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?