Prove that as a $\Bbb Q$ vector space $\Bbb R^n$ is isomorphic to $\Bbb R$ 
Prove that as a $\mathbb{Q}$ vector space, $\mathbb{R}^n$ is isomorphic to $\mathbb{R}.$

I came across this statement somewhere and I have been trying to prove it ever since, but I just don't know how to start or what to define as the map.
 A: The truth of this statement depends on the axiom of choice.1 This is a well-accepted axiom, so you shouldn't worry that maybe someone is pulling your legs. But it also means that you can't quite define an isomorphism.
In order to prove this, you need to note two things:


*

*Two vector spaces over a field $F$ are isomorphic if and only if they have the same dimension.

*If $V$ is a vector space over an infinite field $F$, and $|V|>|F|$, then $\dim V=|V|$.



Footnotes.


*

*In models where every set of real numbers has the Baire Property (at least assuming Dependent Choice), every linear function $T\colon\Bbb R^n\to\Bbb R$ is automatically continuous. 
You can show that no continuous function can be an isomorphism like that. The above assumption is consistent with the failure of the axiom of choice, as shown by Solovay (and Shelah).
A: You can use the statements that both $\Bbb R$ and $\Bbb R^n$ are infinite (uncountable) dimensional extensions over $\Bbb Q$ and two vector spaces of same dimension are isomorphic. And $|\Bbb Q| < |\Bbb R|$. So $dim(\Bbb R)= |\Bbb R|= |\Bbb R^n|=dim(\Bbb R^n)$.
A: Two vector spaces over the same field (in our case $\mathbb Q$) are isomorphic if and only if they have the dimension (this also holds for infinite dimensional vector spaces). Now it is left to show that the $\mathbb Q$ vector space $\mathbb R$ has dimension $|\mathbb R|$. Then you immediately get that the dimension of $\mathbb R^n$ is the same since $\mathbb R$ is a subspace of $\mathbb R^n$ and the dimension is bounded by the number of elements $|\mathbb R^n|=|\mathbb R|$.
