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Proof of Distributive Property of Vectors

  1. Feb 26, 2015 #1

    B18

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    1. The problem statement, all variables and given/known data
    Let u, and v be vectors in Rn, and let c be a scalar.
    c(u+v)=cu+cv

    3. The attempt at a solution
    Proof:
    Let u, v ERn, that is u=(ui)ni=1, and v=(vi)ni=1.
    Therefore c(ui+vi)ni=1

    At this point can I distribute the "c" into the parenthesis? For example:

    =(cui+cvi)ni=1
    =(cui)ni=1+(cvi)ni=1
    =cu+cv.
     
  2. jcsd
  3. Feb 26, 2015 #2

    Dick

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    Sure you can. You know you can distribute over real numbers. The components of vectors in ##R^n## are just real numbers.
     
  4. Feb 26, 2015 #3

    B18

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    Ok I figured. So everything looks in order here? Just was expecting it to be a little more involved!
     
  5. Feb 26, 2015 #4

    Dick

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    No, it's not more involved. This is an easy one.
     
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