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Set Theory Question

  1. Dec 2, 2004 #1
    Hello all

    Set X has x elements and Sset Y has y elements and Set Z consists of all elements are are in either set X or set Y with the exception of the k common elements. I know that the answer is

    x+ y - 2k, however how would I get this? Should I just use a practical example?
     
    Last edited: Dec 2, 2004
  2. jcsd
  3. Dec 2, 2004 #2

    matt grime

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    No, proof by example isn't a proof.

    X sym diff Y = (X \ Y)u(Y\X)

    the union is of disjoint sets so the card of the lhs is the sum of the cards on the rhs. now just find card (X\Y) hint X\Y = X\(XnY)
     
  4. Dec 2, 2004 #3

    HallsofIvy

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    You are told that there are k elements in both X and Y. That means there are x-k elements in x that are NOT in y and y- k elements that are in Y but NOT in X. Z will contain (x- k)+ (y- k)= x+ y- 2k elements.
     
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