Set V with addition and multiplication defined
if u + v = 3n + r. Does that mean u = 3n and v = r?
To prove this is a vector space, you have to use the Axioms of a vector space. I know the axioms but I do not know how to show they are satisfied.
[A1] Closure. The set is closed under addition.
I am confused as how you would show this.
3n is an element of N_0 and r is an element of V. Therefore 3n + r is in V. Is this correct?
Closure is obvious because of the table given in lines 5-7 of the explanations.
This question is strange because in the usual definition of a vector space, the set of scalars is supposed to be a field. Here, $\mathbb N_0 = \mathbb N \cup \{0\}$ is far from being a field because, for instance, nonzero elements do not have any inverse for addition: if $n\in \mathbb N$, there is no element $k\in \mathbb N$ such that $n+k=0$. So, the formal answer to the question (at least with the usual definition of a vector space) is no: V is not a vector space.
In fact, your example is a strange variation on the field with three elements, seen as a vector space of dimension one on itself.