The real numbers form a vector space over the rationals (i.e. with Q as the scalar) How would one go about proving this? I'd like someone to just point in the right direction.
 A: Whenever $E\subseteq F$ is an inclusion of fields, $F$ is naturally a vector space over $E$. The axioms are easily checked.
A: To see that $\mathbb R$ is a vector space over the rational numbers one has to verify that:


*

*The addition of $\mathbb R$ is commutative, and there is a neutral element.

*There is a well-defined operation of multiplying a real number by a rational scalar.

*The scalar multiplication and the vector addition behave as they should. (That is, $(\alpha\beta)v=\alpha(\beta v)$, and both distributive laws.)


But most of those are quite trivial to verify because $\mathbb R$ is a field extending the rational numbers.
Do note that while it is tempting to try and prove by finding a basis for $\mathbb R$ over $\mathbb Q$ (i.e. a collection of real numbers which spans the entire field, and is linearly independent over $\mathbb Q$), such set is impossible to write "by hand". There is no formula expressing it directly, and we can only prove its existence using non-constructive axioms of mathematics.
A: since every field is a vector space over its any subfield Hence F is a field over E which is subfield of F
