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Math Help - |x-1| = 1 - x

  1. #1
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    |x-1| = 1 - x

    Hey forum, I'm dealing with this one

    |x-1| = 1 - x

    I use the definition of the absolute value of an arbitrary argument which gives

    |x-1| = \{ x-1 \ , \te{ \ if \ } x-1 \geq 0 \ \Leftrightarrow \ x \geq 1 \ \ \ \ (1) \\ -(x-1) \ , \te{ \ if \ } x-1 < 0 \ \Leftrightarrow \ x < 1 \ \ \ \ (2) \\

    We see that it is true for the second condition, because

    -(x-1) = 1-x \\ -x + 1 = 1 -x \\ 1 -x = 1- x

    Therefore, the answer must be that x must be smaller than 1. But the equation is also true for x=1 ! What's up with that?
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  2. #2
    Super Member Quacky's Avatar
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    Re: |x-1| = 1 - x

    First case:

    x-1=1-x

    2x=2

    x=1

    Second case:

    -(x-1)=1-x

    -x+1=1-x

    So we get x<{1} from case 2, and x=1 from case 1. Put them together to give the whole range of solutions and you get that x\leq{1}
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  3. #3
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    Re: |x-1| = 1 - x

    Thanks!!!
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  4. #4
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    Re: |x-1| = 1 - x

    Another way of looking at it: |z| is defined as "z if [itex]z\ge 0[/itex], -z if z< 0".
    Obviously 1- x= -(x- 1) so we must have x- 1< 0.
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