How Can You Prove the Commutative Property of Vector Addition in Rn Spaces?

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Discussion Overview

The discussion revolves around the proof of the commutative property of vector addition in Rn spaces, specifically attempting to establish the relationship u + v = v + u using the axioms of a vector space while excluding the commutative axiom itself. The scope includes theoretical reasoning and mathematical proof techniques.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in proving u + v = v + u without using the commutative axiom, questioning the validity of their approach using other axioms.
  • Another participant suggests that attempting to prove the commutative property from the other axioms is misguided, stating that it is an axiom and cannot be derived from the others.
  • A participant provides a representation of vector addition in terms of components, suggesting that u + v can be expressed as {u1 + v1, u2 + v2, ..., un + vn}, and implies that this leads to the conclusion u + v = v + u.
  • Another participant cautions that the previous representation does not constitute a proof and emphasizes the distinction between the addition of vectors and the addition of their components, reiterating that commutativity is an axiom that cannot be derived.
  • A similar representation of vector addition is repeated, with a note that the assumption of the vector space being Rn is necessary for the argument.

Areas of Agreement / Disagreement

Participants generally disagree on whether the commutative property can be proven from other axioms of vector spaces. Some assert that it cannot be derived, while others attempt to provide representations that suggest it can be shown under certain conditions.

Contextual Notes

There are limitations regarding the assumptions made about the vector space, particularly the necessity of specifying that it is Rn for certain arguments to hold. The discussion also highlights the potential confusion between component-wise addition and vector addition as defined in vector spaces.

Shenlong08
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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.
 
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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.
 
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)
 
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 [itex]u+v[/itex] and one for the additions [itex]u_i+v_i[/itex] 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)
 
vdgreat said:
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.
 

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