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Math Help - Commutative Ring

  1. #1
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    Commutative Ring

    Let I be any set and let R be the collection of all subsets of I. Define addition and multiplication of subsets A,B ⊆ I as follows:
    A+B = (A∪B) ∩ [A∩B(with a line over this)] and AB = A∩B. Show that R is a commutative ring under this addition and multiplication.

    I know that to be a commutative ring it has to have the following:
    closure, associativity, commutativity, distribution, identity and inverses.

    Closure - show well-defined
    Assoc - (a+b)+c = a+(b+c) and a(bc) = (ab)c
    Comm - a+b = b+a and ab = ba
    Dist - a(b+c) = ab+ac and (a+b)c = ac+bc


    I am not sure how to write them out using the defined subsets above.

    Please help.
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  2. #2
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    So, A + B = (A\cup B)\cap\overline{(A\cap B)}. When talking about set operations, this is called the symmetric difference and is commonly denoted by A\vartriangle B.

    Closure - show well-defined
    Do you have difficulties with this?

    Assoc - (a+b)+c = a+(b+c) and a(bc) = (ab)c
    The associativity of intersection is pretty obvious. For addition, it's a little messy. Hint: prove that x\in (a+b)+c iff x belongs to exactly one of a, b, c or to all three, and similarly for x\in a+(b+c).

    Comm - a+b = b+a and ab = ba
    Follows directly from the definition and commutativity of union and intersection.

    Dist - a(b+c) = ab+ac and (a+b)c = ac+bc
    Draw a Venn diagram of a, b, c and then prove that x\in a(b+c) iff x\in ab+ac.
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  3. #3
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    Any help with identity or inverses?
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  4. #4
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    Quote Originally Posted by page929 View Post
    Any help with identity or inverses?
    What is A\Delta A=~?

    What is A\Delta \emptyset=~?
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  5. #5
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    Quote Originally Posted by Plato View Post
    What is A\Delta A=~?

    What is A\Delta \emptyset=~?
    emptyset
    A complement
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  6. #6
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    The best way to view this operation is A\Delta B=(A\cap\overline{B})\cup(B\cap\overline{A}).
    That is the same as (A\cup B)\cap (\overline{A\cap B}).

    Now A\Delta \emptyset  = \left( {A \cap \overline \emptyset  } \right) \cup \left( {\emptyset  \cap \overline A } \right) = A
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