$\Bbb{P}$ 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.
Best Answer
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