Why we need the "8 axioms of addition and multiplication" in the definition of a vector space? We know that the reason why we want to introduce a vector space and work with a vector space is that we want to work with a set whose elements can be added and scaled (or a set whose elements are closed under addition and scalar multiplication). Given this motivation, I'm confused by the definition of a vector space:
Definition. A vector Space over field $F$ is a set $V$ such that:
(i). Two operations are defined: vector addition: $V × V → V$ ; scalar multiplication:
$F × V → V$ .
(ii). The set V and these two operations satisfy 8 axioms.
Note that as addition is defined as a mapping $V\times V\rightarrow V$, and scalar multiplication is defined as mapping $F\times V\rightarrow V$, this means part (i) of the definition already gives us a set that is closed under finite addition and scalar multiplication. 
My question: Given that what we want from a vector space is merely that it is a set whose elements can be added and scaled,  what is the primary motivation for imposing the 8 axioms (part (ii) of the definition)? 
 A: Without the axioms of addition and scalar multiplication, we wouldn't know addition was commutative. Let $v,w \in V$ a space with two operations $+ : V \times V \to V$ and $\cdot: F \times V \to V$. To build up the theory of vector space, we need to be able to say things like $v+w = w+v$. But this is only true if we require the axiom of commutativity to hold for the $+$ operation. Similarly we want distribution of scalar multiplication $c(v+w) = cv+cw$. Again this only holds if we enforce axioms.
A: Though we gave a preliminary name 'vector addition'  (and a suggestive notation '$+$') to the operation $V\times V\to V$, it is not assumed that it is indeed 'an addition' operation on some known structures.
In itself it can be any two variable function on $V$.
Instead, we assume the basic and most important properties (the axioms) to try to capture what it means to be an 'addition-like operation'.
Similarly for the scalar multiplication.
Note also that $F$ is already assumed to be a field, i.e. to be equipped with (constants named $0$ and $1$), an addition(-like operation), a substraction(-like operation), a multiplication(-like operation), and a division(-like operation) by any nonzero element.
