What is the probability of two randomly chosen integers being coprime?

Bonus(d4): How does the result relate fo the density

of square-free numbers?

A little research shows that, yes, the coprimality of 2 random integers and the infinite sum of the squares of the positive integers are related. The following is extracted from https://en.wikipedia.org/wiki/Coprime_integers:

Informally, the probability that any number is divisible by a prime (or in fact any integer) p is 1/p. For example, every 7th integer is divisible by 7. Hence the probability that two numbers are both divisible by p is 1/p^2. One is led to guess that the probability that two numbers are coprime is given by a product over all primes:

Prod {over prime p} ( 1- 1/p^2) = Prod {p} (1 / (1 - p^-2)^-1

= 1/*ζ(2)*

<math xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> <mrow> <mstyle displaystyle="true" scriptlevel="0"> <mi>p</mi> </mstyle> </mrow> <annotation encoding="application/x-tex">{displaystyle p}</annotation> </semantics> </math>

Here *ζ* refers to the Riemann Zeta Function which Euler showed in 1735 sums to *π*^{2}/6.

Who woulda thunk it?

*Edited on ***July 16, 2018, 1:25 pm**

*Edited on ***July 17, 2018, 8:10 pm**

*Edited on ***July 17, 2018, 8:11 pm**