# 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, 2023 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, 2023 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, 2023 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, 2023 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, 2023 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.