:52, 13 July 2010 (UTC)

Clarification required

I would appreciate a clarification of WTF this is all about. It looks like petitio principii to me. To get your inequalities (standard looking continued fraction stuff sort thing) you need to assume in the first place that a^2 - 2*b^2 = 0 is not possible (but of course that begs the question) to pass to a^2 - 2*b^2 >= 1 to get the inequalities you arrive at but I don't call that 'direct' (which would be indeed be distinctly disconcerting). To be quite frank I call it extremely (and I suspect wilfully) stupid. I suggest the entire selection is deleted. Meanwhile I've inserted the nearest thing I can find to a WTF template. Rinpoche (talk) 04:54, 10 September 2010 (UTC)

I now understand what the contributor is expressing with his remark about valuations. He is using the Unique Factorisation Theorem to assert that the number of prime factor twos in p^2 is even whereas those in 2*q^2 is odd, a contradiction which establishes the integers are distinct (but that is nevertheless equivalent to saying the SQRT(2) is irrational even from a constructive point of view), and then uses this to prove a result, which in the context of real numbers would generally be regarded as stronger from a constructivist point of view, to show that p/q and SQRT(2) are 'apart'.

The proof still use contradiction and it is wrong to suppose that constructive proofs never use contradiction. Moreover 'apartness' is not a valid distinction comparing integers as there is a decidability criteria for testing equality. These web references discuss the issues 1 proofs that don't depend on contradiction 2 Proof of negation and proof by contradiction. They both assert that it is accepted that the classic proof of the irrationality of the SQRT(2) is in fact constructive.

The section doesn't attempt to describe what a contructive proof is or even link it to the Wiki entry (and which would indeed read very strangely to a newcomer as it describes a constructive proof as one which demonstrates the existence of a thing, but how can that be accomplished in a theorem which asserts the non-existence of a thing?)

Finally the reference provided is a relatively obscure one. While Errett Bishop is a noted mathematician I suspect the contributor has misunderstood the reference.

I suggest the section should be at the very least rewritten to discuss the issue of what a constructive proof of the irrationality of SQRT(2) really amounts to (and addressing the issue that the classical proof is regarded by some at least as constructive anyway) and presents the existing material more transparently with a less obscure reference.

I suggest further that the issue really isn't notable for this article and should be deleted, perhaps transferred to Constructive proof. Rinpoche (talk) 00:48, 11 September 2010 (UTC)

I agree with that, it was more a demonstration of constructive proof than anything new about proving the square root of 2 is irrational. Dmcq (talk) 18:21, 11 September 2010 (UTC)

David Corfield explained the issues there (at the first linke) well enough. Constructively speaking, applying LEM to an efffectively decidable predicate over Z is acceptable. That's why the 2p^2-q^2>0 step is OK. Tkuvho (talk) 20:01, 11 September 2010 (UTC)

but don't you agree that failing to conclude irrationality after demonstrating p^2-2*q^2>0 would be very puzzling for most readers and needs clarification? Is it in fact non-constructive to do so? I can't see that it really can be but the whole constructive thing is not an issue for me and I've never taken much of an interest in it. Most of all, and what is really at issue here, is whether the section (and certainly as it stands) really notable for the article. I think Dmcq is quite right to say it's more about constructive proofs than SQRT(2). I think it should go. Rinpoche (talk) 02:28, 12 September 2010 (UTC)

Is it puzzling to most readers that one does not stop after demonstrating that p^2-2*q^2>0 ? Certainly. As explained in detail at the page you listed above, at this stage you only know that sert{2} is not rational. To conclude that it is irrational, you need an explicit lower bound on the distance from a rational p/q, e.g. the bound 1/3q^2 currently mentioned here. Is it in fact non-constructive to conclude irrationality without the explicit lower bound? That depends on what you mean by "constructive". In Bishopian constructivism, considered the most influential branch of constructivism today, it is in fact not enough. I understand though what you say about this not being an issue for you. However, as a general rule, persecuting alternative viewpoints is unencyclopedic. Tkuvho (talk) 04:57, 12 September 2010 (UTC)

Oh well, I stand properly corrected I'm sure. I am obliged. Perhaps all these remarks would be better off in a page on logic at which I confess I'm indeed all at sixes and sevens and no ambition at all to make it eights and better. All I can say is that as the merest of fancier of things arithmetical I hadn't the faintest idea of what the section was on about at first sight and as a matter of fact still don't and that after some effort honest. If that's what being encylopedic is all about well then persecute and be damned I say. At least the section should surely explain to simple folk other than you like me that a distinction is made between not being rational and irrational. Ever helpful I've offered a provisional edit - by all means feel free to expand on it. Rinpoche (talk) 06:04, 12 September 2010 (UTC)

Before I found this thread on the talk page, I was thoroughly mystified by this section "Constructive proof". It turns out that Rinpoche has articulated some of my concerns, which are:

1. It begins In a constructive proof, one distinguishes between on the one hand not being rational,.... This sounds like all constructive proofs are about rationals and irrationals. Is that true? If not, it should say In a constructive proof of irrationality.... or In this constructive proof....

2. As Rinpoche points out above, the section doesn't attempt to describe what a constructive proof is as the term is used here; it now links to the Wiki entry constructive proof, which as Rinpoche said describes a constructive proof as one which demonstrates the existence of a thing, which is not done here. So as it stands it uses "constructive proof" one way and implicitly defines it (via wikilink) another way.

3. It uses the (to me unfamiliar, despite my often having used the concept) term "valuation" without giving a quick definition right here for the reader who wants to be able to understand with a minimal amount of clicking around. Given that it's easy to quickly define, I think it should be defined here.

4. It says that we are applying the law of trichotomy in the context of an effectively computable predicate over N. While "trichotomy" is linked, "effectively computable predicate" is neither linked (and there is no such article to link to) nor defined. For intelligibility to the reader who doesn't already know the material, I think "effectively computable predicate" should be defined.

5. It says An easy calculation then yields a lower bound of ${\frac {1}{3b^{2}}}$ for the difference $|{\sqrt {2}}-a/b|$. The calculation may be easy, but I don't see what it is -- if it's easy, it should be given here, since this is not a proof without a proof of this assertion. Duoduoduo (talk) 18:03, 15 March 2012 (UTC)

I tried. Tkuvho (talk) 18:16, 15 March 2012 (UTC)

Thanks very much!! I'll put in the simple derivation of the bound, along with a specific section reference in the source. Duoduoduo (talk) 20:31, 15 March 2012 (UTC)

New version

I propose a new version of the constructive proof, which looks clearer to me. See below:

Constructive proof

Let $a$ and $b$ be two positive non-zero integers. The constructive proof shows that there is a finite difference between ${\sqrt {2}}$ and the rational number $a/b$, without using reductio ad absurdum. Since ${\sqrt {2}}<2$ we can assume $a/b<2$, which implies ${\sqrt {2}}+a/b<4$. Moreover, since 2 divides $2b^{2}$ an odd number of times and $a^{2}$ an even number of times, $a^{2}$ and $2b^{2}$ must be distinct, i.e. $|2b^{2}-a^{2}|\geq 1$. It clearly follows that

$|{\sqrt {2}}-a/b|={\frac {|2b^{2}-a^{2}|}{b^{2}({\sqrt {2}}+a/b)}}\geq {\frac {1}{b^{2}({\sqrt {2}}+a/b)}}>{\frac {1}{4b^{2}}}.$

This proof constructively exhibits a discrepancy between ${\sqrt {2}}$ and any rational.

Superhiggs (talk) 15:49, 4 October 2012 (UTC)

When you have shown that $a^{2}$ and $2b^{2}$ must be distinct a simple piece of arithmetic manipulation shows that $(a/b)^{2}!=2$, for arbitrary a and b. Calculating limits on closeness adds nothing to the argument 2A00:23C5:6E00:DA00:2D78:1E49:71DD:E8ED (talk) 17:10, 19 February 2020 (UTC)

much simpler proof

the proof in the article seems overly complex, a much simpler proof would be

1. Assume that √2 is a rational number, meaning that there exists an integer a and an integer b such that a / b = √2.
2. Then √2 can be written as an irreducible fraction (the fraction is reduced as much as possible) a / b such that a and b are coprime integers and (a / b)² = 2.
3. Clearly √2 is not an integer so b > 1
4. If a and b are coprime then a² and b² must also be coprime since squaring introduces no new prime factors
5. therefore a²/b² must be an irriducible fraction with a denominator greater than 1.
6. therefore a²/b² cannot be 2 which contridicts the initial assumption

comments? Plugwash 22:31, 10 July 2007 (UTC)

I came here to make a similar comment. In fact, this approach actually works to prove a much more general result:

   The square root of an integer is rational if and only if that integer is a perfect square.

If a²/b² is an integer and a and b are relatively prime, then we must have b=1. That's it. Yet we are regularly subjected to a more complicated argument for a special case. Am I missing something? DrHow (talk) 18:07, 9 October 2018 (UTC)

Yes you're assuming that if a and b are coprime then a^2 and b^2 are coprime. Proving that takes a lot more work, it requires proving the Fundamental theorem of arithmetic. Assuming this for a more general system where it is in fact wrong led to an embarrassing incorrect proof of Fermat's lat theorem. Dmcq (talk) 23:02, 9 October 2018 (UTC)

Does this argument presuppose uniqueness of prime factorization, rather than only the simpler fact about divisibility by 2? Michael Hardy (talk) 01:05, 17 March 2008 (UTC)

I think you mean contradicts

Irreducible

DarkestMoonlight (talk) 19:37, 20 March 2008 (UTC)

I agree with this. The usual proof involves a discussion about a-squared being even and so a is even, thus a-squared is divisible by 4, so b is even, and so on, by infinite descent. May I offer the following: If a/b=c/d with both fractions irreducible, a and b are coprime, c and d are coprime. Then ad = bc Every factor of a must be a factor of c as it is not a factor of b. Every factor of c is also a factor of a, since c,d are coprime. Therefore a=c and b=d. So every rational has exactly one irreducible form. Proving irrationality of root2 is easy now: 2=2/1 = a-squared/ b-squared, an irreducible fraction. So b=1 and root2 = a, an integer, a contradiction. I cannot see why the Wikipedia gurus resist the inclusion of such a proof, when all the others are more complicated. Bill Dixon, Madrid, Spain BillDixon56 (talk) 06:27, 16 September 2020 (UTC)

May I add that both my attempts to include a proof like this in the main article have been motivated by an honest desire to show the result in a simple way. Editors who reverted them said it was not rigorous, without saying why, or resorted to some kind of procedural rulebook to just throw it out. Im not a Wikipedia expert and Im not a professional mathematician, but it looks OK to me and I'd be glad to know a substantive reason why it is not. Bill Dixon, Madrid, Spain BillDixon56 (talk) 07:23, 16 September 2020 (UTC)

Analytic proof

Shouldn't this section attempt to describe in the first place what an 'analytic' proof is (and for that matter address the nice question of whether the preceding proofs are or are not 'analytic' - whatever they are their dependence on well-ordering ensures they aren't merely algebraic)?

It's quite a substantial and I doubt particularly easy to follow discussion (I mean frankly I can't be arsed myself to read through it). The suspicion must arise it's either a plagiarism or original research and there's no reference provided either by way of peer review.

Who needs it? It's plainly out of place and I'm going to delete it when I return to edit this page unless it's deficiencies have been corrected. Rinpoche (talk) 15:15, 11 September 2010 (UTC)

It certainly looks like the usual idea of an analytic proof to me where they depend on inequalities and limits. The proof looks like a variant of the one described in Liouville number, there must be a proper name around somewhere for that as it is quite important.

I think this proof is an intrinsically interesting one, but I really would like citations like Wikipedia is supposed to provide for sections like that. Dmcq (talk) 18:14, 11 September 2010 (UTC)

Actually another tweak on Liouville's theorem gives a lower bound on the difference between any approximation and the square root of 2 like in the constructive proof. Dmcq (talk) 18:26, 11 September 2010 (UTC)

Well, I'll have a read through then (but it looks desperately tedious ... oh dear). Liouville numbers are about transcendental quanities, surely? Needles to say SQRT(2) isn't transcendental. It certainly needs a reference surely.

I've noticed that the 'geometric proof' is likewise desperately laboured. In general I don't come away with a very good impression of this article. To be plain I think it's rather weak. I do think much of it should be rewritten. However I don't want to tread on any one's toes. If the page is being actively maintained and the contributors are satisfied with their work then it's not for me to do more than point out what I consider it's deficiencies. Rinpoche (talk) 02:17, 12 September 2010 (UTC)

THe theorem provides a bound on how well one can approximate thee solution of an nth degree polynomial and proves the Lioville number is trancendental because it can be approximated better than any any such bound. The n=2 and n=1 is what applies here. Dmcq (talk) 09:02, 12 September 2010 (UTC)

I agree with Rinpoche: the proofs are laboured. 1. Students get confused by infinite descent. 2. The geometric proofs are not distinct since they merely substitute the computation of areas of figures for algebraic multiplication. 3. The analytic proofs depend on all sorts of things which are unnecessarily complicated or difficult to understand, like the Rational Roots theorem, monic numbers, etc. 4. The contradictions arrived at are rather subtle. Once it is proved that a rational has a unique lowest form, we then have a-squared/ b-squared = n/1, a fraction in its lowest form. Thus b= 1 and root-n = a, an integer. For n=2 this is a very clear contradiction. All the other stuff about whether a and b are even, or what their factors are, and the rigmarole of infinite descent becomes redundant. Bill Dixon, Madrid, Spain BillDixon56 (talk) 08:35, 16 September 2020 (UTC)

Proof by missed factorization

Let ${\sqrt {2}}$ be rational.

Then ${\sqrt {2}}={\frac {p}{q}}$, with $p,q$ coprime.

If ${\sqrt {2}}$ is rational so is ${\sqrt {2}}+1$ and ${\sqrt {2}}-1$

But then

$({\sqrt {2}}+1)({\sqrt {2}}-1)=1$

which is

$(p+q)(p-q)=q^{2}$

This is impossible as $p+q$ cannot divide $q^{2}$.

This proof does not require unique factorization. $p+q$ cannot divide $q^{2}$ because $p$ and $q$ are coprime.

The only thing the proof requires is $r|s\wedge (r+t)|s\implies t|s$

$q|q^{2}\wedge (p+q)|q^{2}\implies p|q^{2}$ and this is not possible as $p$ and $q$ are coprime.

For this last we need only Euclid's lemma in the formulation given on

If $n|ab$, and $n$ is relatively prime to $a$, then $n|b$.

If $p|q^{2}=qq$, and $p$ is relatively prime to $q$, then $p|q$, which it does not.

(Adjusting details for the comments bellow that think otherwise, otherwise the text becomes confusing.)

References

Do not delete what other people write on talk pages. What you wrote cannot be used in the article for the reasons I gave and you deleted. The unique factorization theorem is needed to say that if p and q are coprime then p does not divide q^2. Annd noo reliable source has been provided. And it is original research. Please see WP:5P2. Dmcq (talk) 22:35, 28 October 2018 (UTC)

Just for the record I am deleting this as some people think they are very "smart", No, to prove that if p does not divide q it does not divide q^2 you do not need unique factorization only Euclid´s lemma.— Preceding unsigned comment added by 51.175.86.30 (talk • contribs)

Refactored, regardless how smart 51.175.86.30 thinks of herself. Purgy (talk) 07:39, 29 October 2018 (UTC)

Euclid's lemma is the major step in the proof of the fundamental theorem of algebra, which is the unique factorization theorem. And you need the full theorem as your p and q never mind p-q and p+q are not necessarily primes. Please see WP:TPO about not removing other people's comments. Also see WP:TPNO in the same guideline about no personal attacks. In fact just read WP:5P about editing Wikipedia and cooperating with other editors. Dmcq (talk) 11:58, 29 October 2018 (UTC)

Can you just simple stop? Why are you trying to be right when you are not? I gave you the link of extended Euclid lemma that does not require primes? I removed your initial comment because it had nothing to do with the second version of the text? Why are there so many people ready to act as inquisition? Just stop! You are wrong. Period. You cannot defend it by me reading policy of any Web site. Jesus! While I am trying to make something out of the text you are trying to be annoying. And yes it is personal. You exactly you are very annoying.

Wikipedia is an encyclopaedia based on reliable sources, it is not a forum for people to expound their own thoughts. You don't have a WP:reliable source for what you say so it can't go in. There is no point in you continuing with this and you will eventually be blocked from editing Wikipedia if you continue like you're starting. Dmcq (talk) 19:28, 29 October 2018 (UTC)

Not a WP:forum. Dmcq (talk) 17:26, 4 November 2018 (UTC)

Block? First of all I cannot care less. If people from Wikipedia are going to behave like a police, why do you think that I would care? Go on block it. Reliable sources... You must be kidding me. You can say that you are trying, and failing miserably. Wikipedia is not based on reliable sources, neither anything like that is possible. I am not saying anything about trying but being cocky is nowhere close to reality of what Wikipedia truly is. And now what that I said it: Wikipedia is more or less a political tool for spreading a politically correct views at the moment of writing. Thousand articles are written and overwritten as soon as some political change happens in a couple of countries. Science is here only to create a fake moral ground for nothing more than spreading suitable lies, science is just giving it all a background so it all looks "reliable". That is what Wikipedia has become. For Wikipedia, gossiping article is of the same importance as recently published mathematical proof. That is what is making it all look "reliable". It has nothing to do with Wikipedia as it is. Any such global endeavor is doomed and will suffer the same fate. The main problem is that power is given to the people, just like administrators on this site, who have not a clue of what they are really doing. Who are these administrators? Nobody, just like me. Yet they have "powers"... Poor us. — Preceding unsigned comment added by 51.175.86.30 (talk) 11:21, 4 November 2018 (UTC)

I agree with 51.175.86.30 A mathematical proof is either valid or its not. Why does anyone need a 'source'. Wikipedia is just becoming a collection of references of references. The page on root 2 is tedious and dull and the proofs are laboured and arcane. Yawn. BillDixon56 (talk) 12:17, 16 September 2020 (UTC)

Geometric Proof - take 2

The first geo. proof claims (correctly) that the area of the darkest central square is (2b-a)^2. It took me almost a minute to establish this. (I found the bald assertion of facts not in evidence off-putting). IMHO, when I walk thru a proof, I shouldn't have to fill in steps which have been left out. I tried to insert a line which clarified and explained where the assertion came from, but it didn't "take". If we add the line "The middle square has sides of length b - (a - b) or, rearranging, 2b-a." just prior to that problematic assertion, I think the proof would be much clearer. I know it is a matter of how much detail you have to go into for each step but, at least for me, it wasn't obvious the the middle region had that area and was a distraction to have to check it.98.21.247.38 (talk) 14:48, 12 January 2021 (UTC)

I agree that the area of the central square is not immediately obvious and an interim step would be helpful. S Philbrick(Talk) 17:11, 5 February 2021 (UTC)

Cartoon on the subject

You might like this [1] Dmcq (talk) 15:19, 9 December 2016 (UTC)

Dmcq, Heh. S Philbrick(Talk) 17:51, 5 February 2021 (UTC)

"Constructive Proof"

The constructive proof is no proof since the argument is circular. Let me explain why the argument does not yield a proof that ${\sqrt {2}}$ is irrational:

Right in the beginning of the proof the author argues that $2b^{2}$ cannot be equal to $a^{2}$. Given that both $a$ and $b$ are strictly positive integers, the statement $2b^{2}=a^{2}$ is logically equivalent to the statement that ${\sqrt {2}}={\tfrac {a}{b}}$. This in turn means that the statement $2b^{2}\neq a^{2}$ is equivalent to ${\sqrt {2}}\neq {\tfrac {a}{b}}$. Thus, the entire argument relies on the hidden assumption that ${\sqrt {2}}$ is irrational. Hence, the argument is circular and does not yield a proof for the irrationality of ${\sqrt {2}}$. The only thing the argument from this section shows is the following: If one already knows that ${\sqrt {2}}$ is irrational one has some rational number $q\in \mathbb {Q}$ in reduced form $q={\tfrac {a}{b}}$ such that $1<q<{\tfrac {3}{2}}$, then one has the estimate $\vert q-{\sqrt {2}}\vert \geq {\tfrac {1}{3b^{2}}}.$ So either one should remove the section completely or present just the estimate in the section "Properties". Yaddle95 (talk) 16:03, 11 September 2022 (UTC)

EDIT: In did the second of both things. — Preceding unsigned comment added by Yaddle95 (talk • contribs) 16:25, 11 September 2022 (UTC)

Perhaps you should re-read the first sentence of the material you are trying to remove: In a constructive approach, one distinguishes between on the one hand not being rational, and on the other hand being irrational (i.e., being quantifiably apart from every rational), the latter being a stronger property. So the section starts with the already-established point that ${\sqrt {2}}$ is not rational, and proceeds from that point to quantify how far from being rational it is. —David Eppstein (talk) 20:21, 11 September 2022 (UTC)

Unclear scope of "coprime" in Pythagorean Triple proof of irrationality.

The Sierpinski lemma is quoted as

If a, b, and c are coprime positive integers such that a2 + b2 = c2, then c is never even.

As written, it's not clear whether a, b, and c are SETWISE coprime (there exists no natural number k which evenly divides a and b and c) or PAIRWISE coprime (there exists no natural number k which evenly divides more than one of a, b, and c). Reading the sufficiency clause as requiring PAIRWISE coprime numbers causes a serious problem later in the proof when we arrive at b² + b² = a², where the two left-hand side elements, here b and b, are self-evidently not coprime.

One can easily show that some natural number k which evenly divides both a and b (that is, gcd(a,b)=k) must evenly divide c² (see note [1] below), meaning the situation where a and c are pairwise coprime but a and b aren't cannot arise; but this lemma goes unstated, and it certainly isn't so obvious as to escape comment.

I hope what I've written here is clear. My point isn't that the proof is incorrect as-such, just that it's missing one of the strings required to tie the result together.

[1] Rather, you can easily show k²|c²; it's then a bit more complicated to show that k² | c² ⇒ k | c, but it does follow from a comparison of their respective prime decompositions -- if there is a prime q number in the prime decomposition of k not in the prime decomposition of c, there is no square of that prime available in c² to evenly divide it out. 2603:7000:A101:F9F1:3D23:B869:D654:A984 (talk) 22:09, 3 February 2023 (UTC)

Should I nominate this article for GA?

I feel this article can become a GA because,

1. It is well-written and concise to the square root of 2.
2. It is neutral (because it isn’t political)
3. It is relatively stable, and does not break the three revert rule.
4. It is well illustrated

However, I do not want to risk a drive-by nomination like I did at several articles, so that’s why I’m asking whether I should nominate this for GA status. Brachy08 (Never Gonna Give You Up, Never Gonna Let You Down) 02:56, 12 April 2023 (UTC)

There are many unsourced sentences, paragraphs, and even sections. I would likely quickfail it for that, even if it had a nominator who had put sufficiently many hours of work into the article and sufficiently much content to not count as drive-by. The lead does not properly summarize the body of the article, instead having its own content not repeated with more detail elsewhere. Several references appear unreliable and the reference formatting is not consistent. The long sections of formula after formula are also very dubious with respect to WP:GACR 3b, and the applications section looks thin and unbalanced. —David Eppstein (talk) 05:53, 12 April 2023 (UTC)

Thanks for letting me know. Brachy08 (Let’s Have A Kiki, I Wanna Have a Kiki) 00:15, 17 April 2023 (UTC)