Math Help - Proving zero vector is unique - by contradiction?

1. Proving zero vector is unique - by contradiction?

I started a proof by contradiction and would like to know if I'm on the right track, and get some direction please. I know there are solutions out there for this but I would like to try finishing my proof if possible.

Exercise: Prove that the zero vector is unique. This means that if $\vec{z}$ is any other vector in $\mathbb{R}$ with the property that $\vec{z}+\vec{v}=\vec{v}$ for any $\vec{v}\in\mathbb{R}^{n}$, then $\vec{z} = \vec{0}_{n}$.

My Proof:
By contradiction, assume that the zero vector equals some other vector $\vec{z}\in\mathbb{R}^{n}$.
Then by the definition of the zero vector, this means that
(1) $\vec{0} + \vec{z} = \vec{0}$.
Since
$\vec{0}=\vec{z}$, then for some vector $\vec{v}\in\mathbb{R}^{n}-\vec{0}$, we can write
(2) $\vec{z}+\vec{v}=\vec{0}$, therefore
(3) $\vec{z}=\vec{0}-\vec{v}$. By substituting equation (3) into equation (1) we get,
(4) $\vec{0}+\vec{0}-\vec{v}=\vec{0}$, and thus
(5) $\vec{-v}=\vec{0}$.
We have reached a contradiction and therefore proven that the zero vector is unique. QED

2. Your whole methodology is flawed, so far as I can tell. The generally accepted way to show that the zero vector is unique, is to assume there's another vector that satisfies the behavior of the zero vector, and that, therefore, it must equal the usual zero vector. That is, using your notation, you assume both that there is a vector $\vec{z}$ such that $\vec{v}+\vec{z}=\vec{v}$ for all $\vec{v}\in V=\mathbb{R}^{n},$ and that the usual zero vector $\vec{0}$ satisfies $\vec{v}+\vec{0}=\vec{v}$ for all $\vec{v}\in V.$

Now you must SHOW that $\vec{0}=\vec{z}.$ In your proof, at one point, you assumed $\vec{0}=\vec{z}$ and went on to show a contradiction! That logic doesn't work at all. This is not a proof by contradiction. Use the two equations you have assumed to show the required equation is true.

3. I agree with Ackbeet.
To prove the statement "0 is unique" is true by contradiction, we assume to the contrary that "0 is unique" is false. In other words, we assume:

There exist two vectors, a and b NOT EQUAL, such that for all vectors x, a + x = x = b + x.

Then we can conclude that a = b when we contradict some other axiom!

Since we already know that there is one vector satisfying
0 + v = v for all v, we consider what happens if another vector has this same property. i.e.

z + v = v for all v.

Then immediately, we have 0 + v = v = z + v, so 0 + v = z + v. So 0 = z.