Irrational Natural Density Are there any sets of natural numbers with irrational natural density? 
I.e., does there exist a set $A\subset \mathbb{N}$ such that 
$$
\lim_{n->\infty} \frac{|A \cap [1,n]|}{n} \not\in \mathbb{Q}\\
$$
I'd be interested in a proof of existence/nonexistence or an example of such a set
Thanks!
 A: It is not difficult to find a sequence whose natural density is $\alpha$ for any $\alpha \in [0,1]$, rational or irrational. A rough argument goes as follows:
Let $A_n$ be an abbreviation for $\displaystyle\frac{|A \cap [1,n]|}{n}$.
Now construct your sequence $A$ as follows:
Include enough terms to ensure that $A_n$ is greater than $\alpha$, then exclude enough terms to ensure that $A_n$ goes below $\alpha$ etc.
This will always be possible since no matter what the initial segment of $A$ looks like, you can always make $A_n$ tend to zero or 1 by including or excluding all terms from that point on. 
A: Set $A=\{\lfloor \alpha n \rfloor\}_{n=1}^{\infty}$, and $\alpha$ could be any irrational number greater than $1$. Because $\alpha n - 1\leq \lfloor \alpha n \rfloor \leq \alpha n$, then$$\frac{1}{\alpha} \leq \lim \frac{n}{\alpha n}\leq \lim \frac{n}{\lfloor \alpha n \rfloor} \leq \lim \frac{n}{\alpha n-1}=\frac{1}{\alpha}.$$
It's not obvious but the squeeze above really gives the natural density of $A$. 
Proof:
For any infinite subset $A$ of $\mathbb{N}$, its elements can be arranged to a strictly increasing sequence $(a_{k})$. Note that the subscript 'counts' the number of $A$. Then $\forall (big)n \in \mathbb{N}$, there must exists a natural number $k$ such that $$a_{k} \leq n \leq a_{k+1},$$
and of course $$k=|A \cap \{ 1,2,…,n\}|.$$
Then
$$\frac{k+1}{a_{k+1}}-\frac{1}{a_{k+1}} \leq \frac{|A \cap \{ 1,2,…,n\}|}{n} \leq \frac{k}{a_{k}}.$$
Taking limit gives the desired result.
