$\mathbb{R}^3$ as a field I have read in an introduction to quaternions that it had been proven before that one can not structure the Euclidian $\mathbb{R}^3$ as a field. 
Where can one find a proof thereof?
For which n, could we bestow $\mathbb{R}^n$ with a field structure?
 A: Just as a set, $\mathbb{R}^n$ has the same cardinality as $\mathbb{R}$. We can therefore fix a bijection $f : \mathbb{R}^n \to \mathbb{R}$ and define the field operations and constants on $\mathbb{R}^n$ by pulling back the known operations from $\mathbb{R}$ (e.g. for $x, y \in \mathbb{R}^n$, define $x + y = f^{-1}(f(x) + f(y))$). This makes $\mathbb{R}^n$ into a field, for any $n$.
Since you mentioned quaternions, I assume your question might have been: "For which $n$ can we make $\mathbb{R}^n$ into a field which respects the natural structure as an $n$-dimensional $\mathbb{R}$-vector space?" The answer is given by the Frobenius theorem: $n=1, 2$ if we need the field to be commutative, $n=4$ is also possible if we allow non-commutative fields.
Wikipedia: https://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)
A: The only $n$ for which $\mathbb R^n$ is a field are $n=1,2,4$ the case of quaternions being the first example known of non-commutative field. A proof of the impossibility for $n=3$ you can read in "Algèbre linéaire et géométrie élémentaire" of Jean Dieudonné, p. 197, Hermann,Paris, 1964. There is english translation of this book.
