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Homework Help: Simplify set math problem

  1. Jan 11, 2006 #1
    I have no formal training in set theory, and I need to simplify the following:

    [tex](A\cup B - A\cap B)\cup C - (A\cup B - A\cap B)\cap C[/tex]

    Preferably, it should somehow end up as:

    [tex]A\cup (B\cup C - B\cap C) - A\cap (B\cup C - B\cap C)[/tex]
     
  2. jcsd
  3. Jan 11, 2006 #2

    StatusX

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    The union of A and B is the set of points in A or B or both, so subtracting off the part in both A and B leaves the part in either A or B. Call this set D. Then you do the same thing with D and C. So you want the points that are either in C or D (not both), which means they are either in A, or B, or C, but not more than one of these. Does that help at all?
     
  4. Jan 11, 2006 #3

    AKG

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    I don't see how the second thing is a simplification of the first, but the two are the same. You could use:

    (A + B + C) - (AB + BC + AC)

    where + is union, and juxtaposition is intersection.
     
  5. Jan 11, 2006 #4

    CarlB

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    This is basically the XOR relation. That is,

    (A xor B) xor C = A xor (B xor C) = A xor B xor C,

    well familiar to those of us who design integrated circuits.

    Carl
     
  6. Jan 12, 2006 #5
    What are the basic properties of unions and intersections? For example, what is

    [tex](A-B)\cap C[/tex]

    I know how unions and intersections interact with one another, but what about the above case?
     
    Last edited: Jan 12, 2006
  7. Jan 12, 2006 #6

    StatusX

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    We can pretend we are given a base space X containing all the sets, and then we can define a set X-B, the complement of B in X, and then this becomes:

    [tex] (A-B) \cap C = (A \cap (X-B))\cap C = A \cap (X-B)\cap C[/tex]

    Since intersection is associative, and so you can do what you want with this:

    [tex](A-B) \cap C = (C-B) \cap A = (A\cap C) -B [/tex]
     
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