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The purpose of this question is to collect the most outrageous (or ridiculous) conjectures in mathematics.

An outrageous conjecture is qualified ONLY if:

1) It is most likely false

(Being hopeless is NOT enough.)

2) It is not known to be false

3) It was published or made publicly before 2006.

4) It is Important:

(It is based on some appealing heuristic or idea; refuting it will be important etc.)

5) IT IS NOT just the negation of a famous commonly believed conjecture.

As always with big list problems please make one conjecture per answer. (I am not sure this is really a big list question, since I am not aware of many such outrageous conjectures. I am aware of one wonderful example that I hope to post as an answer in a couple of weeks.)

Very important examples where the conjecture was believed as false when it was made but this is no longer the consensus may also qualify!

Shmuel Weinberger described various types of mathematical conjectures. And the type of conjectures the question proposes to collect is of the kind:

On other times, I have conjectured to lay down the gauntlet: “See,

you can’t even disprove this ridiculous idea."

Summary of answers (updated: March, 13, 2017 February 27, 2020 September 2, 2023):

  1. Berkeley Cardinals exist

  2. There are at least as many primes between $2$ to $n+1$ as there are between $k$ to $n+k-1$

  3. P=NP

  4. A super exact (too good to be true) estimate for the number of twin primes below $n$.

  5. Peano Arithmetic is inconsistent.

  6. The set of prime differences has intermediate Turing degree.

  7. Vopěnka's principle.

  8. Siegel zeros exist.

  9. All rationally connected varieties are unirational.

  10. Hall's original conjecture (number theory).

  11. Siegel's disk exists.

  12. The telescope conjecture in homotopy theory. (Disproof by Robert Burklund, Jeremy Hahn, Ishan Levy, and Tomer Schlank announced 2023), preprint

  13. Tarski's monster do not exist (settled by Olshanski)

  14. All zeros of the Riemann zeta functions have rational imaginary part.

  15. The Lusternik-Schnirelmann category of $Sp(n)$ equals $2n-1$.

  16. The finitistic dimension conjecture for finite dimensional algebras.

  17. The implicit graph conjecture (graph theory, theory of computing)

  18. $e+\pi$ is rational.

  19. Zeeman's collapsing conjecture.

  20. All groups are sofic.

  21. The Lovász conjecture.

(From comments, incomplete list) 22. The Jacobian conjecture; 23. The Berman–Hartmanis conjecture 24. The Casas-Alvero conjecture 25. An implausible embedding into $L$ (set theory). 26. There is a gap of at most $\log n$ between threshold and expectation threshold (Update: a slightly weaker version of this conjecture was proved by Keith Frankston, Jeff Kahn, Bhargav Narayanan, and Jinyoung Park!; Further update: the conjecture was fully proved by Jinyoung Park and Huy Tuan Pham ). 27. NEXP-complete problems are solvable by logarithmic depth, polynomial-size circuits consisting entirely of mod 6 gates. 28. Fermat had a marvelous proof for Fermat's last theorem. (History of mathematics).

Tim Campion
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Gil Kalai
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    Does the Jacobian conjecture qualify? – Steve Huntsman Jan 17 '17 at 20:03
  • Hmm, I think it qualifies as a comment but not as an answer. For an actual answer I would like "most likely false" to represent a large consensus and not a personal view of the answerer. But once I asked the question my view about what qualifies is just one view in the crowd... – Gil Kalai Jan 17 '17 at 20:12
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    The Berman–Hartmanis conjecture. – T. Amdeberhan Jan 17 '17 at 20:20
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    "In this paper we try to convince the leader that there is no good reason to believe that the Jacobian Conjecture holds. Although there are several arguments in favor of this conjecture, we show that these arguments haven't got the power to justify the statement that the Jacobian Conjecture holds in general." -van den Essen (1997): http://www.seminariomatematico.unito.it/rendiconti/cartaceo/55-4/283.pdf – Steve Huntsman Jan 17 '17 at 20:20
  • Right! I prefer examples where it is still now commonly believed that the conjecture is false and where the proposer proposes the conjecture genuinely suggesting that it is true. But these two requirements may be too harsh. – Gil Kalai Jan 17 '17 at 20:22
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    There is a fine line between an outrageous conjecture and a bold conjecture. But still I see the spirit of your interesting question. – Joseph O'Rourke Jan 17 '17 at 20:25
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    It seems to me there is a conflict between "you can't even disprove this ridiculous idea" and "the proposer proposes the conjecture genuinely suggesting that it is true", so it's not clear to me what exactly you're after. – Gerry Myerson Jan 17 '17 at 21:59
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    Incidentally, my question at http://mathoverflow.net/q/101821/1946 was asked in the spirit of this question (but it is too recent to qualify for your 2006 requirement). – Joel David Hamkins Jan 17 '17 at 22:01
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    I don't think anyone has disproved the ridiculous ideas that there are only finitely many Mersenne composites, or that all the decimal digits of $\pi$ from some point on are sixes and sevens, or that the partial quotients for continued fractions of real algebraic irrationals are always bounded, but I don't think anyone has proposed any of these ideas genuinely suggesting they are true. – Gerry Myerson Jan 17 '17 at 22:19
  • I don't understand what you mean by point 5. By definition, the negation of any commonly-believed-false statement is a commonly-believed-true statement, isn't it? – Federico Poloni Jan 17 '17 at 22:32
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    The answers below all look of interest, as does the question. And, to boot, this is Community Wiki. Why not keep it open? – Lucia Jan 17 '17 at 22:49
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    It depends on the utility of this question. As a short term diversion to appeal to some of the forum community it serves quite well. As part of a database of questions and answers for future reference by the interested-in-mathematics consumer, I think it is too based in opinion and belongs on a blog. If the intent were to set some challenge questions to spur research, then I think the question should be reworded. As it stands now, it isn't much better than an opinion poll. Gerhard "MathOverflow Isn't Question And Opinion" Paseman, 2017.01.17. – Gerhard Paseman Jan 17 '17 at 23:02
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    @GerhardPaseman I think it is not too uncommon for good mathematicians, working in or near an area, to nevertheless not know about various conjectures. Especially if there has been recent development in tangentially-related areas, this type of list very well might lead to some of these conjectures being refuted. I support keeping the question open. – Theo Johnson-Freyd Jan 17 '17 at 23:36
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    And besides, I will learn things from reading it! – Theo Johnson-Freyd Jan 17 '17 at 23:36
  • @Theo, in which case, let's rewrite the question to fit both a good intent of the asker and the good intent of MathOverflow. As it is currently written, I am not sure either is achieved. Gerhard "Being Ridiculous Can Serve Research" Paseman, 2017.01.17. – Gerhard Paseman Jan 18 '17 at 00:30
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    I feel like a number of famous, "elementary" conjectures, while often believed to be true, have no particular (philosophical) reason to be true, and thus from a cynical perspective might be described as likely false. Examples are the Collatz conjecture (https://en.wikipedia.org/wiki/Collatz_conjecture) and the union-closed sets conjecture (https://en.wikipedia.org/wiki/Union-closed_sets_conjecture) – Sam Hopkins Jan 18 '17 at 04:56
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    What's next - a big list with Trump tweets concerning mathematics? Does this outrageous conjecture of mine count as an example? – Franz Lemmermeyer Jan 18 '17 at 05:39
  • @Sam, I don't know about "philosophical", but I think there are good mathematical reasons for Collatz to be true. – Gerry Myerson Jan 18 '17 at 05:39
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    I agree with Theo on that. E. g. Socrates was, I think, The asker number one. The way I understand it, the question is about conjectures of Socratic quality. If the things Socrates asked would be only pedagogic challenges and not the things he really burningly wanted to know, he would not be Socrates. And recall what happened to him. By the way there is a Socratic badge here on MO. A golden one ;) – მამუკა ჯიბლაძე Jan 18 '17 at 05:40
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    This one doesn't count because it status is settled, but from what I understand of the history of mirror symmetry, when the physicists first proposed it, Yau for one initially thought that it was too outrageous to be true. Along similar lines, I believe that Tao initially thought that the phenomenon of compressed sensing couldn't possibly be true, and that the Candes-Romberg-Tao paper was born out of Tao's attempts to find a disproof. – Timothy Chow Jan 18 '17 at 21:59
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    @TimothyChow The Tao example is a really good one (and with ample documentation), even if it doesn't meet all the conditions of the OP. – Todd Trimble Jan 19 '17 at 02:36
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    The Kahn-Kalai conjecture (the general one for Boolean functions) almost qualifies here, doesn't it – user36212 Jan 19 '17 at 23:58
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    Not an expert on this subject, but "all groups are sofic" might potentially qualify here. – Terry Tao Jan 21 '17 at 22:59
  • I'm far from being an expert on this but perhaps Casas-Alvero conjecture is relevant. https://en.wikipedia.org/wiki/Casas-Alvero_conjecture . Although the most outrageous aspect of it is probably the date it was first conjectured (2001 !!!). – Saal Hardali Jan 21 '17 at 23:40
  • Dear user36212, yes we both regards the conjectures (both for Boolean functions and for graphs) as fairly outrageous and would be very interested in counterexamples. Dear Timothy and Todd, these are good examples and I am certainly not too fussed about meeting all my conditions. – Gil Kalai Jan 22 '17 at 10:42
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    @GilKalai : It's sort of too late now, but I wonder if another (and possibly better) way to phrase your question would be, what conjectures were regarded as impossibly bold or optimistic at the time they were first made? This would allow for both conjectures that have been settled and conjectures that are still open. It would also allow for bold conjectures that we've gotten used to but that were considered outrageous at first. And it would eliminate statements that nobody has ever believed. – Timothy Chow Jan 22 '17 at 21:26
  • This question reached the sidebar and as such has received a large influx of people who usually don't visit this site. Many conjectures in here are described in ways only mathematicians can understand them. I was wondering if the people who posted an answer already could clarify it in a way that people who aren't well versed in math could understand the conjecture? – Nzall Jan 24 '17 at 10:19
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    Dear @Nzall , what does it mean to reach the sidebar? – Gil Kalai Jan 24 '17 at 11:15
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    @GilKalai It means that the question has had enough activity on this site to be viewed as a "hot network question", which in turn means that it's eligible to appear in the list of questions which appear in the sidebar, beneath the meta highlights and the linked questions. This list is effectively a "best of Stack Exchange", and a lot of people tend to check out the questions on that bar, especially if the title is interesting or (like this question) clickbaity. It means that a lot of users from other exchanges on the network will look at the question and potentially give their input. – Nzall Jan 24 '17 at 11:26
  • @GilKalai The biggest consequence is that the question gets a lot of new readers, votes, comments and potentially answers, many of which haven't had much more than high school or maybe first grade college education on the topic. Making the topic a bit more understandable for those users may introduce them to aspects of mathematics they didn't know exist and may encourage them to look further on the site. – Nzall Jan 24 '17 at 11:31
  • @Nzall, certainly it could be a good idea to add elementary explanations for at least some of the answers. It is not so easy but worth the effort. – Gil Kalai Jan 24 '17 at 13:37
  • Dear Terry and Saal, indeed "all groups are sofic" is a famous conjecture which might be suitable. The Casas-Alvero conjecture strikes me as a good example as well but I dont know anything about it. – Gil Kalai Jan 24 '17 at 16:24
  • Why 2006? It seems rather arbitrary. Perhaps instead "is at least 10 years old" which will allow for more recent "outrageous" answers as time goes on. – Damien Jan 25 '17 at 03:36
  • Hi Damien, I agree with that. I also agree with Timothy's comment regarding unsettled conjectures. I am a little worried that adding settled conjectures would have made the question too board (but those can be mentioned in comments and also in other questions.) – Gil Kalai Jan 25 '17 at 10:09
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    [Disclaimer: I haven't really done any serious maths since my degree many moons ago, and yes, I arrived here via the sidebar.] I'm surprised noone mentioned Fermat's famous "I have discovered a truly marvelous proof of this, which this margin is too narrow to contain" quote. I guess it's not technically a conjecture, but there's an implicit conjecture in there that there exists a marvellous proof which would have been attainable before 1621, and this fits all the criteria of the question, as well as being presumably in keeping with the spirit of the question. – Adam Spiers Jan 26 '17 at 16:09
  • Adam, this was indeed a conjecture but it does not fit the question because: a) It was settled (point 2), b) it was not believed to be false (point 1). But I agree that beside the formalities Fermat's conjecture and Fermat's narrow margin claim were quite outrageous! – Gil Kalai Jan 26 '17 at 17:28
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    @GilKalai No, you misunderstood. I was proposing Fermat's quote about his Last Theorem as an answer to the question, not the Theorem itself. The implicit conjecture that there exists a marvellous proof which would have been attainable before 1621 is not settled (point 2), and IIUC is widely believed to be false. – Adam Spiers Jan 29 '17 at 16:06
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    Hmm, I see. This is indeed an outrageous conjecture about mathematics and its history :) – Gil Kalai Jan 29 '17 at 18:55
  • @FranzLemmermeyer I recently made a conjecture that if time goes to infinity trump will make a misspelling on twitter at some point and type abelian instead of a billion. The legend goes that this already happend and that this caused all multiplications at quantum level to be non-commutative, making the universe ridiculously hard to understand. On the other hand there are people thinking that trump tweeting about math is fake news. – Maarten Derickx Jul 29 '18 at 23:47
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    A conjecture from https://www.youtube.com/watch?v=BdHFLfv-ThQ is "$\pi^{\pi^{\pi^\pi}}$ is an integer". I don't know how old this conjecture is, but apart from its age it seems to fit the bill. – Dirk Mar 26 '21 at 08:58

26 Answers26

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A long-standing conjecture in Number Theory is that for each positive integer $n$ there is no stretch of $n$ consecutive integers containing more primes than the stretch from 2 to $n+1$. Just looking at a table of primes and seeing how they thin out is enough to make the conjecture plausible.

But Hensley and Richards (Primes in intervals, Acta Arith 25 (1973/74) 375-391, MR0396440) proved that this conjecture is incompatible with an equally long-standing conjecture, the prime $k$-tuples conjecture.

The current consensus, I believe, is that prime $k$-tuples is true, while the first conjecture is false (but not proved to be false).

Gerry Myerson
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    Does this conjecture have a name? – Brevan Ellefsen Jan 17 '17 at 23:04
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    I've seen it referred to as "the Hardy-Littlewood convexity conjecture". See Section 1.2.4 of the Crandall and Pomerance book, Prime Numbers: A Computational Perspective. – Gerry Myerson Jan 17 '17 at 23:14
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    What makes the consensus favor prime $k$-tuples over the Hardy-Littlewood convexity conjecture? Intuitively the former seems much less plausible than the latter. – orlp Jan 18 '17 at 04:08
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    @orlp, there are heuristic arguments that not only do prime $k$-tuples exist but there are asymptotic formulas for how many of them there are up to any given bound, and these asymptotic formulas are always in close agreement with the numerical evidence. I don't think there are any such arguments for convexity, so it's gotta go. – Gerry Myerson Jan 18 '17 at 05:36
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    This is nowadays known as the second Hardy-Littlewood conjecture. – Wojowu Jan 18 '17 at 13:46
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    Is it the case that Hardy-Littlewood really considered both the k-tuples conjecture and the convexity conjecture to be plausible? Makes me wonder what mutually implausible conjectures are also held to be believed nowadays (2017). – Mark S Jan 19 '17 at 00:53
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    @Mark, it's not even entirely clear where or whether Hardy & Littlewood made the conjectures. See http://mathoverflow.net/questions/54223/whence-the-k-tuple-conjecture and http://mathoverflow.net/questions/30827/where-does-the-hardy-littlewood-conjecture-that-pixy-pix-piy-origin and http://math.stackexchange.com/questions/1072194/source-of-hardy-littlewoods-2nd-conjecture for some discussions. – Gerry Myerson Jan 19 '17 at 01:55
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    http://www.opertech.com/primes/k-tuples.html shows some numerical details. There might be a counter example to the conjecture before 10^1197 - trying to find it with current methods is quite hopeless. @MarkS Quite possible they found both conjectures plausible at the same time; a k-tuple contradicting it is quite hard to find. – gnasher729 Jan 21 '17 at 23:03
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    And I think this conjecture being called "the second Hardy-Littlewood conjecture" - which means that if the first Hardy-Littlewood conjecture is true, the second one is false :-) – gnasher729 Jan 21 '17 at 23:05
  • Godel proved there must be unproveable true statements. One mark of such might be that A and B are both plausible, neither leads to anything outright ridiculous, but that it is proven that A implies not-B, and vice versa. – nigel222 Jan 24 '17 at 11:57
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    @nigel, if 2nd Hardy-Littlewood is false, then there is a counterexample, so it's not unproveable. – Gerry Myerson Jan 24 '17 at 21:54
  • Assuming k-tuples, is the Hensley and Richards' proof effective in disproving convexity? If so with what bound? – Yaakov Baruch Mar 25 '21 at 18:11
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W. Hugh Woodin, at a 1992 seminar in Berkeley at which I was present, proposed a new and ridiculously strong large cardinal concept, now called the Berkeley cardinals, and challenged the seminar audience to refute their existence.

He ridiculed the cardinals as overly strong, stronger than Reinhardt cardinals, and proposed them in a "Refute this!" manner that seems to be in exactly the spirit of your question.

Meanwhile, no-one has yet succeeded in refuting the Berkeley cardinals.

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    Yes, this is the spirit of the question! – Gil Kalai Jan 17 '17 at 20:18
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    Dangit, beat me to it! +1. (Need to get my typing speed up!) – Noah Schweber Jan 17 '17 at 20:25
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    On the other hand, the algebras of elementary embeddings have barely been investigated, and the algebras of elementary embeddings seem like a good spot to obtain an inconsistency (hopefully at a level inconsistent with AC). We know that the algebras generated by a single elementary embedding look like, but barely anything is known about algebras generated by multiple elementary embeddings. I would wait until we better understand the algebras of elementary embeddings before declaring this problem as difficult. – Joseph Van Name Jan 17 '17 at 23:07
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    @JosephVanName Well, it has been an open question for 25 years, and I know of some very smart people who have worked on it. So I think it qualifies as "difficult". But meanwhile, there is a current resurgence of interest in these very strong ZF large cardinals, and so a resolution may be close. Please go for it! – Joel David Hamkins Jan 17 '17 at 23:10
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    @Joel David Hamkins. I will put the consistency of large cardinals beyond AC on my eventual to-do-list. – Joseph Van Name Jan 18 '17 at 00:34
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    Johnson answered [by] striking his foot with mighty force against a large stone, till he rebounded from it -- "I refute it thus." – Max Jan 19 '17 at 10:05
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    The Berkeley cardinal may or may not exist, but the Stanford Cardinal has won prizes. https://en.wikipedia.org/wiki/Stanford_Cardinal – Gerry Myerson Jun 16 '18 at 00:28
  • Maybe there's a problem with the existence notion used in modern mathematics... – Wakem Oct 28 '21 at 17:46
  • I think all existence notions should be somewhat restrictive. I mean we are already talking about the existence of infinite sets! This existence we are talking about now isn't the same notion of existence Cantor himself had. – Wakem Oct 28 '21 at 17:48
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$P=NP$

Let me tick the list:

  1. Most likely false, because, as Scott Aaronson said "If $P = NP$, then the world would be a profoundly different place than we usually assume it to be."

  2. Yes, it's The Open Problem in computational complexity theory

  3. Yes, it's old

  4. It's important, again quoting Scott: "[because if it were true], there would be no special value in "creative leaps," no fundamental gap between solving a problem and recognizing the solution once it's found. Everyone who could appreciate a symphony would be Mozart; everyone who could follow a step-by-step argument would be Gauss..."

  5. It's an equality rather than a negation

Carlo Beenakker
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    Dear Carlo, I wanted to avoid "just the negation of a famous commonly believed conjecture" . (So while an equality it violates my condition 5 as I saw it :) ) But I still +1 it. – Gil Kalai Jan 17 '17 at 20:08
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    Was it ever believed to be true I wonder? I get the impression that nowadays you are hard pressed to find anyone who believes it. Some might hence argue that it is actually the negation of a well-known conjecture, namely that P isn't NP. (posted before I saw Gil's comment) – Kevin Buzzard Jan 17 '17 at 20:08
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    @KevinBuzzard -- "In a 2002 poll of 100 researchers, 9 believed the answer to be yes" (from Wikipedia, with this link to the poll) – Carlo Beenakker Jan 17 '17 at 20:09
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    @KevinBuzzard: Dick Lipton's "P = NP" blog is broadly devoted (or at least, a frequent point of discussion) to speculating that this is true. – Daniel R. Collins Jan 17 '17 at 22:35
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    I am downvoting in disagreement with Aaronson's hyperbolic comments. If every NP algorithm is in P but with much larger time bounds, then the consequences in 4 don't follow. And the world often is profoundly different than people assume: I don't want to declare assumptions likely simply because they're strongly held. –  Jan 17 '17 at 22:51
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    @MattF. -- in all fairness to Scott Aaronson, I took just one of his 10 reasons to believe $P!=NP$, and he calls this particular reason the "philosophical argument"; there are 9 more. – Carlo Beenakker Jan 17 '17 at 22:55
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    Without taking a stand on P vs. NP, I support the hyperbolic comments. They seem to be right on, in the metaphorical sense in which they are offered. The consequences in 4 are meant for those who consider the truly large numbers. If one is small-minded, one might miss the essence of the natural numbers. – Joel David Hamkins Jan 17 '17 at 23:24
  • Indeed, if the answer indicated ideas from Aaronson's reasons 3,6,7 instead I would probably upvote it. –  Jan 17 '17 at 23:41
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    Fails #5. It's the negation of the almost-certainly-true $P \neq NP$ – R.. GitHub STOP HELPING ICE Jan 18 '17 at 03:12
  • @CarloBeenakker Thank God science and mathematics aren't Democratic. – bjd2385 Jan 18 '17 at 06:33
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    There are much better examples from complexity theory if we allow this kind of thing. For example: "NEXP-complete problems are solvable by logarithmic depth, polynomial-size circuits consisting entirely of mod 6 gates." This is a far more dramatic illustration of our inability to prove lower bounds. – Timothy Chow Jan 18 '17 at 16:23
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  • and 4. are completely unsupported by any evidence. They are just hyperbole .
  • – Simd Jan 19 '17 at 22:33
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    @TimothyChow: I think this possibility was ruled out recently by Ryan Williams' breakthrough result. But there are many other, only slightly less ridiculous, possible complexity class equalities to choose from. – Ashley Montanaro Jan 20 '17 at 07:44
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    @AshleyMontanaro : No, I picked my example to be just outside what Williams was able to prove. ACC circuits have constant depth, not logarithmic depth. – Timothy Chow Jan 20 '17 at 18:38
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    Again, quoting Scott: "It is important for it to be false due to the passionate appreciation of my self-proclamation of superiority, and also, others' proclamations I have made too! Jesus, If this is true, I'll have to accept that I and my beloved crushes are memberes of the plebe like anyone else. Scary! :'(" #FirstOrderProblems – Red Banana Apr 23 '17 at 17:55
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    I'd say P=PSPACE is even more outrageous than P=NP. It's still outrageous even if we assume P=NP is true. – none Nov 29 '19 at 08:54
  • Agreed about P=PSPACE being the outrageous one. Essentially, if you're not constrained by memory then you're also not constrained by time :) – Ralph Furman Mar 25 '21 at 18:56