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Homework Help: Set-builder question.

  1. Feb 3, 2013 #1
    I'm asked to sketch the set {(x,y)∈R^2:(y-x)(y+x)=0} on a x-y plane.
    By expanding the rule I get that y=x or that y=-x. In the answer to this question the graph shows two perpendicular lines which cross at the origin and continue in both the positive and negative direction.

    My question is how, can negative values of x and y belong to the set of R^2? To me it seems that the set should contain only positive numbers. Any explanation would be appreciated.
     
  2. jcsd
  3. Feb 3, 2013 #2

    Dick

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    R^2 isn't really the square of anything, it's a notation for the cartesian product of R and R. It's just all pairs (x,y) where x is in R and y is in R.
     
    Last edited: Feb 3, 2013
  4. Feb 3, 2013 #3
    Thank you =)
     
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