In a given vector space V the zero vector is unique

Question

I am trying to attempt the proof for the above theorem from my book. I want to know if my approach is correct.

Attempt:

Let 0 and 0 be additive identities

0+v=v (eq1)

0+v=v (eq 2)

setting eq1=eq2

0+v=0+v

0+v+(-v)=0+v+(-v)

0+0=0+0

0=0

Answer 1

This is actually true for any group, which vector spaces are a particular instance of. The proof is actually a very simple one liner.

If $e_1$ and $e_2$ are zeroes, then $e_1=e_1+e_2=e_2$.

Answer 2

I wouldn't say that what you've done is wrong, but it isn't the best way to go about it. You are actually dangerously close to the regular proof—it's actually hidden inside yours—which would go as follows.

Let $0$ and $\textbf{0}$ be zero vectors, then

\begin{align} 0 &= 0+\textbf{0}\qquad \text{by definition of $\textbf{0}$ being a zero vector} \\ \textbf{0} &= 0+\textbf{0}\qquad \text{by definition of $0$ being a zero vector} \end{align}

So then by transitivity, $0 = \textbf{0}$ so they are indeed the same.