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Showing that the mean of one number and its absolute val..

  1. Oct 4, 2015 #1
    1. The problem statement, all variables and given/known data
    I must prove that the mean of one number and its absolute value is superior or equal to 0 but inferior or equal to the absolute value of the number

    2. Relevant equations
    I must prove this by the method of case.

    3. The attempt at a solution

    I know that the first case is : x>=0 and the second x<0

    But after that, I have no idea on how to do this....
     
  2. jcsd
  3. Oct 4, 2015 #2

    SammyS

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    What are the cases you refer to ?
     
  4. Oct 4, 2015 #3
    You mean you don't understand what the proof by case is ? I mean that to arrive to my result, I must begin x>=0 and the second case is x<0. After, I imagine there is some sort of addition that I Must do to obtain my final result.
     
  5. Oct 4, 2015 #4
    Are you still here ? I think that I must do this : x<0 then absolute(x)=-x

    and x>=0 then absolute(x)=x
     
  6. Oct 4, 2015 #5

    SammyS

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    I misread your post.

    Case 1: x ≥ 0
    If x ≥ 0, what is |x| ?
     
  7. Oct 4, 2015 #6
    Then this means |x|=x

    And if x<0 then this means |x|=-x
     
  8. Oct 4, 2015 #7

    SammyS

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    Yes. That's the way to start each part.
     
  9. Oct 4, 2015 #8
    Yeah, I was able to understand this. Here is what I did after :

    |x|=x add x on each side

    |x|+x=2x

    divide by 2

    (|x|+x)/2=x

    Case 2 :

    |x|=-x

    |x|+x=x-x

    |x|+x=0

    Multiply by 1/2 by each side.

    (|x|+x)/2=0

    Is this correct ?
     
  10. Oct 4, 2015 #9

    SammyS

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    Yes.
     
  11. Oct 4, 2015 #10
    Ok, I guess that I must use my x<=0 inequality and replace the x.

    (|x|+x)/2<=0
     
  12. Oct 5, 2015 #11

    Ray Vickson

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    No. You were asked to prove that
    [tex] 0 \leq \frac{x+|x|}{2} \leq |x| [/tex]
    Note that this is "##\geq 0##", not "##\leq 0##", but, of course, they are the same thing when you actually have "##= 0##".
     
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