From: nikl@mathematik.tu-muenchen.de (Gerhard Niklasch) Subject: Re: Thought on Prime numbers and Random numbers Date: 18 Sep 1999 19:59:28 GMT Newsgroups: sci.math Keywords: How much did Dirichlet prove about primes in progressions? In article <7rqsd4$b3u$1@nnrp1.deja.com>, Bob Silverman writes: |> In article , |> gerry@mpce.mq.edu.au (Gerry Myerson) wrote: |> > Are you sure about this? Dirichlet tells you there are infinitely |> > many primes in both progressions, but that's not enough to get the |> > ratio of the two counting functions to go to 1. Don't you need the |> > Prime Number Theorem for Arithmetic Progressions, as I said in my |> > earlier post? |> |> Ah. I was under the impression that Dirichlet's Theorem included |> what you are calling PNT for AP, i.e. #{x < N; x = a + bk, (a,b) = 1} |> = 1/phi(b) pi(N). |> |> Are they indeed separate theorems? |> |> Doesn't Dirichlet's theorem include this? Since phi(b) is the same |> whether a = -1 or +1, the ratio must indeed go to 1. Dirichlet's Theorem, as stated in most textbooks I'm aware of: If b > 1 is an integer and a is a positive integer coprime to b, the arithmetic progression { a + bk ; k nonnegative integer } contains infinitely many primes. What Dirichlet's proof actually proves is stronger than that: The set of primes p in such an arithmetic progression has Dirichlet density 1/phi(b), that is, the sum of the reciprocals 1/p of primes p < x congruent to a mod b, divided by the sum of the reciprocals of all primes < x, converges to 1/phi(b) as x -> infinity. The PNT for AP is again considerably stronger than that; it is a statement about natural densities: #{p < x congruent a mod b} divided by #{p < x} converges to 1/phi(b) as x -> infinity (or, in view of the plain old PNT, #{p * spam totally unwelcome * http://hasse.mathematik.tu-muenchen.de/~nikl/ ******* all browsers welcome * This .signature now fits into 3 lines and 77 columns * newsreaders welcome