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Cardinality of an interval of rationals

  1. Sep 26, 2010 #1
    For any two rational numbers q1<q2, card ((q1,q2) cap Q) = card Q, right?
    How to prove it, if it's true?
     
    Last edited: Sep 26, 2010
  2. jcsd
  3. Sep 26, 2010 #2

    HallsofIvy

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    Please define your terms. The cardinality of the set of all rationals between any two given numbers, whether rational or irrational is the same as the cardinality of the rationals, yes. The cardinality of interval (q1, q2) where q1 and q2 are rational numbers is the cardinality of R because that is an interval of real numbers.

    I don't know what the "I" in "(q1, q2)-I". The set of integers? So that "(q1, q2)- I" is the set of all numbers that are NOT integers between q1 and q2? If so then the answer is as before: If (q1, q2)- I is the set of all non-integer rational numbers between q1 and q2, the cardinality of that set is the cardinality of Q. If (q1, q2)- I is the set of all non-integer real numbers between q1 and q2, the cardinality of that set is the cardinality of R, the set of all real numbers. That is because the set of integers between any two given rational numbers is finite and so does not affect the cardinality of any infinite set.
     
  4. Sep 26, 2010 #3
    I have edited the original post. Please see.
    ((q1, q2) cap Q) is the set of all non-integer rational numbers between q1 and q2. By intuition I know it has the same cardinality as Q, yet I don't know how to build a 1-1 correspondence between it and Q, please help.
     
  5. Sep 26, 2010 #4
    Okay, I think I got it.

    Let m, n be any positive integers, any positive element q of Q can be written as [tex]m/n[/tex], and any negative element q of Q can be written as [tex]-m/n[/tex].

    Let's define function f as follows:
    [tex]f(q) = \frac{q_{1}+q_{2}}{2} + \frac{q_{2}-q_{1}}{2}\cdot\frac{m}{m+n} (q>0)[/tex]
    [tex]f(q) = \frac{q_{1}+q_{2}}{2} - \frac{q_{2}-q_{1}}{2}\cdot\frac{m}{m+n} (q<0)[/tex]
    [tex]f(q)=0 (q=0)[/tex]
     
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