Riemann Zeta Function: Important Number Relationships (1)

I wish here to correct an earlier error on my “Spectrum of Mathematics” blog with respect to “Remarkable Features of the Number System“ (1 – 4) which was later followed up in “Another Interesting Relationship” and “Further Investigation” and also on the present “Riemann Hypothesis” blog with respect to “Holistic Synchronicity” (1 – 3).

In those entries I suggested that a certain constant relationship (based on π) would hold with respect to the number system as a whole regarding the distribution of those numbers with non-repeating prime factors and repeating prime factors - or structures as I referred to them - respectively.

So based on some initial empirical investigation at various intervals of the number system, I concluded then that the average frequency of those numbers with non-repeating prime factors would be π/(π + 2) = .611 (approx) with the average frequency of the remaining numbers with repeating prime factors 2/(π + 2) = .389 (approx).

However when reading about - what are referred to in the literature as - square-free numbers, I realised that this in fact represented an alternative way of stating my position.

Thus the average frequency of square-free numbers (i.e. numbers that are not divisible by prime factors raised to the power of 2 or higher) would by definition represent those numbers with non-repeating prime factors.

And it is well-known that the average frequency of such numbers (or alternatively the probability of obtaining a square-free number) = 6/π², which of course is the corresponding value of the Riemann zeta function for 1/ζ(s) where s = 2.

Now 1/ζ(2) = 6/π² = .608 (approx) which is very close to π/(π + 2) = .611 (approx).

In fact, in reaching my earlier erroneous conclusion, I had initially considered 6/π² the most likely estimate. However as the sample values that I took at various intervals (to estimate the frequency of numbers with non repeating prime factors) repeatedly averaged out very close to .611, I then changed to an alternative π estimate.

So now in retrospect, it is apparent that the consistency of my estimates represented in fact sample bias over the number intervals that I chose.

In any case it is quite easy to establish the correct result (without resort to empirical data).

The probability that a number is divisible by the square of a prime p = 1/p².

So for example in the simplest case where p = 2, we would expect 1 in 4 numbers to be divisible by 2².

Therefore the probability that a number is square-free with respect to a given prime (i.e. not divisible by the square of a particular prime) = 1 – 1/p².  

However as well as considering divisibility by the square of 2, we equally have to consider all the other primes i.e. 3, 5, 7, ….

So therefore the probability that a number is generally square-free with respect to any prime = ∏(1 – 1/p²)  = 1/ζ(2) = 6/π².And as we have seen, this equally represents the probability that a number is composed of non-repeating prime factors, or alternatively, the average frequency of numbers with non-repeating prime factors.


And this result can be easily generalised as the probability that a number cannot be divided by a prime (raised to the power of n) = 1 – 1/pⁿ.  And then when we allow for divisibility by all the primes we get ∏(1 – 1/pⁿ) = 1/ζ(n).  

Then returning to the case for s = 2, this then entails that the probability that a number is not square-free i.e. is composed of at least one repeating prime factor = 1 – 6/π².

However there is yet another way of stating this important relationship.

The probability that a number is square-free, in fact represents the probability that any two randomly chosen numbers will contain no common factors.

And once again this probability is given as 1/ζ(2) = 6/π².

Let us consider for example the two numbers 6 and 8. These contain 2 as a common prime factor. Alternatively we could express this situation as the fact that the product of 6 and 8 (= 48) is necessarily divisible by the square of 2.

Therefore if two numbers chosen at random contain common prime factors then the product of those numbers cannot be square-free.

However, clearly if the two numbers do not have common prime factors, then the product of these numbers is square-free.