Density of $e^{in \alpha}$ 
How do you prove that, for any $\alpha \in \mathbb R\setminus \mathbb Q[\pi] $,  the set $\{e^{in \alpha}, n\in \mathbb N\}$ is dense in the unit circle?

A classical result proves $2\pi \mathbb Z + \alpha\mathbb Z $ is a dense additive subgroup of $\mathbb R$, but this does not fit the bill, as I need density of $2\pi \mathbb Z + \alpha\mathbb N $...
Am I missing something obvious ? 
 A: Proving that $\{e^{in\alpha}\}_{n\geq 0}$ is dense in the unit circle for any $\alpha\in\mathbb{R}\setminus \pi\mathbb{Q}$ is equivalent to proving that for any $\beta\in\mathbb{R}\setminus\mathbb{Q}$, the sequence given by $\{\beta n\}=\beta n\pmod{1}$ is dense in the interval $[0,1)$.
We may clearly assume $\beta\in(0,1)$ without loss of generality, then consider the continued fraction expansion of $\beta$:
$$\beta=[0;b_1,b_2,b_3,\ldots] \tag{1}$$
with convergents given by $\frac{p_1}{q_1},\frac{p_2}{q_2},\frac{p_3}{q_3},\ldots$ $(1)$ has an infinite number of terms since $\beta\not\in\mathbb{Q}$, and
$$ \left|\beta-\frac{p_n}{q_n}\right|\leq \frac{1}{q_n^2}\tag{2} $$
holds by the general properties of convergents. $(2)$ implies that the distance between $\beta q_n$ and $p_n$ is bounded by $\frac{1}{q_n}$. Since $q_n\to +\infty$ and $x\mapsto e^{ix}$ is a Lipschitz-continuous function, $1$ is a limit point for the original sequence $\{e^{in\alpha}\}_{n\geq 0}$, and since $e^{in\alpha}\cdot e^{im\alpha}=e^{i(m+n)\alpha}$, every element of the sequence is a limit point for the sequence. That leads to density in a straightforward way.
Actually, we may prove through Dirichlet's box principle or Riemann integral criterion that $\{\beta n\pmod{1}\}_{n\geq 0}$ is a equidistributed sequence, and density follows from equidistribution.
A: Let $[x]$ denote the largest integer not exceeding $x.$ 
An elementary proof that $S(x)=\{nx-[nx]:n\in \mathbb N\}$ is dense in $[0,1]$  for irrational real $x>0:$
For real $y, $ define $E(y)=\min \{|y-z|:z\in \mathbb Z\}.$
(A). It suffices to show that $\inf_{y\in S(x)}E(y)=0.$
Because if $\inf_{y\in S(x)}E(y)=0$ then:
(A1). Note that  $ y\in S(x)\implies E(y)\in (0,\frac {1}{2})$ because $x\not \in \mathbb Q.$ 
(A2).  For any $n\in \mathbb N,$ let $m\in \mathbb N$ such that $y=mx-[mx]\in S(x)$ with $0<E(y)<\frac {1}{n+1}.$ Let $k\in \mathbb N$ such that $kE(y)<1<(k+1)E(y).$ (Remark: $kE(y)\ne 1\ne (k+1)E(y)$ as $x\not \in \mathbb Q.$ ) Then $$T=\{jmx-[jmx]:1\leq j\leq k\land j\in \mathbb N\}=\{jmx-j[mx]:1\leq j\leq k\land j\in \mathbb N\}\subset S(x).$$ And for every $z\in [0,1]$ there exists $z'\in T$ with $|z-z'|\leq E(y)<\frac {1}{n+1}.$
(B). To show that $\inf_{y\in S(x)}E(y)=0:$
For brevity let $$A_n=k_nE(nx-[nx])$$ $$\text { and  }\; B_n=(1+k_n)E(nx-[nx])$$ $$\text {  where }\, k_n\in \mathbb N \text {  satisfies }\; A_n<1<B_n.$$
Now let $k'_n=k_n$ if $1-A_n<\frac {1}{2}E(nx-[nx]) .$ And let $k'_n=(1+k_n)$ if $B_n-1<\frac{1}{2}E(nx-[nx]).$ (Remark.Equality cannot occur in either case, and the two cases are disjoint, as $x\not \in \mathbb Q.$) Then  we have $$E(k'_nnx-[k'_nnx])<\frac{1}{2}E(nx-[nx]).$$  So with $f(1)=1 $ and $f(n+1)=k'_{f(n)}f(n)$ we have $$0<E(f(n+1)x-[f(n+1)x])<\frac{1}{2}E(f(n)x-[f(n)x]).$$
