I'm looking for statements that look obviously false but have no disproof (yet).

For example The base-10 digits of $\pi$ eventually only include 0s and 1s.

To make this question a little objective, I'm thinking about the "Vegas gambling odds" I would need to bet on each statement. Or, equivalently, the final answer will be the one statement I'd choose if forced to bet my life against one. I'm hoping voters could try to take one of those approaches too and I'll probably then just side with the biggest vote-getter.

I'm looking for statements that even children would doubt, and that really is the goal of my question, but I'm also a little curious whether the definitions of more advanced math might somehow create an even more laughable but possible statement. So, don't hold back if anything in your mind seems more obvious to you.

EDIT: Most statements probably have an obvious improvement method (e.g., as @bof pointed out, the $\pi$ example can use two different long fixed blocks of digits to fill the tail instead of just 0/1), so I'll simply judge with added style points for making statements "short, sweet, and easy for children to contemplate". The main difference from this prior question is that answers do not need to be "important" (so the focus here in my question is not on advanced mathematics) and that I'm choosing the least believable statement as the answer.

CONCLUSION: If I owned a casino, here's where I'd set the odds. I struck out two statements though because I personally wouldn't take bets on them (like I mentioned in comments, I think the losing gamblers would complain about ambiguous terms even if I handed out some huge book explaining them). I'm curious where the smart money would go with these odds...and wish we had a trusted oracle to settle the bets in the end. I also changed the title of this question to try to keep it open.

$10^{\ \ 3}:1\quad\quad$P = NP

$10^{13}:1\quad\quad$The number $2\uparrow 2\uparrow 2\uparrow 2+3\uparrow 3\uparrow 3\uparrow 3$ is a prime number

$10^{12}:1\quad\quad$The continuum is $\aleph_{37}$

$10^{\ \ 4}:1\quad\quad$Peano arithmetics proves 1=0

$10^{\ \ 5}:1\quad\quad \zeta(5)$ is rational

$10^{\ \ 6}:1\quad\quad e+\pi$ is rational

$10^{\ \ 2}:1\quad\quad$The Riemann Hypothesis is FALSE

$10^{30}:1\quad\quad$The base-10 square root of every prime number eventually only includes 0s and 1s

$10^{14}:1\quad\quad$The number $\Large \pi^{\pi^{\pi^\pi}}$ is an integer

$10^{15}:1\quad\quad$The base-10 digits of π eventually only include 0s and 1s

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    Pretty much same as https://mathoverflow.net/questions/259844/the-most-outrageous-or-ridiculous-conjectures-in-mathematics – Gerry Myerson Sep 17 '18 at 06:51
  • @Gerry The answers in that question are more advanced and even my pi example seems less likely than those. I might need to re-word my question. – bobuhito Sep 17 '18 at 06:58
  • I’m sorry, but did I misread the question? I thought you’re betting your life that the theoretically-possible statement is *not provably true*. So, I answered, “The next statement I make is false.” That’s getting downvoted to oblivion by people asking why that should be considered unlikely. Because I just bet my life that I’m going to falsify it? Although it’s impossible to be sure without seeing the future? Did I misread the challenge? – Davislor Sep 17 '18 at 18:27
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    Unfortunately, I'm voting to close this as off-topic. It's an opinion-based poll question, and not a good format for this site IMHO. –  Sep 17 '18 at 21:44
  • Do you have any reasons for the mentioned odds ? Some of the odds are very different from the odds I would guess. – Peter Sep 19 '18 at 11:58
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    Except for the one which checked for a prime number, it was more of a feeling than math...so, I started a [chat](https://chat.stackexchange.com/rooms/83390/odds-of-statements-that-look-obviously-false-but-cannot-be-disproved) where we can talk about any one in particular. – bobuhito Sep 19 '18 at 15:05
  • Perhaps, bobuhito, you would like to "accept" one of the answers by clicking in the check mark next to it. That may make it harder to delete the question and all of its answers. – Gerry Myerson Sep 30 '18 at 21:59

12 Answers12


$$P = NP$$

is unlikely but still possible.

To explain it to a child, you could ask if it's easier to:

  • check the definition of a word in a wordbook
  • or find a word in the wordbook given its definition

It seems obvious that the former is easier than the latter, $P = NP$ would mean that both are equally easy.

Eric Duminil
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    Similar to my comment for "Peano arithmetics proves 1=0", but less so, I'm a little worried about losing my bet in a loophole here. For the children, you'd have to explain that "easiness" is determined by how the lookup time scales with the dictionary length, not the lookup time alone, and that opens up a can of technical worms IMHO. – bobuhito Sep 17 '18 at 15:42
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    I take your meaning, but "equally easy" is a bit misleading. $n^2$ and $n^200$ are both in $P$, but they're not equally easy. – mephistolotl Sep 17 '18 at 17:24
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    There are even better examples in complexity theory. For instance P = PSPACE is completely ridiculous, but we can't rule it out yet. PSPACE is the class for polynomial space, i.e. "take as much time as you want to solve the problem, but you can only use a polynomial amount of scratch paper". PSPACE is thought to contain the entire infinite polynomial hierarchy, of which P and NP are just the first levels, each level of which is thought to strictly contain the previous one. But we can't prove any of it. – usul Sep 17 '18 at 18:53
  • @usul You meant to say that it's thought to *strictly* contain the entire polynomial hierarchy. That it contains it is pretty clear from the definitions. – Yonatan N Sep 17 '18 at 23:02
  • The [Robertson-Seymour Theorem](https://en.wikipedia.org/wiki/Robertson%E2%80%93Seymour_theorem) is enough to cause me to refrain from saying that $P=NP$ is *obviously* false since it suggests that the intuition that problems in $P$ are tractable is a somewhat misleading intuition. – John Coleman Sep 18 '18 at 02:31

My personal favorite :

The number $$2\uparrow 2\uparrow 2\uparrow 2+3\uparrow 3\uparrow 3\uparrow 3$$ is a prime number.

Since this number is very large (it has $3\ 638\ 334\ 640\ 025$ digits), it is very likely composite. However, according to Enzo Creti's calculation, there is no prime factor below $10^{12}$, so the number might be prime.

Outside mathematics I would vote for the statement : "God exists" and on the second place : "Our universe is not the only one"

To coin another mathematical statement :

Goldbach's conjecture is false

This is almost surely false , but as long Goldbach's conjecture is not proven, it cannot be ruled out.

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    Why is "our universe is not the only one" *most unlikely*? Wouldn't it be unlikely for our universe to *be* the only one? Or did you confuse "most unlikely to be true" with "most unlikely to be provable"? – user541686 Sep 17 '18 at 10:05
  • @Mehrdad I do not care what the physicists say. For me, it does not make sense to assume another universe. If the physicists are honest, they also only speculate about that. – Peter Sep 17 '18 at 10:21
  • Isn't Goldbach's conjecture proven? http://milesmathis.com/gold3 – Aganju Sep 17 '18 at 11:41
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    @Aganju A quick search will tell you why that's a load of nonsense and why you shouldn't take anything else by that same person seriously. – hvd Sep 17 '18 at 12:30
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    The problem with the universes thing is: what even is "a universe"? Before we can talk about whether there is more than one, we ought to sort that one out, since if, say, we define "universe" to be everything there is, there clearly cannot be anything beyond it. If we mean an observable-universe-like "bubble", then it's very likely there are more, given that current estimates on space curvature suggest there should be a lot more space beyond our own bubble. – Wojowu Sep 17 '18 at 12:47
  • What is the actual probably prime you mentioned ? – Overmind Sep 17 '18 at 12:59
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    Overmind: https://en.wikipedia.org/wiki/Knuth's_up-arrow_notation. Once you grasp that you'll understand why the number expressed above cannot reasonably be rewritten out. – Alan Baljeu Sep 17 '18 at 14:34
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    "Outside mathematics I would vote for the statement : 'God exists' and on the second place : 'Our universe is not the only one.'" I don't know that editorializing on philosophical questions is appropriate for a question about mathematics. – Kevin Sep 17 '18 at 17:27
  • Why be so selfish to confine yourself to the narrow perspective that humanity is at the center of something? To say that our universe is the only one would mean that our universe is at the center by default. Yea, to say our *reality* is the only one would mean that our reality is at the center by default. I say humanity must humble itself to the possibility that we may be in the middle of nowhere: just one tiny insignificant grain of sand in all the world's beaches. – Jack G Sep 21 '18 at 22:33
  • @Wojowu If we insist that the universe is only what we can observe, then I agree. But I do not agree the big bang theory claiming that. If we reasonably consider objects/space that are too far away from us to be observable to belong to the universe, then it is unlikely that there is another universe because this universe would not be reachable in principle (if it where, it would belong to our universe). – Peter Sep 22 '18 at 07:13
  • @JackGiffin As Wojowu already pointed out, the problem is the definition of "universe". I do not agree to the mainstream that apparently defines the observable universe to be THE universe. – Peter Sep 22 '18 at 07:23

The continuum is $\aleph_{37}$.

Gerry Myerson
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    Lol! The only issue with this example is, if it is proven or disproven (in ZFC), then we can also prove that ZFC itself is inconsistent, throwing the proof or disproof into doubt. It could still be that ZFC proves itself inconsistent, but actually is consistent, but that would be a very interesting situation... – user21820 Sep 17 '18 at 09:40
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    @user21820: *"It could still be that ZFC proves itself inconsistent, but actually is consistent, but that would be a very interesting situation"*... I'm having trouble wrapping my head around this. How could this work? If it proves itself inconsistent then isn't it inconsistent? – user541686 Sep 17 '18 at 10:07
  • I thought the consistency of ZFC was independent of ZFC. (Though, this does have the strange consequence that it's consistent with ZFC that ZFC is inconsistent.) – Akiva Weinberger Sep 17 '18 at 11:53
  • I don't get it. Don't we know it's $\aleph_{1}$? – Eric Duminil Sep 17 '18 at 12:29
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    @Eric, no. That was what Cantor believed, and it has come down to us as Cantor's Continuum Hypothesis, but we've known for about 55 years now that that question is independent of the usual axioms of set theory. https://en.wikipedia.org/wiki/Continuum_hypothesis will get you started. – Gerry Myerson Sep 17 '18 at 12:51
  • @Mehrdad: You're not the only one who has had trouble understanding that. But it is an easy consequence of the incompleteness theorem. See [this post](https://math.stackexchange.com/q/2486348/21820) for a simple proof and discussion. In short, given any computable formal system S that can reason about program execution, there is some sentence Con(S) about program halting (Q' in the linked post) that S cannot prove or disprove. If S is consistent, then Con(S) is true, and furthermore S' = S+¬Con(S) must be consistent. However, S' proves ¬Con(S) and hence also proves ¬Con(S'). – user21820 Sep 17 '18 at 13:11
  • No it isn't. Incompleteness means there are true statements that cannot be proved. It also can mean we cannot always prove consistency. But if we prove inconsistency, then it is certainly not consistent, as that is the very definition of inconsistent. – Alan Baljeu Sep 17 '18 at 14:41
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    $\aleph_{42}$ ist more likely. – Dirk Sep 17 '18 at 17:11
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    @AlanBaljeu: That's wrong. It is perfectly possible to have a consistent theory that proves its own inconsistency -- take, for example, ZFC extended with the axiom "ZFC is inconsistent" (which has to be at least as consistent as ZFC itself, because if it's inconsistent, then ZFC would prove its own consistency by r.a.a and therefore be inconsistent too). For ZFC itself it is conceivable that it has models but all those models contain non-standard integers and in particular non-standard numbers that the model thinks are proofs of $0=1$. – hmakholm left over Monica Sep 17 '18 at 18:05
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    Henning: Citation needed, because I would dispute your argument as posted: Your system (ZFC+Not ZFC) obviously is inconsistent, though I'm not sure how you would formally define Not ZFC in a usable system. ZFC cannot prove ZFC, but it might (hypothetically) disprove ZFC if it is inconsistent. So we have an inconsistent theory that proves its consistency, but such theories prove every statement. A consistent theory would not disprove itself. – Alan Baljeu Sep 17 '18 at 19:07
  • @user21820 You misinterpret the proofs on that page. The proof starts with "T is consistent" and end with "T is inconsistent", but these are reductio ad absurdum proofs where the premise as an assumption for the sake of the proof. The conclusion rejects the premise, and therefore we insist that the premise is an invalid assumption. T is not, was not, and never can be considered consistent. It is only inconsistent. – Alan Baljeu Sep 17 '18 at 19:12
  • @AlanBaljeu: You are really funny. I wrote the proofs on that page. You clearly do not understand the incompleteness theorems as your above comments demonstrate. If you wish to ask to clarify your misconceptions, feel free to do so. But please don't dispute others when you don't know the subject matter. – user21820 Sep 18 '18 at 05:47
  • @AlanBaljeu: And please, what I and Henning are stating can be found in any **introductory text** to mathematical logic. No citation is needed. Just check any standard text. – user21820 Sep 18 '18 at 05:48
  • @Peter is it possible that from a certain point, the 10-base digits of $\pi$ start to repeat itself? I mean 3.1415...31415...maybe no because otherwise $\pi$ would be a rational number? – Enzo Creti Sep 18 '18 at 07:37
  • @user21820 I have studied logic but it's been a long time. Perhaps my confusion is simply in the terms. Are the words 'consistent' and 'inconsistent' understood as model-theoretic or proof-theoretic or both? – Alan Baljeu Sep 18 '18 at 15:31
  • @AlanBaljeu: I'm afraid it has nothing to do with the terms. I will say again the facts, but I think you should read my linked post thoroughly if you wish to understand the subject matter. Take any formal system S with a proof verifier program V, which given input strings (p,x) will always halt and tell you whether p is a proof of x or not. Suppose S can reason about program execution, so if a program halts on an input and produces an output then S can prove that fact. Then we can write down a sentence Con(S) saying ( V(p,"0=1") = "no" for every string p ). [cont] – user21820 Sep 18 '18 at 17:35
  • [cont] Since V is itself a program, we can ask whether S proves Con(S) or not. Also, if S is consistent then Con(S) is obviously a true sentence about programs. But by the proof in the linked post, S is unable to prove Con(S). If S is a classical/intuitionistic first-order theory, then S' = S+¬Con(S) is consistent, but S' proves ¬Con(S) and so also proves ¬Con(S'). So we have S' is **consistent but proves its own inconsistency** (as expressed by the ¬Con(S')). – user21820 Sep 18 '18 at 17:37
  • I follow the argument above and it makes sense as long as you stay within the system. But when you step outside S', you get the statements Con(S') is true, and S' proves ¬Con(S'), and the syllogism says therefore ¬Con(S') is true. Therefore Con(S') and ¬Con(S'). Therefore contradiction. Anyway, if what I'm saying doesn't make sense, it will take multiplied hours to get to the bottom of the matter, so I invite you to answer either way and the conversation will be left off for a future date. – Alan Baljeu Sep 18 '18 at 18:15
  • @AlanBaljeu I suggest perhaps putting your confusion in the form of a question and posting it on this site. – Nico A Sep 19 '18 at 10:56
  • What is the point of the number $37$. Is it just a joke number (which could be replaced by any other number), or does it have a meaning in this context ? – Peter Sep 19 '18 at 11:54
  • @Peter, it's not a joke number, but it is (more-or-less) an arbitrary number. So far as I know (and I'm am absolutely no expert on this, so maybe someone will correct me), the only values that have ever been seriously put forward for the continuum are $\aleph_1$ and $\aleph_2$. – Gerry Myerson Sep 19 '18 at 13:55
  • @AlanBaljeu: What you said doesn't make sense. S' proves ¬Con(S'), but the naturals N satisfy Con(S'). Con(S') is just an arithmetical sentence and has **no meaning** until it is interpreted. When interpreted about N, it is true. But it is not **proven** by S', and in fact S' was constructed precisely such that it would **disprove** Con(S') although it is in fact consistent. Just as if I say "I am wrong." it does not mean there is a contradiction, but it only means that I am nonsensical! – user21820 Sep 24 '18 at 10:10
  • Thanks for the reply. Food for thought that some day I will plug into a full math.se question to clarify in full. – Alan Baljeu Sep 24 '18 at 18:47
  • @user21820 *"It could still be that ZFC proves itself inconsistent, but actually is consistent, but that would be a very interesting situation"* ...or maybe ZFC's consistent, and there's a metric space on the sentences expressible in ZFC and ZFC proving itself consistent is a limit point thereof. – samerivertwice Oct 08 '18 at 10:31
  • @RobertFrost: That makes absolutely no sense whatsoever, and sorry I'm not interested in continuing. – user21820 Oct 08 '18 at 10:36
  • @user21820 deriving sense from it depends how hard you try. – samerivertwice Oct 08 '18 at 13:42
  • https://math.stackexchange.com/questions/2831642/is-bbb-z-frac12-bbb-z-sim-a-torsion-group-where-x-sim-x2v-2x#comment5842801_2831642 – user21820 Oct 08 '18 at 14:33
  • How much sense something has is independent of how hard one tries to derive sense from it. – user21820 Oct 08 '18 at 16:39
  • Au contraire. The clue's in the phrase to *make sense*. I'm pointing out the work that must be carried out by the interpreter in *making sense* of what is written, that sense only ever follows work. My claim is that there will be others who work harder to do so and do derive sense from it. But I can understand you not wanting to. – samerivertwice Oct 09 '18 at 02:49
  • ...ironically you have linked the perfect example, something which is correct and although relatively succinct and accurate, requires a great deal of work to make sense of. – samerivertwice Oct 09 '18 at 03:05

I think the most unlikely, but still possible statement is:

Peano arithmetics proves $1=0$.

I guess there's not a single mathematician who believes this to be true. Indeed, if it were true, it would basically invalidate almost all current mathematical theories.

Yet we cannot disprove it. Even more, we can prove that we cannot disprove it, unless it is actually true. In other words, by disproving it, you'd actually prove it! (That's essentially what Gödel proved in his incompleteness theorems.)

Now one might object that you can disprove it from e.g. ZFC set theory. But that's only under the assumption that ZFC itself is consistent, which you again cannot prove, except by assuming an even more powerful theory is consistent. So all those proofs really show is that if Peano arithmetics is inconsistent (i.e. proves $1=0$), then all those theories are inconsistent, too.

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    I personally feel like this one is a bit sneaky with definitions (starting with "proves") and I feel like there's a decent chance I'll lose my bet in some loophole. – bobuhito Sep 17 '18 at 15:32
  • I upvoted this for the sheer nonsense value of "We can prove that we cannot disprove it, unless it is actually true." – Wildcard Sep 17 '18 at 16:15
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    I'm not so sure about "not a single mathematician". Edward Nelson famously believed your statement to be true -- he died four years ago, but there are still active ultrafinitists. (However, I have not been able to quickly confirm whether _they_ think PA is inconsistent, rather than, say, merely deny that it is _true_ about platonic integers, or claim the question is not meaningful). – hmakholm left over Monica Sep 17 '18 at 18:18
  • @HenningMakholm: Have you heard the joke that if you ask them whether the natural number $2^n$ exists they will take $Ω(n)$ time to answer "yes"? So asking them whether PA is inconsistent will not get an answer anytime soon (excluding those like Nelson who actually had serious issue with PA). =P – user21820 Sep 18 '18 at 05:52
  • I have a fundamental objection with the notion of truth and false in that the only meaningful truth status is "contradictory". Then if proving it disproves it, it follows that it's contradictory, which is as close to false as it gets for my money anyway, so it's false. – samerivertwice Nov 08 '18 at 14:59

$e+\pi$ is rational

The most that I could find on this was that at least one of $e\pi$ and $e+\pi$ is irrational. I'd bet on both. But it's currently unproven I think.

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  • https://math.stackexchange.com/questions/159350/why-is-it-hard-to-prove-whether-pie-is-an-irrational-number – JollyJoker Sep 17 '18 at 11:04
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    A bit stronger statement that is easily proved is that at least one of $e \pi$ and $e + \pi$ is transcendental. (This is done by considering the polynomial $(x-e)(x-\pi)$ and expanding it.) – Boaz Moerman Sep 19 '18 at 09:29
  • The problem is open. But this conjecture : https://en.wikipedia.org/wiki/Schanuel%27s_conjecture would imply that it is transcendental. – Peter Sep 19 '18 at 12:07

The Riemann Hypothesis is FALSE.

Nilotpal Sinha
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    Some great mathematicians would bet that the RH is false, for example, Littlewood. – bof Sep 17 '18 at 06:42
  • @bof, Littlewood is no onger in a position to make wagers. – Gerry Myerson Sep 17 '18 at 06:50
  • @bof Some other great mathematicians would bet that the RH is true, for example, Reiamnn :) – Nilotpal Sinha Sep 17 '18 at 06:54
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    @GerryMyerson You may be right. Personally, I don't know what Littlewood may or may not be in a position to do. – bof Sep 17 '18 at 06:58
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    You don't have to follow my rules, but I can't believe this gets upvoted...I'd much rather bet on pi not ending with 0s and 1s. – bobuhito Sep 17 '18 at 07:03
  • I'd much rather bet on pi not ending. – Gerry Myerson Sep 17 '18 at 07:04
  • @Gerry I'm sure you're joking, but to keep it real, that's got a proof so not allowed. – bobuhito Sep 17 '18 at 07:07
  • Littlewood does not just use his authority, he actually has some arguments that cannot be just ignored. I read an article about some doubts, the Lehmer phenomenon is the most serious argument. That the mathematical community apparently has decided to ignore these arguments, is another story. – Peter Sep 19 '18 at 11:42
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    @Peter, I doubt the community has decided to ignore Littlewood's arguments. I rather imagine that individuals have looked at the best arguments for both sides and then put their money on whichever outcome they deem more likely. It doesn't matter, anyway, since the truth of RH will not be decided by a vote, but by a proof. – Gerry Myerson Sep 19 '18 at 14:00

This one: $\zeta(5) \in\Bbb Q$.

Nilotpal Sinha
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The base-10 square root of every prime number eventually only includes 0s and 1s.

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My favorite "not disproved, but highly unlikely" statement is:

The number $\Large \pi^{\pi^{\pi^\pi}}$ is an integer.

I don't have a solid source for this being an open problem, but the question is discussed (and considered to be an open problem) at "Why is it so difficult to determine whether or not $\Large \pi^{\pi^{\pi^\pi}}$ is an integer?"

(The reason given there is that the number has over 600,000,000,000,000,000 digits to the left of the decimal point, and has never been computed to that amount of precision.)

Tanner Swett
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    Please add an explanation to support the claim this has not yet been disproven. – hardmath Sep 17 '18 at 18:57
  • "the number has over 600,000,000,000,000,000 decimal places" Sorry but what does that even mean? Remember that the phrase "decimal places" refers to digits to the right of the decimal point, not to the left... – Did Sep 18 '18 at 18:15
  • @Did Whoops! Thanks for pointing that out. I've changed it to "digits to the left of the decimal point". – Tanner Swett Sep 18 '18 at 19:29
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    For those who are interested : The exact number of digits of the integer part of this number is $$666\ 262\ 452\ 970\ 848\ 504$$ which is approximately $6.66\cdot 10^{17}$. – Peter Sep 19 '18 at 11:28

If we denote by $d_n=p_{n+1}-p_n$ the difference of two consecutive primes, then the following holds true:
The two inequalities $$d_{n+2}>d_{n+1}>d_n$$ and $$d_{n+2}<d_{n+1}<d_n$$ occur finitely often. This would mean that the sequence $d_{n+1}-d_n$ will alternate in sign (from some point on).
This is of course something which (surely) does not happen, but it is still I think an open problem.

Konstantinos Gaitanas
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  • I was expecting to see more statements about primes, and do agree this sounds quite unlikely. To be even more specific and unbelievable, is it possible the d sequence alternates 2, N1, 2, N2... (from some point on)? – bobuhito Oct 01 '18 at 04:22

My favorite:

The only primes in the set of Fermat numbers $F_n=2^{2^n}+1$ are $F_0,F_1,F_2,F_3$ and $F_4$.

Martin Hopf
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If Peano Arithmetic is consistent, then by Löb's theorem, Peano Arithmetic is consistent with the sentence $$``1+1=3 \text{ is provable}",$$coded in arithmetic. Therefore, we cannot disprove (in PA) the provability of $1+1=3$.

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