Seeing the plane as a four (or more) dimensional vector space on $\mathbb Q$ As I was trying to answer a question about the enumeration of circuits
one can build with a set of miniature train track elements, I realized
that all plane positions that could be reached had coordinates that
could be both expressed as rational linear combinations of $1$ and
$\sqrt 2$, actually of the form $a+b(\sqrt 2/2)$ with $a,b\in\mathbb
Z$. It occurred to me that the set of relevant coordinates for one
dimension can thus be seen as a two dimensional vector space on the
rationals. Hence the point of the plane where one can find the end of
a circuit starting at the origin may be described by a four
dimensional vector space on $\mathbb Q$ - I am saying $\mathbb Q$ because a field is required for vector spaces, though I am only using coefficient in $\mathbb Z$. I do not know whether there are sophisticated results to be obtained by this view, but it does simplify computations when trying to analyze the shape of circuits, at least to determine the closed ones, for the motivating problem above.
Unless I am mistaken, I guess none of this s very deep, but I have
been wondering whether such views of the plane, or the real line, have known or classical
uses.
Of course, this does not give the whole plane. And,
using other incommensurable irrationals, one can add new
dimensions arbitrarily, while still describing real coordinates.
 A: Absolutely! If you follow the definitions carefully, you'll notice that $\mathbb R$ is itself a vector space over $\mathbb Q$. The set of elements of the form $a+b\sqrt{2}$ for rational $a,b$ is a subspace of this - and, as you note, you can keep adding more irrationals to such a representation, and as long as they are linearly independent over the rationals, you'll get a bigger subspace. All your application is using is that the map $(a,b)\mapsto a+b\sqrt{2}$ is in fact injective - but in the context of linear algebra, this is pretty clear since it is a surjective map between two two dimensional spaces.
A much deeper result, highly relevant to Galois theory, is that the algebraic numbers are a vector space over $\mathbb Q$ - and if you start with the rationals and start adding new elements, you start making field extensions, which are vector spaces over the original field - for instance, one might talk about the field $\mathbb Q[\sqrt{2}]$, which is the set of elements of the form $a+b\sqrt{2}$ under the usual addition and multiplication - meaning, at the least, we can do everything we might want in a field in the elements of the form $a+b\sqrt{2}$. More generally, we can prove that $\mathbb Q[\alpha]$ - which is the field generated by rational combinations of $1$, $\alpha$, $\alpha^2$, $\alpha^3$, $\ldots$ - for any algebraic number $\alpha$ is a vector field over $\mathbb Q$ and its dimension is called the degree of $\alpha$ (and is equal to the degree of the minimal polynomial of which $\alpha$ is a root). The fact that we get a unique representation of elements of the form $\mathbb Q[2]$ by choosing a basis like $\{1,\sqrt{2}\}$ also comes in handy if we want to start describing maps on the field - like, we can quickly see that $a+b\sqrt{2}\mapsto a-b\sqrt{2}$ is, in fact, well-defined and linear - and then we really start to get going when we notice that it's a multiplicative homomorphism too. One can see plenty of other connections between linear algebra and abstract algebra if one wanders down that field of mathematics.
A: You have just described the ring $\mathbb{Z}[\frac{\sqrt 2}{2}]\times \mathbb{Z}[\frac{\sqrt 2}{2}]$ which is a subset of the plane. Of course there are infinitely many other rings which are subsets of the plane. Let $a_1,a_2,\ldots$ and $b_1,b_2,\ldots$ be any algebraic numbers. The one could consider
$$
\mathbb{Z}[a_1,a_2,\ldots]\times \mathbb Z[b_1,b_2,\ldots]\subset \mathbb R^2.
$$
These rings are closely related to number fields, which are studied in algebraic number theory.
