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Notations, non-standard concepts, and definitions used commonly in these investigations are detailed in this post.)

1. Since, by the Prime Number Theorem, the number of primes $\leq \sqrt n$ is $O(\frac{\sqrt n}{log_{_{e}}\sqrt n})$, it would follow that determining a factor of $n$ requires at least one logical operation for each prime $\leq \sqrt n$, and therefore cannot be done in polynomial time—whence $P \neq NP$IF whether or not a prime $p$ divides an integer $n$ were independent of whether or not a prime $q \neq p$ divides the integer $n$.

2. Currently, conventional approaches to determining the computational complexity of Integer Factorising apparently appeal critically to the belief that:

(i) either—explicitly (see here)—that whether or not a prime $p$ divides an integer $n$ is not independent of whether or not a prime $q \neq p$ divides the integer $n$;

(ii) or—implicitly (since the problem is yet open)—that a proof to the contrary must imply that if $P(n\ is\ a\ prime)$ is the probability that $n$ is a prime, then $\sum_{_{i = 1}}^{^{\infty}} P(i\ is\ a\ prime) = 1$.

3. If so, then conventional approaches seem to conflate the two probabilities:

(i) The probability $P(a)$ of selecting a number that has the property of being prime from a given set $S$ of numbers;

Example 1: I have a bag containing $100$ numbers in which there are twice as many composites as primes. What is the probability that the first number you blindly pick from it is a prime. This is the basis for setting odds in games such as roulette.

(ii) The probability $P(b)$ of determining that a given integer $n$ is prime.

Example 2: I give you a $5$-digit combination lock along with a $10$-digit number $n$. The lock only opens if you set the combination to a proper factor of $n$ which is greater than $1$. What is the probability that the first combination you try will open the lock. This is the basis for RSA encryption, which provides the cryptosystem used by many banks for securing their communications.

4. In case 3(i), if the precise proportion of primes to non-primes in $S$ is definable, then clearly $P(a)$ too is definable.

However if $S$ is the set $N$ of all integers, and we cannot define a precise ratio of primes to composites in $N$, but only an order of magnitude such as $O(\frac{1}{log_{_{e}}n})$, then equally obviously $P(a)$ cannot be defined in $N$ (see Chapter 2, p.9, Theorem 2.1, here).

5. In case 3(ii) the following paper proves $P(b) = \frac{1}{\pi(\sqrt{n})}$, since it shows that whether or not a prime $p$ divides a given integer $n$ is independent of whether or not a prime $q \neq p$ divides $n$:

Why Integer Factorising cannot be polynomial time

Not only does it immediately follow that $P \neq NP$ (see here), but we further have that $\pi(n) \approx n.\prod_{_{i = 1}}^{^{\pi(\sqrt{n})}}(1-\frac{1}{p_{_{i}}})$, with a binomial standard deviation. Hence, even though we cannot define the probability $P(n\ is\ a\ prime)$ of selecting a number from the set $N$ of all natural numbers that has the property of being prime, $\prod_{_{i = 1}}^{^{\pi(\sqrt{n})}}(1-\frac{1}{p_{_{i}}})$ can be treated as the de facto probability that a given $n$ is prime, with all its attended consequences for various prime-counting functions and the Riemann Zeta function (see here).

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