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Show a function is unbounded

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Let f be the function defined f(x)=1/x. Prove that f is not bounded on (0,1)


    2. Relevant equations



    3. The attempt at a solution

    I think I should prove by contradiction.

    Assume f is bounded on (0,1).
    Since f is bounded, there exists a real number M such that |f(x)| ≤ M for all x in (0,1)
    f(x) will never be negative since it is on the interval (0,1), hence |f(x)| = f(x)

    This is where I begin to get unclear on where to go next. I want to show that M+1 ≤ M
    Is it correct to use 1/(M+1) and plug it into f(x)?
     
  2. jcsd
  3. Mar 9, 2012 #2

    Dick

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    I don't think you should prove it by contradiction. If n is an number greater than one then 1/n is in (0,1).
     
    Last edited: Mar 9, 2012
  4. Mar 9, 2012 #3
    Can I argue that since 1/n is an infinite sequence, then this function is not bounded?
     
  5. Mar 9, 2012 #4

    Dick

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    You need a better argument than that. What is f(1/n)?
     
  6. Mar 9, 2012 #5
    f(1/n)=n

    Then I could say for all n in the positive integers?
     
  7. Mar 9, 2012 #6

    Dick

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    You could say that, but it doesn't prove f is unbounded until you say why that proves f is unbounded.
     
  8. Mar 9, 2012 #7
    Since f(1/n)=n for all n in N. Since N has an infinite amount of elements, then the function is unbounded on (0,1)?
     
  9. Mar 9, 2012 #8

    Dick

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    Having an infinite number of elements has little to do with being unbounded. What does unbounded mean?
     
  10. Mar 9, 2012 #9
    That there is no lower bound, no upper bound or both.
     
  11. Mar 9, 2012 #10

    Dick

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    Ok, so give me an argument that f has no upper bound.
     
  12. Mar 9, 2012 #11
    There is no n such that 1/n is not in the interval (0,1), so there is no real number M that will satisfy |1/n|≤M.
     
  13. Mar 9, 2012 #12

    Dick

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    You don't want to satisfy |1/n|<=M. You want to show that you can find a number in x in (0,1) such that f(x)>M.
     
  14. Mar 9, 2012 #13
    So if M is greater than one, 1/(M+1) is in (0,1) and M+1 < M is not true?
     
  15. Mar 9, 2012 #14

    Dick

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    I really hope you meant f(1/(M+1))=M+1 > M.
     
  16. Mar 9, 2012 #15
    No, I was still thinking about the contradiction argument. Sorry.
     
  17. Mar 9, 2012 #16

    Dick

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    That's ok. But you've got it now, yes?
     
  18. Mar 9, 2012 #17
    Yep. Thank you again for your help!
     
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