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Math Help - basic question on sets

  1. #1
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    basic question on sets

    How come (A' n B') n [(A u C) n (B' n C')] = (A' n B') n (A u C) n (B' n C') ?

    Is it by the commutative laws ? THese brackets and parantheses are really confusing me .

    Isnt that by distributive

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

    and it doesn't equal to (A' n B') n (A u C) n (B' n C') from the above .

    I really thank you for your time and effort to explain this to me .
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by thereddevils View Post
    How come (A' n B') n [(A u C) n (B' n C')] = (A' n B') n (A u C) n (B' n C') ?

    Is it by the commutative laws ? THese brackets and parantheses are really confusing me .

    Isnt that by distributive

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

    and it doesn't equal to (A' n B') n (A u C) n (B' n C') from the above .

    I really thank you for your time and effort to explain this to me .
    Its because the Intersection operator is associative:

    A\cap\left(B\cap C\right)=A\cap B\cap C=\left(A\cap B\right)\cap C

    You only distribute in cases like A\cap\left(B\cup C\right)
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Its because the Intersection operator is associative:

    A\cap\left(B\cap C\right)=A\cap B\cap C=\left(A\cap B\right)\cap C

    You only distribute in cases like A\cap\left(B\cup C\right)

    Oh thanks a lot !

    So A n (B n C) doesnt equal to (A n B) n (A n C) as distributive laws don work on cases like this ??
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by thereddevils View Post
    Oh thanks a lot !

    So A n (B n C) doesnt equal to (A n B) n (A n C) as distributive laws don work on cases like this ??
    Well it works, but A\cap\left(B\cap C\right)=\left(A\cap B\right)\cap\left(A\cap C\right)=A\cap B\cap A\cap C=A\cap A\cap B\cap C and A\cap A=A so you get \left(A\cap B\right)\cap \left(A\cap C\right)=A\cap B\cap C.

    Its just extra work.
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