# How to show additive subgroup of $R^n$ is not discrete? [closed]

Suppose we have the additive subgroup of reals generated by $$\sqrt{3}$$ and $$\sqrt{5}$$. How would you show you that this subgroup does not form a lattice?

• I’m voting to close this question because as asked it's a math question with no motivation Jan 31 at 20:43
• "How would you show you that this subgroup does not form a lattice?"; are you sure it's not? Isn't the group $\{a\sqrt{3} + b\sqrt{5}\}$ trivially isomorphic to $\mathbb{Z}^2$? Isn't the latter a lattice? Jan 31 at 21:00
• @poncho Okay but it should be discrete (this is the definition). More precisely it should contain a shortest vector. I think, without a proof, one can find arbitrarily small vectors in this group. Jan 31 at 21:05
• "More precisely it should contain a shortest vector"; the meaning of 'shortest' depends on the metric; for the traditional metric on $\mathbb{R}$ ($dist(A, B) = |A-B|$), it's straightforward to show there are arbitrary short vectors. For another metric, say, $dist( A\sqrt{3}+B\sqrt{5}, C\sqrt{3}+D\sqrt{5}) = |A-C| + |B-D|$, there is a shortest nonzero vector Jan 31 at 21:15
• There’s a frustrating divergence between the mathematics and cryptography communities around the meaning of “lattice”. Cryptographers talk about lattices using the Euclidean metric, rather than the more general mathematical definition. Feb 1 at 9:18

It suffices to show that there exists a sequence of arbitrarily short vectors, i.e. integer sequences $$a_n, b_n$$, such that

$$|a_n\sqrt{3}+b_n\sqrt{5}| \to 0$$

For this explicit example, note that an obvious choice of $$a_n$$, $$b_n$$ is

$$a_n\sqrt{3} = -b_n\sqrt{5}\iff -\frac{a_n}{b_n} = \sqrt{5/3}.$$

One can't exactly choose $$a_n, b_n$$ to satisfy this inequality ($$\sqrt{5/3}$$ is irrational). Instead, we choose sequences $$a_n/b_n$$ of increasingly good rational approximations of $$-\sqrt{5/3}$$. One can do this explicitly via continued fractions. For example, wolfram alpha states the relevant continued fraction is $$[1, 3, 2, 3, 2,\dots]$$, meaning $$1 + \frac{1}{3 + \frac{1}{2+\frac{1}{3+\dots}}}$$.

Anyway, the various finite truncations of this continued fraction will be a sequence of increasingly good rational approximations to $$\sqrt{5/3}$$. As the continued fraction is infinite, we can keep repeating this process, leading to $$a_n, b_n$$ such that $$|a_n\sqrt{3}+b_n\sqrt{5}|\to 0$$.

If $$-\sqrt{5/3}$$ were rational, its continued fraction would be finite, and the above argument would eventually fail. Note that this means that something like the subgroup generated by $$\sqrt{3}$$ and $$5\sqrt{3}$$ is a lattice --- it is the lattice $$\sqrt{3}\mathbb{Z}$$. It is important that we didn't consider the subgroup generated by $$\sqrt{3}, 5\sqrt{3}$$ and 1 here --- this is no longer a lattice, by (roughly) the same argument.