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Question on a step in my proof.

  1. Sep 24, 2008 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    I don;t think the actual problem I am trying to prove is relevant. But I have got to a step in my proof where I must show that | a - b | = | -(a - b)|

    So by definition
    |a - b| = +/- (a-b)

    and |-(a - b)| = +/- (- (a - b)) = +/- (a-b)

    Thus the two are equal.

    Is this acceptable?

    Or do I really need to break it down into cases where a - b is Positive negative or 0?

    Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 24, 2008 #2

    Dick

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    |a-b|=+/-(a-b) is NOT the definition of absolute value. The definition of |a-b| is that it is (a-b) if a-b>=0 and -(a-b) if a-b<0. You do need to break it into cases. But it's easier to prove |x|=|-x| (by cases) and then set x=a-b.
     
  4. Sep 24, 2008 #3
    Got it, yes, I realize I can say |a-b| is greater than or equal to +/- (a-b).

    but certainly not equal.

    I showed that |x| = |-x| as follows:

    case x>0:
    If x > 0 then |x| = x by definition. If x>0 then -x<0 since x and -x can't both be in R+. if -x < 0 then |-x| = -(-x) = x

    case x = 0. If x = 0 then -x = 0 and |x| = |0| = 0 and |-x| = |0| = 0.

    case x < 0. If x<0 then |x| = -x by definition. If x<0 then -x>0 since x and -x can both be in R+. since -x>0, |-x| = -x and we are done.
     
  5. Sep 24, 2008 #4

    Dick

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    That proof is just fine. If you know |ab|=|a||b|, you can also prove it by |-x|=|(-1)*x|=|-1|*|x|=|x|.
     
  6. Sep 24, 2008 #5

    tiny-tim

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    Hi Diffy! :smile:

    If your main proof required you to show that |a - b| = |b - a|,

    then I would say that it is enough to say (a - b) = -(b - a), and therefore |a - b| = |b - a|.

    But to show that | a - b | = | -(a - b)|, in that form, doesn't ssem to me to require any proof … |x| = |-x| is part of the definition of ||. :smile:

    (btw, using the symbol +/-, or even ±, would not be acceptable … you should use something more rigorous, like √(x2), or sup{x,-x}, or write the concept out in English :wink:)
     
  7. Sep 24, 2008 #6

    Dick

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    That sounds a little funny. If a<b then -a>-b. That's why x<0 implies -x>0.
     
  8. Sep 24, 2008 #7
    Dick, there was a mistake,

    It should say if x<0 then -x>0 since x and -x can't both be in R+.

    I don't have that if a<b then -a>-b anywhere, but I do have in the my order property that only one of the following is true:
    a in R+
    a = 0
    -a in R+

    I was attempting to use that. Thanks a lot for your help dick!

    Hey TinyTim,

    It wasn't immediately clear (at least to me) from my definition so I felt the need to prove it.
    I had:
    |a| = a if a>0
    |a| = 0 if a = 0
    |a| = -a if a < 0

    Thanks for your help Tiny-Tim!
     
  9. Sep 24, 2008 #8

    Dick

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    Ok. That makes sense.
     
  10. Sep 24, 2008 #9

    tiny-tim

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    Yes, that method's fine! :smile:
     
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