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Do i have to consider the complex case of triangle inequality

  1. Nov 22, 2008 #1
    in order for me to understand this theorim
    do i have to think of the vectors as complex??

    what the values of x1 ,y1 ,x2 ,y2

    [tex]
    \langle x , y \rangle = x_1^* \cdot y_1 + x_2^*\cdot y_2 + \ldots
    [/tex]
    did i get the structure correctly

    x=(x1,y1) y=(y1,y2) each one of x1,x2,y1,y2 is a complex number??
     
  2. jcsd
  3. Nov 22, 2008 #2

    Dick

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    That has not much to do with any triangle inequality, but yes, it looks like the definition of a complex inner product. And I think your understanding of what the objects are is fine.
     
  4. Nov 22, 2008 #3
    i know that the sum of two sides is always bigger then the third one

    but in this formula
    |x+y|=<|x| +|y|

    on both sides i have a sum of two members
    why the third side is |x+y| ??
    and why there is an option for them to be equal??
    (the sum of two sides is always bigger then the third one)
     
  5. Nov 22, 2008 #4

    Dick

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    You are supposed to be thinking of x and y as vectors. If the vector x represents one side and the vector -y represents the other side, x+y represents the third side. The sum of the sides can be equal to the third side if all three vertices lie on the same line.
     
  6. Nov 22, 2008 #5
    in the article they present the case where they use it in a simple way of
    |x+y|=<|x| +|y|

    http://en.wikipedia.org/wiki/Triangle_inequality

    if both the sides lie on the third line then its not a triangle(its not legal)

    on both sides i have a sum of two members
    why the third side is |x+y| ??
    and why there is an option for them to be equal??
    (the sum of two sides is always bigger then the third one)
     
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