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 π2/6.
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