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Proof: |A - B | <

  1. Sep 20, 2004 #1
    There are many questions i can't get. This is just one. Can anyone give me a hint on what to do? It's probably really simple :(

    If A and B are real numbers such that [tex]|A - B| < \frac{1}{2}[/tex], show that:

    [tex]|A^2 - B^2| \leq \frac{1}{2} \{2|B|+
    \frac{1}{2} \}[/tex]
     
    Last edited: Sep 20, 2004
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  3. Sep 20, 2004 #2

    arildno

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    Welcome to PF!
    Try to rewrite [tex]|A^{2}-B^{2}|[/tex] as a product of two expressions..
    (It's a well known identity)
     
  4. Sep 20, 2004 #3
    Is this what you meant?

    [tex] |(A+B) (A-B)| \leq |B| + \frac{1}{4} [/tex]

    If so, i already had that but didn't know where to go from there...already wasted a lot of paper going no where :grumpy:

    Am i supposed to use [tex] -\frac{1}{2} < A - B < \frac{1}{2} [/tex] somewhere?

    thanks
     
    Last edited: Sep 20, 2004
  5. Sep 20, 2004 #4

    arildno

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    First of all:
    You are to PROVE that inequality, not ASSUME it!!
    Hence, what you start with, is the following EQUALITY:
    [tex]|A^{2}-B^{2}|=|A-B||A+B|[/tex]
    (Ok so far?)
    Now, use the inequality you've been given to derive:
    [tex]|A^{2}-B^{2}|=|A-B||A+B|\leq\frac{1}{2}|A+B|[/tex]
    Agreed?
    Now you'll need to use your given inequality to provide an estimate of |A+B|!
     
  6. Sep 20, 2004 #5

    arildno

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    PS:
    You'll need to use the triangle inequality as well..
     
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