Density of the set $\{n^{\frac{1}{q}}:q\in N\}$ in the natural numbers where $n>1$

Question

I know that any natural number greater than $1$, $n>1$, has an uncountably infinite number of roots such that $\{n^{\frac{1}{q}}:q\in N\}$ in the natural numbers and hence no counting function since $\lim_{q\to\infty}{n^{\frac{1}{q}}}=1$.

My first question is, what type of density does the set of roots of any $n>1$ have in the natural numbers?

Does density only apply to sets of integers? My second question is, can this set's density be determined or estimated if it exists?

I am stumped because there is no way to count how many roots of a number lie between $1$ and another arbitrarily large natural number. I understand that the asymptotic density of the set of perfect squares is $0$ since the limit of the number of perfect squares less than or equal to a given value over that value tends to $0$ as the value gets arbitrarily large, but I know that such a limit cannot be applied here.

Answer 1

As a set of real numbers, that set has density zero since it is countable.

It probably is everywhere dense, in the sense that for every $c > 0$ and integer $m$ there are integers $n$ and $q$ such that $|m-n^{1/q}| < c $.

I'll see if I can prove it.

I'll be back if I succeed, otherwise I'll leave it at this.

Answer 2

Proof of the fact that $\{n^{1/q}: n,q \in \mathbb N\}$ is dense in $(1,\infty)$: let us first prove that $\{\frac {\log\, n} q:n,q \in \mathbb N\}$ is dense in $(0,\infty)$. Given $x\in (0,\infty)$ and $0< \epsilon <x$ choose $n$ so large that the length of the interval $(\frac {\log\, n} {x+\epsilon},\frac {\log\, n} {x-\epsilon})$ is greater than $1$ and such that $\frac {\log\, n} {x+\epsilon} >1$ . This interval then contains a positive integer $k$. From the inequalities $\frac {\log\, n} {x+\epsilon} <k <\frac {\log\, n} {x-\epsilon}$ we get $x-\epsilon <\frac {\log\, n} k <x+\epsilon$ or $|\frac {\log\, n} {k}-x|<\epsilon$. We have proved that $\{\frac {\log\, n} q:n,q \in \mathbb N\}$ is dense in $(0,\infty)$. Now given $y>1$ we can find a sequence of the type $\{\frac {\log\, n_j} {q_j}\}$ converging to $\log\, y$ and this implies $n_j^{1/q_j}=e^{\frac {\log\, n_j} {q_j}}$ tends to $e^{\log\, y}=y$.