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Real Analysis Proof

  1. Sep 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that if a>0, there exists n in N such that 1/n < a < n.

    2. Relevant equations



    3. The attempt at a solution

    I am starting with a>0 and trying to manipulate, algebraically, to get n > a > 1/n.

    From a > 0 I can add 1 to both sides to obtain, a+1 > 1. Then I can choose some n such that:

    n > a+1 > 1.

    From there, divide by n to achieve the desired result for one part of the inequality:

    1> (a+1)/n > 1/n. We know from above that n > 1, so I can write:

    n > (a+1)/n > 1/n.


    From here I'm not sure what to do.
     
  2. jcsd
  3. Sep 11, 2012 #2

    jbunniii

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    Try proving the two inequalities separately:

    There exists an integer n in N such that a < n

    There exists an integer m in N such that 1/m < a

    Then think about how to combine these results.
     
  4. Sep 11, 2012 #3

    1. By the Archimedean Property, a < n for some n in N.

    2. Again, by the Archimedean Property, 1 < am, for some m in N. So, 1/m < a.

    Now I have: 1/m < a < n.

    I don't see how I can justify that m=n.
     
  5. Sep 11, 2012 #4

    jbunniii

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    OK, there exists an integer n in N such that a < n.

    If you make n larger, is the inequality still true?

    Similarly, there exists an integer m in N such that 1/m < a.

    Does this remain true if you make m larger?
     
  6. Sep 11, 2012 #5
    yes, the inequalities both hold.

    So could I just say, "We can make n and m large to arbitrary values such that m=n." Would that finish the proof?
     
  7. Sep 11, 2012 #6

    jbunniii

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    Well, the question asks for a specific integer that satisfies both inequalities.

    So if n satisfies the first inequality, and m satisfies the second one, what integer will satisfy both?
     
  8. Sep 11, 2012 #7
    so nm?
     
  9. Sep 11, 2012 #8
    The bigger of the two :D
     
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