# Tag Info

If $a=\frac rs$ (and both are in shortest terms and $s,q>0$) then $\left|\frac rs-\frac pq\right|=\frac{|rq-ps|}{sq}\ge\frac1{sq}$, so we can use $c=\frac 1s$. From $\frac1{sq}\le \left|a-\frac pq\right|\le \frac 1{q^2}$ we get $q\le s$. There are only finitely many fractions $\frac pq$ with denominator $q\le s$ and(!) $|a-\frac pq|\le 1$.
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 ...