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Do primes become more or less frequent as you go further out on the number line? That is, are there more or fewer primes between 1 and 1,000,000 than between 1,000,000 and 2,000,000?

A proof or pointer to a proof would be appreciated.

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Yes (as has been answered): the heuristic reason is that we can list the primes $p_1, p_2, \dots$. Each time you apply the sieve of Eratosnese you exclude the numbers divisible by $p_n$ and weed out "$(1-1/p_n)$" of the natural numbers (i.e., asymptotic density). So in the end you are left with $\prod ( 1 - 1/p_n)$ fraction of the natural numbers; since the sum $\sum \frac{1}{p_n} = \infty$, this is the proportion zero. So we should expect the primes to have density zero. – Akhil Mathew Jul 21 '10 at 0:23
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Check out the Riemann-zeta function if you want something a bit more technical. It gives a pretty precise estimate of the distribution of primes, indirectly. – Noldorin Jul 21 '10 at 7:38
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From the Wikipedia article about the prime number theorem:

Roughly speaking, the prime number theorem states that if a random number nearby some large number N is selected, the chance of it being prime is about 1 / ln(N), where ln(N) denotes the natural logarithm of N. For example, near N = 10,000, about one in nine numbers is prime, whereas near N = 1,000,000,000, only one in every 21 numbers is prime. In other words, the average gap between prime numbers near N is roughly ln(N).

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The Sieve of Eratosthenes is a very intuitive visual representation of why the frequency of prime numbers goes down as you go further out on the number line.

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This is really a less detailed version of Akhil's comment. – BBischof Jul 21 '10 at 4:43
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@BBischof Akhil's comment derives the answer to this question with mathematical rigor. My answer offers a link to a visual representation of the trend for the sake of gaining intuition. Its nice to have both the proof of the result and intuition about the result. – Ami Jul 21 '10 at 5:01
Agreed @Ami, it is a useful answer if not as rigorous. – dsolimano Jul 21 '10 at 13:20
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