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Math Help - Set

  1. #1
    Member SengNee's Avatar
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    Set

    For two sets, P, Q, the difference P-Q is defined to be the set consisting of those elements of P that are not in Q. Prove, with the aid of de Morgan's laws, that
    a) A - (B U C) = (A - B) ∩ (A - C)
    b) A - (B ∩ C) = (A - B) U (A - C)
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  2. #2
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    Here is b.
    \begin{array}{ll}<br />
 A\backslash \left( {B \cap C} \right)  & = A \cap \left( {B \cap C} \right)^c  \\ <br />
  &  = A \cap \left( {B^c  \cup C^c } \right) \\ <br />
  &  = \left( {A \cap B^c } \right) \cup \left( {A \cap C^c } \right) \\ <br />
  &  = \left( {A\backslash B} \right) \cup \left( {A\backslash C} \right) \\ <br />
 \end{array}
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  3. #3
    Member SengNee's Avatar
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    Quote Originally Posted by Plato View Post
    Here is b.
    \begin{array}{ll}<br />
 A\backslash \left( {B \cap C} \right)  & = A \cap \left( {B \cap C} \right)^c  \\ <br />
  &  = A \cap \left( {B^c  \cup C^c } \right) \\ <br />
  &  = \left( {A \cap B^c } \right) \cup \left( {A \cap C^c } \right) \\ <br />
  &  = \left( {A\backslash B} \right) \cup \left( {A\backslash C} \right) \\ <br />
 \end{array}
    What is the symbol, "\" means?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by SengNee View Post
    What is the symbol, "\" means?
    it's the symbol for set subtraction. it's used the same as the minus signs you wrote in your original post

    (a) is very similar, you should be able to do it now

    by the way, Plato utilized DeMorgan's law in the second line of his "proof", the first line was by definition
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  5. #5
    Member SengNee's Avatar
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    a)
    A - (B U C)
    = A ∩ (B U C)'
    = A ∩ (B' ∩ C')


    I only can do until this step...
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by SengNee View Post
    a)
    A - (B U C)
    = A ∩ (B U C)'
    = A ∩ (B' ∩ C')


    I only can do until this step...
    just expand.

    so, A \cap (B' \cap C') = (A \cap B') \cap (A \cap C')

    now what?
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  7. #7
    Member SengNee's Avatar
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    Quote Originally Posted by Jhevon View Post
    just expand.

    so, A \cap (B' \cap C') = (A \cap B') \cap (A \cap C')

    now what?

    That means,

    Distributive law not only for,

    \begin {array}{ll}<br />
A \cap ( B \cup C) = (A \cap B) \cup (A \cap C)   \\   A \cup ( B \cap C) = (A \cup B) \cap (A \cup C)   \\   \end {array}

    But also for,
    \begin {array}{ll}<br />
A \cap ( B \cap C) = (A \cap B) \cap (A \cap C)   \\   A \cup ( B \cup C) = (A \cup B) \cup (A \cup C)   \\   \end {array}


    Right?
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by SengNee View Post
    That means,

    Distributive law not only for,

    \begin {array}{ll}<br />
A \cap ( B \cup C) = (A \cap B) \cup (A \cap C)   \\   A \cup ( B \cap C) = (A \cup B) \cap (A \cup C)   \\   \end {array}

    But also for,
    \begin {array}{ll}<br />
A \cap ( B \cap C) = (A \cap B) \cap (A \cap C)   \\   A \cup ( B \cup C) = (A \cup B) \cup (A \cup C)   \\   \end {array}


    Right?
    yes
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