New answers tagged diophantine-approximation
2
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(!) $|a-\frac pq|\le 1$.
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 ...
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