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$\Bbb{P}$ 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.

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    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. http://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.2012-02-12
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    You don't need anything as powerful as the prime number theorem to prove this. See my comment below my answer.2012-02-12
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    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.2012-02-12

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