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$\mathbb R$ is clearly a $\mathbb Q$-vector space since all axioms needed to be a vector space are verified. Its basis is infinite because it would need to have all roots, $\pi$, $e$...

  1. Is there any quick and "relatively easy" but solid proof that this statement is true ? I don't know how to formally write "the basis needs to contain all (actually not all but still a lot) irrationals and |$\mathbb R$ \ $\mathbb Q$| would be bigger than some uncountable set.
  2. Can we write the basis in a simple form ? What I mean is that, the basis of polynomials $\mathbb R$[t] is infinite but can be written as {$t^k$ , k $\in$ $\mathbb N$}
Chady
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  • Not sure what the question is. The basis for $\mathbb R$ over $\mathbb Q$ must be uncountable as a countable union of countable sets is countable. Is that what you were asking? – lulu Dec 02 '20 at 15:54
  • You can prove that a finite dimensional vector space over a countable field is countable, as there is an injection from a $d$ dimensional $F-$vector space to $F^d$ given by $a_1x_1+\ldots+a_dx_d\mapsto (a_1,\ldots,a_d)$ where $a_1,\ldots, a_d\in F$, $x_1,...,x_d$ are the basis vectors. – Rushabh Mehta Dec 02 '20 at 15:54
  • The problem here is that the proof of existence of such a basis relies on the Axiom of Choice. In other words, such a basis can not be explicitly descrbed. – Crostul Dec 02 '20 at 15:57
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    Every vector space over any field has a basis. The proof of this theorem uses the axiom of choice, and more precisely that every set can be well-ordered. A basis of $\mathbf R$ over $\mathbf Q$ is called a Hamel basis. I don't know of any simple, explicit description. – Bernard Dec 02 '20 at 15:57
  • @Bernard and even if there was a description, it would involve a structure equally complicated as the Hamel basis. – Rushabh Mehta Dec 02 '20 at 15:59
  • I'm asking how to show it's infinite (i thought of showing that I can find sets of linear independant vectors as big as I want) and if the basis could be written as for example {$\sqrt(x),\pi,e,sin(2)...$} maybe – Chady Dec 02 '20 at 15:59
  • Ok perfect, thanks for your answers everybody ! – Chady Dec 02 '20 at 16:02
  • If $\mathbf R$ were finite-dimensional over $\mathbf Q$, it would be an algebraic extension of $\mathbf Q$ (as a field). It is known since that numbers such as $\pi$ or $\mathrm e$ are transcendental. (Lindemann-Weierstrass theorem). – Bernard Dec 02 '20 at 16:06
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    Regarding complexity of describing a basis for $\mathbb{R}$ over $\mathbb{Q}$, see https://mathoverflow.net/questions/46063/explicit-hamel-basis-of-real-numbers. In particular, no such basis can be a Borel set in $\mathbb{R}$, so it has to be fairly complicated. On the other hand, in some universes of set theory there is a closed set in $\mathbb{R}^3$ such that if you project that set to $\mathbb{R}^2$, take the complement of that, and then project that complement into $\mathbb{R}$, you get a basis for $\mathbb{R}$ over $\mathbb{Q}$, so a basis doesn't have to be horrifically complicated. – Chris Eagle Dec 02 '20 at 16:10

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  1. By definition, the powers of $\pi$ are linearly independent over $\mathbb Q$ since $\pi$ is transcendental. Therefore, $\mathbb Q[\pi]$ is a infinite-dimensional subspace of $\mathbb R$.

  2. No description of a basis for $\mathbb R$ over $\mathbb Q$ is known. In most if not all definitions, a description is a string in a language and so there are only countably many descriptions, not enough to describe an uncountable set.

lhf
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