$\Bbb{P}$ is the count of prime numbers in $\Bbb{Z}$
And so, $\Bbb{Z}-\Bbb{P}=NP$ is the count of non-prime numbers in $\Bbb{Z}$ what is the answer of this equation: $\Bbb{P} / NP$

I thought that question and I made that proof, if I'm mistake please correct me.

$E=$Even, $O=$Odd
$1, 2, 3, 4, \cdots \Bbb{Z}$
$O, E, O, E, \cdots$

Clearly there is $\Bbb{Z}/2$ count of Even, and $\Bbb{Z}/2$ count of Odd numbers exist.

If any number in $\Bbb{Z}$ can write as $M \times N$ it is non-prime number, otherwise it's prime number $M \times N$ can be one of that $4$ combinations:

$E \times E = E$
$E \times O = E$
$O \times E = E$
$O \times O = O$

So, $M \times N$ is $\frac{3}{4}$ in ratio of Even numbers, and $\frac{1}{4}$ ratio of Odd.

Even Numbers: $\frac{3}{4}$ * NP
Odd Numbers: $\frac{1}{4}$ * NP

Even Numbers: $0 * P$
Odd Numbers : $P$

There is equal counts of even and odd numbers, so; $\frac{3}{4} * NP + 0 = 1/4 * NP + P$
$\frac{1}{4} * NP = P$
$NP = 2 * P$

If this equation is true, then non-prime numbers are only double-times of prime numbers. Please check my proof.

  • $\begingroup$ You really need to do some number theory... go check this out to begin with, it'll show you why primes are so hard to study. en.wikipedia.org/wiki/Prime_number_theorem And by the way your proof is just wrong. The ratio $\pi(n) / n$ goes to zero as $n$ goes to infinity. The proof uses the prime number theorem in a very elementary way. $\endgroup$ – Patrick Da Silva Feb 12 '12 at 0:08
  • $\begingroup$ You don't need anything as powerful as the prime number theorem to prove this. See my comment below my answer. $\endgroup$ – Michael Hardy Feb 12 '12 at 0:56
  • $\begingroup$ You'll find two distinct types of answers below. One deals with average rates of occurrence of prime numbers (Hardy, NKS). The other deals with comparing sizes of sets using one-to-one maps (Myerson). Either of these could be a valid way of answering your question, because you haven't unambiguously defined your question. $\endgroup$ – Ben Crowell Feb 12 '12 at 1:10

I think the most usual way of thinking about this is to find $$ \lim_{n\to\infty} \frac{\text{number of positive prime numbers}\le n}{\text{number of positive integers}\le n}. $$ The limit is $0$.

This depends on listing the numbers in their usual order. Suppose one writes them in this different order: $$ \text{1st prime number}, \text{first non-prime number}, \text{second prime number}, \text{second non-prime number}, \text{third prime number}, \text{third non-prime number}, \ldots. $$ Then the limit would be $1/2$.

Later note: See this article:

"An Elementary Proof that Primes are Scarce", by E. L. Spitznagel Jr., American Mathematical Monthly, volume 77, number 4, April 1970, pages 396--397. jstor.org/stable/2316153

  • $\begingroup$ I thought I remembered seeing an elementary proof of this fact published in the Monthly in the '70s. But I'm not finding it on Google Scholar now. $\endgroup$ – Michael Hardy Feb 12 '12 at 0:42
  • $\begingroup$ Found it: "An Elementary Proof that Primes are Scarce", by E. L. Spitznagel Jr., American Mathematical Monthly, volume 77, number 4, April 1970, pages 396--397. jstor.org/stable/2316153 $\endgroup$ – Michael Hardy Feb 12 '12 at 0:54
  • 2
    $\begingroup$ The paper's proof fits in a comment: all primes $2^{j-1}<p\le2^j$ divide $\binom{2^j}{2^{j-1}}$, so $$2^{2^j}\ge\binom{2^j}{2^{j-1}}\ge\prod_{2^{j-1}<p\le2^j}p\ge(2^{j-1})^{\pi(2^j)-\pi(2^{j-1})}$$ Comparing exponents, $\pi(2^j)-\pi(2^{j-1})\le2^j/(j-1)$. Letting $j-1$ vary from $i$ to $2i-1$ and adding, $$\pi(2^{2i})-\pi(2^i)\le\sum 2^j/(j-1)\le\sum 2^j/i<2^{2i+1}/i.$$ This with $\pi(2^i)\le2^i$ gives $\pi(2^{2i})/2^{2i}<2^i/2^{2i}+2/i$ which can be made as small as desired. (This is enough because if $2^{2i-2}<n\le2^{2i}$, then $\pi(n)/n<\pi(2^{2i})/2^{2i-2}=4\pi(2^{2i})/2^{2i}$.) $\endgroup$ – ShreevatsaR Jul 22 '13 at 15:50

"$P$ is the number of prime numbers in $Z$" makes no sense until you develop a theory of infinite numbers. Such a theory was developed by Cantor in the late 19th century. When you learn it, you will find that the number of integers, the number of even integers, the number of odd integers, the number of primes, and the number of non-primes are all the same; they are all what is called $\aleph_0$, which we read as "aleph-zero" or "aleph-nought".


I don't exactly follow your logic, and your conclusion is incorrect, but you may be interested to know that this is a historic problem which spurred a lot of developments in analytic number theory up through the nineteenth century, and is intimately related with such famous problems as the Riemann Hypothesis.

Unfortunately, a simple counting argument like yours won't be powerful enough to do the job. You should have a look at the Prime Number Theorem, which is essentially an answer to the question that you posed. It states that the number of primes below $x$, which we write $\pi(x)$, has a growth rate on par with the function $x/\ln x$.

  • 2
    $\begingroup$ A simple counting argument is not powerful enough, but you don't need anything anywhere near as powerful as the prime number theorem to prove that the density of the set of primes is $0$. $\endgroup$ – Michael Hardy Feb 12 '12 at 0:20
  • 1
    $\begingroup$ Fair enough. I was thinking more of addressing the question of the asymptotic behavior of $\pi(x)$. $\endgroup$ – NKS Feb 12 '12 at 0:23

Wow, that was some of the most over garbled and over explained thinking I've seen, no wonder you got lost. I'm thinking this was put here as a troll!

You eventually determined that prime numbers are always an odd number. "1/2" at the most "1/3" at what point is it "1/4" at what point is it "1/5". You didn't flesh out how often an odd number is a prime number. This is the only critical part of determining the "Diminishing" ratio and you simply skipped it.

Please think beyond what you think you've learnt and actually process what you're doing. You can't rely on pre-formulated deductions and then go around asking other people for the answer with out understanding where it came from.

Knowledge is empty without comprehension. And it's obvious you're not comprehending what it is you're asking for.


PS. In determining this ratio, you also determine why bigger machines are less efficient and why quad core processors work better on 4 separated streams as opposed to one task or why 4 wheels of a car can't hold 4 times the weight 1 wheel can.


I believe this is should work. Half of the numbers are even so we are left with 1/2. 1/3 of the remaining possibly prime numbers are not prime because they're divisible by 3.

So 2/3 of the numbers are not prime.

Of the 1/3 of the numbers that are possibly prime, 1/5 of them are not because they're divisible by 5.

So $2/3 + 1/15 = .73333...$ numbers are not prime

Then 1/7 of the remaining .2666... numbers are not prime so we know $0.7714285714285714...$ of the numbers are not prime.

Continuing this logic for all primes, I receive this result for the proporations for all natural numbers less than the values on the right.

$0.96516622547038294206 \times10^7$
$0.95936178982835162343 \times10^6$
$0.95124708214898474554 \times10^5$
$0.93911530754416157215 \times10^4$
$0.91903473649315761166 \times10^3$
$0.87968270952506481155 \times10^2$
$0.77142857142857142858 \times10^1$
$0.000000000000000000 \times10^0$

EDIT: here is the value for 10^8 with 50 decimal precision:


It shows that we are converging towards either the complete density of composites in the natural numbers or a very small ratio.

  • 1
    $\begingroup$ I think you are computing $1-(1-(1/2))(1-(1/3)(1-(1/5))\dots$. The infinite product diverges to zero. $\endgroup$ – Gerry Myerson Jan 3 '15 at 19:11

protected by user21820 Aug 15 '16 at 11:36

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.