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2 Proofs: Does this work?

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove the following:

    (i) ##|x|-|y| \le |x-y|##

    and

    (ii) ##|(|x|-|y|)| \le |x-y|\qquad## (Why does this immediately follow from (i) ?)

    2. Relevant equations

    ##|z| = \sqrt{z^2}##

    3. The attempt at a solution

    (i) ##(|x|-|y|)^2 = |x|^2 - 2|x||y| + |y|^2 = x^2 - 2|x||y| + y^2 \le x^2 - 2xy + y^2= (x-y)^2 \implies \boxed{|x|-|y| \le |x-y|.}##


    (ii) For this part, I looked at the question "Why does this immediately follow from (i)" for inspiration and saw that if I could show that ##|(|x|-|y|)| \le |x-y|## then the proof is complete by transitivity.

    Is it as simple as:

    ##|(|x|-|y|)| = \sqrt{(|(|x|-|y|)|)^2} = \sqrt{(|x|-|y|)^2} = |x|-|y|?##

    I think that it is, but it is getting late :redface:
     
  2. jcsd
  3. Feb 14, 2012 #2

    HallsofIvy

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    If |x|> |y|, (ii) is identical to (i). If |y|< |x|, swap the two.
     
  4. Feb 14, 2012 #3

    LCKurtz

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    No, it isn't. ##\sqrt{(|x|-|y|)^2} = |x|-|y|## only when ##|x|\ge |y|##.
     
  5. Feb 14, 2012 #4

    LCKurtz

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    Remember that ##\sqrt{a^2}=|a|## and if ##0\le a\le b## then ##\sqrt a\le \sqrt b##. So how can you change the above conclusion?
     
  6. Feb 14, 2012 #5
    Isn't this what is called the kaushy swartz?
     
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