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Math Help - Symmetric Difference Proof

  1. #1
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    Symmetric Difference Proof

    We know that

     A \triangle B = (A\backslash B) \cup (B \backslash A)

    and we also know that

     A \triangle B = (A \cup B) \backslash (A \cap B)

    show that

     (A\backslash B) \cup (B \backslash A) = (A \cup B) \backslash (A \cap B)

    using only the following laws: DeMorgan's, Complement, Distributive, Absorption... basically the main set theory laws.

    I have attempted to do this but I'm running into some trouble trying to figure out how to move from the union of the two to the difference of the two.
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  2. #2
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    [(a-b) u b] - [(a-b) u a] ......
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  3. #3
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    not sure how you came to this..

    This is my first experience with set theory so I might be a little slow
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  4. #4
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    Here's most of the work:

    (A\cap B')\cup (B\cap A') = [(A\cap B')\cup B]\cap [(A\cap B']\cup A']=[(A\cup B)\cap (B'\cup B)]\cap[(A\cup A')\cap (B' \cup A')]

    See if you can finish it.
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  5. #5
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    Or you can write (A\cup B)- (A\cap B)=(A\cup B)\cap (A\cap B)^c=(A\cup B)\cap (A^c\cup B^c)=...
    Last edited by dwsmith; January 6th 2011 at 01:09 PM. Reason: replaced C with B
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  6. #6
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    Reply to dwsmith:

    If you're in ZF set theory, set complements aren't allowed. In naive set theory, they are. Just a side note.

    [EDIT]: Spelling of "complements", as per dwsmith's post below. What a nice set!
    Last edited by Ackbeet; January 6th 2011 at 12:47 PM. Reason: Spelling.
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  7. #7
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    so how do I start from the first step of (A-B) union (B-A)?
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  8. #8
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    complements
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  9. #9
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    Quote Originally Posted by Ackbeet View Post
    Reply to dwsmith:

    If you're in ZF set theory, set complements aren't allowed. In naive set theory, they are. Just a side note.

    [EDIT]: Spelling of "complements", as per dwsmith's post below. What a nice set!
    I have to get something right every now and then.
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  10. #10
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    the only complement laws I know are  A \cup (B \backslash A) = A \cup B
    and  A \cap (B \backslash A) = 0
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  11. #11
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    A\setminus B is usually defined as A\cap B' (I'm using ' for complement)

    It literally translates to "the elements in A and not in B"

    The first step in my argument above is just this translation.
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  12. #12
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    Quote Originally Posted by gutnedawg View Post
    the only complement laws I know are  A \cup (B \backslash A) = A \cup B
    and  A \cap (B \backslash A) = 0
    \emptyset \ \mbox{or} \ \{\} not = 0
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  13. #13
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    I understand this however this is not the notation that we are using. I was instructed that the absorption laws would prove to be very useful
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  14. #14
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    Quote Originally Posted by dwsmith View Post
    \emptyset \ \mbox{or} \ \{\} not = 0
    Yea I know I just didn't feel like looking up how to write the null set
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  15. #15
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    (A\cup B)- (A\cap B)=(A\cup B)\cap (A\cap B)^c=(A\cup B)\cap (A^c\cup B^c)=...

    Use distributive law here.

    What will you get?
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