Proof that a subset of $\Bbb R^2$ is not a vector space Here is a subset of $ \Bbb R^2 $ :
$ A = [(x_1, x_2) \in\Bbb R^2 : x_1 - x_2^2 = 0] $
I am trying to prove that it is not a vector subspace.

*

*It is not empty, the $(0,0)$ vector works

*It is closed under scalar multiplication (sign of $x_1$ and $x_2$ are changed but it works)

*I was a bit struggling to prove that it does not work for all $x_1$ and $x_2$ (Is it possible ?)

I have chosen to take an $u = (1, -1)$ and $v = (16, 4)$. Both of them are in $\Bbb R^2$.
When I do $u + v$, I got a vector that is $(17, 3)$. This vector is not in $A$ since $x_1 - x_2^2 \neq 0$
My question : Is there something wrong in my reasoning / solution ?

Following JMoravitz counterexemple for 2) :
Take v = (1, 1) and multiply it by a scalar = 2, it does not fit in A.
On the 3, my counterexemple works and prove that A is not a vector space
