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Does anyone know the current progress in showing the Riemann hypothesis? I was only able to find this paper of Conrey that says at least 40% of the zeros of the Riemann Zeta function lie on the critical line.

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closed as off topic by Joël, Andy Putman, Noah Stein, plusepsilon.de, Benjamin Steinberg May 11 '12 at 14:15

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@David: I am pretty sure that such a strip is not known to exist. Proving that every zero has real part smaller than $\delta < 1$ would give an error term in the prime number theorem of $O(x^{\delta})$, which is certainly not known. The curve bounding the current known zero free region in fact approaches the line $s=1$ as $|t|\to \infty$. –  Daniel Loughran May 11 '12 at 8:08
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A huge number of references are recorded here empslocal.ex.ac.uk/people/staff/mrwatkin//zeta/physics.htm, especially concerning the relation of RH to physics. –  David Corfield May 11 '12 at 8:26
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I am not sure that this question is appropriate for MO: it's ways too wide and unspecific. Before opening it, the title give the impression that someone came up with a proof of RH, and that the PO was asking about the status of that proof. RH is probably the most famous open problem in mathematics. Any serious progress on it would be almost immediately reported on say, the wikipedia page on RH, and on many other places. A more legitimate question would be something more specific, like "is it known that 40% of the zeros are on the critical line, and if known, where can I find a reference?". –  Joël May 11 '12 at 12:31
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Dear Eugene, don't take it personally. You do ask good questions on that forum. I just think that this specific question was too wide. RH is a field of research in itself (with many aspects and angle of attacks) so asking in general about "current progress on RH" is a little bit like asking "Hi. Any current progress in algebra ?", or pushing it a little bit farther like "hey guys, what's up (in mathematics)?". –  Joël May 11 '12 at 18:03
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Eugene, it appears you have been on MO for just 60 days or the like, that is from your profile information. As far as history, some of our most pernicious cranks have asked questions in the same areas you are asking about, precisely because these are some of the biggest unsolved problems. If you would like more detail on that email me (or maybe Yemon). Meanwhile, I'd like to encourage you to see MO as less of an encyclopedia about exciting problems, and more of a resource when you are stuck in your own work. In short, questions on RH and BSD make me nervous. –  Will Jagy May 11 '12 at 19:26

2 Answers 2

up vote 11 down vote accepted

In terms of fraction of zeros on the critical line (which seems to be your question), the best result to date is 41.05% by Bui, Conrey and Young ("More than 41% of the zeros of the zeta function are on the critical line" arXiv:1002.4127). Of course, this is only one measure of progress.

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Thanks for the reference. –  Eugene May 11 '12 at 18:43

It is unsolved as of today.

However, some latest researches are:

Edit:

FWIW, do note that the shortest "proof" of RH by Mark Colyvan as mentioned in IEP here using paraconsistent logic:

As the founders of relevant logic, Anderson and Belnap, urge in their canonical book Entailment, a ‘proof’ submitted to a mathematics journal in which the essential steps fail to provide a reason to believe the conclusion, e.g. a proof by explosion, would be rejected out of hand. Mark Colyvan (2008) illustrates the point by noting that no one has laid claim to a startlingly simple proof of the Riemann hypothesis:

Riemann’s Hypothesis: All the zeros of the zeta function have real part equal to > 1/2.
Proof: Let R stand for the Russell set, the set of all sets that are not members of themselves. It is straightforward to show that this set is both a member of itself and not a member of itself. Therefore, all the zeros of Riemann’s zeta function have real part equal to 1/2.

Needless to say, the Riemann hypothesis remains an open problem at time of writing.

The cited 2008 article by Colyvan however does not use this for RH but Fermat's Last Theorem.

Speaking of non-classical approach, Douglas S. Bridges further writes about the self-referential nature in Reality and Virtual Reality in Mathematics (2006):

There is an even more dramatic example of a proof which might cause the same unease. A famous conjecture of Riemann in the nineteenth century, the Riemann Hypothesis, remains unsolved today despite the efforts of some of the greatest mathematicians in the intervening 150 years. Early last century, the English mathematician J.E. Littlewood produced a theorem whose difficult proof was split into two cases. In the first case, Littlewood assumed that the Riemann Hypothesis was true, and in the second that it was false. Writing R to denote the Riemann Hypothesis, and P to denote the conclusion of Littlewoods theorem, we can express his proof in the schematic form
$(R\bigvee\neg R)\Rightarrow P.$ (1)
Here I have introduced the standard logical symbols $\bigvee$ (or), $\neg$ (not), and $\Rightarrow$ (implies) What is the meaning of Littlewoods proof? Since we are unable at this date to decide whether or not the Riemann Hypothesis is true, we cannot say which of the two cases of his proof actually applies. If, as most mathematicians expect, the Riemann Hypothesis turns out to be provable, then that part of Littlewoods proof that is based on the assumption that the Riemann Hypothesis is false is worthless and can be thrown away. Moreover, in such a proof, if $P$ is an existential statement one that asserts the existence of a certain object $x$ with certain properties then the two cases of a proof of $P$ that follows the Littlewoods schematic form (1) may produce di¤erent objects $x$ with the desired properties (as in our earlier proof involving $\sqrt{2}^{\sqrt{2}}$ ); under such circumstances, we might be unable to tell which of the two possibilities for x was the desired one until we could prove the truth or falsity of the Riemann Hypothesis.
The formalist might attempt to remove our unease about Littlewoods proof as follows. Suppose that the desired conclusion $P$ of Littlewoods theorem is false. Then Littlewoods arguments, schematised in (1), show that neither the Riemann Hypothesis nor its negation can hold (since each of these alternatives leads us to a proof of $P$): In other words, if $P$ is false, then the Riemann Hypothesis is false and it is false that the Riemann Hypothesis is false! This is plainly absurd. Hence we conclude that $P$ cannot be false and is therefore true.

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+1 for your first sentence! –  David Roberts May 11 '12 at 7:30

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