Wikipedia:Reference desk/Archives/Mathematics/2013 May 21

= May 21 =

Twin primes
The recent news articles about "twin primes" confuse me. In the infinite set of all integers, why can't we immediately conclude that there's an infinite number of twin primes, for the same reason we can conclude that there is an infinite number of integers divisible by 7, and an infinite number of integers divisible by 7 billion? Tarcil (talk) 01:48, 21 May 2013 (UTC)


 * Every 7th integer is divisible by 7 so there is obviously infinitely many integers divisible by 7. Similarly for 7 billion. I don't understand why you would think "the same reason" works for twin primes. Lots of things occur no or only a finite number of times among the infinitely many integers. PrimeHunter (talk) 02:03, 21 May 2013 (UTC)


 * For example, there are only finitely many prime numbers divisible by 7 :-) Sławomir Biały  (talk) 01:45, 22 May 2013 (UTC)
 * why can't we immediately conclude — Because maths is based on proofs, not assumptions. — 79.113.209.71 (talk) 03:58, 22 May 2013 (UTC)

Because the summation $$\sum_{\text{twin primes}}\frac{1}{p}$$ is known to be convergent. Count Iblis (talk) 12:07, 22 May 2013 (UTC)
 * Huh!? If you could explain why, or provide some relevant links, it would be very much appreciated! — 79.113.209.71 (talk) 12:35, 22 May 2013 (UTC)
 * See Brun's theorem.  Sławomir Biały  (talk) 12:43, 22 May 2013 (UTC)

If it were proven there are infinitely many twin primes, what would be the value of that knowledge? Could it be put to any practical use? --  Jack of Oz   [Talk]  12:17, 22 May 2013 (UTC)


 * It's the other way around; if the summation were divergent then you could immediately conclude that there are an infinite number of twin primes. This isn't the case and there are probably no such similar shortcuts allowing for an immediate conclusion that there are an infinite number of twin primes. Count Iblis (talk) 12:44, 22 May 2013 (UTC)


 * But why would anyone care either way? We already know there's at least a humongous number of them.  What difference would it make to anything to know for sure they go on forever?  --   Jack of Oz   [Talk]  06:05, 24 May 2013 (UTC)


 * Because at first glance it seems to be a bald fact with no obvious reason for its truth. For example, even assuming the laws of physics, why should there be X number of protons in the universe, rather than X+1 or X-1?  Would it not be terribly interesting if someone were to discover and explanation for that otherwise inexplicable coincidence? Of course there are some people who aren't impressed by the fact that Saussure predicted Hittite would have 'aitches' even before it was deciphered (to oversimplify the matter almost infinitely).  But other humans are.μηδείς (talk) 06:18, 24 May 2013 (UTC)
 * Well, no, there's plenty of reason to believe it's true. If you think of the primes as being essentially randomly distributed, then it really should be true, because their density falls off so slowly that every now and then they really "ought" to bump up against each other as closely as possible.  So it doesn't really need explaining &mdash; an explanation would be needed more if it turned out to be false.  But in response to Jack &mdash; probably no difference, in the short term.  But it would form a part of the tapestry of knowledge, and you just never know when you'll find a use for something in it. --Trovatore (talk) 06:25, 24 May 2013 (UTC)
 * Explainability's not quite the point that I was trying to make. The point is more along the lines that just as understanding the laws of physics doesn't tell you where the galaxies actually are without looking, understanding what primes are doesn't let you map them out ahead of time in the same way that you can map out numbers divisible by three.  For example, given any random range, like 555,000 to 556,000, we know ahead of time how many numbers in that range will be divisible by three, while I am unaware of any formula other than brute calculation that will tell you how many primes and in what pattern. μηδείς (talk) 00:26, 25 May 2013 (UTC)
 * Right, but you can know approximately how many primes there probably are in that range, from the prime number theorem. There should be about $$\frac{1000}{\log 555500}$$, or about 76 primes, in that interval.  Not exactly, but without checking, I would bet 10 to 1 that the number is between 60 and 90. --Trovatore (talk) 01:05, 25 May 2013 (UTC)
 * It's the fact that such estimations are approximate that makes the real truth bald facts and interesting to people who care about such things. μηδείς (talk) 21:38, 25 May 2013 (UTC)
 * Thanks both. It was as I thought, intellectual curiosity.  Nothing wrong with that.  --   Jack of Oz   [Talk]  06:28, 24 May 2013 (UTC)

One can do much better than approximately how many primes. Riemann's Explicit formula - an exact formula for primes in terms of the zeros of the Riemann zeta function, in some sense their duals. Don Zagier's The first 50 million prime numbers, Mathematical Intelligencer 0 (1977), 7-19 is a beautiful introduction. See also. Conjecturally, between x and x+a there are Ca/(log x)^2 twin primes for constant C =1.32... as Zagier describes. Intellectual curiosity about primes has been spectacularly productive - the explicit formula inspired the parallel Selberg trace formula, relating prime geodesics on Riemann surfaces to spectrums of differential operators. Or the generalization of prime to prime ideal in commutative rings A and their geometrization, the scheme Spec A, the spectrum of prime ideals etc., ad infinitum.John Z (talk) 00:02, 26 May 2013 (UTC)
 * I don't know if it's possible to express what you just said in layman's English, John. But it is beyond me. Are you saying that given any arbitrary range of numbers like I did above, Riemann's got a formula that will give you the exact number of primes in that range without actually identifying them or doing some sort of heavy duty calculation, that say, in effect, takes into account all the factors of those numbers to get its result? When you add qualifications (I do not understand) like "in terms of the zeros of the Riemann zeta function, in some sense their duals" it makes me suspect things are a bit complicated. μηδείς (talk) 16:36, 26 May 2013 (UTC)
 * The short answer is "Yes". And well, if YOU can't explain something to an intelligent and interested "layman" like yourself, Medeis, YOU, Mr. Smartypants, don't understand it. And YOU are a stupid fraud who should refund all of your victims=students' hard-earned money. And leave town on the next stagecoach.... But the point is that Riemann, stupendous genius that he was, did come up with an exact formula, with a lot of annoying analytic details that smart people like Zagier or Serge Lang or Harold Davenport understand and explain very well. (Davenport's Multiplicative Number Theory has an excellent, snappy but rigorous presentation of the details of Riemann's classical story and the PNT.).


 * To quote Zagier, "Although Riemann never proved the prime number theorem, he did something which is in many ways more astonishing - he discovered an exact formula for $$\scriptstyle\pi(x)$$." IF you knew all the values of s for which the Riemann zeta function ζ(s) = $$ \sum_{n=1}^\infty\frac{1}{n^s},$$ is uh, 0 (these are called "zeros" or "roots"). THEN Riemann's formula is an exact infinite sum for the Prime-counting function $$\scriptstyle\pi(x)$$, which involves nothing about divisibility and factors etc, but adds up terms using these infinitely many mysterious zeros. Basically, the main term in the formula is the x/log x type stuff of the prime number theorem, which is a decent approximation in itself. But the zeros give exact "error terms"! Convergence of such an infinite sum means that as you use more and more of these zeros, you get a closer and closer approximation to $$\scriptstyle\pi(x)$$.  Look at the last couple graphs in Zagier's paper - showing how taking into account the first 10 or 30 or so zeros of zeta, you get visually obvious, increasingly accurate approximations to the primes.John Z (talk) 23:33, 27 May 2013 (UTC)
 * You do an incredible Willy Wonka impersonation. I am going to take infinite sum to mean brute calculation. That doesn't mean the answer's not interesting. μηδείς (talk) 03:02, 28 May 2013 (UTC)