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Homework Help: (A intersect B)' proof

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data
    I need to prove or disprove:
    (AintersectB)'=A'intersectB'



    2. Relevant equations



    3. The attempt at a solution
    Let x[tex]\in[/tex](A[tex]\cap[/tex]B)'
    Then x[tex]\notin[/tex]A[tex]\cap[/tex]B
    x[tex]\notin[/tex]A or x[tex]\notin[/tex]B
    Then x[tex]\in[/tex]A' or x[tex]\in[/tex]B'
    x[tex]\in[/tex]A'[tex]\cup[/tex]B'
     
  2. jcsd
  3. Nov 16, 2008 #2
    x is not an element of A intersection B means that x is not a common element of A and B. So you want to look at how you went from line 2 to line 3. You might want to try some examples to see what is happening, like A = (0,1), B = [0,1] (as a subset of the reals).
     
  4. Nov 16, 2008 #3
    Ok I typed line 2 and 3 wrong
    Then x[tex]\notin[/tex]A[tex]\cup[/tex]B
    x[tex]\notin[/tex]A or x[tex]\notin[/tex]B
     
  5. Nov 16, 2008 #4
    Let A={1,2,3}
    B={3,4} universe={1,2,3,4,5,6}
    (A intersect B)={3}
    (A intersect B)'={1,2,4,5,6}

    A'={4,5,6}
    B'={1,2,5,6}
    A'intersectB'={5,6}
     
  6. Nov 16, 2008 #5
    You might want to check your 'ors' and 'ands'. If you have the union of (1,2,3) U (4,5,6), then given any x, that would mean that x is in (1,2,3) or (4,5,6), right? What about for intersections? Your proof is almost right..
     
  7. Nov 16, 2008 #6
    So, x[tex]\notin[/tex]A[tex]\cap[/tex]B.
    This implies x[tex]\notin[/tex]{3}
    So,x[tex]\in[/tex]{1,2,4,5,6}
    Would this mean x[tex]\in[/tex]A' or x[tex]\in[/tex]B'?
     
  8. Nov 16, 2008 #7
    Ok so x[tex]\notin[/tex]A and x[tex]\notin[/tex]B.
    Then x[tex]\in[/tex]A' and x[tex]\in[/tex]B'.
    x[tex]\in[/tex]A'[tex]\cup[/tex]B'
     
  9. Nov 16, 2008 #8
    well, no, you had it right using the 'or's. I just wasn't sure if moving from line 2 to line 3 you understood what you were doing or if you were formulating it to make your answer right.
     
  10. Nov 16, 2008 #9
    Ok, then did I go wrong somewhere in my proof? Should I have said x[tex]\notin[/tex]A and x[tex]\notin[/tex]B. Then if I went on from there, I could get a right conclusion?
     
  11. Nov 17, 2008 #10
    If you dont care what method..You can use VENN DIAGRAMS!
     
  12. Nov 17, 2008 #11
    This counterexample should tell you that the statement [tex](A \cap B)' = A' \cap B' [/tex] is not true!
     
    Last edited: Nov 17, 2008
  13. Nov 17, 2008 #12

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    A Venn diagram is not a rigorous mathematical proof.
     
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