6 ms·

What do you mean? The real vector space defined over the rationals is infinite-dimensional; it can't have a finite basis. I think any basis of R over Q would ha

What do you mean? The real vector space defined over the rationals is infinite-dimensional; it can't have a finite basis. I think any basis of R over Q would have to be not only infinite, but uncountably so. Do you mean that any given vector of the space can be represented using a linear combination of a finite subset of the basis elements with rational scalar coefficients?

deleted 6y ago
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giraj 6y ago
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.
- AnHonestComment 6y ago
  What...? Every set {1/n^k | k in N} for n in N is a basis for R over Q, in the sense that any r in R is a Sum[0, inf)[q_i * b_i]. It’s not a finite basis, but it’s definitely countable. It’s not generally the case that a countable union of countable sets is countable — that only holds in the finite case, as we see with power series representations of real numbers.
- a1369209993 6y ago
  > 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 6y 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.)
a1369209993 6y ago
> any given vector of the space can be represented using a linear combination of a finite subset of the basis elements with rational scalar coefficients? FWIW, this is definitely false: x = q₀·√2/1 + q₁·√3/3 + q₂·√5/9 + q₃·√7/27 + q₄·√11/81 + ... = Σᵢ(qᵢ·√πᵢ/3ⁱ) for rational qᵢ. Edit: primes approximate e^i, so 2^i isn't large enough.
- deleted 6y ago
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- fractionalhare 6y ago
  This isn't false; it's a theorem. Using the axiom of choice, all vectors of any space can be represented as finite linear combinations. In the case of infinite-dimensional vector spaces, this means you would need at most n vectors from the infinite basis B, and n scalars from the underlying field F. I'm interested in what you're trying to demonstrate with your example though, or if we misunderstand each other, because this theorem is very well established.

deleted 6y ago

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giraj 6y ago

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.

AnHonestComment 6y ago
What...? Every set {1/n^k | k in N} for n in N is a basis for R over Q, in the sense that any r in R is a Sum[0, inf)[q_i * b_i]. It’s not a finite basis, but it’s definitely countable. It’s not generally the case that a countable union of countable sets is countable — that only holds in the finite case, as we see with power series representations of real numbers.
a1369209993 6y ago
> 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 6y 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.)

a1369209993 6y ago

> any given vector of the space can be represented using a linear combination of a finite subset of the basis elements with rational scalar coefficients? FWIW, this is definitely false: x = q₀·√2/1 + q₁·√3/3 + q₂·√5/9 + q₃·√7/27 + q₄·√11/81 + ... = Σᵢ(qᵢ·√πᵢ/3ⁱ) for rational qᵢ. Edit: primes approximate e^i, so 2^i isn't large enough.

deleted 6y ago
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fractionalhare 6y ago
This isn't false; it's a theorem. Using the axiom of choice, all vectors of any space can be represented as finite linear combinations. In the case of infinite-dimensional vector spaces, this means you would need at most n vectors from the infinite basis B, and n scalars from the underlying field F. I'm interested in what you're trying to demonstrate with your example though, or if we misunderstand each other, because this theorem is very well established.