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Prove |ab|=|a||b|

  1. Feb 14, 2008 #1
    Hi does anyone know a proof for the multiplicative propery of absolute values

    i.e. Prove |ab|=|a||b|
     
  2. jcsd
  3. Feb 14, 2008 #2

    HallsofIvy

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    How about doing exactly what you always do with absolute values: break it into cases.

    1) If [itex]a\ge 0[/itex] and [itex]b\ge 0[/itex]
    Then [itex]ab\ge 0[/itex] so |ab|= ab while |a|= a, |b|= b. ab= (a)(b) so |ab|= |a||b|.

    2) If [itex]a\ge 0[/itex] while b< 0
    The [itex]ab\le 0[/itex] so |ab|= -ab while |a|= a, |b|= -b. -ab= (a)(-b) so |ab|= |a||b|.

    Can you do the other two cases?
     
  4. Feb 14, 2008 #3
    Thanks halls!
     
  5. Feb 15, 2008 #4

    danago

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    My book uses the following proof:

    [tex]
    \left| {ab} \right| = \sqrt {(ab)^2 } = \sqrt {a^2 b^2 } = \sqrt {a^2 } \sqrt {b^2 } = \left| a \right|\left| b \right|
    [/tex]
     
  6. Feb 15, 2008 #5

    HallsofIvy

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    Well, if you want to do it the easy way!
     
  7. Feb 15, 2008 #6
    I still find |a|=-a when a<0 weird! Surely if a = -a, |-a| = a
     
  8. Feb 15, 2008 #7
    When a<0, -a is positive.
     
  9. Feb 15, 2008 #8

    HallsofIvy

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    Yes that's true. Because if a= -a, then a= 0!

    Are you sure that's what you meant to say?
     
  10. Feb 15, 2008 #9
    Yeah I think when you look at the graph of y=|x| it becomes clear (as mud)!
     
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