Rational numbers closed under field and/or vectorspace? The rational numbers $\mathbb Q = \{a/b | a, b \in \mathbb Z, b \neq 0\}$ where $\mathbb Z$ is a set of integers
is a Field. 
But in my textbook notes 
The set $\mathbb Q$ of all rational numbers is not a vector space reason: not closed under scalar multiplication. 
Is there no scalar multiplication in fields? 
 A: The scalar multiplication of a vector space $V$ over a field $k$ is a map
$$k\times V\to V$$
satisfying certain axioms. For example, if you have a field $k$, it comes with a multiplication map
$$k\times k\to k,$$
so every field can be viewed as vector space over itself. (Exercise: Check all the axioms!) In particular, $\mathbb Q$ is a vector space over  $\mathbb Q$. Your textbook probably means that $\mathbb{Q}$ is not a vector space over $\mathbb R$. This is correct, because that would require a scalar multiplication of the form
$$\mathbb R \times \mathbb Q \to
\mathbb
Q.$$
But this cannot be the usual multiplication: If you multiply $\pi$ and $1$, you get $\pi$, which lies outside of $\mathbb Q$. This is what your textbook means when it says that $\mathbb Q$ is not closed under scalar multiplication (with scalars from $\mathbb R$). Now you could try to be clever and find a scalar multiplication which is not the usual multiplication, but it still cannot work: Vector spaces over $\mathbb R$ are either $0$ or uncountable (Exercise: Proof this!), but $\mathbb Q$ is countable.
