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Claimed Proof of Riemann Hypothesis

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.

92 of 345 comments (clear)

  1. Dirty Words by Rik+Sweeney · · Score: 5, Funny

    Me, I use math to write dirty words on my calculator.

    Such as 80085?

    1. Re:Dirty Words by Anonymous Coward · · Score: 5, Funny

      5318008

    2. Re:Dirty Words by UnknowingFool · · Score: 5, Funny

      No, you mean 5318008 or for the slashdot crowd, 55378008

      --
      Well, there's spam egg sausage and spam, that's not got much spam in it.
    3. Re:Dirty Words by Directrix1 · · Score: 4, Funny

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

      --
      Occam's razor is the blind faith in the natural selection of least resistance and in universal oversimplification. -- EF
    4. Re:Dirty Words by Directrix1 · · Score: 5, Funny

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

      --
      Occam's razor is the blind faith in the natural selection of least resistance and in universal oversimplification. -- EF
    5. Re:Dirty Words by Archangel+Michael · · Score: 4, Funny

      On linux, wouldn't it be ...

      host:>man 80085

      ???

      --
      Agent K: A *person* is smart. People are dumb, stupid, panicky animals, and you know it.
    6. Re:Dirty Words by Firehed · · Score: 4, Funny

      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?

      --
      How are sites slashdotted when nobody reads TFAs?
    7. Re:Dirty Words by JayJay.br · · Score: 2, Funny

      Newbie...

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

      Mod me down, I dare you!!!

    8. Re:Dirty Words by StikyPad · · Score: 4, Funny

      You haven't grafted a color TFT screen to your calculator yet?

      Who let these guys in?

    9. Re:Dirty Words by andy19 · · Score: 5, Funny

      Coming from a Slashdotter, are you surprised?

    10. Re:Dirty Words by Anonymous Coward · · Score: 2, Informative

      Hold the calculator upside down, mod-who-didn't-get-the-joke...

    11. Re:Dirty Words by droopycom · · Score: 5, Funny

      You just gave me the best idea for an iPhone app:

      Boobies that bounce according to how the phone is bouncing....

    12. Re:Dirty Words by hyperion454 · · Score: 3, Funny

      One of those guys must have been Gene Simmons.

    13. Re:Dirty Words by DFENS619 · · Score: 5, Funny

      Your ideas are intriguing to me and I wish to subscribe to your newsletter.

    14. Re:Dirty Words by Andor666 · · Score: 3, Funny

      äOEæ--¥é..."ããï¼ï¼

    15. Re:Dirty Words by david.given · · Score: 2, Funny

      Nah, if you really want a dirty word, try 71077345...

    16. Re:Dirty Words by h2k1 · · Score: 3, Interesting

      in portuguese, 50135.50738 (nice breasts).

    17. Re:Dirty Words by PachmanP · · Score: 4, Funny

      Link?

      Does your project have donation page?

      --
      You're thinking small. Why miniaturize the laser, when we could instead enlarge the sharks? -John Searle
  2. I like to describe my workplace with my calculator by InvisblePinkUnicorn · · Score: 2, Informative

    37047734

  3. Yeah but did they point this out? by i_want_you_to_throw_ · · Score: 5, Funny

    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)

    1. Re:Yeah but did they point this out? by rdwald · · Score: 5, Funny

      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...

    2. Re:Yeah but did they point this out? by colonslashslash · · Score: 5, Funny

      Wait... both of you RTFA?

      We have a new /. record!

      --
      She's built like a steak house, but she handles like a bistro....
    3. Re:Yeah but did they point this out? by StikyPad · · Score: 4, Funny

      Not so fast. I read it -2 times.

    4. Re:Yeah but did they point this out? by jd · · Score: 4, Funny

      I imagined I read it, so that's +i.

      --
      It's a small world and it smells funny; I'd buy another if it wasn't for the money; Take back what I paid (SoM)
    5. Re:Yeah but did they point this out? by Anonymous Coward · · Score: 2, Funny

      I read half of TFA...so that's 1/2 + i, which just happens to be on the critical line.

    6. Re:Yeah but did they point this out? by StikyPad · · Score: 4, Funny

      Come on, be real.

  4. Tried to RTFA by multipartmixed · · Score: 5, Funny

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

    --

    Do daemons dream of electric sleep()?
    1. Re:Tried to RTFA by Notquitecajun · · Score: 4, Funny

      Ummm...I think that WAS layman's terms. For you math geeks, try being a history major and looking at all that. It just looks like a cat walked on the keyboard to me...

    2. Re:Tried to RTFA by PlatyPaul · · Score: 5, Informative

      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].

      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 big assumption), this would finally solve a long-standing problem (dating back to 1859).


      Details of the actual solution are a bit heavy. Those actually interested in this sort of number theory might want to start here.

      --
      Misery loves company. Online misery loves unsuspecting random strangers.
    3. Re:Tried to RTFA by stranger_to_himself · · Score: 5, Funny

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

    4. Re:Tried to RTFA by Frankie70 · · Score: 2, Funny

      It's like 50 football fields laid in line from here to Riemann.
      Rieman sounds like a place in Germany.

    5. Re:Tried to RTFA by Notquitecajun · · Score: 2, Funny

      Okay...I would ask WHY this is important, but someone is ponying up a million bucks for the solution. THAT tells me this is important. I'm not sure if I care why...

    6. Re:Tried to RTFA by JohnsonJohnson · · Score: 5, Informative

      It's important because the zeros of the zeta function tell you how the prime numbers are distributed and prime numbers are to number theory as elements are to chemistry, everything you could care about is built out of them. The RH is also related to host of other more esoteric, but no less important conjectures; the truth of a large part of modern mathematics relies on knowing if the RH is true or false.

      Although it's unlikely to impact the storage capacity of a flash drive any time soon the zeta function shows up in high energy physics and thus does have real world consequences.

    7. Re:Tried to RTFA by PlatyPaul · · Score: 2, Interesting

      Here's another easy-to-grasp one: public key encryption (think: credit card purchases online) is dependent upon the use of large primes. Large primes are currently not the easiest/fastest to find - what if you knew better where to look for them?

      --
      Misery loves company. Online misery loves unsuspecting random strangers.
    8. Re:Tried to RTFA by drinkypoo · · Score: 4, Funny

      Thus, archaeologists are as anal about their 1 meter units (or even smaller units) as chemists are about their titrations (or whatever chemists do).

      Last time I tried to get anal with my 1 meter unit, I damned near killed someone.

      --
      "You're right," Fisheye says. "I should have set it on 'whip' or 'chop.'"
    9. Re:Tried to RTFA by Garridan · · Score: 2, Informative

      JSMath.

    10. Re:Tried to RTFA by Walkingshark · · Score: 2, Funny

      Great. Here comes the "intelligent arithmetic" movement.

      --
      The world you experience is only a close approximation of reality.
  5. $1,000,000 prize to be collected then if true by deft · · Score: 4, Informative

    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]

    --

    There's nothing Intelligent about Intelligent Design.
    1. Re:$1,000,000 prize to be collected then if true by rufty_tufty · · Score: 5, Informative

      Good explanation here too:
      http://www.irregularwebcomic.net/1960.html

      --
      "The weirdest thing about a mind, is that every answer that you find, is the basis of a brand new cliche" -
    2. Re:$1,000,000 prize to be collected then if true by Anonymous Coward · · Score: 4, Insightful

      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.

    3. Re:$1,000,000 prize to be collected then if true by UnknowingFool · · Score: 4, Funny

      I looked at the proof and have absolutely no idea what it said. But in the finest slashdot tradition, I WILL have opinions here shortly. Abrasive, loud, and irrefutable ones

      --
      Well, there's spam egg sausage and spam, that's not got much spam in it.
    4. Re:$1,000,000 prize to be collected then if true by Anonymous Coward · · Score: 5, Informative

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

    5. Re:$1,000,000 prize to be collected then if true by A+beautiful+mind · · Score: 2, Funny

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

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

      --
      It takes a man to suffer ignorance and smile
      Be yourself no matter what they say
    6. Re:$1,000,000 prize to be collected then if true by somersault · · Score: 2, Funny

      Do you even know, what a number field is

      It's where they grow new numbers, right?

      PS: Do you, even, know: how to use correct punctuation?1!?!?!?

      --
      which is totally what she said
    7. Re:$1,000,000 prize to be collected then if true by Cowculator · · Score: 2, Informative

      No, you're wrong because you have no idea what you're talking about. Every number field has its own zeta function which roughly describes the distribution of prime ideals in that field, and the Riemann zeta function is the one corresponding to the rational field. The Riemann hypothesis states that the Riemann zeta function (that is, the one for the field of rational numbers) has no zeros whatsover, rational or otherwise, on the critical strip 0 < Re(s) < 1 except along the line Re(s) = 1/2, and this is exactly the statement he's claiming to have proved.

    8. Re:$1,000,000 prize to be collected then if true by MiniMike · · Score: 4, Funny

      Step 1: Find 5-month old baby.
      Step 2: Interrogate baby from step 1, asking questions relevant to the Riemann Hypothesis.
      Step 3: Profit!

      Progress so far:
      Step 1: Complete.
      Step 2: Complete. Reply to question consisted of: "Blah gurgle <splursh> gah hwooo naaae".
      Step 3: Incomplete, but I have reduced the problem from one of Mathematics to one of Linguistics. I expect results soon.

    9. Re:$1,000,000 prize to be collected then if true by Anonymous Coward · · Score: 2, Informative

      Ahem, Xian-Jin Li has a mathematical criterion named after him: http://en.wikipedia.org/wiki/Li%27s_criterion

  6. Reimann? by areusche · · Score: 5, Funny

    Reimann? Like the Noodles right?

    1. Re:Reimann? by Anonymous+Monkey · · Score: 4, Funny

      It's not just noodles. Its a way of life.

      --
      We are the Borg...
  7. Hmmm.... by Otter · · Score: 5, Funny
    The only part of it I understood was:

    The author is grateful to J.-P. Gabardo, L. de Branges, J. Vaaler, B. Conrey, and D. Cardon who have obtained academic positions in that order for him during his difficult times of finding a job.

    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.

    1. Re:Hmmm.... by Anonymous Coward · · Score: 2, Informative

      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.

    2. Re:Hmmm.... by Anonymous Coward · · Score: 5, Funny

      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.

      Well, that's all in the past now.

  8. Math = $$ by RabidMoose · · Score: 5, Funny

    According to the http://en.wikipedia.org/wiki/Riemann_hypothesis wikipedia article, this means $1,000,000 if the proof turns out to be valid. Unfortunately, I didn't understand anything else in that article.

  9. So what? by feijai · · Score: 3, Insightful

    arXiv has become the repository for junk that couldn't pass peer review. Wake me up when we see a published journal article.

    1. Re:So what? by JambisJubilee · · Score: 5, Informative

      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

    2. Re:So what? by TheLink · · Score: 2, Insightful

      I doubt most have even got around to reading the paper. They're too busy thinking of ways to display boobies on their calculators.

      Just look at the above threads.

      --
  10. not so fast by Anonymous Coward · · Score: 5, Informative

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

    1. Re:not so fast by Anonymous Coward · · Score: 5, Funny

      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.

    2. Re:not so fast by Sheafification · · Score: 2, Interesting

      Indeed. Among some mathematicians it is a pleasant diversion to take bets on which of the major unsolved (or unprovable) problems has the most solutions appear on the arXiv this week.

    3. Re:not so fast by Kingrames · · Score: 3, Funny

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

      Someone beat you to the "1" part.

      --
      If you can read this, I forgot to post anonymously.
    4. Re:not so fast by Neil+Strickland · · Score: 3, Interesting

      That's true, but most of them are obvious drivel. I have looked through this one, and it is clearly a real attempt by a genuine mathematician who understands the relevant background. I'd still bet on it being wrong, but not stupidly wrong.

    5. Re:not so fast by bugeaterr · · Score: 2, Funny

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

      Who needs proof!

      Do like Al Gore and declare,

      "The debate on the Riemann Hypothesis is OVER!"

  11. Tough problems by dj245 · · Score: 3, Interesting

    Part of the reason these problems are so tough because to solve them, you have to understand what the problem is first. 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.

    --
    Even those who arrange and design shrubberies are under considerable economic stress at this period in history.
    1. Re:Tough problems by aproposofwhat · · Score: 2

      New math is the only way to go about solving some of these problems.

      You mean like this?

      --
      One swallow does not a fellatrix make
    2. Re:Tough problems by afabbro · · Score: 4, Funny

      ...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 are inventing new math :-)

      --
      Advice: on VPS providers
    3. Re:Tough problems by jd · · Score: 2, Funny

      It's easier to have just one heavy maths function and one trivial maths function than two heavy maths functions, so division is easiest implemented as multiplication with the inverse of one of the two numbers, inverses being relatively trivial in exponential notation. As only computers operate this way, the grandparent poster is obviously an artificial intelligence.

      --
      It's a small world and it smells funny; I'd buy another if it wasn't for the money; Take back what I paid (SoM)
    4. Re:Tough problems by AshtangiMan · · Score: 2, Informative

      Thank god (or Goedle) that they don't have to be both consistent and complete.

    5. Re:Tough problems by mav[LAG] · · Score: 2, Funny

      I carry something else to avoid reproducing.

      A clipboard?

      --
      --- Hot Shot City is particularly good.
  12. Oblig. by JuanCarlosII · · Score: 5, Funny
  13. Re:Apology for the Re by JuanCarlosII · · Score: 2, Interesting

    Not really, the kind of person who would solve a problem of this nature is probably going to be the Andrew Wiles reclusive genius type - a lot like the Russian gent whose name escapes me who solved the Poincare Conjecture. Thus he's not necessarily going to be too keen to teach/lecture/supervise and so would possibly not be too attractive to prospective employers.

    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.

  14. Re:Congratulations! by Anonymous Coward · · Score: 3, Funny

    Solving the energy crisis is easy.

    Use less energy.

    Kthxbye.

  15. Re:The continuum hypothesis will be next... by sm62704 · · Score: 3, Insightful

    First Fermat, now this. Is nothing sacred?!

    Money. Not much else these days.

    --
    mcgrew's razor: Never attribute to stupidity that which can be explained by greedy self-interest
  16. typo by Ungrounded+Lightning · · Score: 4, Informative

    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 ..."

    --
    Bantam Dominique roosters crow a four-note song. Once you've heard it as "Happy BIRTHday" you can't NOT hear it that way
    1. Re:typo by mcrbids · · Score: 5, Funny

      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 ..."

      You have a slight typo. It should be: "You have a slight typo. It should be: ..."

      --
      I have no problem with your religion until you decide it's reason to deprive others of the truth.
  17. Numb3rs by netsavior · · Score: 5, Funny

    Charles Eppes: Imagine you have an infinite number of plot holes, and you want to test how they compare to imaginary numbers. The Riemann Hypothesis states that I can use the zeros in this formula to predict how bullets will bounce off of concrete to a degree of statistical accuracy that it will actually give me the social security number of the guilty shooter.

  18. Re:Try this. by Anonymous Coward · · Score: 5, Funny

    your mother?

  19. Re:Coulda told us more... by Weaselmancer · · Score: 2, Insightful

    If you tried, you'd miss by 1/2.

    --
    Weaselmancer
    rediculous.
  20. His Advisor Also Claimed Proof by Sirius00 · · Score: 2, Interesting

    This guys advisor, according to the Math Genealogy Project, is Louis deBranges. DeBranges also claimed to have proven this a few years back, but his proof was not accepted (for reasons unknown to me). The $1M might still be safe.

  21. Re:Congratulations! by danzona · · Score: 2, Interesting

    it has huge effects on prime number distribution

    Prime 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.

  22. Re:The continuum hypothesis will be next... by hansraj · · Score: 5, Insightful

    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.

  23. The REAL importance is Primes by Anonymous Coward · · Score: 5, Interesting

    Section two of the wiki article (http://en.wikipedia.org/wiki/Riemann_hypothesis) is the great importance here. If indeed there is a proof of Riemann's Hypothesis, then there is a similar proof of the Generalized Riemann Hypothesis, which is in turn a big step in finding the exact distribution of prime numbers.

    Finding the distribution of prime numbers has epic consequences, like breaking most encryption, for starters.

    1. Re:The REAL importance is Primes by payola · · Score: 5, Informative

      The Riemann Hypothesis and RSA encryption both have to do with prime numbers, but the relationship between the two pretty much ends there. To break RSA you need to know how to factor large numbers quickly. RH, on the other hand, pretains to the distribution of prime numbers. It's pretty unlikely that a proof would make computers any faster at factorizing.

      So this begs the question that a lot of people have been asking on this thread: why should you care? There tongue-in-cheek answer is that a solution is worth $1,000,000. While that response may suffice for non-mathematicians, mathematicians would have another, more important reason to celebrate. RH and its generalization, the Grand Riemann Hypothesis, have an absolutely enormous number of profound impliations in number theory, and it is difficult to overstate how critical a proof of either would be. (The implications are too technical to write about here, but you can read about them in most good survey books on analytic number theory; for example, see section 5.8 of Iwaniec & Kowalski). A successful proof probably won't affect your life in any meaningful way (unless you work with analytic number theory for a living), but it would be monumental in the world of math - indeed, this is precisely why there's a reward for solving it. If that's not enough for you, just remember that many mathematicians are motivated not by fame or money but by the beauty and elegance of mathematics, and any proof of RH would establish a truly beautiful and amazing result.

      Of course, there's also the question: is Li's proof correct? I certainily don't know, and I doubt anyone will for quite some time, but there's an interesting story. Li's Ph.D. adviser was Louis de Branges who, as noted on this very website, claimed to prove RH in 2004. His proof has not been accepted by the mathematical community and is widely considered to be incorrect, in large part because the method he wclaims to use was shown, in a 2000 paper co-authored by none other than Xian-Jin Li, to have holes in it.

  24. DOOOOOMED!!!!!!! by Anonymous Coward · · Score: 2, Funny

    I can't believe they are brazenly going forward with research into this subject without knowing if it could possibly lead to the creation of a black hole that will swallow the earth.

  25. Re:Try this. by Anonymous Coward · · Score: 2, Funny

    Tired?

  26. Wrong by InvisblePinkUnicorn · · Score: 5, Funny

    "hellhole - nice."

    No, it's elohlleh, pronounced "elO'-heh-luh", which in the Primitive Quendian proto-language used by the early Elves after their awakening by Eru Ilúvatar, roughly translates to "a dreary, oppressive, or unpleasant place".

    Totally different.

  27. Re:Numb3rs by multipartmixed · · Score: 4, Funny

    Dude, you owe me a monitor.

    Note to self: Do not drink coke while reading /.

    --

    Do daemons dream of electric sleep()?
  28. Or, in layman's terms... by CarpetShark · · Score: 4, Informative

    I just finally found a simple explanation of complex numbers, and just heard of this Riemann Hypothesis, so I may be way off, but let me try to put what (I think) I've figured out so far in layman's terms for the rest of the lost souls:

    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.

    Basically, 10 trillian calculations have been done involving certain complex numbers, which all show a clear pattern: if you get an answer of 0, the real part of the number given to the function always seems to be 0.5. As yet, no one has proven this, and so, presumably, no one truly understands why that's the case yet. Also, presumably, when we do understand it, we'll have forward (either in a a step or a leap) in our ability to use complex numbers (and the multi-dimensional calculations they represent.

  29. Re:Not quite right? by JuanCarlosII · · Score: 3, Funny

    k is an arbitrary constant

    Or a 'variable' as it is also known.

  30. Re:Reimann hypothesis by tobiah · · Score: 2, Interesting
    page 4 for equations 3.2-3.4 he assumes g0 can be bounded because it has compact support on (0, Inf). This is false, a function that is continuous on all of (0, Inf) also has compact support on (0, Inf). That sort of thing is why functions with compact support are only interesting on a bounded domain.

    If you let g0(x)=1/x, then the integral at the bottom of page 38 blows up to Inf.

    I don't see a way to fix that, Theorem 8.6 is pretty important to this proof, and probably false. Those bits represent Li's major contribution to the problem, the rest of it is restating previous results.

    --
    "The ability to delude yourself may be an important survival tool" - Jane Wagner -
  31. Disproof by tobiah · · Score: 2, Interesting

    Ah well, not quite right. But let g0(x)=x works, because there's no integrability condition. Thm 8.6 then falls apart because h0 is no longer in L^2(C), or V(h) is not an operator, take your pick.

    --
    "The ability to delude yourself may be an important survival tool" - Jane Wagner -
  32. Re:Reimann hypothesis by payola · · Score: 2, Informative

    Extremely highly-regarded mathematician Terence Tao has said, in response to a comment left on his blog, that the proof is probably incorrect.