|
|
|
|
|
|
by giraj
2055 days ago
|
|
|
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.) |
|
|