Are $\mathbb{R}$ and $\mathbb{R}^n$ isomorphic for $n > 1$? My linear algebra class recently covered isomorphisms between vector spaces. Much emphasis was placed on the fact that $\mathbb{R}^m$ is isomorphic to particular subsets of $\mathbb{R}^n$ for positive $m < n$. For example, $\mathbb{R}^2$ is isomorphic to every plane in $\mathbb{R}^3$ passing through the origin.
My question is whether $\mathbb{R}$ is isomorphic to $\mathbb{R}^n$ for all $n > 1$. It follows from the fact that $\lvert \mathbb{R} \rvert = \lvert \mathbb{R}^n \rvert$ that there exist bijections $\phi : \mathbb{R} \rightarrow \mathbb{R}^n$ for all $n > 1$, but showing that these sets are isomorphic as vector spaces would require showing that such $\phi$ are linear. The standard constructions don't appear to give linear maps, but that doesn't preclude the existence of some linear mapping.
I'm rather surprised that my book didn't answer this question, because it seems like a natural question to ask. That, or the answer is obvious and I'm overlooking some important fact.
 A: I have absolutely no doubt that your book covers this point. Two vector spaces are isomorphic if and only if they have the same dimension. So $\mathbb{R}$ (dimension $1$) cannot be isomorphic to any $\mathbb{R}^n$ (dimension $n$) if $n>1$.
A: No, they're not. In fact, the following holds

Theorem
Two vector spaces $V,W$ (over a common field) are isomorphic iff $\dim V = \dim W.$
Proof. 
Your linear algebra course will probably cover this in a few weeks, or you can look it up in any linear algebra book.

Same cardinality is a necessary but not sufficient condition, you need more than a bijection, you need a bijective linear transformation.
Consider the case of $\Bbb R, \Bbb R^2$. 
Suppose $f:\Bbb R^2\to \Bbb R$. Let $\{e_1,e_2\}$ be the standard basis of $\Bbb R^2$, then
$$f(e_1)=a, f(e_2)=b$$
Neither of $a,b$ can be $0$ (why?).
But from here, $$\frac b af(e_1)-f(e_2)=f\left(\frac b a e_1-e_2\right)=0\qquad $$
As $f$ is a bijective linear transformation, we must have that $\frac b a e_1-e_2=0$, but from the linear independence of $e_1,e_2$ this is impossible. 
Thus, no such $f$ exists.

Addendum.
An important observation is that if $n\leq m$, then $\Bbb R^n$ is isomorphic to a subspace of $\Bbb R^m$, just let $$f:\Bbb R^n \to S\subseteq\Bbb R^m\\
f(x_1,\dots ,x_n)=(x_1,\dots,x_n,0,\dots,0)$$
For some appropiate subspace $S$ of $\Bbb R^m$.
