BIG LIST: Statements that look obviously false but cannot be disproved

Question

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

Answer 1

The continuum is $\aleph_{37}$.

Answer 2

The Riemann Hypothesis is FALSE.

Answer 3

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

Answer 4

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.

Answer 5

$$P = NP$$

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.

Answer 6

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.

Answer 7

$e+\pi$ 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.

Answer 8

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

Answer 9

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.)

Answer 10

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.

Answer 11

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 $F_4$.

Answer 12

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 $1+1=3$.