Kronecker's theorem - converse I don't know how to prove that Kronecker's theorem is false if $\alpha_{1}$, $\alpha_{2}$, $\dots$, $\alpha_{M}$
are not independent.

Kronecker's theorem
  Suppose
  that $\alpha_{1}$, $\alpha_{2}$, $\dots$, $\alpha_{M}$
  are independent real numbers. Then given real
  numbers $\beta_{1}$, $\beta_{2}$, $\dots$, $\beta_{M}$
  and $\epsilon>0$ we can find integers
  $N$, $m_{1}$, $m_{2}$, $\dots$, $m_{M}$ such that
  $$|N\alpha_{j}-\beta_{j}-m_{j}|<\epsilon$$
  for each $1\leq j\leq M$.

Where if $\alpha_{1}$, $\alpha_{2}$, $\dots$, $\alpha_{M}$
satisfy 
$\sum_{j=1}^{M} n_{j}\alpha_{j}\notin{\mathbb Z}$
for integers $n_{j}$ not all zero
then we say that they are independent real numbers.
It's easy if $M=1$, since then $\alpha_1$ is rational and $N\alpha_1$ only takes finitely many values modulo 1. But I can't get anything done even for $M=2$. I've tried drawing a picture but I'm struggling to get anywhere.

If $\alpha_1,\cdots, \alpha_M$ are not independent, then they satisfy
$$
n_0+\sum_{j=1}^{M} n_{j}\alpha_{j}=0,$$
for integers $n_i$ not all zero. 
Then the set $\{( N\alpha_1,\cdots, N\alpha_M)|N\in \mathbb{N}\}$ will be a subset of 
$$
\{(x_1,\cdots, x_n)| \sum_{j=1}^{M} n_{j}x_j =m \textrm{ for some } m\in\mathbb{Z}\}.
$$
The above set is not a dense subset of $\mathbb{R}^M$. 
