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Sets and their Cardinality?

  1. Aug 4, 2010 #1
    Hey guys, this is my first post, (Hi) was just wondering if i could get your help. I'm studying for my repeats and you guys can save me.

    If X = {1,2,3,4}, Y = {2,4,6} what is the cardinality of the following sets?

    (i) A = {x|x mod 2 = 0 and 0 <=x<=20}
    (ii) B = X * X * Y
    (iii) C = {(x,y)|x ≠ y and x,y ∈ X}

    Please explain your train of thought in solving this. I am trying hard to understand the right way to approach this question quickly, Thank you for your time guys...
     
    Last edited: Aug 4, 2010
  2. jcsd
  3. Aug 4, 2010 #2
    I've tried them out, with the following answers:-

    (i) A = {2,4,6,8,10,12,14,16,18,20}, Therefore A has a cardinality of 10 (elements).
    (ii) B = Cartesian Product of 'X times X times Y' or better yet '4 by 4 by 3' elements each to give a total of 48 in cardinality?!
    (iii) C = UNSOLVED!!!

    I wanna make sure someone agrees with me having the right answers since you're the pros

    I really wanna know what the '|' symbol stands for or means, as in 'x|x'. Hard to specifically search for in a book.
     
  4. Aug 4, 2010 #3
    Figured it out. I will post the full answer for future questioneers

    (i) A = {2,4,6,8,10,12,14,16,18,20}, Therefore A has a cardinality of 10 (elements).
    (ii) B = Cartesian Product of 'X times X times Y' or better yet '4 by 4 by 3' elements each to give a total of 48 in cardinality?!

    (iii) C = {(x,y)|x ≠ y and x,y ∈ X}

    pairs x,y {such as (1,1),(1,2),etc...} drawn from set X with a cardinality of '4 by 4 = 16' as in the question "x,y ∈ X". Due to the statement 'x ≠ y' pairs can't come in equals, discarding the following four sets (1,1),(2,2),(3,3),(4,4). The end product is 16-4 giving a cardinality of 12 for set 'C'.
     
  5. Aug 4, 2010 #4
    How much is 0 (mod 2)? In the set A, the possible values of x include 0, don't they? Anyway, I'm sure you remember that any number that is 0(mod2) is even and vice-versa.
     
  6. Aug 5, 2010 #5
    Yes i do, and thanks for pointing that out in any case, it's always the little things that count >_<
     
  7. Aug 5, 2010 #6
    C has the cardinality|X*X| - |x|.

    Because for every x in X, there is a pair (x,x), which are exactly the ones not in {(x,y) : x != y /\ x,y in X}
     
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