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(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:

    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'
  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}

  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}
    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'.
  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


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    Staff Emeritus
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    A Venn diagram is not a rigorous mathematical proof.
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