# Probability that any two numbers are prime

July 3, 2019 tags: mathTODO…. fix this

Two numbers are relatively prime if they have no common divisors except 1. It also means that the numbers have no prime factor in common.

The probability that a number is divisible by 2 is 1/2. The probability that two numbers are both divisible by 2 is (12).12=122. And thus, probability that neither of the two numbers are divisible by 2 is (1−122)

Similarly, probability that neither of the two numbers are divisible by 3 is (1−132)

Thus the probability that neither of the two numbers are divisible by any of the prime numbers is P(RelativelyPrime)=(1−122).(1−132).(1−152).(1−172)…

However, this is the minimum probability limit for any two numbers being relatively prime.

“Wait, but that’s an infinite product. What am I gonna do with that big product?”

We’ll see.

Let’s intruduce an infinite sum, S=112+122+132+142+152+… Multiplying both sides by 122

,

122.S=122.112+122.122+122.132+122.142+122.152+… 122.S=122+142+162+182+1102+…

Subtracting the last term from S , (1−122).S=1+(122−122)+132+(142−142)+152+…

S1=(1−122).S=1+132+152+172+192…

All the multiples of 2 are gone.

Now, multiply S1 by 132. 132.S1=132.1+132.132+132.152+132.172…

Subtracting the last term from S1 . (1−132).S1=1+152+172+(192−192)+… (1−132).S1=1+152+172+1112+…+1232+1252+… (1−132).(1−122).S=1+152+172+1112+…+1232+1252+…

All the remaining multiples of 3 are gone.

If we continue this with all the prime numbers, all the terms on the right are gone except 1. S.(1−122).(1−132)(1−152)(1−172)(1−1112)…=1 So, S=1(1−122).(1−132)(1−152)(1−172)(1−1112)… (1−122)(1−132)(1−152)(1−172)(1−1112)…=1S At the beginning, we defined S=112+122+132+142+152+… This is a very famous sum defined as, ζ(2)=112+122+132+142+152+… The very beautiful symbol ζ is called Reimann Zeta Function named after the mathematician Bernhard Riemann. It is defined for all complex numbers as, ζ(s)=11s+12s+13s+14s+15s+… where s

is a complex number.

It might be surprising, but it is true that ζ(2)=π26

The proof to this fact is here.

So, going back to our original question, P(RelativelyPrime)=(1−122).(1−132).(1−152).(1−172)… P(RelativelyPrime)=1ζ(2) P(RelativelyPrime)=1π26 P(RelativelyPrime)=6π2

There we go!! Probability that any two numbers are relatively prime to each other is not a very complicated expression.