70
$\begingroup$

I've recently been reading about the Millennium Prize problems, specifically the Riemann Hypothesis. I'm not near qualified to even fully grasp the problem, but seeing the hypothesis and the other problems I wonder: what practical use will a solution have?

Many researchers have spent a lot of time on it, trying to prove it, but why is it important to solve the problem?

I've tried relating the situation to problems in my field. For instance, solving the $P \ vs. NP$ problem has important implications if $P = NP$ is shown, and important implications if $P \neq NP$ is shown. For instance, there would be implications regarding the robustness or security of cryptographic protocols and algorithms. However, it's hard to say WHY the Riemann Hypothesis is important.

Given that the Poincaré Conjecture has been resolved, perhaps a hint about what to expect if and when the Riemann Hypothesis is resolved could be obtained by seeing what a proof of the Poincaré Conjecture has led to.

$\endgroup$
12
  • 6
    $\begingroup$ Define "practical use". And it is important because it is there and it is fun to know stuff, though this would probably not fit within the "practical use" tag... $\endgroup$
    – DonAntonio
    May 28, 2013 at 9:51
  • $\begingroup$ It is very likely to be a duplicate isn't it? $\endgroup$
    – Dominic Michaelis
    May 28, 2013 at 9:52
  • $\begingroup$ @DominicMichaelis; I don't follow, a duplicate of what? $\endgroup$
    – Mythio
    May 28, 2013 at 10:00
  • 4
    $\begingroup$ Well it is a very famous hypothesis so you won't be the first one to ask it, partial answers are for example here $\endgroup$
    – Dominic Michaelis
    May 28, 2013 at 10:14
  • 3
    $\begingroup$ immortality ... in the sense of Pythagoras ($a^2 + b^2 = c^2$) and Einstein ($E=mc^2$) $\endgroup$
    – zerosofthezeta
    Sep 14, 2013 at 7:43

5 Answers 5

Reset to default
95
$\begingroup$

Proving the Riemann Hypothesis will get you tenure, pretty much anywhere you want it.

$\endgroup$
5
  • 15
    $\begingroup$ Worth the downvotes :) $\endgroup$
    – nbubis
    Dec 29, 2013 at 2:51
  • 14
    $\begingroup$ Not to mention the 1M on your bank account... $\endgroup$
    – Matemáticos Chibchas
    Dec 29, 2013 at 3:44
  • 6
    $\begingroup$ I've already proven it, I just need to check my calculations... ;) $\endgroup$
    – King Squirrel
    May 4, 2014 at 17:26
  • 37
    $\begingroup$ @KingSquirrel If only that damn margin were to be a bit wider... ;-) $\endgroup$
    – triple_sec
    Aug 26, 2014 at 7:46
  • $\begingroup$ @KingSquirrel I have proved it too, just waiting for the prize to be increased to $10M.;) $\endgroup$
    – ishandutta2007
    Aug 12, 2022 at 8:55
52
$\begingroup$

The Millennium problems are not necessarily problems whose solution will lead to curing cancer. These are problems in mathematics and were chosen for their importance in mathematics rather for their potential in applications.

There are plenty of important open problems in mathematics, and the Clay Institute had to narrow it down to seven. Whatever the reasons may be, it is clear such a short list is incomplete and does not claim to be a comprehensive list of the most important problems to solve. However, each of the problems solved is extremely central, important, interesting, and hard. Some of these problems have direct consequences, for instance the Riemann hypothesis. There are many (many many) theorems in number theory that go like "if the Riemann hypothesis is true, then blah blah", so knowing it is true will immediately validate the consequences in these theorems as true.

In contrast, a solution to some of the other Millennium problems is (highly likely) not going to lead to anything dramatic. For instance, the $P$ vs. $NP$ problem. I personally doubt it is probable that $P=NP$. The reason it's an important question is not because we don't (philosophically) already know the answer, but rather that we don't have a bloody clue how to prove it. It means that there are fundamental issues in computability (which is a hell of an important subject these days) that we just don't understand. Solving $P \ne NP$ will be important not for the result but for the techniques that will be used. (Of course, in the unlikely event that $P=NP$, enormous consequences will follow. But that is about as likely as it is that the Hitchhiker's Guide to the Galaxy is based on true events.)

The Poincaré conjecture is an extremely basic problem about three-dimensional space. I think three-dimensional space is very important, so if we can't answer a very fundamental question about it, then we don't understand it well. I'm not an expert on Perelman's solution, nor the field to which it belongs, so I can't tell what consequences his techniques have for better understanding three-dimensional space, but I'm sure there are.

$\endgroup$
8
  • 16
    $\begingroup$ about Perelman's solution and the Poincaré conjecture : in theoretical physics, more precisely in general relativity, there are equations relating the distribution of mass in the universe to the curvature of space-time. Perelman's solution used curvature (via the Ricci flow) to get topological information, i.e. the shape of the compact simply connected manifold (homeomorphic to a sphere). these kind of techniques are related to determnining the shape of the universe (space-time) which is a 4-manifold, depending on the curvature (which is determined by mass repartition, which we can observe) $\endgroup$
    – Albert
    May 28, 2013 at 13:22
  • 2
    $\begingroup$ @Glougloubarbaki - it would be great if you could expound your comment as an answer. $\endgroup$
    – nbubis
    Dec 29, 2013 at 2:45
  • 2
    $\begingroup$ Is it really obvious that "enormous consequences will follow" from a proof of P = NP? $\endgroup$
    – bof
    Nov 14, 2014 at 4:53
  • $\begingroup$ @bof highly likely, yes. At the very least lots and lots of expert computer scientists will look at the proof in disbelief as it shatters something they really felt very confident about. And, unless the proof will be completely unusable computationally, some really hard to solve problems will be found polynomial solutions. $\endgroup$
    – Ittay Weiss
    Nov 14, 2014 at 4:57
  • 1
    $\begingroup$ P=NP. I'd love an opportunity to demonstrate my NTM. Nonetheless, the argument that no-one believes P=NP is just patently false. P=NP via BPP. $\endgroup$
    – Jonathan Charlton
    Jan 18, 2016 at 6:44
27
$\begingroup$

Explaining the true mathematics behind the Riemann Hypothesis requires more text that I'm allotted (took most of my undergraduate degree in mathematics to even touch the surface; required all of graduate school to fully appreciate the beauty).

In very simple terms, the Riemann Hypothesis is mostly about the distribution of prime numbers. The idea is that mathematicians have some very good approximations (emphasis on approximate) for the density of the primes (so you give me an integer, and I can use these approximate functions to tell you roughly how many primes are between 0 [really 2] and that integer). The reason we use these approximations is that no [known] function exists that efficiently and perfectly computes the number of primes less than a given integer (we're talking numbers with literally millions of zeros). Since we can't determine the exact values (again, I'm simplifying a lot of this) the problem mathematicians want to know is exactly HOW good are these approximations.

This is where the Riemann Hypothesis comes in to play. For well over a century, mathematicians have known that a special form of the polylogarithm function (again, more fun math if you're bored) is a really great approximation for the prime counting function (and it's way easier to compute). The Riemann Hypothesis, if true, would guarantee a far greater bound on the difference between this approximation and the real value. In other words, the importance of the Riemann Hypothesis is that it tells us a lot about how chaotic the primes numbers really are. That's an incredibly high-level explanation and the Riemann Hypothesis deals with literally hundreds of other concepts, but the main point is understanding the distribution of the primes.

$\endgroup$
8
$\begingroup$

The Riemann hypothesis is a conjecture about the Riemann zeta function $$\zeta(s)=\sum_{n=1}^{\infty}\dfrac{1}{n^s}$$ This is a function $\mathbb{C} \rightarrow \mathbb{C}$. With the definition I have provided the zeta function is only defined for $\Re(s)\gt1$. With some complex analysis you can proof that there is a continuous (actually holomorphic if you know what it means) extension of the function so that it is defined in whole $\mathbb{C}$. The Riemann zeta function has some trivial zero points like $-2,-4,-6.$ The hypothesis says that the other zero points lie on the critical line $\Re(s)=\dfrac12$. This hypothesis had many application in analysis and number theory. The first proof of the prime number theorem used this conjecture.

In order to give an anwer to your question a would like to refer to this website, where you can find tons of applications of the Riemann hypothesis.

$\endgroup$
3
  • 37
    $\begingroup$ This is not an answer to the question, but a (very) brief introduction to what the RH is. OP asked about what applications proving RH would have. $\endgroup$
    – Lord_Farin
    May 28, 2013 at 9:58
  • 2
    $\begingroup$ the link was interesting (+1) $\endgroup$
    – zerosofthezeta
    Sep 14, 2013 at 7:46
  • $\begingroup$ And we want the zeros of $\zeta(s)$ because $\frac{\zeta'(s)}{s\zeta(s)}$ is the Laplace transform of $\Psi(u) = \sum_{p^k < e^u} \log p$ a function showing us the distribution of primes numbers. If the RH is true, then $\frac{\Psi(u) -e^u}{e^{u/2}}= o(e^{-u/2})+\log 2\pi + \sum_\gamma \frac{e^{i \gamma u}}{1/2+i\gamma}$ where $\gamma$ are the imaginary parts of the non-trivial zeros $\endgroup$
    – reuns
    Dec 23, 2016 at 7:33
5
$\begingroup$

The techniques used in the proofs of some of the most difficult theorems are used to prove so many other theorems. A proof of 1 of these theorems will give us access to an incredible amount of new techniques that will definitely make mathematics shorter,simpler and easier to understand.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .