Number theory during covid break -1

Suppose 2 natural numbers are chosen at random, what is the probability that they are relatively prime? if this is a π b \dfrac{a}{\pi^b} for integers a , b a,b find a + b a+b


generalize to n n natural numbers.


The answer is 8.

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1 solution

Aareyan Manzoor
Apr 1, 2020

If two numbers are coprime, it means that they have no common prime factors. That means that for each p p , it cannot divide both numbers. The probability a number chosen at random is divisible by p p is just 1 p \frac{1}{p} , so the probability that p p divides both is p 2 p^{-2} and hence its complement is 1 p 2 1-p^{-2} . We take the product over all p p and this is just p ( 1 p 2 ) = 6 π 2 \prod_p (1-p^{-2}) = \frac{6}{\pi^2} Similarly the general case is 1 ζ ( n ) \frac{1}{\zeta(n)}


There is an interesting way to do the case of two numbers by considering that the probability something is coprime to a a is just ϕ ( a ) a \frac{\phi(a)}{a} , i leave it as an interesting exercise to the reader to do that.

Heyyy I'm back too, albeit maybe only for today

Julian Poon - 1 year, 2 months ago

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oh hey long time, yeah i only occasionally come back here when im bored.

Aareyan Manzoor - 1 year, 2 months ago

Can you explain how is the product related to the Riemann Zeta function?

Piro Manco - 1 year, 1 month ago

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The exact name of the identity is the Euler Product. There's a really well written proof here based on the Sieve of Eratosthenes.

Julian Poon - 1 year, 1 month ago

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