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Set theory homework - Theoretic reasoning

  1. Jul 14, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove where X and Y are both sets use theoretic reasoning

    i) Z \ (X [itex]\cap[/itex] Y) = (Z \ X) [itex]\cup[/itex] (Z \ Y)
    ii)(Y \ X) [itex]\cup[/itex] Z = (Y [itex]\cup[/itex] Z) \ (X \ Z)
    iii) Z \ (Y \ X) = (X [itex]\cap[/itex]Z) [itex]\cup[/itex](Z \ Y)

    2. Relevant equations
    \ = set difference



    3. The attempt at a solution
    i know you dont do other peoples work for them but i am very clueless on this subject a guiding hand with step by step help would be greatly appreciated
     
  2. jcsd
  3. Jul 15, 2011 #2

    gb7nash

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    What is "theoretic reasoning"?
     
  4. Jul 15, 2011 #3

    HallsofIvy

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    If you mean just regular "proofs", each of those is basically to prove "A= B" where A and B are sets. And the standard way to do that is to prove "[itex]A\subset B[/itex]" and then "[itex]B\subset A[/itex]". Finally, the way to prove [itex]A\subset B[/itex] is to start by saying "let x be a member of A" and prove, using the definitions of A and B, that x must be a member of B.

    For example, you want to prove, in (i), that
    [tex]Z\(A\cap B)= (Z\A)\cup (Z\B)[/tex]
    so you start by proving
    [tex]Z\(A\cap B)\subset (Z\A)\cup (Z\B)[/tex]

    And you do that by saying "let x be in [itex]Z\(A\cap B)[/itex]" which means, of course, that x is in Z but not in [itex]A\cap B[/itex]. Saying that x is not in [itex]A\cap B[/itex] means x is not in both A and B. So there are two possibilities- x is in Z and A but not in b or x is in Z and B but not in A. Do those two possibilities as two cases:
    1) x is in Z and A but not in B. Then, since x is in Z but not in B, x is in Z/B. Do you see why that means x must be in [itex](Z/A)\cup(Z/B)[/itex]?
    2) x is in Z and B but not in A. Then, since x is in Z but not in A, ...
     
  5. Jul 15, 2011 #4
    ok i understand 1)

    heres my attempt at 2)
    (Y \ X) ∪ Z = (Y ∪ Z) \ (X \ Z)

    let x be in (Y \ X) ∪ Z
    -x is in Y and Z but not in X, so x is in X\Z
    also (Y ∪ Z) \ X, because x is in Y and Z but not in X

    hence (Y \ X) ∪ Z = (Y ∪ Z) \ (X \ Z)

    THIS OK?
     
  6. Jul 16, 2011 #5

    HallsofIvy

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    No, but it is a start. That shows that
    [tex](Y\X)\cup Z\subset (Y\cup Z)\(X\Z)[/tex]
    You still need to show the other way.
     
  7. Jul 16, 2011 #6
    what do you mean when you say show the other way?

    is the 'other way' this?
    -x is in X and Z but not in Y, so x is in Y\Z

    so overall
    (Y \ X) ∪ Z = (Y ∪ Z) \ (X \ Z)

    let x be in (Y \ X) ∪ Z
    -x is in Y and Z but not in X, so x is in X\Z
    -x is in X and Z but not in Y, so x is in Y\Z

    also (Y ∪ Z) \ X, because x is in Y and Z but not in X

    hence (Y \ X) ∪ Z = (Y ∪ Z) \ (X \ Z)
     
  8. Jul 16, 2011 #7

    gb7nash

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    A = B if and only if A ⊂ B and B ⊂ A.

    You showed that (Y \ X) ∪ Z ⊂ (Y ∪ Z) \ (X \ Z). You still need to show that (Y ∪ Z) \ (X \ Z) ⊂ (Y \ X) ∪ Z.
     
  9. Jul 17, 2011 #8
    ok lol kinda obvious isnt it

    ok heres a revised version of 2
    (Y \ X) ∪ Z = (Y ∪ Z) \ (X \ Z)

    let x be in (Y \ X) ∪ Z
    -x is in Y and Z but not in X, so x is in X\Z
    -x is in Y and Z but not in X, so (Y ∪ Z) \ X

    let x be in (Y ∪ Z) \ (X \ Z)
    -x is in Y and Z, so x is in Y∪Z
    -x is in Z but not in X, so x is in A\C
    -x is in Y and Z but not in X, so x is in (Y∪Z)\X

    hence (Y \ X) ∪ Z = (Y ∪ Z) \ (X \ Z)

    good?
     
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