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November 18, 2009

Unsolved math problem turns 150

Posted: 12:14 PM ET

Happy 150th anniversary to the Riemann Hypothesis, one of the most important math problems ever!

Proposed by Bernhard Riemann in 1859, the Riemann Hypothesis deals with prime numbers. You may recall that a prime number is a positive whole number that has only two positive whole number divisors: one and itself. The first of them are 2, 3, 5, 7, 11, 13, in order.

This hypothesis would be able to provide a better estimate than ever before of a special function denoted as Pi(x). Pi(x) represents the number of prime numbers that are no bigger than x, where x is a positive number. For example, Pi(14) would be 6, because there are six prime numbers (2, 3, 5, 7, 11, 13) no bigger than 14. That's probably the most understandable explanation you're going to get that doesn't involve "zeta functions" and other technical terms.

Given that many of the best mathematicians have tried and failed to provide a solution, the proof is probably not easy or obvious, says Peter Sarnak, professor of mathematics at Princeton University and an authority on the subject. “Most experts expect that a proof will require a major new insight into the structure of whole numbers and the prime numbers,” he said.

But if you can solve it, the Clay Mathematics Institute will give you $1 million.

A proof would have implications not only for mathematics, but also for cryptography and computer science, says Ramin Takloo-Bighash, associate professor at the University of Illinois at Chicago. Internet security protocols, after all, are largely based on prime numbers. Experimental and theoretical evidence has supported the truth of the Riemann Hypothesis, although there are a small number of naysayers who say it can’t be proven, Takloo-Bighash said.

Still, there’s enough confidence in the truth of the Riemann Hypothesis that mathematicians have established “conditional” theorems, which can never be validated until someone proves the 150-year-old problem, says Kenneth Ribet, professor of mathematics at the University of California, Berkeley.

Riemann's paper on the subject was first published in November 1859, but no one knows the day. So, the American Institute of Mathematics picked a Wednesday in the middle of November to celebrate the 150th anniversary, said Brian Conrey, executive director.

Intrigued? Stop by one of these lectures today.

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Filed under: Mathematics


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Michael Harris   November 18th, 2009 1:05 pm ET

Very interesting proposition, however, you did not include how to follow up about winning the prize??? Where is this information? I checked at the Claymath website and there is nothing about this topic?

Please let me know.

Thanks.


Mom   November 18th, 2009 1:23 pm ET

It should be more specific, Can' t understand what the problem is ?


User   November 18th, 2009 5:17 pm ET

Prize info located here:

http://www.claymath.org/millennium/


Raman   November 18th, 2009 5:19 pm ET

Here is the link to the seven millennium problems, each worth $1m prize: http://www.claymath.org/millennium/

Note that Poincare conjecture has already been proven by Perelman. You can have a stab at the remaining six. And here is the link to a description of Riemann hypothesis: http://www.claymath.org/millennium/Riemann_Hypothesis/


boo   November 18th, 2009 5:51 pm ET

even if you solve the problem if there is a solution that depicts a prime whole as the number you have two years in question to waite to see if your answer is even consisdered as being correct or close to solving the problem then its up the the scolars to pick it apart and then considerif it be even up to standards for the institute seems like you do all the work they get all the rewards ???


Fun with numbers   November 18th, 2009 6:44 pm ET

Dear Mom, if you don't understand what the problem is any further explination will fly completely over your head. As for Michael Harris, the reason there is no information on how to follow up with winning the prize is simple. No one has solved it in 150 years, therefore the chances that a blog on CNN will draw out the solution are effectivly zero.


Ubik   November 18th, 2009 7:06 pm ET

Actually, I have discovered a truly marvelous proof of this, which this text box is too narrow to contain. Sorry! ;)


Katerina   November 18th, 2009 7:22 pm ET

http://en.wikipedia.org/wiki/Riemann_hypothesis


jc   November 18th, 2009 7:41 pm ET

I have a solution, but there isn't enough room in the margins for me to write it out.


Amy   November 18th, 2009 7:59 pm ET

Read more about the Riemann Hypothesis here:

http://en.wikipedia.org/wiki/Riemann_hypothesis

More information about the Millennium Prize Problems can be found here:

http://www.claymath.org/millennium/


Roy   November 18th, 2009 8:20 pm ET

Please show the actual initial problem.


A Friend   November 18th, 2009 8:30 pm ET

I got dizzy just reading the first couple of lines...Not smart beware!


Elena P   November 18th, 2009 8:43 pm ET

Here is the link on the claymath website:
http://www.claymath.org/millennium/


Todd Timberlake   November 18th, 2009 9:53 pm ET

Here are some more specifics. There is a function known as the "Riemann zeta function" that was introduced by Bernhard Riemann in his 1859 paper. Riemann showed that the "zeros" of this function are intimately connected with how the prime numbers are distributed among the whole numbers (whole numbers are just the normal counting numbers 1, 2, 3, 4, ...). A "zero" of a function is just a number that you plug into that function and the result is zero. The Riemann zeta function has what are called "trivial zeros" at all of the negative even integers (-2, -4, -6, etc). But it also has lots of other zeros. These other zeros are complex numbers, which means they involve the imaginary number i (the square root of negative one). The Riemann Hypothesis states that all of the "nontrivial zeros" of the Riemann zeta function have the form of a real number plus (1/2)i. So far every nontrivial zero that has been calculated has turned out to satisfy the Hypothesis (and we're talking a LOT of them). But that doesn't constitute a proof of the Hypothesis, because maybe there is a zero that doesn't fit that pattern and we just haven't found it yet. The Hypothesis has lots of interesting connections to other areas of math and even to areas of physics.

Hope that helps! For more just Google "Riemann Hypothesis" and check out the Wikipedia or MathWorld articles. But be ready to put your math hat on!


Dr. Mathq   November 18th, 2009 10:21 pm ET

If youre smart enough to solve the problem, you should have no problem figuring out how to contact them


Steve   November 18th, 2009 11:28 pm ET

0 is number 1, move everything 1 digit and theirs your answer. (0) is always number 1! Ask the banks!


MikeHunt   November 18th, 2009 11:35 pm ET

i know the answer, but i'm not telling.


Richard Ashcraft   November 18th, 2009 11:45 pm ET

http://www.claymath.org/millennium/Riemann_Hypothesis/riemann.pdf


Mike in NYC   November 19th, 2009 12:17 am ET

So what exactly is the Riemann hypothesis? A little oversight there, Liz.

That pic is pretty funny, BTW.


Thomas   November 19th, 2009 2:38 am ET

This problem is not insolvable, rather it is intractable. The problem can easily be solved by computers using a brute force algorithm, but not in a reasonable amount of time, especially when the number one is calculating to is arbitrarily large.

As long as one doesn't want to calculate pi(x) for an arbitrarily large number, the algorithm for solving the problem is actually rather simple, you count the number of prime numbers from 1 to n (non inclusive), n being the number you are checking up to. How one would check whether the numbers are prime? Simply determine the modulus between that number and all the numbers smaller than it, and if at any given time the modulus is 0, the number is composite.


http://twitter.com/whybs   November 19th, 2009 3:32 am ET

Here are useful sites

1. http://bit.ly/3KnBV0
2. http://bit.ly/1YqjkE
3. http://bit.ly/1JDDyV (7 problems @ prizes of $1M / each!)
4. http://bit.ly/WR1Me (email inquiries to prize.problems@claymath.org )

Hopefully, your math skills surpass my Googling skills! :)

Good luck!


Greg   November 19th, 2009 3:57 am ET

There is more information about the Millenium Prizes (one of which a solution to the Riemann Hypothesis would earn) at the following Clay Math Institute page:

http://www.claymath.org/millennium/

There are lots of descriptions of the Riemann Hypothesis, of various lengths and level of detail, all over the web. The Wikipedia page is as good a place to start as any:

http://en.wikipedia.org/wiki/Riemann_hypothesis


Jason   November 19th, 2009 4:54 am ET

Mom: You didn't understand what the problem is because this article never explained what it is. And then they complement themselves for giving such an "understandable explanation"! I expect more from CNN; that this kind of sloppiness is offered and accepted feels like a symptom of meager math and science education in this country.

The Reimann hypothesis, unlike Fermat's last theorem, takes a little bit of time to explain carefully. (Although nowhere as much time as it takes to explain some of the other major mathematical propositions, like the recently solved Poincare conjecture.) Here's a link to an explanation that may be appropriate: http://www.donotrememberthisaddress.com/mathematics/whatisriemannhypothesis.php


Stephen H   November 19th, 2009 8:21 am ET

I believe the answer is 42.


JT   November 19th, 2009 8:37 am ET

great, but what is the problem? you never say what it is, only an example of what Pi(x) is.


Charlie   November 19th, 2009 8:52 am ET

M's Landau obviously did not get her degree in Math. Her explanation of Riemann's Hypothesis indicates a complete lack of understanding of the problem. Those who are interested in a better explanation might try this link: http://www.math.hmc.edu/funfacts/ffiles/30002.5.shtml


Bobby G.   November 19th, 2009 8:57 am ET

All you need to understand is that 1+1=2 and 1-1=0.

Also understand that just because you have checks doesn't mean you have money.


H A Helfgott   November 19th, 2009 9:18 am ET

The problem is as follows. We've got an approximate formula for
Pi(x) (it's roughly of size x/(log x), or more precisely the integral of
1/(log t) for t from 2 to x). We know that this is a pretty accurate formula.
Experiments show that it is actually extremely accurate – to a much greater extent than it has be proven to be.

The Riemann hypothesis states that this formula (without being exact) is essentially as accurate as it could possibly be (given that one can show that is isn't perfect).

All the talk about the Riemann zeta function is very important, but it's just a way to rewrap what I've said above.


David   November 19th, 2009 9:51 am ET

Ha! Yet another problem that Obama hasn't been able to solve... how predictable...


DMR   November 19th, 2009 9:54 am ET

Jason, it's very typical for CNN–I can usually find one grammatical error a week with minimal effort–they are only a touch better than fox, and far, far behind bbc...


Prof. Gerald Lambeau   November 19th, 2009 9:59 am ET

Quick, someone email Will Hunting and get the answer ASAP. He is probably roaming around the southside of Boston with his buddy Ben trying to stir up trouble with Math geeks at Cambridge.

For the love of God my head hurts from not knowing the answer to ridiculous theories in math. Get his number now!!


Justin   November 19th, 2009 10:03 am ET

A few mistakes up there. Actually, the hypothesis conjectures all non-trivial zeros of the zeta function have the form 1/2+iy but even that much is way too much for most non-mathematicians. I agree though: it's not going to be solved by any known method. Something new has to happen to prove it.


Will Hunting   November 19th, 2009 10:17 am ET

Hey guys I saw this problem up on the chaulk board last night when I was mopping the floor and I wanted to let you all in on the answer. No thanks necessary and feel free to keep the prize. Here goes:

enter the cubed root of 150,399,696,179,997,056,512 on your calculator and turn it upside down.

How ya like them apples?


Brian   November 19th, 2009 10:34 am ET

They randomly picked Wednesday? How unimaginative.
Should have gone with Nov. 13th at 17:19 and 23 seconds.


required   November 19th, 2009 10:50 am ET

Doesn't work for 2.

Pi(2) = 0

0 is not positive


Brian   November 19th, 2009 11:28 am ET

I like the whole idea about this but I don't even get what the problem is either. Pi(14) would be 6 it definately makes sense. The higher number x is, the more prime numbers there are. So what is the problem they want to solve?


specter   November 19th, 2009 12:04 pm ET

Where do black holes go? What is dark matter? How did we get here? What is the meaning of it all? Ignorance is truly bliss. I should have payed more attention in school. To borrow from a fellow ignoramis,"there are known known's and known unknown's...." I know I don't know the answer to a lot of these questions. /shrug


S. M.   November 19th, 2009 1:43 pm ET

I know the answer to this but I don't need a million dollars right now.


Dan   November 19th, 2009 1:54 pm ET

2+2=5


shutthehellup   November 19th, 2009 2:37 pm ET

How bout you write it up first, then find out how to claim your prize.


Martin C. Winer   November 19th, 2009 2:45 pm ET

Here is the solution:
http://www.rankyouragent.com/primes/primes_simple.htm
and based on these formulae you can predict the presence of prime twins and other constellations all through the number line:
http://www.martincwiner.com/wp-content/uploads/2009/02/primeconstellations2.pdf
http://www.martincwiner.com/martin-winers-math/
Simple really.


ioan   November 19th, 2009 2:52 pm ET

Took me 5 minutes to solve it :-)
There you go:

function Pi(x: integer): integer;
var
i: integer;
J: Integer;
isPrime: boolean;
begin
if x >= 2 then
result := 1
else
result := 0;
for i := 3 to x do
begin
isPrime := true;
for J := 2 to I – 1 do
begin
if (i mod j) = 0 then
begin
isPrime := false;
break;
end;
end;
if isPrime then
result := result + 1;
end;
end;


sam   November 19th, 2009 3:00 pm ET

Simple (English) Wikipedia might be easier to understand for some folks http://simple.wikipedia.org/wiki/Riemann_hypothesis


MathPhd   November 19th, 2009 3:21 pm ET

Yeah, I completely agree with FunwithnNumbers. As a PhD student at Penn, I've worked on solving the Riemann Hypothesis with other ivy-league students and Professors on the East Coast for a year. Something tells me a middle aged Mom who posts on CNN that she doesn't "understand the problem" wouldn't provide too much help in the solution to the one of the most cryptic mathematical equations of all time.


Jorgen   November 19th, 2009 4:18 pm ET

Anything past basic math be moot. Its the same as a language. People make up languages all the time such as elvish and Klingon, and they make sense. The same thing applies to math, it was invented, people began to follow it and as you get into things like imaginary numbers it just becomes stupid. Stop the madness, admit that this problems have no real application on anything useful and use math just enough to balance your check book, because lets face it, it has to be done.


onlycontent   November 19th, 2009 7:41 pm ET

5318008


Todd Timberlake   November 19th, 2009 8:09 pm ET

Thanks Justin. Definitely an error in my earlier post – not sure how that happened, I must have been half asleep when I wrote it. The REAL part of the nontrivial zeros is 1/2, the imaginary part can be lots of different things (Google Andrew Odlyzko to find datasets with hordes of nontrivial zeta zeros, all of which fit the hypothesis).

It's nice that CNN is even including a link to a story about the Riemann Hypothesis – but I agree with several other posters that the article could give a little more explanation. It's a difficult subject, but if we can't describe difficult science and math then we can't talk about almost anything that is a current topic of research. For anyone who wants to learn more about this fascinating topic I recommend Derbyshire's "Prime Obsession" (especially if you like some history with your math).


dtc   November 19th, 2009 10:03 pm ET

why is there a picture of a black guy trying to solve a math problem.


Hard Tymes   November 20th, 2009 10:51 pm ET

Right about now I wish we had Mr Data from Star Trek!
I'd make book that android could solve it!
One other item, folks.
Why don't we give the problem to that monster computer in Brussels, Belgium? Then we could go on to solve the reason why we can't balance a checkbook! :)

Hard Tymes ~~ HardTymes.Com


Conic   November 22nd, 2009 8:50 pm ET

I thought I'd save you all some time. I've already solved all of the remaining six problems. So you shouldn't even bother to try. I'm smart! It's all in my head.

Now to get the prize money. Let's see; how much am I going to get? New problem... what is six times one million. Gonna have to think on this one for a few years. ;)


Vinod   November 27th, 2009 10:51 pm ET

I am sorry if I sound bad I think you are saying P(infinity) = 2, 3, 5, 7, 11, 13, ..........,....,....., infinity. So that means you are trying to make me found infinity also with that question.


Third Stooge   November 28th, 2009 10:29 am ET

I'm pretty sure the final answer will have "Made in China" in it somewhere...this whole scenario and problem shows us why. Maybe one out of every 1000 people (or more) who read this article have any notion or interest in the actual thrill of solving such a problem. The rest of the readers only have the intelligence to make snide comments. It is proof why this country has gone down the tubes and allowed other countries to take away the pride of being educated that Americans used to have. I'm not surprised to know that the majority of people can't perform simple math without the use of a calculator.


Mark C   November 29th, 2009 8:29 pm ET

**** The same thing applies to math, it was invented, people began to follow it and as you get into things like imaginary numbers it just becomes stupid. Stop the madness, admit that this problems have no real application on anything useful and use math just enough to balance your check book, because lets face it, it has to be done. ****

Wow, could you possibly be a bigger moron?

Kindly explain how you get a spaceship to Saturn using "basic math" that you use to balance your checkbook. Or how you do the fluid dynamics necessary to design an airplane. Or understand semiconductors (which involve quantum mechanics) well enough to build the computer you posted this drivel from.


Louise   November 30th, 2009 1:08 pm ET

If you think this is hard; just try explaining the federal tax code.


Jessica*   December 2nd, 2009 1:05 am ET

It cracks me up to see how many people post on the CNN blog that they have the answer – and even funnier that people are quoting Wikipedia – since when is Wikipedia a reliable source of information?


RingerMe   December 5th, 2009 9:00 am ET

FTAA proof would have implications not only for mathematics, but also for cryptography and computer science,


Okie in Exile   December 6th, 2009 8:44 pm ET

Jessica,
Wikipedia is usually pretty good on mathematical topics. Geeks take stuff pretty seriously.


Double Action   December 8th, 2009 1:55 am ET

Wolfram Alpha is my internet math professor.


ebrown2112   December 11th, 2009 1:53 am ET

The answer is 42, duh.


mrwords   December 13th, 2009 9:25 am ET

Only if x is between 182 and 190, inclusive.


brooke   November 5th, 2010 1:59 pm ET

@required: 0 is not x in the equation. 2 is x. Pi(x) Pi(2)=0. I'm pretty sure 2 is a positive number.


anirban saha   April 6th, 2011 10:41 am ET

this hypothisis is greatest beauty in mathematics. i want to solve this. i want the e-mail id of the institute to contact.


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Obama on Internet Security – Riemann Hypothesis
[youtube=http://www.youtube.com/watch?v=tdhQSlUNUf4&w=640&h=390]


Peter L. Griffiths   August 30th, 2013 2:43 pm ET

In discussing the Riemann hypothesis, why does nobody mention that the non-trivial zeta evaluations are all terms in the cotangent series whose first term is 1/2 which seems to have been interpreted as being a critical line.


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Peter L. Griffiths   November 2nd, 2013 1:48 pm ET

Euler's product formula for primes quoted in Riemann's 1859 paper is not quite correct, the Harmonic series needs to be divided by the composite fractions including 1 followed by 1/4 contained in the Harmonic series.


Peter L. Griffiths   November 3rd, 2013 12:36 pm ET

Further to my comment of 2 November 2013, this division of the Harmonic series by the composite series has up to now been overlooked because of the initial transfinite location of the finite composite fractions. These composite fractions need to be brought together in a finite location to divide the Harmonic series.


Peter L. Griffiths   November 8th, 2013 12:24 pm ET

Further to my comment of 3 November 2013, I have come to the conclusion that it is not only the primes but also the composites which are deducted from the original Harmonic series (H), leaving 1 equal to H. product (p-1)/p which is Euler's formula. The great attraction of this formula is that the fractions in both
H and product (p-1)/p can be raised to the same power or root and still equal 1. I have however come to the conclusion that Riemann was on the wrong track for identifying primes. To identify primes, the number of factors in the composites needs to be studied. Nobody seems to have done this.


Peter L. Griffiths   November 25th, 2013 11:56 am ET

This comment I hope delivers the fatal thrust. Near the beginning of his 1859 paper Riemann incorrectly assumes that the complex variable s =1/2 +ti is a Zeta power. Riemann fails to recognise that an expression containing an imaginary number such as 1/2 +ti cannot be a power unless the base is a log base such as e. The best known example of this is Cotes's formula
cosu + isinu equals e^(iu) where it is not possible for e to be meanfully replaced by other values. This means that Riemann is badly wrong in applying as a power s =1/2 +ti. It also means that practically all the arguments in his 1859 paper are fallacious.


Peter L. Griffiths   December 17th, 2013 1:18 pm ET

I wish to clarify the observation in my comment of 25 November 2013 that it is not possible for e to be replaced by other values. I now consider that it is possible for e to be replaced by other values but only because the general formula for any variable base n and power u should be the constant n^(u/logn). When n equals e, log n will be clearly 1. The fact that n^(t/logn) has to be a constant particularly when applied to i that is n^(ti/logn) undermines virtually all the arguments in Riemann's 1859 paper.


Peter L. Griffiths   December 20th, 2013 2:10 pm ET

The e number mentioned in my previous comments has an equivalent n^(1/logn) where n can be any number. This does not affect my conclusions.


Peter L. Griffiths   August 23rd, 2014 11:54 am ET

One possible approach to identifying primes is to apply Euler's discovery H(1-[1/2])(1-[2/3])(1-[4/5])(1-[6/7]) ... =1, where H is the harmonic series 1+(1/2) +(1/3) +(1/4)...., particularly if neither series is taken to infinity.


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