Proving commutativity of addition for vector spaces I'm trying to prove commutativity of addition for vector spaces, using the axioms for vector spaces. Apparently commutativity can be proven! Im having trouble getting a good feel for what is allowed and what is not. Here's my work so far:
$u+v+u+v = 2(u+v) = 2u + 2v = u+u+v+v = u+(u+v)+v$
Here I just wanna claim that $u+(v+u)+v = u+(u+v)+v$ 
$\Rightarrow -u+u+(v+u)+v+(-v) = -u+u+(u+v)+v+(-v)$ : here im just adding -u to the right, and -v to the left. Question: is this "adding to both sides" really legit in this context? Why? 
Quick help proof: $-v+v = (-1)v+(1)v = (-1+1)v = 0v = 0 = v-v$
And another: $ 0+v = v+(-v) + v = (1)v + (-1)v + v = (1-1)v + v = v = v+0$
We have $0 + (u+v) + 0 = 0+(v+u)+0 \Rightarrow u+v = v+u$
This feels ugly and not at all elegant, especially the great leap "add -u to both sides" feels completely out of place. Do I need more lemmas? Is there a more elegant way? 
//not homework or anything, just for my own pleasure, feel free to provide theory, as it is more insightful than solutions. :)
Thanks! 
EDIT: corrected notation a little.
 A: Normally commutativity is taken as an axiom, but you can deduce it from associativity, distributivity and from the existence of inverses as follows:
$(u + v) - (v + u) = (u + v) - v - u$ (by distributivity) 
$ = u + (v - v) - u$ (by associativity) 
$ = u + 0 - u = (u + 0) - u = u - u = 0$
So $u + v = v + u$
A: Your first equation essentially answers the question:
$$u+v+u+v=2(u+v)=2u+2v=u+u+v+v$$
From here, because we know that a vector space is a group under addition, add on the left by $-u$ and on the right by $-v$ to get
$$-u+u+v+u+v-v=-u+u+u+v+v-v$$
$$v+u=u+v$$
This is legitimate because by definition a vector space is a group under addition.  If your definition doesn't have this as part of it, I'd recommend adding your definition.
A: The proof of this fact almost goes as follows (after generalizing associativity to omit parentheses):
$$u + u + v + v = (u + v)(1 + 1) = u + v + u + v$$
Then cancel $u$ on the left, and cancel $v$ on the right, to obtain $u + v = v + u$, "QED."
The issue, though, is that left and right cancellation laws, although not so tough to prove, are generally done by assuming commutativity. Let us suppose we have only proved right cancellation, and now we will prove left cancellation without overassuming, i.e., without assuming $0$ is a left identity and without assuming that the inverse on the right also works on the left.
Proposition 1. Given that $x + y = 0$ we can conclude that $y + x = 0$.
Proof. $y + x + y + x = y + 0 + x = y + x$. Right cancellation on $y+x$ yields the desired result.
Proposition 2. Given that $x + 0 = x$ we can conclude that $0 + x = x$.
Proof. Suppose $y$ is the (right and left, cf. Prop 1) inverse of $x$: $0 + x = x + y + x = x + 0 = x$.
Now that we have right and left cancellation laws, the scare quotation marks can be removed from the QED above.
A: I agree that commutativity is an axiom for Vector Spaces, so what you might need to show is that the ordinary vector spaces you know about satisfy the axioms of Vector Spaces.  Then you can forget about my capital letters.
You know numbers commute.  Use that to show vectors commute.
$\begin{align}u+v&=&(u_1,u_2,...,v_n)+(v_1,v_2,...,v_n)\\
&=&(u_1+v_1,u_2+v_2,...,u_n+v_n)\\
&=&(v_1+u_1,v_2+u_2,...,v_n+u_n)\\
&=&v+u
\end{align}$
A: I hate to be a bearer of  bad news but you cannot prove commutativity of addition using the other axioms.
Proving $(u+v) - (v+u) = 0$ does not necessarily mean $-(v + u)$ is the additive inverse of $u+v$
unless you assume  the uniqueness of the additive inverse.
To prove uniqueness of the additive inverse you need commutativity.
Not one axiom in the definition of a vector space is redundant.
