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Math Help - Proving ||z|-|w|| <= |z-w|

  1. #1
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    Proving ||z|-|w|| <= |z-w|

    hi, I stumbled upon a simple demonstration but i can´t solve it:

    ||z|-|w|| <= |z-w|

    Can someone help me?

    Pedro
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    Quote Originally Posted by Pedro View Post
    hi, I stumbled upon a simple demonstration but i can´t solve it:

    ||z|-|w|| <= |z-w|

    Can someone help me?

    Pedro
    This is the reverse triangle inequality.
    Last edited by mr fantastic; November 24th 2010 at 05:52 PM.
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    ok, thank you!
    Last edited by mr fantastic; November 24th 2010 at 05:52 PM.
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    This proof relies on this fact: a \geqslant 0\;\& \, - a \leqslant b \leqslant a\, \Leftrightarrow \,\left| b \right| \leqslant \left| a \right|.

    So \left| u \right| \leqslant \left| {u - w} \right| + \left| w \right|\, \Rightarrow \,\left| u \right| - \left| w \right| \leqslant \left| {u - w} \right|

    Likewise \left| w \right| - \left| u \right| \leqslant \left| {w - u} \right| = \left| {u - w} \right|

    Thus we now have  - \left( {\left| {u - w} \right|} \right) \leqslant \left( {\left| u \right| - \left| w \right|} \right) \leqslant \left( {\left| {u - w} \right|} \right)

    Can you use the fact to finish?
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  5. #5
    rcs
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    i understood what is discussed above... but i would like know how to explain and admit that
    -|a| ≤ a ≤ |a|

    and

    -|b| ≤ b ≤ |b|

    can anybody help me to prove this?
    all i know is that a negative is of course less than a positive number
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    Please start a new thread for any new question.

    You said that you understand that  a \geqslant 0\;\& \, - a \leqslant b \leqslant a\, \Leftrightarrow \,\left| b \right| \leqslant \left| a \right| .

    If you do, can we say that |a|\le ||a||~?

    If so then is it not true that -|a|\le a \le |a|~?
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    The fact is, that any nonnegative number is equal to its absolute value. So if \displaystyle a \geq 0, then \displaystyle a = |a|.

    However, any negative number is less than its absolute value (since the absolute value is always nonnegative). So if \displaystyle a < 0, then \displaystyle a < |a|.

    So that means for any \displaystyle a that \displaystyle a \leq |a|.


    A similar argument works the other way to show \displaystyle -|a| \leq a.

    So that means \displaystyle -|a| \leq a \leq |a|.
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  8. #8
    rcs
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    thanks you! so same in the -|b| ≤ b ≤ |b|.
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