Who would have thought that area and circumference of a circle were related to factorising numbers and to infinite series?

Yet the probability that two integers m and n picked at random are co-prime (have no common factors) is:

ζ(2)^{(-1)}=6/(π^{2})=0.60792…

ζ(z) is the Riemann zeta function, which for z=2 is

sum from r=1 to infinity of 1/r^{2}.

Let N be a positive whole number, and p

_{1}, p

_{2}, p

_{3}…. p

_{r}be the primes less than N

Let Π(N) be the number of coprime pairs of numbers ≤N

⌊x⌋ is the greatest whole number ≤x, e.g. ⌊2.9⌋=2

Then divide up all the N^{2} pairs of numbers≤N into the Π(N) which have no common divisor, the (⌊N/p_{1}⌋)^{2} which have p_{1} as a common divisor, the (⌊N/p_{2}⌋)^{2} which have p_{2} as a common divisor, etc.

We will have double-counted those which have p_{1}p_{2} (etc.) as a common divisor. Subtract (⌊N/p_{1}p_{2}⌋)^{2} (etc.) Then we will have double-subtracted those which have p_{1}p_{2}p_{3} as a common divisor. And so on…

With all the corrections

N^{2} = Π(N) + sum over all primes≤N of (⌊N/p⌋)^{2} −

sum over all pairs of primes≤N of (⌊N/p_{i}p_{j}⌋)^{2} +

sum over all groups of three primes≤N of (⌊N/p_{i}p_{j}p_{k}⌋)^{2} −…

so

Π(N)/N^{2} = 1 − [sum (1/N)⌊(N/p)⌋]^{2} + sum [(1/N)⌊N/p_{i}p_{j}⌋]^{2} − sum [(1/N)⌊N/p_{i}p_{j}p_{k}⌋]^{2} + …

Take the limit as N tends to infinity

Then the probability of two numbers≤N being coprime approaches

1 − sum (1/p)^{2} + sum (1/p_{i}p_{j})^{2} − ….

which is the infinite product over all the primes of

(1−(1/2)^{2}).(1−(1/3)^{2}).(1−(1/5)^{2}).(1−(1/7)^{2})…. **[1]**

But ζ(2) is defined as the sum from r=1 to infinity of 1/r^{2}. Factorising r into prime factors, that is:

the sum over all powers r_{1}, r_{3}, r_{3}… ≥0 of 1/[2^{2r1}.3^{2r2}.5^{2r3}….]

and that is this infinite product of infinite sums:

(sum for all r_{1}≥0 of 1/2^{2r1}).(sum for all r_{2}≥0 of 1/3^{2r2})…

The infinite sums are geometric series adding to

1/[1−(1/2)^{2}], 1/[1−(1/3)^{2}], 1/[1−(1/5)^{2}], ….

so their infinite product, ζ(2), is equal to 1/(infinite product **[1]**), which is equal to the probability of any two numbers, however big, being coprime.

**It remains to prove that ζ(2)=π ^{2}/6**

Euler proved it this way:

sin x = x − x^{3}/3! + x^{5}/5! − x^{7}/7! − …. (Maclaurin series)

But the roots of sin x = 0 are 0, ±2π, ±3π, ±π…

So (this is a daring step, but possible to justify rigorously) for some constant A

x − x^{3}/3! + x^{5}/5! − x^{7}/7! −…. = A x(1−x^{2}/π^{2})(1−x^{2}/(2π)^{2})(1−x^{2}/(3π)^{2})…

Equating the coefficients of x, A=1

Equating the coefficients of x^{3}:

−(1/3!) = −1/π^{2}−1/(2π)^{2}−1/(3π)^{2}…

and so

1/1^{2} + 1/2^{2} + 1/3^{2} +… = π^{2}/6 ▇

Thanks for all that to Julian Havil’s brilliant book Gamma.

For more about primes, watch Terry Tao’s video:

And for more about the zeta (ζ) function and the Riemann hypothesis, watch Ed Frenkel’s video: