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Homework Help: Please check this

  1. Mar 25, 2005 #1
    Verify [itex]A\cap(B-C)=(A\cap B)-(A\cap C)[/itex].

    This is how I did it:

    From [itex]x\in[A\cap(B-C)][/itex] (lhs) we have that
    1.) [itex]x\in A[/itex]
    2.) [itex]x\in B[/itex]
    3.) [itex]x\notin C[/itex]
    From #1 and #2 we have that
    4.) [itex]x\in(A\cap B)[/itex]
    Now, since [itex]x\in(A\cap C)[/itex] means that [itex]x\in A[/itex] and [itex]x\in C[/itex], but we have [itex]x\notin C[/itex], it follows that
    5.) [itex]x\notin(A\cap C)[/itex]
    Remembering that [itex]x\in(A-B)[/itex] means that [itex]x\in A[/itex] and [itex]x\notin B[/itex] it follows from #4 and #5 that
    [itex]x\in[(A\cap B)-(A\cap C)][/itex] (rhs)

    Is this right?

    - Kamataat
     
  2. jcsd
  3. Mar 25, 2005 #2
    Yes, but it is only half of the proof. You have shown

    [tex](A \cap B) - (A \cap C) \supseteq A \cap (B-C)[/tex]

    To finish the proof you must also go in "reverse" and show

    [tex](A \cap B) - (A \cap C) \subseteq A \cap (B-C)[/tex]
     
  4. Mar 25, 2005 #3
    I see. My mistake was to assume the thing I was trying to prove... and I forgot, that [itex]A=B[/itex] iff [itex]A\subseteq B\wedge B\subseteq A[/itex].

    I'll post the other half in the morning. Got to sleep now. Thanks!

    - Kamataat
     
  5. Mar 26, 2005 #4
    Is this it?:

    From [itex]x\in[(A\cap B)-(A\cap C)][/itex] (rhs) we have
    1.) [itex]x\in(A\cap B)[/itex]
    2.) [itex]x\notin(A\cap C)[/itex]
    From #1 it follows that
    3.) [itex]x\in A[/itex] and [itex]x\in B[/itex]
    From #2 and #3 it follows that
    4.) [itex]x\in A[/itex] and [itex]x\notin C[/itex] (not the other way around because #3 shows that [itex]x\in A[/itex])
    Now we have
    5.) [itex]x\in A[/itex] and [itex]x\in B[/itex] and [itex]x\notin C[/itex]
    From #5 it follows that
    6.) [itex]x\in A[/itex] and [itex]x\in(B-C)[/itex] from which we get
    7.) [itex]x\in[A\cap(B-C)][/itex] (lhs)

    The one question I have is about #5-->#6. As far as I understand it, #5 could also result in "[itex]x\in B[/itex] and [itex]x\in(A-C)[/itex]", or could it not because of steps #3 and #4?

    - Kamataat
     
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