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Help with (a-b)'=b'-a'

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove or disprove:


    2. Relevant equations

    3. The attempt at a solution
    Let x[tex]\in[/tex](A-B)'
    Then x[tex]\notin[/tex](A-B)
    I'm not sure where to go from here...
  2. jcsd
  3. Nov 16, 2008 #2


    Staff: Mentor

    Re: (a-b)'=b'-a'

    What's the context here? What are A and B? What does (A - B)' mean?
  4. Nov 16, 2008 #3
    Re: (a-b)'=b'-a'

    A-B is the set of elements in A that are not in B
    So x is not in A-B means that x is in A but is not in B

    You may want to try to think of some counterexamples before trying to show inclusion both ways.
  5. Nov 16, 2008 #4
    Re: (a-b)'=b'-a'

    Ok so:
    x[tex]\in[/tex]A and x[tex]\notin[/tex]B

    Ok, so suppose A={1,2,3,4,5} B={3,4,6}
    Then A-B={1,2,5}
    So, (A-B)'={3,4,6}
    so, (A-B)'=B
  6. Nov 16, 2008 #5
    Re: (a-b)'=b'-a'

    If you are working in R, then the complement of the set A-B would be R - {1, 2, 5}
    You might want to try some simpler examples like A= (0,1) or {1, 2, 3} and B = [0,1] or {3, 4}
  7. Nov 16, 2008 #6
    Re: (a-b)'=b'-a'

    universe ={1,2,3,4,5,6,7}
  8. Nov 16, 2008 #7
    Re: (a-b)'=b'-a'

  9. Nov 16, 2008 #8
    Re: (a-b)'=b'-a'

    So, if x is not an element of A-B, then x is not an element of {1,2}
  10. Nov 16, 2008 #9
    Re: (a-b)'=b'-a'

    Ok, so

    So x[tex]\notin[/tex]A and x[tex]\in[/tex]B
  11. Nov 17, 2008 #10
    Re: (a-b)'=b'-a'

    Keep going! What are A' , B' and B'-A' ?
  12. Nov 17, 2008 #11
    Re: (a-b)'=b'-a'

  13. Dec 2, 2008 #12
    Re: (a-b)'=b'-a'

    Still not quite sure
    Let x[tex]\in[/tex](A-B)'
    Can I say now x[tex]\in[/tex]B? this is the part that confuses me...
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