Let $V\_2=\{(x,y)\in \mathbb{Q}^2 \mid x^2=-y^2\}$ with the addition and the scalar multiplication of $\mathbb{Q}^2$. 

To check if $V\_2$ is a $\mathbb{Q}$-vector space, we must check the axioms, right? 

We have the following: 
$(x\_1, y\_1), (x\_2, y\_2)\in V\_2 : x\_1^2=-y\_1^2, x\_2^2=-y\_2^2$ 

$x\_1^2x\_2^2=y\_1^2y\_2^2 \Rightarrow x\_1x\_2=\pm y\_1y\_2$ 

$(x\_1,y\_1)+(x\_2,y\_2)=(x\_1+x\_2,y\_1+y\_2) : (x\_1+x\_2)^2=x\_1^2+2x\_1x\_2+x\_2^2=-y\_1\pm 2y\_1y\_2-y\_2^2$ 
it doesn't imply that $(x\_1+x\_2)^2=-(y\_1+y\_2)^2$, right? 

Does it follow from that that $V\_2$ is not closed under the addition, and so it is not a $\mathbb{Q}$-vector space?