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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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    $\begingroup$ Pretty much same as mathoverflow.net/questions/259844/… $\endgroup$
    – Gerry Myerson
    Commented Sep 17, 2018 at 6:51
  • $\begingroup$ @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. $\endgroup$
    – bobuhito
    Commented Sep 17, 2018 at 6:58
  • $\begingroup$ 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? $\endgroup$
    – Davislor
    Commented Sep 17, 2018 at 18:27
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    $\begingroup$ 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. $\endgroup$
    – user296602
    Commented Sep 17, 2018 at 21:44
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    $\begingroup$ Except for the one which checked for a prime number, it was more of a feeling than math...so, I started a chat where we can talk about any one in particular. $\endgroup$
    – bobuhito
    Commented Sep 19, 2018 at 15:05

12 Answers 12

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$$P = NP$$

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.

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    $\begingroup$ 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. $\endgroup$
    – bobuhito
    Commented Sep 17, 2018 at 15:42
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    $\begingroup$ 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. $\endgroup$
    – mephistolotl
    Commented Sep 17, 2018 at 17:24
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    $\begingroup$ 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. $\endgroup$
    – usul
    Commented Sep 17, 2018 at 18:53
  • $\begingroup$ @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. $\endgroup$
    – Yonatan N
    Commented Sep 17, 2018 at 23:02
  • $\begingroup$ The Robertson-Seymour 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. $\endgroup$
    – John Coleman
    Commented Sep 18, 2018 at 2:31
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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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    $\begingroup$ 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"? $\endgroup$
    – user541686
    Commented Sep 17, 2018 at 10:05
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    $\begingroup$ @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. $\endgroup$
    – hvd
    Commented Sep 17, 2018 at 12:30
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    $\begingroup$ 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. $\endgroup$
    – Wojowu
    Commented Sep 17, 2018 at 12:47
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    $\begingroup$ Overmind: 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. $\endgroup$
    – Alan Baljeu
    Commented Sep 17, 2018 at 14:34
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    $\begingroup$ "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. $\endgroup$
    – Kevin
    Commented Sep 17, 2018 at 17:27
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The continuum is $\aleph_{37}$.

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    $\begingroup$ 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... $\endgroup$
    – user21820
    Commented Sep 17, 2018 at 9:40
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    $\begingroup$ @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? $\endgroup$
    – user541686
    Commented Sep 17, 2018 at 10:07
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    $\begingroup$ @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. en.wikipedia.org/wiki/Continuum_hypothesis will get you started. $\endgroup$
    – Gerry Myerson
    Commented Sep 17, 2018 at 12:51
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    $\begingroup$ $\aleph_{42}$ ist more likely. $\endgroup$
    – Dirk
    Commented Sep 17, 2018 at 17:11
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    $\begingroup$ @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 $0=1$. $\endgroup$
    – hmakholm left over Monica
    Commented Sep 17, 2018 at 18:05
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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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    $\begingroup$ 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. $\endgroup$
    – bobuhito
    Commented Sep 17, 2018 at 15:32
  • $\begingroup$ I upvoted this for the sheer nonsense value of "We can prove that we cannot disprove it, unless it is actually true." $\endgroup$
    – Wildcard
    Commented Sep 17, 2018 at 16:15
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    $\begingroup$ 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). $\endgroup$
    – hmakholm left over Monica
    Commented Sep 17, 2018 at 18:18
  • $\begingroup$ @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 $\endgroup$
    – user21820
    Commented Sep 18, 2018 at 5:52
  • $\begingroup$ 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. $\endgroup$
    – it's a hire car baby
    Commented Nov 8, 2018 at 14:59
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$e+\pi$ 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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  • $\begingroup$ math.stackexchange.com/questions/159350/… $\endgroup$
    – JollyJoker
    Commented Sep 17, 2018 at 11:04
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    $\begingroup$ 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.) $\endgroup$
    – Boaz Moerman
    Commented Sep 19, 2018 at 9:29
  • $\begingroup$ The problem is open. But this conjecture : en.wikipedia.org/wiki/Schanuel%27s_conjecture would imply that it is transcendental. $\endgroup$
    – Peter
    Commented Sep 19, 2018 at 12:07
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This one: $\zeta(5) \in\Bbb Q$.

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The Riemann Hypothesis is FALSE.

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    $\begingroup$ Some great mathematicians would bet that the RH is false, for example, Littlewood. $\endgroup$
    – bof
    Commented Sep 17, 2018 at 6:42
  • $\begingroup$ @bof, Littlewood is no onger in a position to make wagers. $\endgroup$
    – Gerry Myerson
    Commented Sep 17, 2018 at 6:50
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    $\begingroup$ @GerryMyerson You may be right. Personally, I don't know what Littlewood may or may not be in a position to do. $\endgroup$
    – bof
    Commented Sep 17, 2018 at 6:58
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    $\begingroup$ 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. $\endgroup$
    – bobuhito
    Commented Sep 17, 2018 at 7:03
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    $\begingroup$ @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. $\endgroup$
    – Gerry Myerson
    Commented Sep 19, 2018 at 14:00
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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.)

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    $\begingroup$ Please add an explanation to support the claim this has not yet been disproven. $\endgroup$
    – hardmath
    Commented Sep 17, 2018 at 18:57
  • $\begingroup$ "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... $\endgroup$
    – Did
    Commented Sep 18, 2018 at 18:15
  • $\begingroup$ @Did Whoops! Thanks for pointing that out. I've changed it to "digits to the left of the decimal point". $\endgroup$
    – Tanner Swett
    Commented Sep 18, 2018 at 19:29
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    $\begingroup$ 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 $6.66\cdot 10^{17}$. $\endgroup$
    – Peter
    Commented Sep 19, 2018 at 11:28
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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:

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 $F_4$.

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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.

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  • $\begingroup$ 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)? $\endgroup$
    – bobuhito
    Commented Oct 1, 2018 at 4:22
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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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