proving a set V is a vector space (in one of the axioms) If the set $V$ is defined by the points that go through the origin in $\mathbb{R}^2$  that satisfy the equation $ax+by=0$ then show $V$ is a vector space.
Resolution:Proving that $V$ is closed under addition and scalar multiplication, I know how to do this.
How do I prove that addition is commutative $A+B=B+A$ , what resolution is correct?
Define $A=(x_1,y_1), B=(x_2,y_2)$
$1) \: A+B=(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2)=(x_2+x_1,y_2+y_1)=(x_2,y_2)+(x_1,y_1)=B+A$
$2) \: A+B=(x_1+x_2,y_1+y_2)$, subs in to the line equation $a(x_1+x_2)+b(y_1+y_2)=a(x_2+x_1)+b(y_2+y_1)$ 
$ B+A=(x_2+x_1,y_2+y_1)$ sub that in to the line equation gives $a(x_2+x_1)+b(y_2+y_1)$ and that is what we have above so
A+B=B+A
Note: to prove that $v$ is closed under addition and scalar multiplication the second method is used.
What the right way to prove this?
(As an aside do I have to prove to each of the sets to prove that are vector spaces besides proving closed under addition and scalar multiplication do i have to prove 8 more ,like the one above and i.e. $c(u+v)=cu+cv$ being $u,v$ vectors and $c$ a scalar)
(And hence memorinzing 8 more axioms)
 A: To show that $V$ is closed under addition, we need to show that $A+B$ lies in $V$ (i.e. satisfies your linear equation) for any $A,B \in V$. Letting $A = (x_1,y_1)$ and $B = (x_2,y_2)$, we have to prove that $A+B = (x_1+x_2,y_1+y_2)$ satisfies the equation, i.e. $a(x_1+x_2) + b(y_1+y_2) = 0$. This is easy with our assumptions, namely we get $a(x_1+x_2) + b(y_1+y_2) = (ax_1 + by_1) + (ax_2 + by_2) = 0$. Proving that $V$ is closed under scalar multiplication is done in a similar way.
For your second question, you won't have to check all the axioms one by one. The reason is that $V$ is a subset of $\mathbb{R^2}$, which we already know is a vector space. You should know the following definition of a subspace.
Let $W$ be a vector space. Then a subset $V \subset W$ is a vector space (subspace of $W$) if and only if the following conditions hold:


*

*$0 \in V$

*$V$ is closed under addition

*$V$ is closed under multiplication


Applying this with $W = \mathbb{R}^2$ should solve the problem much quicker than checking each of the axioms.
A: Both 1) and 2) are correct, just differen ways of showing u+v=v+u being u,v vectors
