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If we're talking about a finite set of the natural numbers, like those between 1 and 500 or 1 and a million, it seems to me that the fraction of numbers in that finite set that have a factor of 5 approaches $1/5$ as the set increases in size. Like roughly $1/2$ of all numbers in such a set have a factor of 2, roughly $1/3$ have a factor of 3, and so on; and this approximation grows less "rough" and more exact as the size of the set increases.

So, can we say that out of the entire set of the natural numbers, exactly $1/5$ are divisible by 5? Or perhaps that the limit of the fraction of the natural numbers less than or equal to a given n divisible by a given integer approaches 1/that integer as n approaches infinity?

(I would love to know how to ask this question with proper notation.)

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    $\begingroup$ There was a similar thread some time ago, see math.stackexchange.com/questions/101768/… and especially the first answer. $\endgroup$ Commented Aug 26, 2012 at 21:12
  • $\begingroup$ Am I correct in assuming that this question has already been answered in enough forms that it isn't a helpful addition to the site? I don't mind deleting if so; the idea of "asymptotic density" is something I didn't know about until today, and now that I can learn about that I at least have something to work with. How do I delete, if I ought to? $\endgroup$
    – Annick
    Commented Aug 26, 2012 at 21:14
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    $\begingroup$ Yours is certainly a good question and it doesn't seem a duplicate to me. If I didn't knew the title of the linked question I probably wouldn't have been able to find it. I was merely referencing so that you can benefit from the answers there. Please keep asking. $\endgroup$ Commented Aug 26, 2012 at 21:27
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    $\begingroup$ @Annick Now that there are several answers, I'd just leave it. If several people want to close it as a duplicate, we can, but it'll still exist for others to see. $\endgroup$
    – GeoffDS
    Commented Aug 26, 2012 at 21:27
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    $\begingroup$ @Annick I would just like to add that this is a very well-asked question, and that I'm sure you will get positive responses in the future if you continue to ask questions in this manner. $\endgroup$ Commented Aug 26, 2012 at 21:31

4 Answers 4

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This can indeed be made formal. To formalize the statement "$x$ fraction of natural numbers satisfy the property $P$", we define the function $$f(n)=\text{ number of natural numbers }\leq n\text{ which satisfy }P$$ and write $\lim\limits_{n\to \infty} \frac{f(n)}{n}=x$. In your first case, the function $f$ is given by $f(n)=\lfloor n/3\rfloor$ and the statement becomes $$\lim\limits_{n\to\infty} \frac{\lfloor n/3\rfloor}{n}=\frac{1}{3}$$ which is easily seen to be true, since $\frac{1}{3}-\frac{1}{n}\leq \frac{\lfloor n/3\rfloor}{n}\leq \frac{1}{3}$. Similar results hold for any natural number in place of $k$.

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    $\begingroup$ Be careful, however: mathematically speaking, the cardinality of numbers divisible by 3 is equal to the cardinality of all numbers! These are two very different counting mechanisms being used. $\endgroup$
    – akkkk
    Commented Aug 26, 2012 at 21:39
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    $\begingroup$ holy god, what the :/ $\endgroup$
    – orokusaki
    Commented Aug 27, 2012 at 1:13
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    $\begingroup$ This formal statement is exactly the definition of natuarl density or asymptotic density. $\endgroup$
    – Ken Bloom
    Commented Aug 27, 2012 at 2:54
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Look up natural density or asymptotic density.

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Here are the positive integers divisible by $3$: $$ 3,6,9,12,15,18,21,\ldots $$ Here are those not divisible by $3$: $$ 1,2,4,5,7,8,10,11,\ldots $$ Now suppose we just alternate between the first list and the second: $$ \begin{array}{} 3 & & & & 6 & & & & 9 & & & & 12 & & & & 15 & & & & \cdots\cdots \\ & \searrow & & \nearrow & & \searrow & & \nearrow & & \searrow & & \nearrow & & \searrow & & \nearrow & & \searrow & & \nearrow \\ & & 1 & & & & 2 & & & & 4 & & & & 5 & & & & 7 \end{array} $$

Then we could argue in the same way that half of all positive integers are divisible by $3$.

It does make sense to say $1/3$ of them are divisible by $3$, understanding that statement in a certain context, but defining that context is something that will bear examination.

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  • $\begingroup$ Why does it not make sense? It doesn't make sense in the sense of cardinality, but it makes perfect sense in the sense used in the other answers. $\endgroup$
    – joriki
    Commented Aug 26, 2012 at 23:00
  • $\begingroup$ Sorry: typo. I've fixed it. $\endgroup$ Commented Aug 26, 2012 at 23:16
  • $\begingroup$ I think a slightly more formal way to say this would be, “the cardinality of the natural numbers divisible by 3 is the same as the cardinality of the natural numbers indivisible by 3.” $\endgroup$
    – bdesham
    Commented Aug 27, 2012 at 3:03
  • $\begingroup$ @bdesham : That may be more formal but it's not exactly the same thing. Cardinality is only one related concept; there are others. $\endgroup$ Commented Aug 27, 2012 at 3:21
  • $\begingroup$ ....in particular; I didn't intend the arrows to indicate a one-to-one correspondence, but only the order in which the numbers appear in a sequence. $\endgroup$ Commented Aug 27, 2012 at 3:22
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If one computes the fraction of positive integers from $1$ to $m$ that are a multiple of $n$, we get a sequence whose limit is $\frac{1}{n}$ as $m\to \infty$. If we ask about the size of the set of positive integers which are multiples of $n$ compared to that of all positive integers, the answer is that they are the same in some sense as they are both countably infinite.

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