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Intersection of Rationals and (0 to Infinity)?

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Let A = [Q[itex]\bigcap[/itex](0,[itex]\infty[/itex])] [itex]\bigcup[/itex] {-1} [itex]\bigcup[/itex](-3, -2]


    2. Relevant equations
    So A = (0,[itex]\infty[/itex]) [itex]\bigcup[/itex]{-1} [itex]\bigcup[/itex](-3,-2]


    3. The attempt at a solution
    I understand that the Rational numbers are cardinally equivalent to (0,[itex]\infty[/itex]),

    but why isn't the intersection of Rationals and (0,[itex]\infty[/itex]) =>(0,[itex]\infty[/itex])\Irrationals ?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 13, 2012 #2
    Quite the contrary, the Irrationals are cardinally equivalent to (0,[itex]\infty[/itex])
     
  4. Mar 13, 2012 #3

    HallsofIvy

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    The set of all rational numbers is countable, unlike [itex][0, \infty)[/itex].
     
  5. Mar 14, 2012 #4
    Then back to my original question why is the intersection of rationals and (0,∞) = (0,∞)

    in other words, why don't irrationals come out of this intersection?
     
  6. Mar 14, 2012 #5
    If x belongs to the intersection of A and B, then x belongs to A and x belongs to B. The intersection of rationals and (0,∞) is the set of numbers which are both rationals and positive real numbers.
     
  7. Mar 14, 2012 #6

    HallsofIvy

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    It isn't. The intersection of the set of all rational numbers with all positive real numbers is all positive rational numbers.
     
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