Probability and number theory, part II

In this post, we will explore the following question: If we choose two integers uniformly at random, what is the probability that they are relatively prime?

We start with a preliminary result. The Riemann zeta function, \( \zeta(s) \), is ubiquitous in number theory. Although the most interesting version of it is the complex-valued one, today we only need the real version. It is defined as follows: for any real number \( s \) greater than \( 1, \)

\( \zeta(s) = \displaystyle\sum_{n=1}^{\infty} n^{-s} \).

The Euler product  for the zeta function is the following alternative expression:

\( \zeta(s) = \displaystyle\prod_{p} \frac{1}{1 – p^{-s}}, \)

where the product is taken over every prime \( p \). To see why this formula is true, we start with the right-hand side and use a geometric series expansion to obtain

\(\displaystyle\prod_{p}\frac{1}{1-p^{-s}}=\displaystyle\prod_{p}(1+p^{-s}+p^{-2s}+\cdots). \)

By the Fundamental Theorem of Arithmetic, every positive integer can be expressed as a unique product of primes, up to order. (We consider \( 1 \) to be the product of no primes; i.e., the empty product.) Thus, there is a one-to-one correspondence between the terms in the sum \( \sum_{n=1}^{\infty} n^{-s} \) and the terms in the expansion of the product \( \prod_{p}(1+p^{-s}+p^{-2s}+\cdots).\) Hence,

\( \displaystyle\prod_{p}\frac{1}{1-p^{-s}}=\displaystyle\sum_{n=1}^{\infty}n^{-s}=\zeta(s).\)

Now back to our original problem! Suppose we choose \( a, b \in \mathbb{Z} \) uniformly at random. Fix an arbitrary prime \( p \). The probability that both \( a \) and \( b \) are divisible by \( p \) is \( \left(\frac{1}{p}\right)\left(\frac{1}{p}\right) = p^{-2} \). Hence, the probability that they are not both divisible by \( p \) is \( 1 – p^{-2} \).

The condition that \( a \) and \( b \) are relatively prime is equivalent to the condition that, for every prime \( p \), \( a \) and \( b \) are not both divisible by \( p \). Furthermore, for distinct primes \( p \) and \( q \), the events that \( a \) and \( b \) are not both divisible by \( p \) and \( a \) and \( b \) are not both divisible by \( q \) are mutually independent. (In fact, more precisely, the events that \( a \) and \( b \) are not both divisible by \( m \) and \( a \) and \( b \) are not both divisible by \( n \), for relatively prime integers \( m \) and \( n \), are mutually independent.)

Thus, the probability that \( a \) and \( b \) are relatively prime is equal to

\( \displaystyle\prod_{p}(1-p^{-2}), \)

where the product is taken over every prime \( p \). But

\( \displaystyle\prod_{p}(1-p^{-2}) = \displaystyle\prod_{p}(\frac{1}{1-p^{-2}})^{-1}=(\displaystyle\prod_{p}\frac{1}{1-p^{-2}})^{-1} = \frac{1}{\zeta(2)}.\)

Hence the answer is \( \frac{1}{\zeta(2)}\approx 0.608. \, \spadesuit \)

The closed-form expression for \( \zeta(2) \) is actually well-known (it equals \( \frac{\pi^{2}}{6}) \), but that’s another story for another day. The result also generalizes to more than two integers, a nice exercise for the reader.

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