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Why is my proof of this set identity incorrect?

  1. Feb 7, 2014 #1
    1. The problem statement, all variables and given/known data

    jaLL28y.png

    2. Relevant equations



    3. The attempt at a solution

    $$A-(A\cap B)=A-B\\ A\cap (A\cap B)^{ C }=A\cap B^{ C }\quad (set\quad difference\quad law)\\ A\cup [A\cap (A\cap B)^{ C }]=A\cup [A\cap B^{ C }]\quad (applied\quad A\cup \quad to\quad both\quad sides)\\ A=A\quad (absorption \quad law)$$
     
    Last edited: Feb 7, 2014
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  3. Feb 7, 2014 #2

    LCKurtz

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    Among other things, it looks like you have started with what you wanted to prove and drew the conclusion that A = A. I'm guessing that the problem wasn't to prove that A = A.
     
  4. Feb 7, 2014 #3

    LCKurtz

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    Also, lets take an example where A and B are any two subsets if U. Here's a proof that A = B, using your argument:

    ##A = B##. Union U with both sides: ##A\cup U = B\cup U##. Therefore ##U = U##. End of proof.
     
  5. Feb 7, 2014 #4
    Well, I can just reverse the steps to fix that.

    Hmm... How come in algebra I can do operations on both sides but I can't do that in set algebra?

    For example in algebra, I can do this:

    4 = 4
    4 + 5 = 4 + 5
    9 = 9

    But I can't seem to do that with set algebra?
     
  6. Feb 7, 2014 #5

    LCKurtz

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    That's right. You can't because it doesn't work that way. And even in ordinary algebra, you don't start with something you are trying to prove by assuming it is true. Then you would get arguments like this: To prove 3 = 5 start with 3 = 5. Multiply both sides by 0 and get 0 = 0. That doesn't prove 3 = 5.

    In your current problem you could start with the left side and use your rules on it. You already have ##A - (A\cap B) = A \cap (A\cap B)^c##. Just keep going with it using the rules until you get an expression identical to the right side of the equation.
     
  7. Feb 7, 2014 #6
    I was told that you can't do that because there's no way to reverse that step, i.e. it is impossible to go from 0=0 to 3=5.

    Alright. The textbook already has the answer since it's an example but I just figured I'd try to solve it like I do algebraic proofs. Is there a specific property that regular algebra has that set algebra is missing?
     
  8. Feb 7, 2014 #7

    LCKurtz

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    When you work backwards like that, you need reversible steps alright. In my argument:

    ## A = B##, ##A\cup U = B\cup U##, ##U = U##, you can't reverse the steps because given that ##A\cup U = B\cup U## that doesn't imply ##A=B##. Your argument has a similar error.
     
  9. Feb 7, 2014 #8
    I see. Are there any operations that I can apply in set algebra that are reversible?
     
  10. Feb 7, 2014 #9

    LCKurtz

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    To add another point to this discussion: In algebra if you have ##a+c = b+c## you can conclude ##a=b##, and if you have ##ac = bc## you can conclude ##a=b## as long as ##c \ne 0##. You don't have similar equations for sets. That is, if you have ##A\cup C = B\cup C## you can't conclude ##A=B## and if you have ##A\cap C = B\cap C## you can't conclude ##A=B##. So cupping or capping something with both sides of an equation isn't going to give you a reversible argument.
     
    Last edited: Feb 7, 2014
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