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How to open an innequality

  1. Jan 5, 2009 #1
    how they get from this expression

    |An+1 - An - L|<e


    L-e<An+1 - An<L+e

    cant understand this transition
  2. jcsd
  3. Jan 5, 2009 #2


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    Try to initially simplify your expression so it is easier to handle. Let v = An+1-An. Somehow, you probably mean something like An+1-An.
    So now, you may write:

    |v - L|<e

    You can show that on a number line, knowing also that e>0, because it is greater than an absolute value, a positive number. You can also represent on the number line, -e, which is on the opposite side of zero, equadistant from zero as e.

    You have two situations possible, but ultimately you may see that they form conjointedness.
    Either v-L is positive, or v-L is negative.

    Can you now work with the original statement more successfully?
  4. Jan 5, 2009 #3


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    |x|< a means -a< x< a. In particular, |v-L|< e means -e< v-L< e and, adding L to each part, L-e< v< L+ e.
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