Determine whether a set is a vector space, problem with axioms 4 & 5 Vector spaces have been confusing me for a while, I haven't really grasped a clear image of the idea yet.
The question is simple though

Determine whether the set of real numbers (x,y) with operations
  $$(x_1,y_1)\oplus(x_2,y_2) = (x_1+x_2+1, y_1+y_2+3)$$  $$ k  \cdot(x,y) =  (kx,y) $$ is a vector space or not. Check all axioms.

The axioms that passed were 1,6,2,3,9 & 10.
The axioms that failed were 7,8
The axioms that confused me were 4 & 5.
In axiom 4 the rule is that there should exist $\vec 0$ such that $$ \vec0 + \vec u =  \vec u$$  
Here's where it confused me, is $\vec 0$ always $= (0,0)$?
Because in this case it wouldn't work, since
$$\vec0 + \vec u = (0,0) + (x_1,y_1)$$
$$= (x_1+1, y_1+3)$$
Unless $\vec 0$ = (-1,-3) then this axiom would be true.
But then when I checked axiom 5, where the rule is
$$-\vec u + \vec u = \vec 0$$
If we apply it we get
$$-\vec u + \vec u = (-x_1,-y_1) + (x_1,y_1)$$
$$= (1,3)$$
So first I don't know if the zero vector could be something different and in this case it doesn't help. Second, in axiom 5, is the negative in $-\vec u$ considered as a scalar and should be treated it as such? So is $-\vec u$ actually $(-x_1, y_1)$ ? Or is it correct as I previously did?
 A: 
Axiom 4. For each element $x$ in $V$ there is an element $y$ in $V$ s.t.
$$x+y=0_V.$$

In your case $0_V = (-1, -3),$ so if $u=(a,b), v=(-a-2,-b-6)$, clearly
$$(a,b)\oplus(-a-2,-b-6)=(-1,-3)=0_V.$$
Your error is that you directly said $\large-\vec{u}=-1\color{blue}{\cdot_R}\vec{u},$ then you're using an operation not defined! And even if you correctly use the defined operation,
$$\large-1\cdot \vec{u}=(-x_1,y_1)\color{red}{=_?}-\vec{u},$$
you haven't prove it!
In short, $\Large-\vec{u}$ means nothing more than it's the inverse of $\Large\vec{u}.$
A: Well first, if you get one axiom to fail then it is not a vector space . Second ,O^-->  does not have to be (0,0) ,it can be any vector (pair) with U satisfying U circled addition V=V for every vector (pair) ( .Third given U=(a,b) ,you must find a vector  W =(c,d) which ,when added to U (add means using your circled addition).
(p,q)+(r,s)=(p+q,r+s) is NOT the addition used for this problem .
  In the definition and list of axioms for a vector space the addition is always denoted by '+' .but you must change this to circled '+'. 
