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Math Help - Set Algebra

  1. #1
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    Question Set Algebra

    Hi using set algebra I have to simplify the following:

    ( (A n B n C) n (A u B u C') u (A u B') )

    the answer I have been given is A n (B' n C) but I get A u B'

    can anybody verify and break down the steps used?

    Many thanks
    Last edited by barc0de; January 3rd 2009 at 07:39 AM. Reason: clarity
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  2. #2
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    Quote Originally Posted by barc0de View Post
    Hi using set algebra I have to simplify the following:

    (A n B n C) n (A u B u C') u (A u B')

    the answer I have been given is A n (B' n C) but I get A u B'

    can anybody verify and break down the steps used?

    Many thanks
    What is B'?
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  3. #3
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    Quote Originally Posted by Mush View Post
    What is B'?
    B' is the complement of B
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  4. #4
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    Quote Originally Posted by barc0de View Post
    I have to simplify the following: (A u B u C) n (A u (B n C))
    Here is a start:
    \left[ {A \cup B \cup C} \right] \cap \left[ {A \cup \left( {B \cap C} \right)} \right] = A \cup \left[ {\left( {B \cup C} \right) \cap \left( {B \cap C} \right)} \right]
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  5. #5
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    Hello, barc0de!

    Simplify: ., (A \cup B \cup C) \cap [A \cup (B \cap C)]

    . . . \begin{array}{cccc}\bigg[A \cup B \cup C\bigg] \cap \bigg[A \cup (B \cap C)\bigg] & &\text{Given} \\ \bigg[ A \cup B \cup C \bigg] \cap \bigg[ (A \cup B) \cap (A \cup C)\bigg] & & \text{Distributive} \\ \\[-4mm] \bigg[(A \cup B \cup C) \cap (A \cup B)\bigg] \cap \bigg[A \cup C\bigg] && \text{Associative} \\ \bigg[A \cup B\bigg] \cap \bigg[A \cup C\bigg] && \text{Subset} \\ \\[-4mm]  A \cup (B \cap C) && \text{Distributive}\end{array}

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  6. #6
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    You need to edit the post by adding some grouping symbols.
    Because \left( {X \cap Y} \right) \cup Z \ne X \cap \left( {Y \cup Z} \right).
    Which one is your problem?
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  7. #7
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    Your edit did nothing to clear away the confusion.
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  8. #8
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    Quote Originally Posted by Plato View Post
    Your edit did nothing to clear away the confusion.
    Sorry I'm not sure what you mean.

    I have to algebraically simplify:

    (A n B n C) n (A u B u C') u (A u B')

    that is all I have
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  9. #9
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    Thanks Soroban. Could you take a look at the following please:

    (A n B n C) n (A u B u C') u (A u B')

    the answer I have been given is A n (B' n C) but I get A u B'

    can you verify and break down the steps used
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  10. #10
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    Which way are they grouped?
    \begin{gathered}<br />
  \left[ {\left( {A \cap B \cap C} \right) \cap \left( {A \cup B \cup C'} \right)} \right] \cup \left( {A \cup B'} \right) \hfill \\<br />
  \left( {A \cap B \cap C} \right) \cap \left[ {\left( {A \cup B \cup C'} \right) \cup \left( {A \cup B'} \right)} \right] \hfill \\ <br />
\end{gathered} <br />
    It makes a huge difference.
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  11. #11
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    This may be where the problem lies in getting 2 different answers. I have typed the problem in as I have been given it. Would it be too much trouble for you to do both?? I would really appreciate it.

    Many thanks
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  12. #12
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    Quote Originally Posted by barc0de View Post
    This may be where the problem lies in getting 2 different answers. I have typed the problem in as I have been given it. Would it be too much trouble for you to do both?? I would really appreciate it.
    \begin{gathered}<br />
  \left[ {\left( {A \cap B \cap C} \right) \cap \left( {A \cup B \cup C'} \right)} \right] \cup \left( {A \cup B'} \right) = A \cup B' \hfill \\<br />
  \left( {A \cap B \cap C} \right) \cap \left[ {\left( {A \cup B \cup C'} \right) \cup \left( {A \cup B'} \right)} \right] = \left( {A \cap B \cap C} \right) \hfill \\ <br />
\end{gathered}
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  13. #13
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    Could you please break down the steps used in finding the first solution?
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