Re: [PrimeNumbers] proving the Riemann hypothesis
>> If the RH is false, the gap between prime P1 and theThe original poster might have intended to say that there is no -constant-
>> next prime P2 could be anywhere from 1 to infinitely
> False. Your attempt at a proof ends here.
upper bound on the gap between two consecutive primes -- which is
Regardless of that, this statement:
>> I can prove that the gap must be smaller than P1xP1-P1. If P1 is 5,is a consequence of Bertrand's postulate (proved by Chebyshev, or
>> this means that the gap between 5 and its next prime must be smaller
>> than 5x5 - 5 = 20
Tchebycheff, or some other proper transcription :-) ), which says that the
gap after prime P is at most of size P.
[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
- For example 2 adjacent gaps cannot be equal if they aren't multiple of
6. For example the gap between 2 pairs of twins is at least 4. For
example each prime number has the form 2n+/-1, 3n+/-1, 4n+/-1, 6n+/-1.
Each pair of twins has the form 12n+-1, there are approximate formulas
for the nth prime and the number of primes < x and so on. You cannot
call all this random ore unpredictable. Of course the prime numbers are
distributed as regularly as possible, that's a tautology. In
mathematics everything is as regular as possible. Is pi random? Build
P=2,357111317192329 , and you have the same case as pi. Consider the
primes to be an irrational number, and there are no problems. If you
mean there is no formula f(n) which produces primes for each n, then
you are right. In this sense primes are random. (I am not quite sure
there is a formula p=[k^n^3] (H.W.Mills) which is said to produce only
prime numbers). If you define "formula" as an algorithm, as a
calculation instruction such as the sieve of Eratosthenes, then the
primes are not random but simply what they are. Perhaps the compound
numbers are random? Or are they only non-transparently complicated?