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Math Help - [SOLVED] Finding the union and intersection of indexed sets.

  1. #16
    Member ilikedmath's Avatar
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    Quote Originally Posted by Plato View Post
    Here is a strong hint on the intersection in #4.
    \left( {\forall x \in (0,1)} \right)\left[ { - \frac{1}<br />
{x} <  - 1 < 1 < \frac{1}<br />
{x}} \right]<br />
    So based on that interval, it looks like (-1, 1) is the intersection.
    Did I get it? or not
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  2. #17
    Member ilikedmath's Avatar
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    Exclamation Revised Answers

    Here are my revised answers after all the help I got:


    (1)
    = (-1, 1)
    = [0, 1)

    --------------------------

    (2)
    = Q
    =
    --------------------------

    (3)



    --------------------------
    (4)
    = (- \infty, \infty)
    = (-1,1)
    --------------------------

    (5)
    = [0,1)
    = {0}

    --------------------------

    (6)



    -----

    (Do they all look correct now? ) This is for a boardwork in class assignment (we present one of these in front of the class ).
    We don't know which problem we'll present until just minutes before we present, so I really wanted to make sure I got them all okay so I can do well in the boardwork presentation in class.
    Thanks!
    Last edited by ilikedmath; October 14th 2008 at 07:59 PM. Reason: Forgot to type in the infinity symbol
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  3. #18
    Member ilikedmath's Avatar
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    Question

    Quote Originally Posted by vincisonfire View Post
    The notation R\Q means all real numbers EXCEPT rational. You thus know that z in never rational.
    You know that for set An, z is smaller than n that is a natural number.
    Eg A10 contains all real number smaller than 10 that are not rational.
    I would say that union contains R\Q and the intersection should be the empty set.
    does this refer to #5 or #6? They both have 'z' but #5 has the A_n and #6 has A_z
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  4. #19
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    The union in #5 ought to be \left( {0,1} \right) \cap \left( {\mathbb{R}\backslash \mathbb{Q}} \right) = \left\{ {x \in :\left( {\mathbb{R}\backslash \mathbb{Q}} \right):0 < x < 1} \right\}.
    Do you see that in #5, \left( {\forall n \in \mathbb{N}} \right)\left[ {A_n  \subseteq \left( {\mathbb{R}\backslash \mathbb{Q}} \right)} \right]?
    Thus, the intersection is \emptyset.

    On the other hand, in #6, \left( {\forall z \in \left( {\mathbb{R}\backslash \mathbb{Q}} \right)} \right)\left[ {A_z  \subseteq \mathbb{N}} \right].
    Thus you need to redo your answers.
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  5. #20
    Member ilikedmath's Avatar
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    Question Rethinking

    Quote Originally Posted by Plato View Post
    The union in #5 ought to be \left( {0,1} \right) \cap \left( {\mathbb{R}\backslash \mathbb{Q}} \right) = \left\{ {x \in :\left( {\mathbb{R}\backslash \mathbb{Q}} \right):0 < x < 1} \right\}.
    Do you see that in #5, \left( {\forall n \in \mathbb{N}} \right)\left[ {A_n \subseteq \left( {\mathbb{R}\backslash \mathbb{Q}} \right)} \right]?
    Thus, the intersection is \emptyset.

    On the other hand, in #6, \left( {\forall z \in \left( {\mathbb{R}\backslash \mathbb{Q}} \right)} \right)\left[ {A_z \subseteq \mathbb{N}} \right].
    Thus you need to redo your answers.
    So I guess vince's answer was referring to #5 then. So are all the rest correct? (besides my #5 and 6).

    For #6, the notation is throwing me off. I'm still confused how to read it. I see that z is any irrational number and the sets in the A_z will be all irrationals less than all the naturals.
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  6. #21
    Member ilikedmath's Avatar
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    Thanks, everyone, for all your help. I ended up getting #2 which I was very pleased with getting. I think I did well. We find out grades a week from today. Thanks again. I will not mark this thread as SOLVED.
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