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Absolute value

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Show the following for every d>0:
    For every real number x with |x-1|<d it follows that |1+x|<2+d

    2. The attempt at a solution
    If x-1>0, then |x-1|=x-1<d. Hence x+2 = |x+2| < 2+d.
    If x-1<0, then |x-1|=-(x-1)<d. Hence x-1>-d => x+1 > 2-d ...??

    Is this really possible to do without putting any other restrictions on d?
     
  2. jcsd
  3. Feb 14, 2010 #2

    CompuChip

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    Homework Helper

    Let me answer your last question first.

    |x - 1| < d means, that the distance of x to 1 is no more than d. Now what is the distance of x to -1? Well, if x is "to the left" of 1, then it is definitely closer to -1 than d. For example, if x is at the maximal distance from 1, namely 1 - d, then it has distance 1 - (1 - d) = d to -1.
    If x is on the other side of 1, its distance to -1 is equal to two units plus the distance to 1.

    If you use the idea that you can measure the distance of x to -1 by going "to 1" first, you hopefully see how the triangle inequality comes in useful.
     
  4. Feb 14, 2010 #3
    Ah of course, the triangle inequality in one dimension :) (not really a triangle)
    [tex]|a-c|\leq |a-b|+|b-c|[/tex]

    If I choose: a=1, b=-1, c=x

    [tex]|1-x|\leq|1+1|+|-1-x|=2+|1+x|<d[/tex]
    So surely:
    [tex]|1+x|<d+2[/tex]

    Thank you!
     
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