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Prove -|x|</x</|x|

  1. Mar 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove -|x|</x</|x|


    2. Relevant equations



    3. The attempt at a solution

    We are trying to prove -|x|</x</|x| (less than or equal to)

    This means that we need to show that x>/ -|x| and x</ |x|.

    Let x be a real number.

    Assume |x|=x if x>0 and -x if x<0

    I'm not sure were to go from here, as I cannot use actual values to prove it. Could someone please show me what to do?

    Thank you very much
     
  2. jcsd
  3. Mar 30, 2008 #2
    Case 1:
    [tex]-|x| \leq x[/tex]

    [tex]-x \leq -x \leq x[/tex]

    So it's obvious that case 1 holds..

    Can you see where to go form there?
     
  4. Mar 30, 2008 #3
    Thank you very much

    Could you please explain to me how you got that? Would I do the same thing for |x|?

    Thank you
     
  5. Mar 30, 2008 #4
    It's just basic a absolute value identity, and yes you would^^
     
  6. Mar 31, 2008 #5
    Thank you

    After I do that, would that complete the proof?

    |x|>x

    x>x or -x>x

    In order to prove that |x|</a if and only if -a</x</a where a>/0, would I do the same thing as in the previous problem?

    Thank you
     
    Last edited: Mar 31, 2008
  7. Mar 31, 2008 #6

    HallsofIvy

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    The simple way to do it is to use the definition: |x|= x if [itex]x\ge 0[/itex], -x is x< 0.

    If x[itex]\ge[/itex] 0, then [itex]-|x|\le x\le |x|[/itex] becomes [itex]-x\le x\le x[/itex]. Isn't that obvious?

    If x> 0, then [itex]-|x|\le x\le |x|[/itex] becomes [tex]x\le x\le -x[/itex]. Isn't that obvious?
     
  8. Mar 31, 2008 #7
    Thank you very much

    In order to prove that |x|</a if and only if -a</x</a where a>/0, would I do the same thing as in the previous problem?

    Thank you
     
  9. Mar 31, 2008 #8
    there is a typo here, it should read i guess x<0. I just wanted to point it out.
     
  10. Mar 31, 2008 #9
    Thank you very much

    Regards
     
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