Exploring the Commutative Property of Vector Addition in Rn Spaces

[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.

 

[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.

 

[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)
therefore, u+v=v+u (vector addition)

 

[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)

 

[HallsofIvy]

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)
therefore, u+v=v+u (vector addition)

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