Is $\frac{lnk}{\pi}$ irrational or rational?Here $k$ is a positive integer. I ask the question because I want to prove $\{\frac{lnn}{2\pi}\}$ （fractional part)  is dense in $[0,1)$.
If we take $n=2^{k}$.It will become $\{\frac{ln2}{2\pi}k\}$.So we need to prove $\frac{ln2}{\pi}$ is irrational.It's difficult for me.But we can take $n=3^{k}.$ So one of $\frac{ln2}{\pi}$ and $\frac{ln3}{\pi}$ must be irrational. If not, $\frac{ln3}{ln2}$ is rational. This means $3^{p}=2^{q}$ and causes a contradiction.
So is there direct way to show $\frac{lnk}{\pi}$ is irrational? Any reference will be thanked. 

You don't know which of $\frac{\ln 2}{\pi}$ or $\frac{\ln 3}{\pi}$ are irrational. But you know that at least one of them is. That's enough. You are allowed to pick "the irrational one", even though you don't know which one it is, as long as you know it exists.
(Small note: If it happens that they are both irrational, which is the most likely result, then "the irrational one" doesn't make sense. So you have to phrase yourself in a way that makes sense no matter which ones are actually irrational.)
Alternatively, and if you want to be a bit more explicit, you don't actually need irrationality for what you want to prove. You can show that for any $N$, on the set $[e^{2\pi N}, e^{2\pi(N+1)})\cap \Bbb N$, the function $f$ given by
$$
f(n) = \left\{\frac{\ln n}{2\pi}\right\}
$$
is increasing, and the difference $f(n)-f(n-1) = \frac{\ln n - \ln(n-1)}{2\pi}$ between consecutive values becomes small as $N$ grows large. This means that picking $N$ large enough, you can get the values of $f$ as close together as you want. So the collection of all values of $f$, across all natural numbers $N$, must be dense.
