Equidistribution theorem in two dimension If $\alpha\in \mathbb R$ is an irrational, then $\{ \langle k\alpha \rangle \}_{k \ge 1}$ is dense in $[0,1]$, where $\langle x \rangle$ denote the fractional part of $x$. Moreover, $\{ \langle k\alpha \rangle \}_{k \ge 1}$ is equidistributed in $[0,1]$, which means that for $0 \le a < b \le 1$,
$$ \lim_{n \to \infty} \frac{\# \{ 1\le k \le n:  a \le \langle k \alpha \rangle \le b \} }{n} = b-a. $$
Does the result holds for higher dimension?
Let $\alpha,\beta\in \mathbb R$ are two irrationals such that $1,\alpha,\beta$ are linearly independent over $\mathbb{Q}$.
How to prove that $\{(\langle k\alpha \rangle, \langle k\beta \rangle)\}_{k \ge 1}$ is dense in $[0,1]^2$ and is equidistributed in $[0,1]^2$?
 A: As a generalization of Weyl's criterion, $\{u_k\}_{k\geqslant 1}$ is equidistributed in $[0,1]^2$ if and only if
$$ \forall \ell\in\mathbb{Z}^2\setminus\{0,0\},\lim\limits_{n\rightarrow +\infty}\frac{1}{n}\sum_{k=0}^{n-1}e^{2i\pi\ell\cdot u_k}=0 $$
But for all $\ell\in\mathbb{Z}^2\setminus\{0,0\}$, we have
$$ \sum_{k=0}^{n-1}e^{2i\pi\ell\cdot(\langle k\alpha\rangle,\langle k\beta\rangle)}=\sum_{k=0}^{n-1}e^{2i\pi\ell_1\langle k\alpha\rangle}e^{2i\pi\ell_2\langle k\beta\rangle}=\sum_{k=0}^{n-1}\left(e^{2i\pi\ell_1\alpha+2i\pi\ell_2\beta}\right)^k $$
Thus
$$ \left|\frac{1}{n}\sum_{k=0}^{n-1}e^{2i\pi\ell\cdot(\langle k\alpha\rangle,\langle k\beta\rangle)}\right|\leqslant\frac{2}{n|1-e^{2i\pi\ell_1\alpha+2i\pi\ell_2\beta}|}\underset{n\rightarrow +\infty}{\longrightarrow}0 $$
because $\ell_1\alpha+\ell_2\beta\notin\mathbb{Z}$ since $1$, $\alpha$ and $\beta$ are linearly independent. As for the density, it follows from the fact that $\alpha\mathbb{Z}+\mathbb{Z}$ and $\beta\mathbb{Z}+\mathbb{Z}$ are dense in $\mathbb{R}$ since $\alpha,\beta\notin\mathbb{Q}$
