Is the LLL algorithm only applicable with integers?

Question

Why cannot it apply with a basis defined with irrational numbers? For example, such a basis:

$(1, \sqrt{2}, \sqrt{3}) $, $(1, \sqrt{5}, \sqrt{7 })$ and $(1, \sqrt{11}, \sqrt{13 })$.

Answer 1

It can be applied to any basis in $\mathbb{R}^n$, including your example.
The original paper of the LLL-algorithm (Factoring Polynomials with Rational Coefficients) defines the algorithm on any lattice, and defines lattice as $L := \left\{ \sum_{i=1}^n r_i b_i \;|\; r_i \in \mathbb{Z} \right\}$, where $b_1, b_2, \ldots, b_n$ is any basis in $\mathbb{R}^n$.
Most of the applications indeed use vectors in $\mathbb{Z}^n$, but not all, for example the article On multidimensional Diophantine approximation of algebraic numbers uses the LLL-algorithm for vectors with algebraic number coordinates (e.g. $\sqrt{2}$ is an algebraic number, but $\pi$ is not).

Answer 2

The input and the output of the LLL algorithm is a basis for a lattice, not a basis for a vector space. The span of your vectors isn't a lattice, and so it doesn't fit the conceptual scheme of the algorithm.

Note that in some restricted cases you can have irrationals. For example $(1,2\pi),(2,\pi)$ form a lattice and so the algorithm can be used with those two inputs.

Edit: Here's an argument that might convince you that your set doesn't form a lattice. Instead of your example, we'll look at $(1,1),(\sqrt{2},\sqrt{3})$. The set of $\mathbb{Z}$ combinations takes the form $\{(a+b\sqrt{2},a+b\sqrt{3}):a,b\in\mathbb{Z}\}$.

By the theorem in the link, the numbers of the form $a+br$ are dense in $(0,1)$ for every irrational $r$, and by increasing $a$ we can translate the set and see that it's dense in $\mathbb{R}$. This tells us that we can find a set of values for $a$ and $b$ such that the first coordinate is dense in $\mathbb{R}$ and that we can do it separately for the second. However, the defining property of a lattice is that it's discrete, and so the set generated by these two vectors doesn't form a lattice.

Now, your set doesn't have a vector of all ones, but we can fix that. If you take one dimension and scale all the vectors by a constant factor, you preserve the fact that it's a lattice. So when you have $(1,\sqrt{2},\sqrt{3}),(1,\sqrt{5},\sqrt{7})$ we can scale the second coordinate by $\sqrt{2}$ and the third by $\sqrt{3}$ to get $(1,1,1),(1,\sqrt{5/2},\sqrt{7/3})$. By the same argument, linear combinations of these two vectors don't form a lattice, and so the original vectors cannot! Adding in the third vector doesn't make things better, because adding in additional vectors can only increase the number of points we can reach