Determining whether a set is Vector Space or Not

Question

Q: Determine whether the set $V = \{(x,\,y):x\ge 0,y\in\mathbb{R}\}$ with the standard operations in $\mathbb{R}^2$, is a vector space.

A:For the each $u\in V$ there exists a $-u\in V$. If we apply this rule to our $V$, we see that it does not apply.

My question is, is the answer I have given above is a correct, applicable one?

Answer 1

Perhaps try to be clear by giving an example.

$(1,0) \in V$ because $1\ge 0$ and $0 \in \mathbb{R}$. However $-(1,0)=(-1,0) \notin V$ since $-1<0$.

Answer 2

Perfect answer, thought in a bit more formal manner you can say, that some elements of $V$ doesn't have an additive inverse, and hence it's enought to give only a single counterexample.

Answer 3

Yes of course it suffices just to exhibit a counter example which doesn’t satisfy the definition that is

$$v=(1,y)\in V$$

but $-v \not \in V$.

Answer 4

Your idea is correct, but the execution needs work. How do you know the rule does not apply?

Remember: if you are saying that the statement "for all $u$, this thing is true", is false, you need to show that the statement "there exists some $u$ for which this thing is false" is true.