Is this $0v=0$ proof correct? I'm taking my first linear algebra course at university and recently I've been introduced to vector spaces. Now, the teacher has asked us to prove that $0v=0$ for any $v$ using only the definition of a vector space. I've came up with a proof, but I'm unsure whether it's correct or not, as I'm not familiar with proofs in Math.
$$
\begin{align}(1+1)v &= 2v \\
(1+1)v-2v &= 2v-2v\\
[(1+1)-2]v &= 2(v-v) \\
0v &= 0
\end{align}
$$
If this is correct I can go on proving other properties of vector spaces. Any feedback appreciated!
EDIT: Ok, taking into account José's answer, I came up with this one
$$
(1+0)v=v \\
1v+0v=v \\
-v+v+0v=-v+v \\
0v=0
$$
 A: Your proof has some problems. What is $-2v$? Is it $-(2v)$ or is it $(-2)v$? If it is $-(2v)$, then you could jump directly from $2v-2v$ to $0$, but then how do you know that $(1+1)v-2v=\bigl((1+1)-2\bigr)v$? And if $-2v$ is $(-2)v$, then how do you know that $2v-2v=2(v-v)$?
A: You can write a proof that doesn't require $(-1)v=-v$:
$$
0v=(0+0)v\\
0v=0v+0v\\
0v-(0v)=0v+0v-(0v)\\
0=0v
$$
Note that $x-y$ actually means $x+(-y)$; with $-(0v)$ I denote the negative of $0v$.
Where does your attempt go wrong? With the same convention as before, you can do $(1+1)v-(2v)=2v-(2v)$. The right hand side is $0$ by definition of $-(2v)$, but you cannot go from $(1+1)v-(2v)$ to $((1+1)-2)v$ unless you prove that
$$
-(2v)=(-2)v
$$
which is true but unfortunately depends on $0v=0$. Indeed, for any scalar $a$,
$$
av+(-a)v=(a+(-a))v=0v=0
$$
so $(-a)v$ summed to $av$ is $0$, which means $(-a)v$ is the negative of $av$:
$$
(-a)v=-(av)
$$
Note that in all of this the property $1v=v$ has not been used. With it we can also state
$$
(-1)v=-(1v)=-v
$$
A: Here another method : 
  where $θ$ is the zero vector and $a$ is a scaler 
 We know that $av$ is a scaled vector and $v + θ = v$ 
So : $$av + θ = av $$
Hence :$$θ = (a-a)v$$
$$θ = 0v$$
