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Math Help - Power Set question.

  1. #1
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    Power Set question.

    Hello all =)

    i am strugling (?) here with some old homework...

    lets see...

    A=P({a,c,d,e,g}) , B=P({c,d,e,f,g}) . ( P(X) is the power set of X) .

    Calculate |A U B| .

    My thoughts are:

    A= 2^5= 32
    B= 2^5= 32

    |A U B| i believe that, it should be for example a new set that contains {a,c,d,e,f,g}
    so thats 2^6=64

    Right?

    If its right, is this a good way to type as an answer?

    Thanks for your time, and assistance =)
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  2. #2
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    Quote Originally Posted by primeimplicant View Post
    Hello all =)

    i am strugling (?) here with some old homework...

    lets see...

    A=P({a,c,d,e,g}) , B=P({c,d,e,f,g}) . ( P(X) is the power set of X) .

    Calculate |A U B| .

    My thoughts are:

    A= 2^5= 32
    B= 2^5= 32

    |A U B| i believe that, it should be for example a new set that contains {a,c,d,e,f,g}
    so thats 2^6=64

    Right?

    If its right, is this a good way to type as an answer?

    Thanks for your time, and assistance =)
    You will need to rethink your answer because A U B does not contain {a,c,d,e,f,g}.

    I recommend considering P({a,c,d,e,f,g}) and removing those sets that are supersets of {a,f}.
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  3. #3
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    \left| {A \cup B} \right| = \left| A \right| + \left| B \right| - \left| {A \cap B} \right|.

    But is that 2^5+2^5-2^4?
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  4. #4
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    Quote Originally Posted by Plato View Post
    \left| {A \cup B} \right| = \left| A \right| + \left| B \right| - \left| {A \cap B} \right|.

    But is that 2^5+2^5-2^4?
    I'm not sure if your question was directed at me but both our methods yield the same answer.

    2^5+2^5-2^4 = 2^6 - 2^4

    Edit: I misread something so the wording of this post came out kind of funny, carry on..
    Last edited by undefined; September 3rd 2010 at 01:01 PM.
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  5. #5
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    Quote Originally Posted by undefined View Post
    I'm not sure if your question was directed at me but both our methods yield the same answer.
    Actually no, was just commenting on the usual formula.

    \displaystyle |P(J)\cup P(K)|=2^{|J|}+2^{|K|}-2^{|J\cap K|}}.
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  6. #6
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    hah,

    thanks a lot , now it makes sense !

    thanks again.
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  7. #7
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    Quote Originally Posted by primeimplicant View Post
    hah,

    thanks a lot , now it makes sense !

    thanks again.
    Well in case you didn't understand my first post and would like to: A U B contains precisely those elements of P({a,c,d,e,f,g}) that do not contain {a,f}, since no element of A contains {a,f} and no element of B contains {a,f}. (The first "contains" means set membership and the other "contain(s)" mean subset.)
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