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Uniqueness of value of Riemann Integral(proof)

  1. May 16, 2013 #1
    The proof is in the document.

    I highlighted the main points that I am questioning in the document.

    I am questioning the fact that A = B...
    (The following is in the document)
    |A-B|=ε where they define the value of ε to be a positive arbitrary real number (ε>0).

    And for A = B that means ε must be equal to zero however ε>0.

    My Reasoning
    For A≥B we have
    A - B = ε
    And for B≥A we have
    A - B = -ε

    And after solving the system of two eq. A - B = ε & A - B = -ε
    we get


    A - B = 0
    ∴ A = B.
    Is this how they proved A is the same as B?
     

    Attached Files:

  2. jcsd
  3. May 16, 2013 #2
    This is a standard move in analysis. If you want to show two things are equal, then you show that their difference is less than epsilon for an arbitrary epsilon. Since you can get the difference as close as you like to zero, it might as well be zero. Have you learned about things like supremum and infimum of a set?
     
  4. May 16, 2013 #3
    Can you explain this mathematically? It doesn't make sense to me how you can say two things are equal just because they differ by an arbitrary value ε where we choose ε to → 0.
    You can say two things are approximate though.

    And I have not taken an analysis course.
     
    Last edited: May 16, 2013
  5. May 16, 2013 #4
    Deleted comment
     
    Last edited: May 16, 2013
  6. May 16, 2013 #5

    Office_Shredder

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    This is not what they have. If |A-B| = ε and ε was nonzero, then A and B would not be equal. What they have is |A-B|≤ ε.
     
  7. May 16, 2013 #6
    Thank you. It makes sense now.
     
  8. May 16, 2013 #7

    HallsofIvy

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    I don't like the wording "it might as well be 0". That makes it sound like we can choose whether to have this number 0 or not. If, for any [itex]\epsilon> 0[/itex], [itex]z< \epsilon[/itex], z cannot be positive since then we could take [itex]\epsilon[/itex] (which can be any positive number) to be z itself and conclude that "z< z"/

     
  9. May 17, 2013 #8
    Good call. I concur.
     
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