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Commutitive law for vectors

  1. Mar 6, 2008 #1
    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. jcsd
  3. Mar 7, 2008 #2
    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.
  4. Mar 24, 2008 #3
    let u={u1, u2....,un)
    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)
  5. Mar 25, 2008 #4
    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)
  6. Mar 25, 2008 #5


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