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How high is the percentage of primes in $\mathbb{N}$?

($\mathbb{N} := \lbrace { 1, 2, 3, \ldots \rbrace }$ ; a prime is only divisible by itself and 1 in $\mathbb{N}$)

The percentage has to be lower than 50% as all even numbers (except for 2) aren't primes. So the percentage has to be lower than $1 - (\frac{1}{2} + (\frac{1}{3} - \frac{1}{6})) = 1 - (\frac{1}{2} + \frac{1}{6}) = 1 - (\frac{1}{2} + \frac{1}{2 \cdot 3}) = \frac{1}{3}$

I guess it will be something like that:

$\frac{\text{primes in } \mathbb{N}}{\text{numbers in } \mathbb{N}} = 1 - \sum_{i=\text{first Prime}}^\text{primes} \frac{1}{\prod_{j=\text{first prime}}^{i\text{-th prime}} j}$

But calculating this (exact) value goes definitely beyond my math skills. Can somebody help me?

  • 0
    See also: [Prove that $\lim_{x\to\infty} \pi(x)/x=0$](http://math.stackexchange.com/questions/28540/prove-that-lim-x-to-infty-pix-x-0)2013-11-20

6 Answers 6

-4

Not completely sure about this but I am working on proving it:

6.598827715%

That is, e raised to the negative e (expressed as a percentage)

The number of primes is the likelihood of zero for a Poisson Distribution with Lambda of e.

  • 4
    This is a very nice number -- it appears in infinite power towers, among other applications, see https://oeis.org/A073230 -- but it's not the answer to this question. Some numerical evidence, if the theory wasn't making sense: $\pi(10^{25})/10^{25}\approx0.0177$, so 6.6% is way too high.2014-04-29
70

People are quoting the Prime Number Theorem here, but that is pretty serious overkill. Even Chebyshev's theorem is a level above, in my opinion.

In my undergraduate number theory course, I prove that the prime numbers have density zero in the positive integers (including a careful statement of what this means). The proof is given as Theorem 6 in these notes. In fact it builds on the OP's observation that the density must be at most $\frac{1}{2}$ because half of all numbers are divisible by $2$. Similarly the density must be at most $\frac{1}{3}$ because every prime number $p > 5$ is relatively prime to $6$, and $\frac{\varphi(6)}{6} = \frac{1}{3}$. One can get a complete proof by showing that for each $\epsilon > 0$, there exists a positive integer $d$ such that $\frac{\varphi(d)}{d} < \epsilon$. This was proved earlier in the course: it is Proposition 6 in this set of notes. The proof doesn't use any analytic fact deeper than the divergence of the harmonic series.

  • 4
    Dear Pete, I agree that this is the natural $p$roof of the de$n$sity statement that the OP is asking about. Regards,2011-11-14
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I will denote the set of all primes by $\mathbb P$.

It is not exactly clear what one can understand under percentage of primes, but one possible interpretation is the limit $\lim\limits_{n\to\infty} \frac{\pi(n)}n = \lim\limits_{n\to\infty} \frac{|\{p\in\mathbb P; p\le n\}|}n$ of the ratio of primes and all numbers in the interval $[1,n]$. This is precisely the asymptotic density of the set $\mathbb P$ and using Prime number theorem one can show that it is equal to zero.


The asymptotic density of the set $A$ is defined as $d(A)=\lim\limits_{n\to\infty} \frac{|\{a\in A; a\le n\}|}n.$ A possible way to show $d(\mathbb P)=0$ without using PNT is to use the result from the paper Ivan Niven. The asymptotic density of sequences. Bull. Amer. Math. Soc., 57(6):420-434, 1951.

Corollary 1. If for a set of primes $\{p_i\}$ we have $d(A_{p_i})=0$ for every $i$, and if $\sum p_i^{-1}=\infty$ then $d(A)=0$.

Here, for any $A\subseteq\mathbb N$ and a prime $p$, the set $A_p$ is defined as $A_p=\{n\in A; p\mid n, p^2\nmid n\}.$

Applying the above result with $A=\mathbb P$ and $\{p_i\}=\mathbb P$ gives $d(\mathbb P)=0$.

We are using the fact that $\sum_{p\in\mathbb P} \frac1p=\infty$. A very nice proof of this fact was given by Erdős, see Proofs from The Book by Martin Aigner, Günter M. Ziegler p.5.


Another short proof of $d(\mathbb P)=0$ is given in this paper:

S. E. Mamangakis. Shorter notes: Remark on $\pi(x) = o(x)$. Proc. Amer. Math. Soc., 13(4):664--665, 1962

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According to the Prime Number Theorem $ \lim_{n\to\infty}\frac{\pi(n)}{n/\log(n)}=1\tag{1} $ where $\pi(n)$ is the number of primes less than or equal to $n$. Therefore, there is an $N$ so that if $n\ge N$, $ \frac{\pi(n)}{n}\le\frac{2}{\log(n)}\tag{2} $ Since there are infinitely many primes and infinitely many positive integers, there is no way to divide the number of primes by the number of positive integers to get a proportion. The next best attempt at a proportion would be the limit of the proportion in $[1,n]$: $ \lim_{n\to\infty}\frac{\pi(n)}{n}\tag{3} $ Inequality $(2)$ says that the limit in $(3)$ is $0$.

Addendum:

Pete Clark mentions that the Prime Number Theorem is overkill for this question, and I agree. In light of this, let's consider first the density of positive integers not divisible by a particular prime $p$. Define $ F_p=\{n\in\mathbb{Z}^+:p\nmid n\}\tag{4} $ It is pretty clear that $ \lim_{n\to\infty}\frac{|F_p\cap[1,n]|}{n}=1-\frac{1}{p}\tag{5} $ Divisibility by a prime $p$ and a prime $q$ are independent; that is, $p\mid n$ and $q\mid n$ if and only if $pq\mid n$. Thus, the density of $F_p\cap F_q$ is $ \lim_{n\to\infty}\frac{|F_p\cap F_q\cap[1,n]|}{n}=\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right)\tag{6} $ Continuing in this manner, we get that the density of positive integers not divisible by any in a set of the primes $\{p_i\}_{i=1}^k$ would be $ \lim_{n\to\infty}\frac{|\bigcap_{i=1}^kF_{p_i}\cap[1,n]|}{n}=\prod_{i=1}^k\left(1-\frac{1}{p_i}\right)\tag{7} $ Therefore, the density of the primes not in $\{p_i\}_{i=1}^k$ is no greater than $\prod_{i=1}^k\left(1-\frac{1}{p_i}\right)$. Since the density of primes in $\{p_i\}_{i=1}^k$ is $0$, the density of all primes is no greater than $\prod_{i=1}^k\left(1-\frac{1}{p_i}\right)$.

If we enumerate all primes as $\{p_i\}_{i=1}^\infty$, using the Fundamental Theorem of Arithmetic, we get that $ \begin{align} \frac{1}{\prod_{i=1}^\infty\left(1-\frac{1}{p_i^\alpha}\right)} &=\prod_{i=1}^\infty\left(1+\frac{1}{p_i^\alpha}+\frac{1}{p_i^{2\alpha}}+\cdots\right)\\ &=\sum_{k=1}^\infty\frac{1}{k^\alpha}\tag{8} \end{align} $ As Pete mentions, since the harmonic series diverges, as $\alpha\to1^+$, $(8)$ implies that $ \prod_{i=1}^\infty\left(1-\frac{1}{p_i}\right)=0\tag{9} $ Thus, the density of primes is $0$.

  • 0
    +1 The technique used in the addendum is simply wonderful. I saw the first time a variant of it [here](http://math.stackexchange.com/questions/427910/a-simple-way-to-obtain-prod-p-in-mathbbp-frac11-p-s-sum-n-1-in)2014-11-02
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Almost all of the natural numbers are composite; that is, the percentage of primes in $\mathbb{N}$ is $0$.

We know that the number of primes less than or equal to a natural number $x$ is asymptotic to $x/\log(x)$; that is, it has the same behavior for large $x$. Since this grows strictly slower than $x$, we see that the proportion of primes goes to $0$.

From this formula, we can also find that the proportion of primes in the natural numbers less than $x$ is $1/\log(x)$.

-3

While the density of primes as a percentage of all natural numbers is zero, that is merely a corollary to some of Kantor's theories about different levels of infinity: there are infinitely more natural numbers than prime numbers, so that as $N\to\infty$, density$\to 0$.

What I found more interesting was how the density of a bounded $N$ decreases as $N$ increases and remains bounded:

$25\%$ of the numbers from $1$ through $100$ inclusive are prime

$21\%$ of the numbers from $101$ through $200$ inclusive are prime

$14\%$ of the numbers from $901$ through $1,000$ inclusive are prime

$9\%$ of the numbers from $9,001$ through $10,000$ inclusive are prime

etc.

  • 5
    You are, unfortunately, sorely misguided here. This has nothing to do with Cantor's theorem; that was about _cardinality_, and you can map the natural numbers one-on-one to the primes and vice versa, so that there are exactly as many naturals as primes, just as there are exactly as many naturals as even numbers. Don't confuse cardinality with density.2014-11-02