Is the statement "1/3 of the natural numbers are divisible by 3" true? Is anything similar to it true? 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.)
 A: 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.
A: 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.
A: 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 $k$.
A: Look up natural density or asymptotic density. 
