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Absolute value

  1. Oct 6, 2008 #1
    Prove the following: if [tex] |a| \leq b [/tex] then [tex] -b \leq a \leq b [/tex] (where [tex] b \geq 0 [/tex]).

    So [tex] a \leq b [/tex] and [tex] -a \leq b [/tex]. Then [tex] -b \leq a [/tex] so that [tex] -b \leq a \leq b [/tex].

    Suppose that [tex] -b \leq a \leq b [/tex]. Then [tex] a \leq b [/tex] and [tex] -a \leq b [/tex] so that [tex] |a| \leq b [/tex].

    Is this a correct proof? You don't have to consider cases (e.g. [tex] a <0, \ a = 0, \ a > 0 [/tex])?
     
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  3. Oct 6, 2008 #2

    Dick

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    It's fine if you already know |a|<=b implies a<=b and -a<=b. If you don't know that you might have to use cases to prove that.
     
  4. Oct 6, 2008 #3
    So if you know that already, then it is ok to deduce from [tex] -b \leq a \leq b [/tex] that [tex] a \leq b [/tex] and [tex] -a \leq b [/tex]? The sign does not matter?
     
  5. Oct 6, 2008 #4

    Dick

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    That's definitely ok. That's what -b<=a<=b MEANS. I'm talking about saying |a|<=b implies -a<=b and a<=b.
     
  6. Oct 6, 2008 #5
    So if we have [tex] -b \leq a \leq b [/tex] and [tex] a = -5 [/tex] then we can say that [tex] -a \leq b [/tex]?

    Similarly, if [tex] a = 5 [/tex] then [tex] -a \leq b [/tex]?

    It doesn't matter what the sign is?

    Thanks
     
  7. Oct 6, 2008 #6

    Dick

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    I'm saying I think you should PROVE |a|<=b implies -a<=b and a<=b by splitting it into cases.
     
  8. Oct 6, 2008 #7
    but if you already know that then post #5 doesn't depend on the sign of [tex] a [/tex]?

    thanks
     
  9. Oct 6, 2008 #8

    Dick

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    If you already know that |a|<=b implies -a<=b and a<=b then this whole thread should have been over at post 2. I already said the proof is fine in that case.
     
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