dimension of $\mathbb{R}$ over $\mathbb{R}$ I wonder if the set of real numbers $\mathbb{R}$ could be a vector space over $\mathbb{R}$ with dimension 2. Which then would be the proper "sum" and "product"?
Thanks.
 A: Let $f : \mathbb{R} \to \mathbb{R}^2$ be any bijection.
Define the operations $\oplus : \mathbb{R}\times\mathbb{R}\to \mathbb{R}$ and $\odot : \mathbb{R}\times\mathbb{R}\to \mathbb{R}$ like this:
$$x \oplus y = f^{-1}(f(x) + f(y)), \quad\forall x,y \in \mathbb{R}$$
$$\lambda \odot x = f^{-1}(\lambda \cdot f(x)), \quad\forall \lambda,x \in \mathbb{R}$$
You can check that $\left(\mathbb{R}, \oplus, \odot\right)$ is indeed a vector space over $\mathbb{R}$.
Futhermore, notice that $f$ is an isomorphism of $(\mathbb{R}, \oplus, \odot)$ and $(\mathbb{R}^2, +, \cdot)$, so $$\dim \,(\mathbb{R}, \oplus, \odot) = \dim\,(\mathbb{R}^2, +, \cdot) = 2$$
Indeed:
$$f(x \oplus y) = f\big(f^{-1}(f(x) + f(y))\big) = f(x) + f(y), \quad\forall x,y \in \mathbb{R}$$
$$f(\lambda \odot y) = f\big(f^{-1}(\lambda \cdot f(x))\big) = \lambda \cdot f(x), \quad\forall \lambda,x \in \mathbb{R}$$
A: Yes, there is a vector space structure on $(\mathbb R,{+})$ that has dimension $2$ over $\mathbb R$. But of course the scalar multiplication of this vector space structure will not agree with ordinary multiplication of real numbers.
First, $\mathbb R$ and $\mathbb R^2$ both have dimension $2^{\aleph_0}$ as vector spaces over $\mathbb Q$.  (Simply considering cardinalities tells us it cannot be otherwise). This means they are isomorphic as $\mathbb Q$-vector spaces, and in particular the additive groups $(\mathbb R,{+})$ and $(\mathbb R^2,{+})$ are isomorphic.
We can use this latter isomorphism to pull the scalar multiplication of $\mathbb R^2$ back to $(\mathbb R,{+})$ to get a vector space that is isomorphic to $\mathbb R^2$, and therefore has dimension $2$.
Unfortunately, this construction does not produce a description of how the scalar multiplication would work.  It depends on the axiom of choice, and this dependence feels essential (if you want the addition to agree with ordinary real addition), so it is likely that there is no solution with an explicit description.
