Additive identity verification How could I test the additive identity for the following vector space with the vector addition defined as: (x1, x2)+(y1, y2) = (x1+y1+2, x2+y2)?
This is what I came up with until now but I don't feel that it's the correct answer
(x1, x2)+(0)=(x1+0+2, x2)
The set V is not a vector space since the additive identity property fails.
 A: For there to be an additive identity you need $id=(id_x,id_y)$ such that for any $(x,y)$:
$(x,y) + (id_x,id_y) = (x,y)$
and
$(id_x,id_y) + (x,y) = (x,y)$
From the first, we get:
$(x,y) + (id_x,id_y) = (x+id_x+2,y+id_y) = (x,y)$
hence 
$x +id_x+2=x$ and $y + id_y = y$
and therefore: $id_x=-2$ and $id_y=0$
OK, but does that work for the second as well? Let's see:
$(id_x,id_y) + (x,y) = (id_x + x + 2, id_y+y) = (-2+x+2,0+y) = (x,y)$
So yes, that works. So, the additive identity is $(-2,0)$
A: Your argument is not quite right. 
Showing there is no additive identity is one place to start, but this
$$(x1, x2)+(0)=(x1+0+2, x2)$$
is faulty, for two reasons.
First, $(0)$ is not a vector in the space (which seems to be $\mathbb{R}^2$). I think you mean $(0,0)$.
Second, even though your argument (when corrected) would show that $(0,0)$ is not an additive identity, you need to show that nothing could be an additive identity. So you need to show that there is no vector $ (a,b)$ that satisfies 
$$
(x1, x2)+(a,b) = (x1 , x2 ).
$$
The answer from @Bram28 shows there is something that works.
So you should check the other vector space axioms. What about associativity of addition? Are there additive inverses? How is scalar multiplication defined?
