Proof of -(-v)=v in a vector space The exercise is: Prove that $-(-v)=v$ for every $v \in V$

Proof
Suppose  $v \in V$ and $V$ is a vector space.
Then $-(-v) \in V$ as result of the scalar multiplication property and
$-(-v)=-(-1\cdot v)=-1 \cdot(-1\cdot v) =(-1 \cdot-1)\cdot v = 1 \cdot v = v $
The desired result $-(-v)=v$ holds.

The solution manual gives the proof:

Proof


I just wanted to make sure that the way I did the proof isn't missing anything?Thanks
 A: Your proof is fine. Although this proof is a long winded version of the proof in the solutions manual, it is a little easier to grasp.
I like your proof, but to be pedantic, one would have to prove the commutativity of scalar multiplication with multiple scalars to a vector for your third equality. This is not presented in the book. So limiting ourselves to a subset of all the axioms and theorems until the point this problem shows up, I believe this is a not so elegant but straightforward proof.
Let $$ -v = w.\tag{1}\label{1}$$
Then we can write $-(-v)=-w$.
Since $-w$ is an additive inverse for $w$ we can write
$$w + (-w) = 0 .\tag{2}\label{2}$$We can also write $$v + (-v) = 0.\tag{3}\label{3}$$ from $\eqref{1}$ and $\eqref{3}$ it is clear that $$v+w=0.\tag{4}\label{4}$$ which means that $$v = -w.$$ Substituting in the value of $w=-v$ we get $$v=-(-v)$$ which concludes this proof.
A: Your solution seems good, acknowledging the fact that V is a vector space, and therefore satisfies the axioms pertaining to the proof.
