Vector spaces and scalar multiplication Say I have the set $V$ of rational numbers as vectors and the field $F$ of reals as scalars. Does $V$ form a vector space over $F$? I ask this because $V$ isn't closed under scalar multiplication.
On Wikipedia it doesn't state that being closed under scalar multiplication is an axiom, but on WolframAlpha it says "A vector space $V$ is a set that is closed under finite vector addition and scalar multiplication." If WolframAlpha were correct, then surely the rationals couldn't form a vector space over the reals?
Thanks for any replies.
 A: The sentence 'A vector space over a field $F$ is a set $V$ together with two binary operations that satisfy the eight axioms listed below' from the Wikipedia article seems slightly vague.  The scalar multiplication is not exactly a binary operation on $V$ but rather a function $\mu:F\times V\rightarrow V$, which implies that vector spaces are closed under scalar multiplication.
A: No, if you define Vector Spaces this way and you require closure of $V$ under scalar multiplication, then no, $\mathbb{Q}$ is not a vector space over the field $\mathbb{R}$. In general, I think you cannot consider a field as $F$ and a subfield of $F$ as $V$: for example, you cannot consider $\mathbb{Q}$ as a Vector Space over $\mathbb{R}$, or you cannot consider $\mathbb{R}$ as a vector space over $\mathbb{C}$.
However, you can prove that the contrary is doable: every field $F$ is a Vector Space over a subfield $K \subset F$. For instance, you can think of $\mathbb{C}$ as Vector Space over $\mathbb{R}$, you can think of $\mathbb{R}$ as Vector Space over $\mathbb{Q}$.
