I am just learning about discrete subgroups of $(\mathbb{R}^n,+)$ and I read that every $\mathbb{Z}$basis of such a subgroup is even linearly independent over $\mathbb{R}$. Conversely, this implies that the ring of integers of an algebraic number field often has accumulation points. My question is: Is there a method/theory of explicitely finding such accumulation points together with the corresponding sequences? To be more concrete, consider $\mathbb{Z}[\sqrt{2}]$. We know that there are arbitrarily small (i.e. close to zero) real numbers of the form $n+m\sqrt{2}$ with $m,n\in\mathbb{Z}$. How can we find these $m, n\in\mathbb{Z}$ so that $n+m\sqrt{2}<\epsilon$ for given $\epsilon>0$?
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3Note that $\frac nm$ is a rational approximation of $\sqrt 2$ which can be found, for example, by continuous fraction expansion of $\sqrt 2$. – Fabio Lucchini Jun 05 '18 at 11:51

1The answers [here](https://math.stackexchange.com/questions/73262/provingthatmnsqrt2isdenseinr/73268) also give some construction metjhods. – Dietrich Burde Jun 05 '18 at 11:54

Note that you can still recover a discrete subgroup, in fact a lattice, from the ring of integers in a number field $K$ considering *all* embeddings of $K$ into $\Bbb{R}$ or $\Bbb{C}$. For instance, if you map $\Bbb{Q}(\sqrt{2})$ into $\Bbb{R}^2$ by sending $a+b\sqrt{2}$ to $(a+b\sqrt{2},ab\sqrt{2})$ the image of $\Bbb{Z}[\sqrt{2}]$ is a lattice. – Andrea Mori Jun 05 '18 at 12:10

Thanks! In particular the answer that Dieter pointed to is very easy and helpful! – Hans Jun 05 '18 at 12:32