Does there exist a defined expression that describes the amount of prime numbers over the amount of natural numbers?
Is such an expression meaningless?
Also, if it helps, just assume the Riemann Hypothesis is true.
Primes vs. Natural Numbers
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Re: Primes vs. Natural Numbers
Check out the Prime number theorem.
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Re: Primes vs. Natural Numbers
There is a famous ratio of the number of prime numbers less than n to n. Is that what you are looking for?
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Re: Primes vs. Natural Numbers
No, more like Cardinality of primes/Cardinality of Natural Numbers.
Re: Primes vs. Natural Numbers
The cardinalities are obviously the same. You can't divide infinite cardinalities.

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Re: Primes vs. Natural Numbers
It's pretty easy to see (using the Prime Number Theorem) that the Schnirelmann density of the primes is 0. Also, the natural density (also called asymptotic density) of the primes is 0.
Re: Primes vs. Natural Numbers
One possible way to try to make more precise the idea of the ratio (Amount of primes)/(Amount of natural numbers) is to look at what proportion of the first n natural numbers are prime, and then take the limit of that proportion as n approaches infinity.
http://en.wikipedia.org/wiki/Natural_density
For the primes, it turns out that you get 0. (And you don't need the full strength of the prime number theorem if you just want that fact.) But this answer by itself might seem a bit unsatisfying  by itself, it doesn't tell you a whole lot about the primes and how they compare to other "thin" sets.
Edit: ninja'd!
http://en.wikipedia.org/wiki/Natural_density
For the primes, it turns out that you get 0. (And you don't need the full strength of the prime number theorem if you just want that fact.) But this answer by itself might seem a bit unsatisfying  by itself, it doesn't tell you a whole lot about the primes and how they compare to other "thin" sets.
Edit: ninja'd!
Re: Primes vs. Natural Numbers
skullturf wrote:One possible way to try to make more precise the idea of the ratio (Amount of primes)/(Amount of natural numbers) is to look at what proportion of the first n natural numbers are prime, and then take the limit of that proportion as n approaches infinity.
http://en.wikipedia.org/wiki/Natural_density
For the primes, it turns out that you get 0. (And you don't need the full strength of the prime number theorem if you just want that fact.) But this answer by itself might seem a bit unsatisfying  by itself, it doesn't tell you a whole lot about the primes and how they compare to other "thin" sets.
Edit: ninja'd!
DENSITIES. That's the thing I was looking for. Yeah, I expected it to be infinitesimal (aka 0).

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Re: Primes vs. Natural Numbers
Yeah, I expected it to be infinitesimal (aka 0).
The density doesn't tell the whole story. There are still more primes than some other 0 density subsets. Here's something to consider. Denote the Schnirelmann density of a set A by d(A), and if A and B are two sequences, denote their sumset by A+B. A set A is called an essential component if for any set B that has 0<d(B)<1, we have d(A+B) > d(B). It turns out the primes are an essential component. That is, even though the primes have 0 density, you strictly increase the density of any positive density subset by adding the primes to it. Also, the primes form a basis for the natural numbers, which means if you add the primes to themselves some finite number of times, you can get any natural number. (This is very related to Golbach's conjecture.)
Just to convince yourself that these properties are special, if A is the set of powers of 2 (including 1), try to show that A is not a basis for the natural numbers, nor is it an essential component.

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Re: Primes vs. Natural Numbers
For yet another way to think about "how many" primes there are among the natural numbers, recall that [imath]\sum 1/p[/imath] diverges. So in this sense there are many more primes than other densityzero subsets.
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