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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?

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up vote 4 down vote accepted

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:

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 $F (=\mathbb Q)$.

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Beat me to it with a nice concise answer. +1 – Rick Decker Nov 28 '12 at 21:33
I think you confused F and V – Belgi Nov 28 '12 at 21:37
@Belgi: Where do you think I confused F and V? F=Q and V=R. I'm pretty sure it's correct. – Brusko651 Nov 28 '12 at 21:40
ok this helps a lot more, thank you!! – Andriy Lysak Nov 28 '12 at 22:00
@Brusko651 - my mistake, sorry! – Belgi Nov 28 '12 at 23:23

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...

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I'm really not that proficient in proofs, but thank you all the same:) – Andriy Lysak Nov 28 '12 at 20:49
You only need $K$ to be a ring. So, for instance, $F[X,Y]$ is a vector space over $F$. – lhf Nov 28 '12 at 22:07

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