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Math Help - Vin diagrams? is there any other way..

  1. #1
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    Vin diagrams? is there any other way..

    Let a & b be sets:

    A u (B-A)U(C-A)=(B u C)-A

    How do I solve?

    The other ones I have are:
    (A n B) E A
    A E (A U B)
    A-B E A
    A n (B - A) 0
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  2. #2
    Super Member Gamma's Avatar
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    be careful

    your first one is not even true.
    A=(0,1)
    B=(1,2)
    C=(2,3)

    Your LHS yields (0,1)\cup (1,2) \cup (2,3)
    but the RHS is (1,2) \cup (2,3)

    you should learn a few TeX codes so set theory makes sense.

    \cup is union
    \cap is intersection
    \subset is subset
    \supset goes the other way
    \in is element of
    \emptyset is the nullset

    I have no idea what your E is.

    Maybe type your question again correctly so we can help?
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  3. #3
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    [tex] \cup [/tex] gives  \cup is union

    [tex] \cap [/tex] gives  \cap is intersection

    [tex] \subseteq [/tex] gives  \subseteq is subset

    [tex] \in [/tex] gives  \in is element

    [tex] \emptyset [/tex] gives  \emptyset is emptyset
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  4. #4
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    Quote Originally Posted by Gamma View Post
    I have no idea what your E is.
    Unfortunately I have learned to speak some ASCII... E is apparently often equivalent to \in; although, rarely, it is sometimes seen outside of its usual habitat as being equivalent to \subset or \subseteq, which I suspect is what's meant in this case.

    Sorry for the mixed metaphor, there.
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  5. #5
    Super Member Gamma's Avatar
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    Quote Originally Posted by orendacl View Post
    Let a & b be sets:

    A u (B-A)U(C-A)=(B u C)-A

    How do I solve?

    The other ones I have are:
    1)(A n B) E A
    2)A E (A U B)
    3)A-B E A
    4)A n (B - A) 0

    Okay so I will show you how to do these without vinn diagrams. I will take Alephzero's suggestion and take E as
    \subset
    1)Let x\in A \cap B then x is by definition in both A and B, therefore x is in A, so x\in A \Rightarrow A \cap B \subset A as desired

    2) Let x\in A, then x is certainly in A or B since it is in A. Thus x\in A \cup B \Rightarrow A \subset (A\cup B) as desired

    3) Let x\in A-B. Then x is in A, but not in B. But this means x\in A so A-B \subset A as desired.

    4) I think this is supposed to say A \cap (B-A)=\emptyset

    So you show both containments.

    Clearly A \cap (B-A)\supset \emptyset.

    So let x\in A \cap (B-A). Then x\in A and x\in B-A. But this means x is in A, and x is in B, but not in A. So clearly no x can both be in A and not be in A, so this is precisely the empty set, so we have the desired A \cap (B-A)=\emptyset


    QED.

    It might be worth at least noting what happens if you cannot chose an x in the set, like for instance if you have the case where like A-B=\emptyset or something, then you just have the containment being trivially true.

    As for the first one, you are gonna have to get the correct statement before I can show you how to prove it because as it stands it is not true as my counterexample demonstrates.
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