For any prime, what percentage of the square-free numbers has that prime as a prime factor?
Question about primes in square-free numbers
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5There are infinitely many square-free numbers; what do you mean by a percentage of an infinite set? – 2012-04-07
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0There are infinitely many primes. A square-free number is a product of any finite subset of these. The proportion of these which are divisible by any particular prime (say, $2$) is not well-defined, since it would be $\frac{\infty}{\infty}$. Of course, you might think that for each even square-free number $2n$ there is a corresponding odd one, $n$, so that the proportion should be $\frac12$. But this correspondence is not necessary; others are also possible, leading to other (equally fallacious) ratios. This is why @BrianM.Scott asks you for clarification. – 2012-04-07
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2On the number line we know that 2 divides half, 3 divides one third, etc. If I pick a prime, what part of the square-free numbers would have it as a factor? – 2012-04-07
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2@BrianM.Scott : I would assume that what is meant is a density $\lim\limits_{n\to\infty} |A\cap B\cap\{1,\ldots,n\}|/|B\cap\{1,\ldots,n\}|$, where $A$ is the set of all square-free numbers having the particular prime as a factor and $B$ is the set of all square-free numbers. – 2012-04-07
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2@Michael: So would I, but I wanted it explicitly confirmed. – 2012-04-07
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1@Michael Hardy: that's what I needed. Thanks. – 2012-04-07
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0Postscript: I have a self-evident proof that $\frac{1}{2}$ of the square-free numbers will contain a specific prime. Would this be of interest to anyone? – 2012-04-07
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1@Rudy: That makes me wonder whether you really do mean what Michael wrote. You may have in mind the obvious bijection between the square-free numbers with and without a specific prime, but that doesn't imply a fraction $1/2$ under any well-known definition of relative density, and certainly not under the one Michael suggested. I believe that under the latter the fraction should be $1/(p+1)$ for a prime $p$. – 2012-04-07
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1I could not find this info in the literature so I asked here. If I didn't get the $\frac{1}{2}$ for an answer, then I would publish my little postulate. I don't think that this is relative density, but I've been wrong before. I show that the square-frees are countable. – 2012-04-07
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1@RudyToody: It is obvious that the squarefree numbers are a countable set: they can't be *un*countable (they are a subset of a countable set), and there are infinitely many of them (at least as many as there are prime). Showing that the squarefree numbers are countable seems like a rather trivial thing. – 2012-04-08
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0@Arturo Magidin: I guess I'm misusing the term. What I mean is that I can show the count of square-free numbers is always a power of 2. And each prime is a factor of half the numbers. And using this info, I hope to be able to determine a limit on how much the sum of the Moebius Mu can vary. – 2012-04-08
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2@RudyToody: You are definitely misusing terminology. It is simply not true that "the couny of squarefree numbers is always a power of 2." For example, the number of squarefree numbers smaller than $8$ is $6$, which is not a power of $2$. I have no idea what it is you are doing, but what you have said so far is, at face value, either trivially true or trivially false. – 2012-04-08
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0I added a link to an old proof. – 2012-04-08
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4@RudyToody: Yeah; sorry, but what you have does not prove what you think you prove. Your "counts" don't count density, don't count limiting quantity. In the end, you are just taking an infinite set and dividing it into two sets that can be bijected. This is trivial, and proves nothing. – 2012-04-08
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0@ArturoMagidin: thanks for the info. I will remove the link and then select the answer to finish this topic. It's been fun. – 2012-04-08
1 Answers
Let $A(n)=\{\mathrm{squarefree~numbers~\le n}\}$ and $B_p(n)=\{x\in A(n); p\mid x\}$.
Then the asymptotic density of $B_p$ in $A$ is $b_p = \lim_{n\rightarrow \infty} |B_p(n)|/|A(n)|$. (It seems from the comments that this is not what @RudyToody is looking for, but I thought it's worth writing up anyway.) Let the density of $A$ in $\mathbb{N}$ be $a = \lim_{n\rightarrow \infty} |A|/n$.
Observe $B_p(pn) = \{px; x\in A(n),p\nmid x\}$, so for $N$ large $b_p$ must satisfy $$ \begin{align} b_p a (pN) & \simeq (1-b_p)aN \\ b_p &= \frac{1}{p+1} \end{align} $$ as @joriki already noted.
To illustrate, here are some counts for squarefree numbers $<10^7$. $|A(10^7)|=6079291$. $$ \begin{array}{c|c|c|c} p & |B_p(10^7)| & |A(10^7)|/(p+1) & \Delta=(p+1)|B|/|A|-1 \\ \hline \\ 2 & 2026416 & 2026430.3 & -7\times 10^{-6} \\ 3 & 1519813 & 1519822.8 & -6\times 10^{-6} \\ 5 & 1013234 & 1013215.2 & 1.9\times 10^{-5} \\ 7 & 759911 & 759911.4 & -5\times 10^{-7} \\ 71 & 84438 & 84434.6 & 4\times 10^{-5} \\ 173 & 34938 & 34938.5 & -1.3 \times 10^{-5} \end{array} $$