# Commutitive law for vectors

1. Mar 6, 2008

### Shenlong08

I'm having trouble with a proof : u+v=v+u
using the definition of a vector space (excluding the commutitive axiom of course), thus I have the other 9 axioms to work with.

I'm not even sure if I'm using one of the axioms right; since I cannot say that u+v=v+u, I don't think that I should say that u+(-u)=(-u)+u=0...which is what I've been trying to use.

2. Mar 7, 2008

### mrandersdk

not sure what you are trying to do. Are you trying to prove the commutitive axiom from all the others?

Because if you are don't, it is an axiom for a reason, that is it cannot not be proven from the others, if it could it would not be an axiom, it would be a theorem and the commutitive axiom would not be in the definition.

3. Mar 24, 2008

### vdgreat

let u={u1, u2....,un)
v={v1,v2....,vn)
then u+v= {u1+v1, u2+v2,....,un+vn} (vector addition)
or, u+v= {v1+v1, v2+u2,....,vn+un} (commutative property of addition on R)

4. Mar 25, 2008

### trambolin

Please, be careful about these, this is not a proof, this is exactly what mrandersdk has stated before. Note that, the two addition signs "+", (one for the addition $u+v$ and one for the additions $u_i+v_i$ that you used are different). We define the vector addition as such, it does not follow from the other axioms.

I can define the scalar addition as subtraction, then you don't have commutativity in that weird vector set (Since I destroyed the commutativity, it is not a vector space anymore, because commutativity is an axiom)

5. Mar 25, 2008

### HallsofIvy

This is assuming that the vector space is Rn which was not given.