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Show that if a < b + ε for every ε>0 then a ≤ b

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that if a < b + ε for every ε>0 then a ≤ b


    2. Relevant equations

    I am not sure if this is a right way to do it? I just want to know if it does make sense

    3. The attempt at a solution
    proof.
    a < b + ε → if a is bounded above by b+ε then b is the least upper bound for a.
    which means a ≤ b.
    ε is the upper bound of a since b≤ε.
     
  2. jcsd
  3. Oct 10, 2011 #2

    Dick

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    Re: proof.

    No, that doesn't work. Try formulating it as a proof by contradiction.
     
  4. Oct 11, 2011 #3
    Re: proof.

    ok! so i tried it this way...

    Suppose a > b+ε for every ε>0,then a≤b
    let ε= 1/2a-1/2b since a-b>ε→→→a-b >1/2a-1/2b (true)

    so a> b+ 1/2a-1/2b
    1/2a>1/2b
    a>b which contradict from the first agreement ∅
     
  5. Oct 11, 2011 #4

    gb7nash

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    Re: proof.

    Unfortunately, no. To do a proof by contradiction, you need to assume the hypothesis and the negation of the conclusion.
     
  6. Oct 11, 2011 #5
    Re: proof.

    ok..another try...i won't stop until i get this ...

    suppose a>b and let ε = 1/2 (a-b) ....(i think you know where i get that)

    so a < b+ε → a< b+(1/2a-1/2b)
    then a-1/2 < b-1/2 → 1/2a<1/2b→ a<b CONTRADICTION ∅
     
  7. Oct 11, 2011 #6
    Re: proof.

    I think thats is the answer...what do you think
     
  8. Oct 11, 2011 #7

    gb7nash

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    Re: proof.

    It looks fine, just fixed a minor typo on your part.
     
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