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Discrete Math - question about sets

  1. Nov 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Use set builder notation to give a description of each of these sets.

    a) { 0,3,6,9,12 }

    b) { -3, -2, -1,0, 1, 2, 3 }

    c) { m,n,o,p }








    3. The attempt at a solution

    X={x l x is an odd possitive multiplier of 3 less than 12 }



    X is supposed to be the set. Can I just name it randomly? Also, can I say it like this? Is that ok? Not really sure about b and c.
     
  2. jcsd
  3. Nov 2, 2009 #2
    You can name your set what you like.

    For b, notice the elements are integers from -3 to 3; that is, x[tex]\in[/tex]Z AND -3 [tex]\leq[/tex] x [tex]\leq[/tex] 3

    For a, x does not have to be odd since 6 and 12 are in the set. You are right that they are positive multiples of 3 and less than 12 but can you describe that in mathematical notation?
     
  4. Nov 2, 2009 #3

    Oh yeah.. nevermind. They are not all odd.

    And no, I don't know. I'd just simply say : X= { x[tex]\in[/tex]Z l x is positive x*3 less than 13 }
     
  5. Nov 2, 2009 #4
    If you denote 3Z as the set multiples of three, then the set in (a) consists of elements x [tex]\in[/tex]3Z such that 0 [tex]\leq[/tex] x [tex]\leq[/tex] 12
     
  6. Nov 2, 2009 #5
    Alternatively, if you've seen quantifiers before you can write the set as

    [tex] \left\{0,3,6,9,12\right\} = \left\{ x \in \mathbb{Z} \mid \exists y \in \mathbb{Z} \left(x=3y\right), 0 \leq x \leq 12 \right\}[/tex].

    The first predicate essentially says that x is included in the set if and only if there exists an integer y such that x is three times y.

    For example, the number 4 would not be included in the set because there is no integer that satisfies [tex]4=3y[/tex]. If you haven't seen quantifiers yet, then nevermind. :smile:
     
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