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Proving vector spaces where (a1a2 < equal to 0)

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data
    For the vector set<a1,a2>, where (a1a2 < equal to 0)


    2. Relevant equations



    3. The attempt at a solution

    I'm not sure why this set is close under scalar multiplication and not in vector addition. Some hints would be nice :D
     
  2. jcsd
  3. Mar 13, 2012 #2

    Fredrik

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    Is <a1,a2> your notation for ordered pairs? In other words, are you trying to define a subset of ℝ2? In that case, it should be easy to find a counterexample.

    I don't think it's possible to give you a hint without completely solving the problem for you.
     
  4. Mar 13, 2012 #3
    Well i think for vector addition, it's open because there's no negative vector of a1 or a2 since a1a1<0? am I correct? and I'm assuming a1 is a vector, and a2 is another vector.
     
  5. Mar 13, 2012 #4

    Fredrik

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    Then I don't understand the definition of the set. What vector space is this supposed to be a subset of? If a1 and a2 are vectors, what does <a1,a2> and a1a2 mean? I also don't understand the sentence "there's no negative vector of a1 or a2".
     
  6. Mar 13, 2012 #5
    um the set is a notation for ordered pairs, I was trying to refer to the axiom for vector addition that states" For each x in V, there exist a vector -x such that x+(-x)=(-x)+x=0" is not satisfied
     
  7. Mar 13, 2012 #6

    Fredrik

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    You said that a1 and a2 are vectors. In what vector space? Or would you like to change that and say that they are real numbers instead? Because if they are vectors, I don't know what a1a2 means.
     
  8. Mar 13, 2012 #7

    Fredrik

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    Why don't you start by just writing down the sum of two arbitrary members of this subset?
     
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