real numbers a vector space over rational numbers? Let $V$ be set of real numbers and $K$ the field of rational numbers.
Is $V$ a vector space over $K$, with ordinary addition of real numbers
and multiplication by rational numbers?
 A: Yes, $V$ is a vector space over $K$, but it is of infinite dimension over $K$.  Choosing a basis for $V$ over $K$ is a non-trivial matter that requires non-constructive foundational material such as the axiom of choice, see http://en.wikipedia.org/wiki/Axiom_of_choice
A: A vector space over a field $K$ is any set $V$ equipped with functions $V\times V\to V$ and $K\times V\to V$ satisfying the axioms (I sometimes call them "computation rules").
So, since $\mathbb{R}$ with the two functions you mention satisfies the axioms, it is a vector space over the rationals. It is infinite dimensional, by reasons of cardinality: a finite dimensional vector space over an infinite field $K$ has the same cardinality as $K$.
In the same vein, $\mathbb{C}$ is a vector space over $\mathbb{Q}$ as well. This has important applications in the theory of algebraic numbers, because one can use the powerful machinery of vector spaces. Note that $\mathbb{C}$ and $\mathbb{R}$ are vector spaces over any of their subfields. One of the most useful results in this respect is the formula
$$
[M:L][L:K]=[M:K]
$$
where $M$, $L$ and $K$ are subfields of $\mathbb{C}$ (any field would do, actually) with $K\subseteq L\subseteq M$ and with $[L:K]$ we denote the dimension of $L$ as a vector space over $K$.
A: Anytime you have a field extension $V/K$ you can consider $V$ as a vector space over $K$.
