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Write with at least one less abs value symbol

  1. Feb 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Here we go. Write the following with at least one less absolute value symbol:

    (ii) |(|a+b| - |a| - |b|)|

    (iii) |(|a + b| + |c| - |a + b + c|)|

    (iv) |x2 - 2xy + y2|


    2. Relevant equations

    12 Properties of Numbers.

    3. The attempt at a solution

    Let's just look at (ii) for now since I am getting all flustered with it:

    |(|a+b| - |a| - |b|)|

    There are 2 potential abs value symbols that could be dealt with:

    1) I could try to show that the quantity |a+b| - |a| - |b| ≥ 0 and the outermost symbols could be dropped.

    OR

    2) Try to show that -|a|-|b| = -|a+b| and re-write the original expression as |(|a+b| - |a + b|)|

    Either way is a pain. It seems like there are so many 'cases' to test. Am I correct in saying that I would need to test all of the following cases?

    I. a = b > 0
    II. 0 < b < a
    III. 0 < a < b
    IV. a < 0 < b where |a| < |b|
    V. b < 0 < a where |b| < |a|
    VI. a < 0 < b where |b| < |a|
    VII. b < 0 < a where |a| < |b|
    VIII. a = b = 0
    IX. a = b < 0

    Am I overkilling here? Are any of these equivalent? I am thinking "no" because its subtraction involved ....

    :mad:
     
  2. jcsd
  3. Feb 3, 2012 #2

    SammyS

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    For (ii) What does the triangle inequality tell you about how |a+b| and |a|+|b| are related?

    For (iv) Factor x2 - 2xy + y2 .
     
  4. Feb 4, 2012 #3
    From the triangle inequality: |a+b| ≥ |a| + |b| → |a+b| -|a| - |b| ≥ 0 ... jeesh.....nice catch SammyS!

    EDIT: oof! that's backwards! Now I'm screwed. I need sleep. Back in morning!


    For (iv)

    I see that it is |(x+y)(x-y)| but I am not sure what to do with that? Again there are many cases to test, right?

    I was originally thinking that since x2 and y2 are always > 0, then I would want to try to show that: x2 + y2 ≥ 2xy for all x,y.
     
    Last edited: Feb 4, 2012
  5. Feb 4, 2012 #4

    SammyS

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    Not quite right!
    (x+y)(x-y) = x2 - y2

    Try again.

    For (iii): Try the triangle inequality here too.
     
  6. Feb 19, 2012 #5
    So I have come back to this one finally: I am still unsure if part (ii) ... I think I am missing the obvious.

    (ii) ##|(|a+b| - |a| - |b|)|##

    From the triangle inequality I have that ##|a+b|\le|a|+|b|\implies |a+b| - |a| - |b|\le 0## so for sure I cannot remove the outer most abs value signs. This means that I am looking at removing (at least) one of the inner sets. As a guess, I am choosing to remove the signs around ##|a+b|## which means that I need to show that for any values a,b, that ##|a+b| - |a| - |b| = a+b - |a| - |b|.##

    But I don't know if this is correct. I am pretty sure it is not.
     
  7. Feb 19, 2012 #6

    LCKurtz

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    Are you really sure? What is ##|x|## if ##x\le 0##?
     
  8. Feb 19, 2012 #7
    I'm sorry LCKurtz, I don't think I understand what you're asking? |x|=-x if x≤0. But what is your point :redface:?
     
  9. Feb 20, 2012 #8

    LCKurtz

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    Think about what you said above which I colored red. What if that is x?
     
  10. Feb 20, 2012 #9
    Hi LKCurtz! I don't think I had enough foresight to realize that I could remove an abs value sign and rearrange the expression :redface:. How about this?

    ##|(|a+b|-|a|-|b|)|=|b|+|a|-|a+b|##
     
  11. Feb 20, 2012 #10

    LCKurtz

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    Yes. That's the idea.
     
  12. Feb 21, 2012 #11
    Just to finish up:

    (iii) ##|(|a + b| + |c| - |a + b + c|)| = |a + b| + |c| - |a + b + c|##

    since ##|a + b + c| \le |a + b| + |c|## by triangle inequality.

    and
    (iv) ##|x^2 - 2xy +y^2| =x^2 - 2xy +y^2. ##

    Since, ##x^2 - 2xy +y^2 = (x-y)^2 \ge 0 \text{ for all x.} ##
     
  13. Feb 21, 2012 #12

    SammyS

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    Yes. Those look good !
     
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