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Natural numbers, integars

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Let a,b [tex]\in[/tex]R with a < b. and let n [tex]\in[/tex]N where n(b-a) > 1.

    a) How do you know that such an n must exist?
    b) Show that there exists m [tex]\in[/tex]Z where a < m/n < b
    c) Show that there exists some irrational c where a < c < b (Hint:rational + irrational = irrational.)

    2. Relevant equations

    see above btw N is for natural numbers, Z for integers and R for real numbers

    3. The attempt at a solution

    a) Since n(b-a) > 1 , n > 1/(b-a) and since b does not equal a there is an n which exists OR can we say that nb > na and the n's cancel out to give us the condition given in the question.

    b) we already know that n is natural and N[tex]\subset[/tex]Q and its safe to assume that Z[tex]\subset[/tex]R is always true but how do I exactly show that m lies in Z.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Feb 22, 2009
  2. jcsd
  3. Feb 22, 2009 #2

    Mark44

    Staff: Mentor

    For a) you can't start off with
    You have to show that such an n exists to make that true.

    For b, your task is not to show that "m lies in Z." You have to show that there exists an integer m such that m/n is between a and b.
     
  4. Feb 22, 2009 #3
    so for part a) has something to do with the archimedean property?
     
    Last edited: Feb 22, 2009
  5. Feb 22, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, Mark44's point is that you cannot start by asserting that n(b-a)> 1. What you can do is start with your second statement: since b-a> 0, 1/(b-a) is a positive real number and, by the Archimedean property, ...
    For b) Show that there exists m Z where a < m/n < b
    note that since n(b-a)= nb- na> 1, there must exist an integer between nb and na.
     
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