$\mathbb{R}^2$ diagonally cut in half through point of origin, vector space?

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I'm starting my math course at uni in a few days and I have a relatively simple question for most of you. I'd like to know whether $\mathbb{R}^2$ diagonally cut in half through the point of origin is a vector space (the space below the cut I mean). I'd say it's not a v-space because scalar multiplication leads out of the set if you take $-1$ as a scalar for the vector $(0,-x)$ so you end up with $(0,x)$ above the cut. I'm new to this type of task so I'd appreciate if someone can verify this for me. Also, how do you show this in a mathematically correct fashion?

Thanks in advance!

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You are right. We say $U$ is a vector subspace (or linear subspace) of some vector space $V$ if:

(1) (contains the zero vector) $0 \in U$

(2) (Closure under addition) If $u_1, u_2 \in U$ implies $u_1 + u_2 \in U$

(3) (Closure under scalar mult.) if $u \in U$, then $au \in U$ for some constant $a$

In your case $V = \mathbb{R}^2$. To show that your set is not a vector subspace you only need to show one of the above conditions does not hold.