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Math Help - Set Questions

  1. #1
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    Set Questions

    Help me solve the question please? They make me headache.

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    i. (A' \cup B') \cap (A' \cup B) \cap (A \cup B)=A' \cap B

    ii. (A \cup B') \cap (A \cup B)=A

    iii. (A \cup B' \cup C') \cap [A \cup (B \cap C)]=A

    iv. A \cup (A' \cup B)'=A

    2. A certain sixth form science class consists of 90 pupils, all of whom take mathematics, 45 study chemistry, 42 study biology, and 52 study physics. Of those taking three subjects, 18 study physics and chemistry, 22 study physics and biology, and 16 study chemistry and biology while 5 take all four subjects. Find the number of pupils who study mathematics only.
    Answer: 2

    Note: I'd like a Venn Diagram and the formula for question 2.
    Last edited by cloud5; June 18th 2009 at 04:33 AM. Reason: Additional Questions
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  2. #2
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    Quote Originally Posted by cloud5 View Post
    Help me solve the question please? They make me headache.

    1. Show
    i. (A' \cup B') \cap (A' \cup B) \cap (A \cup B)=A' \cap B

    (A' u B') n [( A' n A) u B ]

    =( A' u B' ) n ( nullset u B )

    =(A' u B' ) n B

    = (A' n B ) u ( B' n B )

    = ( A' n B ) u null set

    = you got ur prove .
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  3. #3
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    Quote Originally Posted by cloud5 View Post
    Help me solve the question please? They make me headache.

    1. Show

    ii. (A \cup B') \cap (A \cup B)=A

    iii. (A \cup B' \cup C') \cap [A \cup (B \cap C)]=A
    (2) A u ( B' n B ) = A u null set - distributive

    (3) A u [( B' u C' ) n (B n C )] - distributive

    = A u [( B n C )' n ( B n C )] -- de morgans

    = A u null set

    = A

    (4) This is not clear . Its A u (A' u B)' or [A u (A' u B)]'
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  4. #4
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    Quote Originally Posted by mathaddict View Post
    (2) A u ( B' n B ) = A u null set - distributive

    (4) This is not clear . Its A u (A' u B)' or [A u (A' u B)]'
    Opss... made a mistake there... it is

    A \cup (A' \cup B)'=A
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  5. #5
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    Quote Originally Posted by cloud5 View Post
    Opss... made a mistake there... it is

    A \cup (A' \cup B)'=A

    ( A n universal set ) u ( A n B' ) = A n ( universal set u B )

    = A n universal set

    =A
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  6. #6
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    Hello, cloud5!

    I'll do the first two.

    We need a couple of Theorems:

    . . \begin{array}{cc}<br />
[1] & P \cap P' \:=\:\emptyset \\ \\[-3mm]<br />
[2] & P \cup \emptyset \:=\:P \end{array}



    (ii)\;\;(A \cup B') \cap (A \cup B) \;=\;A

    (A \cup B') \cap (A \cup B)

    . . \begin{array}{ccc}= & A \cup (B' \cap B) & \text{Distr.} \\ \\<br />
= &A \cup \emptyset & [1] \\ \\<br />
=& A & [2] \end{array}




    (i)\;\;(A' \cup B') \cap (A' \cup B) \cap (A \cup B) \:=\:A' \cap B

    (A' \cup B') \cap (A' \cup B) \cap (A \cup B)

    . . \begin{array}{cccc}= & \bigg[A' \cup (B' \cap B)\bigg] \cap (A \cup B) & \text{Distr.} \\ \\<br />
= & \bigg[A' \cup \emptyset\bigg] \cap (A \cup B) & [1] \\ \\<br />
= & A' \cap (A \cup B) & [2] \\ \\<br />
= & (A' \cap A) \cup (A' \cap B) & \text{Distr.} \\ \\<br />
= & \emptyset \cup (A' \cap B) & [1] \end{array}

    . . \begin{array}{cccc}<br />
= & \qquad\qquad A' \cap B & \qquad\qquad\quad\; [2]<br />
\end{array}

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