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.

$$
\begin{align}(1+1)v &amp;= 2v \\
(1+1)v-2v &amp;= 2v-2v\\
[(1+1)-2]v &amp;= 2(v-v) \\
0v &amp;= 0
\end{align}
$$

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
$$
(1+0)v=v \\
1v+0v=v \\
-v+v+0v=-v+v \\
0v=0
$$
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 $2v-2v=2(v-v)$?
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 :$$θ = (a-a)v$$
       $$θ = 0v$$