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AunionB=AunionC, B=C

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    I need to decide whether the theorem is correct and decide where the proof fails.
    AunionB=AunionC, B=C

    2. Relevant equations

    3. The attempt at a solution
    We will prove by contradiction. Suppose that A[tex]\cup[/tex]B and A[tex]\cup[/tex]C are not equal. Then there is some object x that is in one and not the other. We proceed by looking at 2 cases:
    First look at the case where x[tex]\in[/tex]A[tex]\cup[/tex]B and x[tex]\notin[/tex]A[tex]\cup[/tex]C. Then x[tex]\notin[/tex]A. So x[tex]\in[/tex]B. Also x[tex]\notin[/tex]C. Therefore x is in B and not in C, which contradicts the condition B=C.

    I thought the theorme was correct, but I can't find where the proof goes wrong.
  2. jcsd
  3. Nov 17, 2008 #2


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    Staff Emeritus
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    That's NOT the way proof by contradiction works! To prove "if X then Y" by contradiction, you assume Y is not true. Here your theorem is "if [tex]A\cup B= A\cup C[/tex] then B= C. Proof by contradiction would start "suppose B is not equal to C".

  4. Nov 17, 2008 #3
    Oh yeah I forgot about that.
  5. Nov 18, 2008 #4
    I think it's not true. You take A={1,2,3}. B={1,0}, C={2,0}. Then B not = C but A union B= A union C
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