Proof for linear-algebra The real numbers form a vector space over the rationals (with $\mathbb{Q}$ as the scalar
field).
Is there a proof out there I can study or can someone please prove it?
 A: Try to prove the more general:
Lemma: If $\,F\subset K\,$ are fields, then $\,K\,$ is a vector space over $\,F\,$
Check that the sum in $\,K\,$ is the vectorial sum, and the multiplication by scalars happens all it within $\,K\,$, so...
A: Do you know the axioms for a vector space? They are a list of rules which a set $V$ with operations $+$ and $\cdot$ has to satisfy to be considered a vector space over F.
For example:
$a+b=b+a \:$ for all $a,b \in V$
$\lambda \cdot(a+b) = \lambda \cdot a + \lambda \cdot b $ for all $a,b \in V$ and $\lambda \in F$.
Here's the complete list: http://en.wikipedia.org/wiki/Vector_space#Definition
All those axioms are completely obvious when $F$ is $\mathbb Q$ and $V$ is $ \mathbb R$ with the normal addition (of real numbers) and multiplication (of a rational number with a real number) that you are used to. So there's really nothing to prove. You only need to look through the axioms and note that they are true for this particular set $V$ and field $F$ and therefore we can say that $V$ is a vector space over $F (=\mathbb Q)$.
