Is it possible to define the vertex coordinates of a dodecahedron as integer multiples of basis vectors in an affine space?

Question

I am trying to think of an affine space defined by cells (whose edges may differ in length), such that the coordinates of all the vertices in a regular dodecahedron can be expressed as integer multiples of the basis vector.

Coordinates:

(±1, ±1, ±1)
(0, ±1/ϕ, ±ϕ)
(±1/ϕ, ±ϕ, 0)
(±ϕ, 0, ±1/ϕ)

I don't think it would be possible as both 1 and ϕ would have to be integer multiples of the cell's basis vector, is my thinking correct?

Answer 1

You're almost correct. Stating that

both 1 and ϕ would have to be integer multiples of the cell's basis vector

you write that a number would be an integer multiple of some vector. This mismatch is indeed not just a language problem, but a mathematical problem.

One way to rewrite this would be by formulating it in terms of one coordinate entry. But concentrating on a given coordinate representation can be misleading, so that approach might lead you to wrong results, as seen e.g. in the tiling of the plane by regular triangles which has a $\sqrt3$ in its coordinates but can be written as a lattice nonetheless.

So the right way to approach this is by speaking about vectors all the way through. If all corners of your dodecahedron are integer combinations of some basis vectors, then the differences between them have to be integer multiples of these as well. This eliminates the question of whether the origin coincides with the center of the dodecahedron. So what you can do is choose any three linearily independent (over $\mathbb R$) vectors as a basis, and express the others as linear combinations of these. If, at some point, you find that one of the other vectors can not be represented as a rational linear combination of your chosen basis vectors, then you know that the set of all corner vectors forms a three-dimensional vector space over $\mathbb R$, but not over $\mathbb Q$. (In fact, you'd have a higher-dimensional $\mathbb Q$-vectorspace, but that's a tricky thing to imagine.) Since any relation between integer multiples would imply rational ratios, that's the point where you have the proof that you can't find a basis which leads to all-integer combinations for all your corners.

So how could you choose your first three basis vectors? Choose all three as differences from the $(\pm1,\pm1,\pm1)$ template, then you know that all rational combinations of these will have rational coordinates, and therefore you can't construct vectors containing $\phi$ as $\mathbb Q$-combinations of these. Which is very close to your original argument, but now placed on the right foundation.