How to find a basis for the vector space of real numbers over the field Q?

  • Thread starter Arian.D
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  • #1
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So the title says everything. Let's assume R is a set equipped with vector addition the same way we add real numbers and has a scalar multiplication that the scalars come from the field Q. I believe the dimension of this vector space is infinite, and the reason is we have transcendental numbers that are not algebraic. On the other hand we know from the axiom of choice that any vector space has a basis, so is there a way to find a basis for this interesting one?

I hope my question isn't wrong.
 

Answers and Replies

  • #2
Erland
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Probably, it is not possible to find such a basis without using at least some weaker form of the Axiom of Choice.
 
  • #3
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I might be way off here, but this is just a first thought. Have you heard of "the" (quotes because there is more than one) Vitali Set? The one I am thinking about is built like this: consider the interval [0,1]. Now, make an equivilence relation x~y iff x-y is rational. Now, pick one element from each equvilence class (you have to use the axiom of choice here.) This seems like it *might* form a basis for R over Q. But, like I said, this is one of the first things that popped in my mind so it might be wayyy off.
 
  • #4
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I think it might work. For example, if [itex][r]_{\alpha}[/itex] is the collection of equivilence classes, and if [itex]z \in [0,1][/itex] and [itex]z \in [r][/itex] for some [itex]r[/itex] then [itex]r-z = q \in \mathbb{Q}[/itex]. So that [itex]z = 1r + q1[/itex]. So if we require that 1 be one of the numbers from the equivilency classes, then this set certainly spans [itex][0,1][/itex]. And by taking [itex]q[/itex] to be an integer + q it seems like this set will span the whole real line. Now, are they linearly independent?
 

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