# Claimed Proof of Riemann Hypothesis

#### CmdrTaco posted about 6 years ago | from the two-scoops-of-math-please dept.

345
An anonymous reader writes *"Xian-Jin Li claims to have proven the Riemann hypothesis in this preprint on the arXiv."* We've mentioned recent advances in the search for a proof but if true, I'm told this is important stuff. Me, I use math to write dirty words on my calculator.

## Dirty Words (5, Funny)

## Rik Sweeney (471717) | about 6 years ago | (#24031621)

Me, I use math to write dirty words on my calculator.Such as 80085?

## Re:Dirty Words (5, Funny)

## Anonymous Coward | about 6 years ago | (#24031671)

5318008

## Re:Dirty Words (1, Offtopic)

## G-forze (1169271) | about 6 years ago | (#24031681)

## Re:Dirty Words (1, Offtopic)

## bigstrat2003 (1058574) | about 6 years ago | (#24031761)

## Re:Dirty Words (0, Offtopic)

## Trails (629752) | about 6 years ago | (#24032537)

## Re:Dirty Words (1)

## rubah (1197475) | about 6 years ago | (#24033151)

Luckily the numerals 0, 8 and 5 are three of five that look exactly the same even upside down in liquid crystal notation.

You can continue to turn your calculator 180 degrees, but why bother when you can be reaching into your pants. . . Or wherever mathnerds reach :O

(inside of gabriel's horn moarlike)

## Re:Dirty Words (5, Funny)

## UnknowingFool (672806) | about 6 years ago | (#24031763)

5318008or for the slashdot crowd,55378008## Re:Dirty Words (4, Funny)

## Directrix1 (157787) | about 6 years ago | (#24031985)

No for the slashdot crowd it would be: 58008uÉÉ . Because obviously we all have calculators that support unicode text entry.

## Re:Dirty Words (5, Funny)

## Directrix1 (157787) | about 6 years ago | (#24032019)

That would've been a lot cooler if Slashdot supported Unicode.

## Re:Dirty Words (4, Funny)

## Firehed (942385) | about 6 years ago | (#24032637)

At that point, isn't it safe to assume that our calculators can just draw a pair of boobs in 2-bit greyscale?

And that we've written apps that simulate what we assume bouncing would look like given our collective lack of experience outside of the pornographic realm?

## Re:Dirty Words (4, Funny)

## Archangel Michael (180766) | about 6 years ago | (#24032037)

On linux, wouldn't it be ...

host:>man 80085

???

## Try this. (-1, Offtopic)

## Anonymous Coward | about 6 years ago | (#24031965)

timeseach day. What was she after that?## Re:Try this. (0, Offtopic)

## chaboud (231590) | about 6 years ago | (#24032197)

There was 1 girl, who was 16, she 69'ed 3

times.What was she?

## Re:Try this. (5, Funny)

## Anonymous Coward | about 6 years ago | (#24032555)

your mother?

## Re:Dirty Words (1, Troll)

## Smidge207 (1278042) | about 6 years ago | (#24031969)

Your post advocates a

(X) technical ( ) legislative ( ) market-based ( ) vigilante

approach to fighting spam. Your idea will not work. Here is why it won't work. (One or more of the following may apply to your particular idea, and it may have other flaws which used to vary from state to state before a bad federal law was passed.)

( ) Spammers can easily use it to harvest email addresses

( ) Mailing lists and other legitimate email uses would be affected

(X) No one will be able to find the guy or collect the money

( ) It is defenseless against brute force attacks

( ) It will stop spam for two weeks and then we'll be stuck with it

( ) Users of email will not put up with it

(X) Microsoft will not put up with it

(X) The police will not put up with it

(X) Requires too much cooperation from spammers

( ) Requires immediate total cooperation from everybody at once

( ) Many email users cannot afford to lose business or alienate potential employers

( ) Spammers don't care about invalid addresses in their lists

( ) Anyone could anonymously destroy anyone else's career or business

Specifically, your plan fails to account for

(X) Laws expressly prohibiting it

( ) Lack of centrally controlling authority for email

(X) Open relays in foreign countries

( ) Ease of searching tiny alphanumeric address space of all email addresses

(X) Asshats

(X) Jurisdictional problems

( ) Unpopularity of weird new taxes

( ) Public reluctance to accept weird new forms of money

( ) Huge existing software investment in SMTP

( ) Susceptibility of protocols other than SMTP to attack

( ) Willingness of users to install OS patches received by email

( ) Armies of worm riddled broadband-connected Windows boxes

( ) Eternal arms race involved in all filtering approaches

( ) Extreme profitability of spam

( ) Joe jobs and/or identity theft

( ) Technically illiterate politicians

( ) Extreme stupidity on the part of people who do business with spammers

( ) Dishonesty on the part of spammers themselves

( ) Bandwidth costs that are unaffected by client filtering

( ) Outlook

and the following philosophical objections may also apply:

(X) Ideas similar to yours are easy to come up with, yet none have ever been shown practical

( ) Any scheme based on opt-out is unacceptable

(X) SMTP headers should not be the subject of legislation

( ) Blacklists suck

( ) Whitelists suck

(X) We should be able to talk about Viagra without being censored

( ) Countermeasures should not involve wire fraud or credit card fraud

( ) Countermeasures should not involve sabotage of public networks

( ) Countermeasures must work if phased in gradually

(X) Sending email should be free

( ) Why should we have to trust you and your servers?

( ) Incompatiblity with open source or open source licenses

( ) Feel-good measures do nothing to solve the problem

( ) Temporary/one-time email addresses are cumbersome

( ) I don't want the government reading my email

( ) Killing them that way is not slow and painful enough

Furthermore, this is what I think about you:

( ) Sorry dude, but I don't think it would work.

(X) This is a stupid idea, and you're a stupid person for suggesting it.

( ) Nice try, assh0le! I'm going to find out where you live and burn your house down!

## Re:Dirty Words (-1, Offtopic)

## Anonymous Coward | about 6 years ago | (#24032673)

Why don't you

1134206?## Re:Dirty Words (0, Offtopic)

## 8ball629 (963244) | about 6 years ago | (#24032695)

80084 + 1 = 80085

## Re:Dirty Words (2, Funny)

## JayJay.br (206867) | about 6 years ago | (#24033107)

Newbie...

correct spelling is "5318008" and you have to look at the calculator "umop apisdn"

Mod me down, I dare you!!!

## I like to describe my workplace with my calculator (2, Informative)

## InvisblePinkUnicorn (1126837) | about 6 years ago | (#24031635)

## Re:I like to describe my workplace with my calcula (0, Offtopic)

## theskipper (461997) | about 6 years ago | (#24032059)

5318008

The

trueking of beghilos, and one that I think about at least 50 times a day.## Yeah but did they point this out? (5, Funny)

## i_want_you_to_throw_ (559379) | about 6 years ago | (#24031645)

## Re:Yeah but did they point this out? (5, Funny)

## rdwald (831442) | about 6 years ago | (#24032393)

By using Fourier analysis on number fields, we prove in this paper E. Bombieri's refinement of A. Weil's positivity condition, which implies the Riemann hypothesis for the Riemann zeta function in the spirit of A. Connes' approach to the Riemann hypothesis.

Weather permitting of course. (Just looking on the positivity side)

I thought you were randomly babbling, but then I RTFA and realized you were just quoting it...

## Re:Yeah but did they point this out? (5, Funny)

## colonslashslash (762464) | about 6 years ago | (#24033039)

bothof you RTFA?We have a new

## Tried to RTFA (5, Funny)

## multipartmixed (163409) | about 6 years ago | (#24031657)

Man, where's Charles Eppes when you need something explained to you in layman's terms?

## Re:Tried to RTFA (4, Funny)

## Notquitecajun (1073646) | about 6 years ago | (#24032073)

## Re:Tried to RTFA (5, Funny)

## stranger_to_himself (1132241) | about 6 years ago | (#24032439)

Ummm...I think that WAS layman's terms. For you math geeks, try being a

history majorand looking at all that. It just looks like a cat walked on the keyboard to me...Are you reading slashdot as some kind of anthropological study?

## Re:Tried to RTFA (5, Informative)

## PlatyPaul (690601) | about 6 years ago | (#24032369)

\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}}[written for LaTeX], or "the sum of 1/(n^s) as n goes from 0 to infinity (increasing by 1 repeatedly)" [in more human-readable form].Riemann was interested in the zeros to this function, where s is a complex number. He conjectured that all zeros (aside from those of the form s = -2c, where c is a positive integer) would have to be of the form (1/2) + ki, where k is a constant and i is the square root of -1.

This paper is saying that they've found a way to verify this intuition by patching a hole in a previous attempt.

Assuming that everything is correct (a

bigassumption), this would finally solve a long-standing problem (dating back to 1859).Details of the actual solution

area bit heavy. Those actually interested in this sort of number theory might want to start here [amazon.com] .## typo (4, Informative)

## Ungrounded Lightning (62228) | about 6 years ago | (#24032481)

The Riemann zeta function is \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} [written for LaTeX], or "the sum of 1/(n^s) as n goes from 0 to infinity (increasing by 1 repeatedly)" [in more human-readable form].You have a slight typo. Should be: "... as n goes from 1 to infinity ..."

## Numb3rs (5, Funny)

## netsavior (627338) | about 6 years ago | (#24032541)

## Dolly parton bought a size 69 bra (1, Funny)

## larry bagina (561269) | about 6 years ago | (#24031669)

## Re:Dolly parton bought a size 69 bra (-1, Offtopic)

## Anonymous Coward | about 6 years ago | (#24031865)

## Re:Dolly parton bought a size 69 bra (1)

## u38cg (607297) | about 6 years ago | (#24032251)

The version I learnt as a kid: There was a girl of 13, who had a bust of 84. She wanted to make it 45, so she went to the doctor. 0, he said. Take these pills 2 times a day - instead she took them four. Of course, she ended up...

## Re:Dolly parton bought a size 69 bra (1)

## Anonymous Cowpat (788193) | about 6 years ago | (#24032507)

Does that make me the only person who remembered how "boobless" is spelt and just typed 55318008 into the calculator when they wanted something to snigger at?

## Re:Dolly parton bought a size 69 bra (1)

## bigstrat2003 (1058574) | about 6 years ago | (#24032969)

## $1,000,000 prize to be collected then if true (4, Informative)

## deft (253558) | about 6 years ago | (#24031779)

Was reading wikipedia because I have no idea why this is important, but need to know enough to impress my friends (and by that I mean, alienate).

But I noticed this is such a big deal, theres a cool million waiting for the person that proves it. John Nash in "beautiful Mind" tries to prove this one too. Sorry gladiator... not today!

So yeah, Check it out, notice the offer at the end, after all the completely unintelligible mathematicrap:

Riemann hypothesis

The Riemann hypothesis (also called the Riemann zeta-hypothesis), first formulated by Bernhard Riemann in 1859, is one of the most famous and important unsolved problems in mathematics. It has been an open question for almost 150 years, despite attracting concentrated efforts from many outstanding mathematicians. Unlike some other celebrated problems, it is more attractive to professionals in the field than to amateurs.

The Riemann hypothesis (RH) is a conjecture about the distribution of the zeros of the Riemann zeta-function (s). The Riemann zeta-function is defined for all complex numbers s 1. It has zeros at the negative even integers (i.e. at s = 2, s = 4, s = 6, ...). These are called the trivial zeros. The Riemann hypothesis is concerned with the non-trivial zeros, and states that:

The real part of any non-trivial zero of the Riemann zeta function is ½.

Thus the non-trivial zeros should lie on the so-called critical line, ½ + it, where t is a real number and i is the imaginary unit. The Riemann zeta-function along the critical line is sometimes studied in terms of the Z-function, whose real zeros correspond to the zeros of the zeta-function on the critical line.

The Riemann hypothesis is one of the most important open problems of contemporary mathematics, mainly because a large number of deep and important other results have been proven under the condition that it holds. Most mathematicians believe the Riemann hypothesis to be true.[1] A $1,000,000 prize has been offered by the Clay Mathematics Institute for the first correct proof.[2]

## Re:$1,000,000 prize to be collected then if true (5, Informative)

## rufty_tufty (888596) | about 6 years ago | (#24031953)

Good explanation here too:

http://www.irregularwebcomic.net/1960.html [irregularwebcomic.net]

## Re:$1,000,000 prize to be collected then if true (1)

## nwf (25607) | about 6 years ago | (#24032491)

## Re:$1,000,000 prize to be collected then if true (0)

## Anonymous Coward | about 6 years ago | (#24032511)

Hmmm... This is the start of a very bad trend for me as a

## Re:$1,000,000 prize to be collected then if true (1)

## camperdave (969942) | about 6 years ago | (#24033043)

http://www.irregularwebcomic.net/1960.htmlGreat! Now how am I supposed to get any work done.

## Tough problems (3, Interesting)

## dj245 (732906) | about 6 years ago | (#24031975)

isfirst. I studied the Riemann hypothesis in college for a good week and I'm still not sure where you might begin solving it. Like the Navier-Stokes equations (another big problem with a big prize) solving it will probably require the invention of some new mathematics. Its not simply a matter of dividing by 3 and carrying the 2. I don't know about you but I haven't the slightest idea about how to go about inventing new math. That's the realm of Newton and Einstein, and few others.New math is the only way to go about solving some of these problems.

## Re:Tough problems (2)

## aproposofwhat (1019098) | about 6 years ago | (#24032239)

You mean like this? [aol.com]

## Re:Tough problems (4, Funny)

## afabbro (33948) | about 6 years ago | (#24032717)

...solving it will probably require the invention of some new mathematics. Its not simply a matter of dividing by 3 and carrying the 2.If you're carrying numbers when dividing, I guess you :-)

areinventing new math## Re:$1,000,000 prize to be collected then if true (4, Insightful)

## Anonymous Coward | about 6 years ago | (#24032131)

The Riemann hypothesis is considered the most important unsolved problem in math. But, considering the source here (random paper on ArXiv by complete unknown), there's no real reason to believe this paper is correct. The number of incorrect proofs to major mathematics problems every year is staggering.

## Re:$1,000,000 prize to be collected then if true (0)

## Anonymous Coward | about 6 years ago | (#24032269)

Oh, and this summary of the Riemann hypothesis somewhat obscures its true importance: it is equivalent to a very strong statement about the distribution of prime numbers. For example, the CS paper from a few years ago that gave a deterministic algorithm to check if a number was prime depended on the Riemann hypothesis for some of its results.

## Re:$1,000,000 prize to be collected then if true (1, Informative)

## Mr. Sketch (111112) | about 6 years ago | (#24032291)

I briefly looked through the proof, and it only claims to be a proof in the rational number field, not for all real numbers. It's still a step in the right direction, but not a full proof.

## Re:$1,000,000 prize to be collected then if true (4, Funny)

## UnknowingFool (672806) | about 6 years ago | (#24032751)

## Re:$1,000,000 prize to be collected then if true (5, Informative)

## Anonymous Coward | about 6 years ago | (#24032805)

No. Every number field has its own zeta function. The standard Riemann hypothesis concerns that of the rationals.

## Re:$1,000,000 prize to be collected then if true (0)

## Anonymous Coward | about 6 years ago | (#24032815)

The Riemann Zeta function is the zeta function corresponding to the rational number field, thus it would really prove RH. Do you even know, what a number field is?

## Re:$1,000,000 prize to be collected then if true (-1, Redundant)

## Anonymous Coward | about 6 years ago | (#24032827)

Sounds suspicious when his abstract claims to prove the Riemann hypothesis, then on page two he says:

"To avoid complication I have only considered the rational number field but I feel that the techniques of this paper can be extended..."

## Re:$1,000,000 prize to be collected then if true (-1, Redundant)

## Anonymous Coward | about 6 years ago | (#24032889)

From the paper:

"To avoid the complication of writings, I only considered the rational number ïeld in this paper. But, I feel that techniques of this paper can be adopted to any algebraic number ïeld without much difficulty to give a proof of the Riemann hypothesis for Dedeking zeta functions."

So no, you're right it's not a full proof, but the claim is that this is pretty all you need, and any proof using a different number field would just look the same. (Though admittedly that claim doesn't seem to be backed up..?)

## Re:$1,000,000 prize to be collected then if true (0)

## Anonymous Coward | about 6 years ago | (#24032413)

Basically the hypothesis tries to explain where all prime numbers lie along the number line, all the way to infinity. So it would be pretty important if there is a proof for it.

## Re:$1,000,000 prize to be collected then if true (1)

## olyar (591892) | about 6 years ago | (#24032461)

John Nash in "Beautiful Mind" tries to prove this one too.

One of the things I remember from the book is that he and his wife had a running joke that all babies know the solution to this problem and then forget it when they learn to talk. Maybe Xian-Jin Li had a flashback.

## Re:$1,000,000 prize to be collected then if true (1)

## olyar (591892) | about 6 years ago | (#24032713)

Found that here [wordpress.com]

## Re:$1,000,000 prize to be collected then if true (0)

## Anonymous Coward | about 6 years ago | (#24032613)

Why is the Riemann hypothesis important?

The Riemann hypothesis has a deep connection with prime numbers and proving it will help prove other theorems related to the distribution of prime numbers. This will help make finding the prime factors of large numbers practical and make many widely used cryptographic techniques almost trivial to break. So if the theorem has been proved (unlikely) we could be in for a bit of trouble...

## Re:$1,000,000 prize to be collected then if true (1)

## Lundse (1036754) | about 6 years ago | (#24033019)

Was reading wikipedia because I have no idea why this is important, but need to know enough to impress my friends (and by that I mean, alienate).

Story of my life...

## Re:$1,000,000 prize to be collected then if true (2, Funny)

## A beautiful mind (821714) | about 6 years ago | (#24033181)

And I would have succeeded if it weren't for these meddling kids! What do you mean you can't see them?!

## Reimann? (5, Funny)

## areusche (1297613) | about 6 years ago | (#24031783)

## Re:Reimann? (4, Funny)

## Anonymous Monkey (795756) | about 6 years ago | (#24031951)

## Re:Reimann? (0)

## Anonymous Coward | about 6 years ago | (#24032921)

Damn you, OPEC!

## Re:Reimann? (0, Flamebait)

## PlatyPaul (690601) | about 6 years ago | (#24032437)

That joke is nearly a year old [haloscan.com] online, and about 1000 if you've spent any time in a university math department.

## Re:Reimann? (1)

## areusche (1297613) | about 6 years ago | (#24032467)

## Hmmm.... (5, Funny)

## Otter (3800) | about 6 years ago | (#24031791)

Sounds about par for the course for academic hiring, and it sounds like he's still pretty traumatized from it. I hope this works out for him and he can go around flipping off all the hiring committees who turned him down.

## Re:Hmmm.... (2, Informative)

## Anonymous Coward | about 6 years ago | (#24032469)

It's brutal trying to try to get into academia in a field that doesn't produce money. The sad thing is that departments want to hire more people but there is never any money or open positions and tenured professors hang onto their positions until they die. Things are a little better in physics than math, but not much (I am an experimental physicist).

I had an undergraduate professor tell us endlessly to NOT go into physics, as it would make us miserable careerwise. I'm still in physics, but most of my friends are not, and I totally understand his point now. I had a history professor tell me that if he knew how hard it would be to get to where he was, he never would have been a history major.

## Re:Hmmm.... (1)

## Henry V .009 (518000) | about 6 years ago | (#24032665)

## Math = $$ (5, Funny)

## RabidMoose (746680) | about 6 years ago | (#24031805)

## So what? (3, Insightful)

## feijai (898706) | about 6 years ago | (#24031815)

## Re:So what? (1)

## 192939495969798999 (58312) | about 6 years ago | (#24032263)

Also, the proof of something that complicated is likely so complicated that only the very best minds would even be able to prove that the proof was wrong.

## Re:So what? (5, Informative)

## JambisJubilee (784493) | about 6 years ago | (#24032335)

I think you misunderstand the scope and purpose of arXiv. arXiv is a repository for *preprints*.

By uploading the file to arXiv before submitting it, not only do you ensure that those that can't afford $10,000+ subscription fees can access the article, but you open up your findings to a much wider international audience.

The lack of peer review is not necessarily a liability in this situation

## Re:So what? (1, Interesting)

## Anonymous Coward | about 6 years ago | (#24032645)

I would hardly consider Perelman's preprints to be "junk that couldn't pass peer review"

## Previous proofs (4, Interesting)

## RockMFR (1022315) | about 6 years ago | (#24031827)

## not so fast (5, Informative)

## Anonymous Coward | about 6 years ago | (#24031845)

there are "proofs" of the Riemann hypothesis on the arXiv every few weeks. Don't believe it 'til it's vetted.

## Re:not so fast (5, Funny)

## Anonymous Coward | about 6 years ago | (#24032185)

Yeah. arXiv once published my paper that shows cases where P = NP; I proved it conclusively for the cases where P = 0 and/or N = 1, but so far I haven't gotten my $1,000,000.00 check from the Clay Math Institute.

## Re:not so fast (0)

## Anonymous Coward | about 6 years ago | (#24032785)

That could be the funniest thing I've read here in a long time. It just clicked the right humor circuits in the ol gray matter.

If I bothered to have an account (why sign up when you can lurk for 7 years?) I would give you points.

Congrats to you, fellow AC.

## Re:not so fast (3, Funny)

## Kingrames (858416) | about 6 years ago | (#24032931)

They sent you your checks for cases where you are equal to 0.

Someone beat you to the "1" part.

## Re:not so fast (2, Interesting)

## Sheafification (1205046) | about 6 years ago | (#24032207)

## Dolly Parton (1)

## jav1231 (539129) | about 6 years ago | (#24031903)

6922251x8=55378008

## The continuum hypothesis will be next... (1)

## JuanCarlosII (1086993) | about 6 years ago | (#24031907)

This is seriously disappointing news though. I've always appreciated the romance of such "theories", and now there's one less in the world. That and my planned deal with the devil to save my soul has now hit the rocks.

## Re:The continuum hypothesis will be next... (3, Insightful)

## sm62704 (957197) | about 6 years ago | (#24032383)

First Fermat, now this. Is nothing sacred?!Money. Not much else these days.

## Re:The continuum hypothesis will be next... (5, Insightful)

## hansraj (458504) | about 6 years ago | (#24033123)

The Continuum Hypothesis is known to be neither provable nor disprovable in the standard axiomatic set theory ZF, enriched with the axiom of choice (ZFC). So I wouldn't really count on someone settling that one either way any time soon. Of course one could come up with a new set of axioms for the set theory and *then* prove or disprove CH but you would be hardpressed to find anyone showing interest in that result. After all, I could just add CH or not(CH) to ZFC and trivially prove or disprove it. So anything in that line first needs to even define what a sensible problem is.

For those who have no clue what I said above:

Continuum hypothesis:There is no set strictly larger than the set of natural numbers and at the same time strictly smaller than the set of real numbers. The size of a set in relation to other is defined in terms of mapping. Positive integers are the same number as even numbers because you can define a bijection between the two. Reals are strictly more than naturals.ZF:Set theory made axiomatic. Few axioms (like empty set exists, supersets are larger than original sets etc) that you need to believe and most of the set theory believed to follow.Axiom of Choice:Given a set of sets, one can make a set containing one element from each set. Looks obviously true but in some equivalent but different sounding formulations looks obviously false. Known to be independent to ZF.Y Independent to axioms X:Believing that Y is true does not yield contradiction together with X unless X itself yield contradictions. Same holds for believing that Y is false.PS: Apologies for not including links. I am feeling lazy. Wikipedia has nice articles about all of the above. Articles on ZF, CH or Axiom of Choice are the place to start for a fun reading.

## Oblig. (5, Funny)

## JuanCarlosII (1086993) | about 6 years ago | (#24031983)

## Coulda told us more... (-1, Offtopic)

## MilesAttacca (1016569) | about 6 years ago | (#24032027)

worthR'ing-TFA. Otherwise Slashdot would just be a collection of titled links and not link, summary, discussion. Just a little gripe.## Re:Coulda told us more... (0)

## Anonymous Coward | about 6 years ago | (#24032141)

## Re:Coulda told us more... (2, Insightful)

## Weaselmancer (533834) | about 6 years ago | (#24032737)

## Re:Coulda told us more... (1)

## Vectronic (1221470) | about 6 years ago | (#24032765)

Thats ok, the article doesnt say anything either...

In its entirety:

A proof of the Riemann hypothesis

Xian-Jin Li

(Submitted on 1 Jul 2008 (v1), last revised 2 Jul 2008 (this version, v2))

By using Fourier analysis on number fields, we prove in this paper E. Bombieri's refinement of A. Weil's positivity condition, which implies the Riemann hypothesis for the Riemann zeta function in the spirit of A. Connes' approach to the Riemann hypothesis. Subjects: Number Theory (math.NT)

MSC classes: 11M26

Cite as: arXiv:0807.0090v2 [math.NT]

Submission history

From: Xian-Jin Li [view email]

[v1] Tue, 1 Jul 2008 19:43:13 GMT (20kb)

[v2] Wed, 2 Jul 2008 11:05:52 GMT (20kb)

So Unless you are some encyclopedia of theorems and proofs, you will have to look it all up anyways.

## Re:Coulda told us more... (0)

## Anonymous Coward | about 6 years ago | (#24032891)

But that's what /. editors are for...

## Apology for the Re (1)

## hackus (159037) | about 6 years ago | (#24032125)

Ok, so many have tried, all have failed.

It may take a decade to test the assertions that this so called proof attempts to demonstrate.

Perhaps we could give the guy a consolation prize, wait for the work to be "proven" wrong and then off course, issue an Apology:

http://www.math.purdue.edu/~branges/apology.pdf [purdue.edu] :-)

-Hack

PS: Does anyone find it STRANGE that the guy who can solve this problem has issues finding a job?

WTF?

## Re:Apology for the Re (2, Interesting)

## JuanCarlosII (1086993) | about 6 years ago | (#24032209)

I doubt too many Maths faculties in the world have people working full-time on the Riemann Hypotheses.

Of course I echo your sentiments that his proof is almost certainly flawed though.

## Re:Apology for the Re (1)

## 1729 (581437) | about 6 years ago | (#24032973)

Interestingly, DeBranges was Xian-Jin Li's advisor:

http://www.genealogy.math.ndsu.nodak.edu/id.php?id=16641 [nodak.edu]

## Congratulations! (-1, Troll)

## areusche (1297613) | about 6 years ago | (#24032221)

## Re:Congratulations! (3, Funny)

## Anonymous Coward | about 6 years ago | (#24032299)

Solving the energy crisis is easy.

Use less energy.

Kthxbye.

## Re:Congratulations! (1)

## StarReaver (1070668) | about 6 years ago | (#24032457)

## Re:Congratulations! (2, Interesting)

## danzona (779560) | about 6 years ago | (#24032935)

it has huge effects on prime number distributionPrime numbers are distributed in pretty much the same way as they were before the proof.

The proof is mathematics for the sake of mathematics. The Riemann Hypothesis has been accepted as true true for over a hundred years, so practical applications that derive from it already exist.

## Numb3rs (0)

## Anonymous Coward | about 6 years ago | (#24032233)

Riemann's was featured prominently in the 5th episode of the first season.

## 1134 (1)

## vingilot (218702) | about 6 years ago | (#24032277)

hEll

## FrisT 4sot (-1, Offtopic)

## Anonymous Coward | about 6 years ago | (#24032287)

## 53188008? (1)

## ya really (1257084) | about 6 years ago | (#24032405)

## In English? (1)

## Codex_of_Wisdom (1222836) | about 6 years ago | (#24032797)

So... could someone explain this theorem in simple(r) terms, please?

## His Advisor Also Claimed Proof (2, Interesting)

## Sirius00 (1318579) | about 6 years ago | (#24032923)

## fracking Mormons. (-1, Flamebait)

## Anonymous Coward | about 6 years ago | (#24033047)

If you'll notice, the author is from Brigham Young University, which is (quite literally) owned by the Moron, er I mean...Mormon cult, er I mean..."church". If there's one thing Mormons are definitely NOT known for, it's their use of rigorous proof.

## You mean that... (1)

## misterhypno (978442) | about 6 years ago | (#24033137)

Simple Simon actually met a Riemann, after all?!

I thought that was just a hypothesis!