Probability that any two numbers are prime

TODO…. 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.