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Author
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Topic: They may have disproven the Riemann hypothesis
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500_Apples
rabble-rouser
Babbler # 12684
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posted 23 May 2007 06:08 AM
Paper Here quote:
Thursday, March 22, 2007
Riemann wrong?
Someone claims to have disproved the Riemann hypothesis!This is one of the most famous unsolved problems in mathematics, first posed by Bernhard Riemann (pictured) in 1859. Tribikram Pati has posted a paper on the ArXiv which shows that an assumption of the truth of the hypothesis leads to a contradiction. Therefore, the hypothesis must be false. His disproof could be worth $1,000,000 as the Riemann hypothesis is one of the Millennium Prize Problems. It is also the only one of Hilbert's problems that made it onto the list. It's worth pointing out that there have been several attempts at proving the hypothesis, many listed here, but will this latest effort finally settle it? I dunno! It's completely out of my area of knowledge so it could be utterly wrong for all I know... I Feel I should urge caution, especially after I complained about people jumping to the wrong conclusions earlier this week... More discussion on the claim here, here and here. I don't think I've wrote a post every day during the week for a long time... I'd better go to uni now, cobordism is calling me!
Link
From: Montreal, Quebec | Registered: Jun 2006
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unionist
rabble-rouser
Babbler # 11323
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posted 23 May 2007 07:02 AM
Well, who knows for sure. It'll take quite some time. But I hear Pati hasn't published much in almost 40 years (and most breakthroughs in math are made by the young, unfortunately) - and it was only a pre-print, not peer-reviewed. So we get lots of questions and stuff like this, which you have to be a math fiend to even want to begin to start verifying or refuting: quote:
The Pati paper is incorrect, due to a circular choice of parameter definitions. On page 10 equations (10), (11), t is defined in terms of delta, and delta defined in terms of A_*, where A_* is to be defined "in the sequel". On page 13 equation (20), A_* is defined in terms of t. One cannot make all of these definitions simultaneously.
I'll wait for the MAA or AMS or someone of like authority to make the announcement before I crack out my Riemann-era vintage champagne. ETA: I had inadvertently conflated the MAA and AMS into AMA. The Unionist Conjecture: quote:
For all x such that 1 < x < n, where n grows exponentially, MAA + AMS is never equal to AMA.
[ 23 May 2007: Message edited by: unionist ]
From: Vote QS! | Registered: Dec 2005
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