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Math Help - New to discrete and in need of help!! Set question.

  1. #1
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    Question New to discrete and in need of help!! Set question.

    Hello everyone,
    I am really new to discrete and totally dont get it!!

    I have a question that I have no idea how to get started or really approach it.
    It is a Set theory question. I really appreciate it if someone can give me the step by step procedure of how to solve it and explanations so I can use it as a learning guide.

    THANK YOU!!!


    If the statement is true, give a proof. If it is false then write its negation and prove it.

    Assume all sets are subsets of a universal set U.

    For all sets A, B,and C, A-(B-C) = (A-B)-C
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  2. #2
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    Suppose that you want to prove two sets A and B to be equal. There are two main ways to do that:
    1) Take one of them, say A, and performing algebraic transformations try to turn it into B.
    2) Prove that and . To do this, first let and try to show that (), and then let and try to show that ().
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  3. #3
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    Quote Originally Posted by Kitizhi View Post
    Hello everyone,
    I am really new to discrete and totally dont get it!!

    I have a question that I have no idea how to get started or really approach it.
    It is a Set theory question. I really appreciate it if someone can give me the step by step procedure of how to solve it and explanations so I can use it as a learning guide.

    THANK YOU!!!


    If the statement is true, give a proof. If it is false then write its negation and prove it.

    Assume all sets are subsets of a universal set U.

    For all sets A, B,and C, A-(B-C) = (A-B)-C
    Take A={1,2,3,4},B={1,2,3},C={3}.

    Now A-(B-C)= {1,2,3,4}-[ {1,2,3}-{3}]= {1,2,3,4}-{1,2}={3,4}

    But (A-B)-C = [{1,2,3,4}-{1,2,3}]-{3}= {4}-{3} = {4}

    Hence A-(B-C)\neq (A-B)-C
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  4. #4
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    Awesome guys, you've both been a great help.
    However I got another question..

    How would I prove this statement is true?
    The previous one wasnt too bad cause it was false and I would just provide a counter-example, but this one is true.

    For all sets A, B and C, A x (B-C) = (A x B) - (A X C)

    Would I assume the negation is true and provide a counter example to prove the negation is false and conclude that the original statement is true?
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  5. #5
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    Quote Originally Posted by Kitizhi View Post
    Awesome guys, you've both been a great help.
    However I got another question..

    How would I prove this statement is true?
    The previous one wasnt too bad cause it was false and I would just provide a counter-example, but this one is true.

    For all sets A, B and C, A x (B-C) = (A x B) - (A X C)

    Would I assume the negation is true and provide a counter example to prove the negation is false and conclude that the original statement is true?
    No because like that is like proving the above by using only one example .We want to give a general proof

    Anyway if the above is true why not try to prove it >

    So let xε[Ax(B-C)] THAT implies xεA AND xε(B-C)====> xεA AND xεB AND ~xεC(= x does not belong to C) and that implies:

    (xεA AND xεB)AND(~xεC v ~xεA)====> xε(AxB) AND ~xε(ΑxC) ===>..........................BY using de morgan..........

    ====>xε[(AxB)-(AxC)] ====> Ax(B-C)\subseteq (AxB)-(AxC)


    You try the converse
    Last edited by benes; February 25th 2009 at 07:19 PM. Reason: WRONG SENTENCE
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  6. #6
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    Quote Originally Posted by benes View Post
    No because like that is like proving the above by using only one example .We want to give a general proof

    Anyway if the above is true why not try to prove it >

    So let xε[Ax(B-C)] THAT implies xεA AND xε(B-C)====> xεA AND xεB AND ~xεC(= x does not belong to C) and that implies:

    (xεA AND xεB)AND(~xεC v ~xεA)====> xε(AxB) AND ~xε(ΑxC) ===>..........................BY using de morgan..........

    ====>xε[(AxB)-(AxC)] ====> Ax(B-C)\subseteq (AxB)-(AxC)


    You try the converse

    Ahh perfect thanks alot. However to complete the proof I have to do the other way around where (AxB)-(AxC) = Ax(B-C) as well right?
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  7. #7
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    Quote Originally Posted by Kitizhi View Post
    Ahh perfect thanks alot. However to complete the proof I have to do the other way around where (AxB)-(AxC) = Ax(B-C) as well right?
    O.K

    x\in[(AxB)-(AxC)]\Longrightarrow[(x\in A\wedge x\in B)\wedge(\neg x\in A\vee\neg x\in C)]\Longrightarrow [(x\in A\wedge x\in B)\wedge(x\in A\rightarrow \neg x\in C)] \Longrightarrow[x\in A\wedge x\in B\wedge\neg x\in C]\Longrightarrow[x\in A\wedge(x\in B\wedge\neg x\in C)]\Longrightarrow x\in[Ax(B-C)]

    Hence  [(AxB)-(AxC)]\subseteq Ax(B-C)
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  8. #8
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    Thanks alot!!
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