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There are many unproven conjectures that if you heuristically took a probability of it being true, it would basically be almost $100\%$ true.

I can see why mathematics must be rigorous as many conjectures thought to be true were proven false, but these conjectures did not heuristically have a $100\%$ chance of being true. Many of these conjectures proven false have an issue from some condition holding true for an infinite number of integers, many times unjustifiably.

So what we would lose from saying things like $\pi^{{\pi}^{\pi^\pi}}$ is not an integer and $e + \pi$ is irrational just because we didn't prove it?

MarianD
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Matthew Liu
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    Because sometimes even highly plausible conjectures are wrong. – Travis Willse Mar 04 '19 at 20:16
  • Pretty much the only thing you could lose is everything if the conjecture is eventually disproven. That being said, in maths you are free to assume any unproven statement, but you should make it clear whenever you do. – Wojowu Mar 04 '19 at 20:16
  • "these conjectures did not heuristically have a 100% chance of being true." Are you sure about that? And are you sure that this will continue to be the case forever? – Noah Schweber Mar 04 '19 at 20:17
  • Because if you use this result later, suddenly you don't know what is only approximately true and what is true exactly. Moreover, some theorems depend on something happening infinitely many times. – enedil Mar 04 '19 at 20:17
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    For example, take the relativized $P\not= NP$ conjecture: that $P^A\not=NP^A$ for any oracle $A$. Baker-Gill-Solovay proved a pair of interesting results: measure-one many oracles $A$ satisfy $P^A\not=NP^A$, but there are oracles $A$ satisfying $P^A=NP^A$ as well. So the relativized $P\not=NP$ conjecture has a "probability 1" justification, but is false. And that's just one I thought of off the top of my head. – Noah Schweber Mar 04 '19 at 20:19
  • I don't understand your use of the word "heuristically" in this context. How are you putting a probablity on a mathematical statement being true or not? – Malcolm Mar 04 '19 at 20:19
  • @Noah Schweber That seems to be like saying all real numbers in [0,1] are not integers just because the measure is 1. For a specific number the probability is 0 but if you add up the probabilities for all real numbers then it is no longer 0. For the examples I've given, the probability of them being true is essentially 100%. – Matthew Liu Mar 04 '19 at 20:50
  • @MatthewLiu Exactly, that's sort of my point (although I chose to give a plausible example that people actually conjectured): what precisely do you mean by "probability one"? At the very least you have to give some reason to expect your notion to never run into errors. If you want to change the way math is done, the onus is on you to justify why that won't be a disaster. – Noah Schweber Mar 04 '19 at 20:51
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    "If you heuristically took a probability of it being true, it would basically be almost 100% true" is what they call not even wrong. There is no sample space, no probability measure, "there is no scientific basis on which to form any calculable probability whatever", as Keynes put it. And given the classical law of explosion, the effects of even a single false conjecture will be global. So we have everything to lose, and no tangible way to assess risk. – Conifold Mar 04 '19 at 21:33
  • @MatthewLiu "For the examples I've given, the probability of them being true is essentially 100%." Based on what exactly? – Noah Schweber Mar 06 '19 at 00:47

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Let me give one catastrophic example: You may be familiar with the Moebius function $\mu(n)$ which is defined as $0$ if $n$ has any factor which is the square of a prime, and for square-free number is $+1$ if $n$ has an even number of prime factors, and $-1$ otherwise.

Now let's look at a an integer-valued function function $M(n)$ the sum $$ M(n) = \sum_{k=1}^n \mu(k)$$ (this is known as the Merten's function) and observe (without proving it) that while $M(1) = 1$, for all $n>1$, $|M(n)| < \sqrt{n}$. We try this out for up to $n=1,000,000$ and it always works, so we call this Merten's conjecture.

Being a bit OCD, we try it out for all $n < 10^{16}$ and not only does it always work, $|M(n)|$ is always less than $\frac35$ for all those cases. So now we decide, let's assume the conjecture is true; let's add it to our body of axioms that we can use in future proofs.

Well, pretty soon we are happy, because now we can prove the Riemann Hypothesis (a much more important conjecture, but one that is perhaps tougher to convince yourself of by numerical examples). And we continue to develop a bunch of other math, so much so that a century from now we forget which theorems depend on the Mertens axiom and which don't.

Along the way, somebody manages to prove the $3n+1$ conjecture (that if you start with any integer and if odd, change it to $3n+1$ or if even, divide by $2$, the conjecture is that eventually you end up with $1$). I can assure you that this is possible starting from theorems derived assuming the Mertens conjecture. And $50$ years later, some student discovers (using the processor on her smartphone) a number which instead of reaching $1$, goes into a cycle that repeats forever.

Well now the world is in the unpleasant position of knowing that the math that people have been working on contains contradictions, meaning that anything can be proven. Good luck trying to unravel which pieces of 21st century math depend on the Mertens axiom and which don't. What a mess.

(By the way, the Mertens conjecture is known to be false, although we may never have computers strong enough to find the smallest counterexample.)

Mark Fischler
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  • There is no way mertens conjecture can possibly be heuristically be 100% true when summing all n towards infinity. Showing cases for n < 10^16 is actually meaningless unless a conjecture has already shown that if a counter example exists, it must be below some number. 10^16 is still 0% of infinity. – Matthew Liu Mar 05 '19 at 03:51