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I was perusing this textbook on algorithmic number theory, where I came across this page where they appear to prove that the Riemann Hypothesis holds almost surely.

This seems like an odd statement for something that is not random (a Theorem). Yet they derive a probability measure for this and proceed forward...

I've heard of the "probabilistic method" ala Erdos, but this seems different. It's not establishing the existence of something, but making a statement about the truth value of a theorem.

Note: This question is similar to one I asked some time back under a now-defunct user name. That also referenced a work (Goldbach's conjecture) that used probability in number theory, but non in an the manner of Erdos.

Question For the Riemann Hypothesis, how would I interpret the statement that this hypothesis is "almost surely true"...would this have any weight in the mathematical community or is it mainly a useful heuristic for producing algorithms (aka...we can assume the hypothesis is true due to the high "probability" of it being true)...?

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    $\begingroup$ We have used computers to search for zeros and every (non-trivial) zero that has been found lies on the critical line. But, mathematics is a not a court of law and "preponderance of the evidence" is not proof. $\endgroup$
    – Doug M
    Apr 20, 2016 at 21:11
  • $\begingroup$ @DougM I don't understand your comment. The authors made a probabilistic argument not a statistical one. $\endgroup$
    – user237392
    Apr 20, 2016 at 21:18
  • $\begingroup$ The FAQ has instruction in how to merge your accounts. I am about to put something on the old question. $\endgroup$
    – Ross Millikan
    Dec 19, 2016 at 20:43
  • $\begingroup$ This made it seem plausible that RH might not be true... $\endgroup$
    – Goldbug
    Jul 16, 2019 at 3:03

2 Answers 2

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The book says "We now explain why the Riemann hypothesis is plausible on probabilistic grounds". That answers the question. We cannot assume that RH is true "due to high probability of being true". It is just one more indication that we should believe in it. A proof, however, is something different.

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  • $\begingroup$ +1 Thanks..:but why is probability vaild ground for belief in this context? $\endgroup$
    – user237392
    Apr 20, 2016 at 21:11
  • $\begingroup$ Well, you are right. There are better reasons to believe in RH, but nonetheless it is something. $\endgroup$
    – Dietrich Burde
    Apr 20, 2016 at 21:15
  • $\begingroup$ Thanks for your responses. There were some other comments on here about Cramer's method that I also found interesting, but the poster took them away... $\endgroup$
    – user237392
    Apr 20, 2016 at 23:32
  • $\begingroup$ @user237392 : the point of the probabilistic argument is that if RH were to be false, then something truly unexpected must hold, so believing it is true sounds more plausible than believing it is false. But hey, people believe in religions sometimes... $\endgroup$
    – Patrick Da Silva
    Dec 19, 2016 at 20:42
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They did not prove that the Riemann hypothesis is almost surely true. For this to be done you would need to classify all zeta functions, find the class that Riemann zeta function belongs to, then prove that for almost all zeta functions in this class the Riemann hypothesis is valid.

From there it would follow that the hypothesis is almost surely true for the Riemann zeta function itself, because it belongs to a class of zeta functions that have negligible many exceptions.

This would not prove that Riemann hypothesis is true for the Riemann zeta function, because it could be that exactly the Riemann zeta function is that negligible exception, but this would be an enormous achievement that would lead to all theorems related to Riemann hypothesis being almost surely true and additionally it would give an enormous boost since almost all zeta functions from a certain class would have Riemann hypothesis valid.

We are not there yet.

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  • $\begingroup$ Also the RH (for $\zeta(s)$ or $\pi(x)$) is true for $\Im(s)$ and $x$ not too large. And if you assume the random model for the primes, then the RH for the random $\Pi(x)$ is almost surely true. $\endgroup$
    – reuns
    Sep 4, 2017 at 18:19

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