"The most natural conjecture is that the prime numbers appear randomly..
If the twin prime conjecture is correct, then we may conclude that the
[primes] are not constructed randomly"
why does he say a random distribution of primes would imply the twin prime conjecture is false?
My intuition is the opposite: a random distribution of primes implies the twin prime conjecture is true. Because suppose there are finite twin primes, set them aside; then test pairs of numbers above the largest of them. Since we assume the twin prime conjecture is false, the chances that random pairs above that number are both prime is exacty 0; but that would mean the primes are not distributed randomly...
I thought about this for a second, and it seems that you are right: if the primes were distributed randomly, the twim prime conjecture would be true (see below). Are there any analytic number theorists in the audience to explain what the quote in the parent post is supposed to mean and/or fix my mistakes?
The prime number theorem [1] states that the probability that a natural number N is prime is approximately 1/log(N). So, the probability that N & N+2 are both prime is about 1/log(N)^2. If they were distributed randomly, there would be about
integral(1/log(x)^2 dx) from x=2 to x=N
twin primes between 1 and N. This integral converges to infinity as N grows large[2] (wish I remembered enough caclulus to do it by hand). So, there is some N for which you'd expect to have 10^100 twin primes less than N, etc.
So it seems that if you decided by a coin toss whether a number should be prime (where you'd make it prime with probability 1/log(N)), there would be infinitely many twin primes.
Wikipedia seems to confirm it's a reasonable estimate [3].
UPDATE: By the way, the same argument implies that if primeness was decided by a coin toss, there would be infinitely many N such that N and N+1 are prime. There is, however, only one such N (namely 2). So, primeness is not decided by a coin toss, and the above is not a proof of the twin primes conjecture :).
I a cousin comment, Retric proves that the integral of 1/log(x)^2 diverges. The point is that log(n)^2 < n for x > 2 (you can prove this by calculus: the only maximum of the function log(x)^2/x is at x=e). So, 1/log(x)^2 > 1/x, and the integral of 1/x diverges.
The most fun way to prove the last statement is that the integral of 1/x from 1 to infinity is greater than 1/2 + 1/3 + 1/4 + 1/5 + ... > 1/2 + 1/4 + 1/4 + 1/8 + ... = 1/2 + 2 * 1/4 + 4 * 1/8 + ... = 1/2 + 1/2 + 1/2 + ....
yes, very good Update. I had also thought of this and was wondering what the hell "randomly distributed" even means rigorously. I wrote the other respondent this by email (before your Update :) ) since I wasn't sure of my reasoning -- http://rifers.org/paste/show/3314
If the twin prime conjecture is correct, then we may conclude that the [prime] integers are not constructed randomly.
Another interesting tid-bit.. Looks to me that mathematics research is getting overwhelmingly complex...
For the next two semesters, I organized a seminar with five graduate
students (including Yitang) on Prof. Hironaka’s monumental papers on
2
the theory of resolutions of singularities. I believed that we doubled the
world population of those who had studied the papers after we finished two
semesters. Prof. Grothendick once described those papers as among the
most complicated thesises in the human history.
Of course it is. By 'randomly' he refers to a maximum entropy ("uniform") distribution conforming Gauss' Prime Number Theorem. Since by that theorem the gap rises, the probability of twin primes would go to 0 and, crucially, the expected number of twin primes would be finite (or any finite gap really).
That's an interesting reason for Yitang's proof being so significant without actually reaching the small gap -- it shows primes violate eloquently the random distribution in some ways.
If the probably of a number being prime is ~1 / ln (N)
The probably of x and x + 2 being prime is ~ (1 / ln (N)) ^2. For large N, (ln (N)) ^2 < N so there are infinitely many twin primes. Because the Sum of 1/(f(x)) as x -> inifinity = infinity if f(x) < (x).
So, the real issue is if the number of twin primes significantly larger than (1/ln(x)) ^ 2.
Ah yes you're right I just naively assumed the gap going to infinity implied the number of small gaps has to be finite, thanks. I believe your argument argument sightly miscounts the twin primes, since you're giving a new "chance" after you had already drawn a prime/not-prime. A lower bound assuming the probabilistic model would be E[Twin Primes] > Sum ( 1/( ln(4k) ln(4k+2) ), which is still divergent so your point stands.
Note the twin prime conjecture is not yet proven (your argument does not follow directly from PNT, but from the approximate probabilistic model that (almost surely?) displays PNT-like count).
> Because the Sum of 1/(f(x)) as x -> inifinity = infinity if f(x) < (x).
Here's a cute counterexample: x/(2*sign(sin(x))) < x, but the sum < infinity (you're missing an absolute value |f(x)| in your statement).
To expound on Retric's comment (I expounded much more in a cousin thread): I also first assumed that what you said must be true. When I tried to check with a calculation, it turned out that the opposite is the case.
My intuition is the opposite: a random distribution of primes implies the twin prime conjecture is true. Because suppose there are finite twin primes, set them aside; then test pairs of numbers above the largest of them. Since we assume the twin prime conjecture is false, the chances that random pairs above that number are both prime is exacty 0; but that would mean the primes are not distributed randomly...