- #1

Adel Makram

- 636

- 15

The probability that two randomly chosen integers to be coprimes is known to be equal to ## \prod_{2}^{\infty}(1-\frac{1}{p^2})=6/\pi^2##

I tried to conceptualize the problem in the following way but got different results.Suppose that we pick up an integer at random which could be either prime or composite. Let's pick up another integer smaller than the first one and wonder about the probability of both integers being coprimes.

##\rho=(P_p)(cop|P_p)+(not P_p)(cop|not P_p)##

Where ##\rho## is the probability that the two integers are coprimes which can be expressed as the sum of the probability that the first integer is a prime multiplied by the conditional probability of the second integer to be coprime given that the first one is prime plus the probability that the first integer being a non-prime multiplied by the conditional probability of the second number to be coprime given the first one is non-prime.

Since we know the frequency of the prime from the PNT (prime number theorem) we can substitute that for the probability of the prime.

Also, from the Euler Phi function, we can substitute it for the last term. ##\varphi=\prod_{p}^{} n(1-\frac{1}{p})##

we get:

##\rho=\frac{1}{Log(n)}+(1-\frac{1}{Log(n)})(\frac{\prod_{p}^{} n(1-\frac{1}{p})}{n})##

n cancel and we left up with

After factoring out we get

For a large n ##\frac{1}{Log(n)}## goes to zero

and the final result becomes

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