Is the set of rational numbers a vector space? Is the set of all rational numbers $\mathbb{Q}$ a vectorspace?
I assumed no, because if
$$\vec{x} \in \mathbb{Q}$$ 
$$\pi \vec{x} \notin \mathbb{Q}$$ failing scalar mult closure
This was a question out of my linear algebra book, looking at the solutions says that it is a vectorspace. Am I making an incorrect assumption?
 A: Any question that asks if this or that set is a vector space is a wrong question. To be a vector space, one needs to specify a set, a field, an addition operation and a scalar product operation. Then one needs to check the axioms. Without that, the question is meaningless. 
The intention was probably to ask whether $\mathbb Q$ is a vector space over the field $\mathbb Q$ with the ordinary notions of addition and multiplication, which it is. Your argument in the question shows that $\mathbb Q$ is not a vector space over $\mathbb R$ with the usual addition and multiplication. But, if the field is not specified, and the operations are not specified, then really the question is just meaningless. In particular, one can endow $\mathbb Q$ with infinitely many different vector space structures over infinitely many different fields. For that matter, the set $\mathbb N$ can also be endowed with infinitely many such vector space structures. 
A: Any field is a vector space over itself.  So, yes, the rational numbers are a vector space over $\mathbb{Q}$.
