Probability of Two Random Integers having no Common Divisor

Theorem
Let $a$ and $b$ be integers chosen at random.

The probability that $a$ and $b$ are coprime is given by:
 * $\map \Pr {a \perp b} = \dfrac 1 {\map \zeta 2} = \dfrac 6 {\pi^2}$

where $\zeta$ denotes the zeta function.

The decimal expansion of $\dfrac 1 {\map \zeta 2}$ starts:
 * $\dfrac 1 {\map \zeta 2} = 0 \cdotp 60792 \, 71018 \, 54026 \, 6 \ldots$

Proof
Let $a$ and $b$ be two integers chosen at random.

For $a$ and $b$ to be coprime, it is necessary and sufficient that no prime number divides both of them.

The probability that any particular integer is divisible by a prime number $p$ is $\dfrac 1 p$.

Hence the probability that both $a$ and $b$ are divisible by $p$ is $\dfrac 1 {p^2}$.

The probability that either $a$ or $b$ or both is not divisible by $p$ is therefore $1 - \dfrac 1 {p^2}$.

Whether or not $a$ is divisible by $p$ or divisible by another prime number $q$ is independent of both $p$ and $q$.

Thus by the Product Rule for Probabilities, the probability that $a$ and $b$ are not both divisible by either $p$ or $q$ is $\paren {1 - \dfrac 1 {p^2} } \paren {1 - \dfrac 1 {q^2} }$.

This independence extends to all prime numbers.

That is, the probability that $a$ and $b$ are not both divisible by any prime number is equal to the product of $1 - \dfrac 1 {p^2}$ over all prime numbers:


 * $\map \Pr {a \perp b} = \ds \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$

From Sum of Reciprocals of Powers as Euler Product:


 * $\ds \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$

from which:


 * $\ds \dfrac 1 {\map \zeta 2} = \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$

where $\map \zeta 2$ is the Riemann $\zeta$ (zeta) function evaluated at $2$.

The result follows from Riemann Zeta Function of 2.

Also see

 * Probability of Random Integer being Square-Free‎, which is the same probability as this