# Every prime $p_{n}$ is a prime factor of $\frac{1}{2}$ of all the square-free numbers. [closed]

I have removed my bold claims and the naive question so I can link to this post from another.

Edit

With each prime, we construct square-free numbers that have that prime as the greatest prime factor and merge those with the square-free number that have the previous primes as factors.

Example: after we have processed $p_{2}$ our square-free numbers are:

$\{1,2,3,6\}$ where you can see that 2 is a factor of half the numbers, as is 3. Since we double the count of square-free number for each prime, the proportionality is absolute (thank-you, Euclid).

We next apply $p_{3}$ to the previous square-frees and get:

$\{1,2,3,6,5,10,15,30\}$ where you can see that the primes $2,3,5$ are each factors of $\frac{1}{2}$ of the square-free numbers. This proportionality holds to infinity.

Edit 2

The reason I asked about the critical line is because you can consider the $p_{n} \propto \textit{all-square-free-numbers} = \frac{1}{2}$ as the state of the data. It remains the same whether you are calculating the first non-trivial zero or the last. It is a constant.

Edit 3

Square-free numbers with even number of factors in the numerators and those with odd number of factors in the denominators:

$\left\{\frac{1}{2}\times\frac{6}{3}\times\frac{10}{5}\times\frac{15}{30}\right\}=1$

Because the proportionality of the primes to the square-free numbers is always $\frac{1}{2}$, this holds to $\infty$.

Coda
The series identified in Edit 3 is the infinite product equivalent to the infinite Merten's sum $= 0$. However, when we restructure the product series into the real-world sequence, we can show that neither sum nor product can converge.

$\left\{1\times\frac{1}{2}\times\frac{1}{3}\times\frac{1}{5}\times6\times\frac{1}{7}\times10\times\frac{1}{11}\times\frac{1}{13}\times14\right\}$

We can see that there will always be one or more uncancelled primes in the denominator at all times, thus no convergence.

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## closed as not a real question by Jacob Black, anon, Micah, Asaf Karagila, Andrés CaicedoFeb 26 '13 at 0:35

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

What exactly do you mean, "1/2 of the square-free numbers"? There are infinitely many square-free numbers. Do you mean natural density? – Alex Becker Feb 25 '13 at 23:30
@AlexBecker, no. I mean that this proportionality is fixed as soon as the prime is encoutered. – Fred Kline Feb 25 '13 at 23:33
I have no idea what "this proportionality is fixed as soon as the prime is encoutered" means. Perhaps you could give an example? – Alex Becker Feb 25 '13 at 23:35
I do not understand the comment above, so do not understand the meaning of "$1/2$ of the natural numbers." – André Nicolas Feb 25 '13 at 23:37
Let's try it for $p=7$. The square free-numbers from $p$ upwards are 7 8 10 11 13 14 15 17 19 21 22 23 26 29 30 31 33 34 35 37 38 39 41 42 43 46 47 51 ..., where the bold ones are multiples of $7$. Doesn't look like half of them "as soon as 7 is encountered" -- so if you don't mean a limiting denstity, then what do you mean? – Henning Makholm Feb 25 '13 at 23:42

I see. So what you mean, rigorously, is for any $n\in\mathbb{N}$, let $S_n$ be the set of squarefree products of primes $p_1,\ldots,p_n$. Then the fraction of $a\in S_n$ for which $p_n$ divides $a$ is $1/2$.

Let's view this in a different way. Let's just take a set of objects $a_1,\ldots, a_n$ and choose a subset of them randomly. For each $a_i$, we have that $a_i$ can be included or not included in the set. Since these are independent, we see that the number of subsets of $S$ is $2^n$. (Note, by the way, that this is the proof for $\sum_{k=0}^n {n \choose k}=2^n$.) Out of those, $2^{n-1}$ (or exactly half) do not include $a_i$, for whichever given $a_i$ you choose.

Do you see how this relates to your problem?

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Yes, finally someone understands where I'm going with this. I break it down into even number of factors and odd number of factors to show that at $\infty$ (supposing we get there) that these two sets are equal and thus the sums net to zero. – Fred Kline Feb 26 '13 at 0:25
It's actually a pretty neat observation. I don't know anything about the Riemann hypothesis though, if there is a connection somebody else will have to explain it. – Alexander Gruber Feb 26 '13 at 0:32

"If proved true, does anyone think this could explain why the critical line is $1/2$?"

It is true, and trivial. Perhaps someone, somewhere, thinks this could explain why the critical line is $1/2$, but it doesn't, any more than $\cos^2x=(1/2)(\cos2x+1)$ explains why the critical line is $1/2$, or $x=(1/2){-b\pm\sqrt{b^2-4ac}\over a}$ explains why the critical line is $1/2$. Read up a little on the zeta function, and you will see why the critical line is at $1/2$.

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