[giraj]

You're right, any basis of R over Q has to be uncountably infinite. If a countable basis existed, we could for example write R as the union of the subspaces spanned by the n first basis elements, indexed by n. That would mean R is countable, since a countably-indexed union is countable.

[a1369209993]

> any basis of R over Q has to be uncountably infinite.

It's actually trivial to give a (infinite but) countable set of (non-orthogonal) basis vectors: 2^i for integer i. 0 and 1 are both rationals (scalars), and every real number has a (possibly infinite) binary expansion, eg e = 10.1011011111100001... = 1·2^1 + 0·2^0 + 1·2^-1 + 0·2^-2 + ...

[giraj]

That's not a basis of R at all. If you want to create a basis of R you better choose irrational elements in your basis, otherwise the Q-span is contained in Q.

What I think you want to say is that "any real number has a binary expansion". Which is true, but the binary sequences don't form a vector space over R, but a Z/2-module. And as a Z/2-module, your { 2^i for integer i } isn't even a basis because you need infinite expansions to express most real numbers. The span of a basis are only the finite linear combinations.

(FWIW, I think you've given a description of the dyadic rationals.)