What does "Consider R as an vector space over Q" mean? As a bonus question in an exam, I had to prove that R is infinite dimensional as a vector space over Q. I would have probably tried cardinality to show it, but I don't know what it means.. 
 A: You may be familiar with the idea that it's possible to select a set $B$ of polynomials so that every polynomial $p$ can be uniquely represented as a finite sum of the form $$\sum c_ib_i$$ where the $c_i$ are numbers and the $b_i$ are elements of $B$; we say that $B$ is a basis for the space of all polynomials. A typical choice for $B$ is the set $\{1, x, x^2, \ldots\}$, but other bases are sometimes used.  Any such basis must be countably infinite, but only a finite subfamily is needed to represent any single polynomial.
Analogously, it's possible to select a set $B$ of real numbers so that every real $r$ can be uniquely expressed as a finite sum $$\sum  q_ib_i$$  for some rational numbers $q_i \in \Bbb Q$ and for some finite subset of $b_i\in B$.  Then we say that $B$ is a "basis" for $\Bbb R$ over $\Bbb Q$.  It's not hard to show that any such $B$ must be uncountable. Your idea about using cardinality considerations is the right one.
A: A vector space over a field $k$ is a set of vectors $V$ with an addition operation and a scalar multiplication operation (subject to some axioms). Consider $V=\mathbb{R}$ and $k=\mathbb{Q}$, and take vector addition to just be real number addition, and scalar multiplication to just be real number multiplication. It's not hard to see (from the field axioms for $\mathbb{R}$) that all the axioms of a $\mathbb{Q}$-vector space are satisfied.
As far as the dimensionality of $\mathbb{R}$ as a $\mathbb{Q}$-vector space, think about the cardinality of an $n$-dimensional $\mathbb{Q}$-vector space for $n$ finite, and then ask whether $\mathbb{R}$ has that cardinality.
A: You need to show any basis of $\mathbb{R}$ over $\mathbb{Q}$ is infinite.  In fact such a basis must be uncountably infinite.
A: We can encode the structure of a vector space as a quadruple $(V,\mathbb{K},s,m)$, where $s: V \times V \to V$ and $m: \mathbb{K} \times V \to V$ satisfy the hypothesis in the axioms.
You are used to seeing $\mathbb{R}$, in its vector space structure, as $(\mathbb{R},\mathbb{R},+,\cdot)$, where $+,\cdot$ are the usual operations in $\mathbb{R}$.
The exercise is requiring you to see it as $(\mathbb{R},\mathbb{Q},+,\cdot|_{\mathbb{Q} \times \mathbb{R}})$.
The point, formalism aside, is that things like linear combinations can now only take rational inputs as scalars. For example: under this last structure, the span of $\{1\}$ is $\mathbb{Q}$.
