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Math Help - Set Theory problem

  1. #1
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    Set Theory problem

    Hey, can anyone please help me with this problem:

    For all sets A, B, C, prove
    If A-C = \emptyset then (A \cup B) - (B \cap C) = (A-B) \cup (B-C)

    This is what I have done so far, but I seem to be stuck:

    (A \cup B) - (B \cap C) Thm A-B=A \cap B'
    (A \cup B) \cap (B \cap C)' DeMorgans
    (A \cup B) \cap (B' \cup C')

    I can't figure it out from there. I'm guessing it might have something to do with distribution but i'm not sure. Any help is appreciated
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  2. #2
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    \left( {A \cup B} \right) \cap \left( {B' \cup C'} \right) = \left( {A \cap B'} \right) \cup \left[ {\left( {A \cap C'} \right) \cup \left( {B \cap C'} \right)} \right]<br />
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  3. #3
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    Hello, Foink!

    You're right . . . the Distributive Property will finish it.


    For all sets A, B, C, prove:
    If A-C = \emptyset then (A \cup B) - (B \cap C) = (A-B) \cup (B-C)

    This is what I have done so far:

    \begin{array}{cc}(A \cup B) - (B \cap C) & \text{Given} \\<br />
(A \cup B) \cap (B \cap C)' & \text{d{e}f. subtraction} \\<br />
(A \cup B) \cap (B' \cup C\,') & \text{DeMorgan} \end{array}

    . . . Good!

    We know: . \begin{array}{cc}S \cap S' \:=\:\emptyset & {\color{blue}[1]} \\ S \cup \emptyset \:=\:S & {\color{blue}[2]}\end{array}

    We are also given: . A - C \:=\:\emptyset\quad\Rightarrow\quad A \cap C\,' \:=\:\emptyset\;\;{\color{blue}[3]}


    \begin{array}{cc}<br />
[(A \cup B) \cap B'] \cup [(A \cup B) \cap C\,'] & \text{Distr.} \\ \\<br /> <br />
[(A \cap B') \cup (B \cap B')] \cup [(A \cap C\,') \cup (B \cap C\,')] & \text{Distr.} \\ \\<br /> <br />
[(A \cap B') \cup \emptyset\:] \cup[\:\emptyset \cup (B \cap C\,')] & [1], [3] \\ \\<br /> <br />
(A \cap B') \cup (B \cap C\,') & [2] \\ \\<br /> <br />
(A - B) \cup (B - C) & \text{d{e}f. subtraction}<br />
\end{array}

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  4. #4
    CPR
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    What is B' or C' in your problem?

    Just wondering if the B' meant elements not in set B or elements not in set C.
    Please let me know
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CPR View Post
    Just wondering if the B' meant elements not in set B or elements not in set C.
    Please let me know
    correct. B' and C' means B-compliment and C-compliment respectively
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  6. #6
    CPR
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    Question on A' U B

    A= { 2, 4, 5}
    B = {1,2,4,5,6}

    so what will be A' U B

    and let say that there was an U={ 1,2, 3, 4, 5, 6,7}

    What would B'?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CPR View Post
    A= { 2, 4, 5}
    B = {1,2,4,5,6}

    so what will be A' U B

    and let say that there was an U={ 1,2, 3, 4, 5, 6,7}

    What would B'?
    given that U, A' = {1, 3, 6, 7} and B' = {3,7}

    you can find A' U B from that
    Last edited by Jhevon; February 26th 2008 at 05:53 PM.
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  8. #8
    CPR
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    Thanks

    It would be {1,6}.
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CPR View Post
    It would be {1,6}.
    yes
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