## Primes vs. Natural Numbers

For the discussion of math. Duh.

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xepher
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### Primes vs. Natural Numbers

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.

Blatm
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### Re: Primes vs. Natural Numbers

Check out the Prime number theorem.

Yakk
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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?
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

xepher
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### Re: Primes vs. Natural Numbers

No, more like Cardinality of primes/Cardinality of Natural Numbers.

++\$_
Mo' Money
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### Re: Primes vs. Natural Numbers

The cardinalities are obviously the same. You can't divide infinite cardinalities.

theorigamist
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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.

skullturf
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### 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!

xepher
Posts: 83
Joined: Tue Mar 09, 2010 1:42 am UTC

### 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).

theorigamist
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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.

forgetful functor
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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 density-zero subsets.