How to show that the additive inverse axiom of a vector space can be replaced with 0v = 0?

Question

I am working on a problem that asks to show that in the definition of a vector space, show that that additive inverse can be replaced with the condition such that: 0v = 0 for all v belonging to V.

My reasoning is that since we're assuming that the axiom "There exists a w such that v + w = 0 for all v, w belonging to F (the set of all real or complex numbers)" can be replaced with 0v = 0, I can begin the proof with:

v + w = 0v                     by supposition
(-v) v + w = 0v (-v)           property of equivalence
w = -v                      

I don't think this is sufficient though; as someone told me I shouldn't be making the assumption that there exists a (-v). Any help here would be greatly appreciated.

Answer 1

Suppose that $(\forall v\in V):0v=0$. Then, given $v\in V$,\begin{align}0&=0v\\&=\bigl(1+(-1)\bigr)v\\&=1v+(-1)v\\&=v+(-1)v\end{align}and therefore $v$ has an additive inverse.

And if each $v$ has an additive inverse $v'$, then\begin{align}0v&=(0+0)v\\&=0v+0v\end{align}and therefore\begin{align}0&=0v+(0v)'\\&=(0v+0v)+(0v)'\\&=0v+(0v+(0v)')\\&=0v+0\\&=0v.\end{align}

Answer 2

Define $u:=(-1)v$ so $u+v=v+u=(-1+1)v=0v=0$. Therefore, $u$ is $v$'s additive inverse.