Which are Bases?

Algebra Level 2

Which of the following are "bases" in the way described?

$(2,\,3)$ and $(-1,\,1)$ generate $\mathbb{R}^2$ over the real numbers .
$\dots,\,2^{-2},\,2^{-1},\,2^0,\,2^1,\,2^2,\,\dots$ generate real numbers by using their binary representation over $\mathbb{F}_2$ .
$1$ and $\sqrt{2}$ generate the algebraic numbers over the rationals.

1 and 3 1 and 2 1, 2, and 3 2 1

1 solution

Henry Maltby
Jul 1, 2016

Only $1$ works.

$(2,\,3)$ and $(1,\,-1)$ could work as a basis, since any element of $\mathbb{R}^2$ can be expressed as a sum of (real) multiples of them in precisely one way. They would therefore work as "coordinates".
This is a more interesting question, because it showcases a subtlety of linear sums. In particular, a binary representation for an irrational (or rational recurring) number is not a finite sum. Even if we were to allow non-finite sums, they in general would not make sense (in this case, a notion of convergence in real numbers) and here would allow for non-uniqueness. For instance, 0.999.. = 1 . But when only finite sums are permitted, not all real numbers can be expressed in such away; in particular, we only get the rational numbers whose denominators are a power of $2$ (possibly $1$ ). So there is no way the prescribed numbers could be a basis for the real numbers over the finite field of two elements.
For instance, there is no rational linear combination of $1$ and $\sqrt{2}$ to give $\sqrt{3}$ . Try proving this yourself!