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Math Help - Proof By Contradiction

  1. #1
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    Proof By Contradiction

    Hello,

    Can someone please help me with this proof?

    (A-C)∩(C-B) = 0.

    Do I start off by....

    suppose: ~[(A-C)∩(C-B) = 0.] ?

    Also, does this (0 ∩ B') equal the empty set?

    Thanks...

    PS. ' indicates the complement.
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  2. #2
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    Why not start by the mere definition? Suppose x\in\left(A-C\right)\cap\left(C-B\right)\Longrightarrow x\in A-C\,\,\,and\,\,\,also\,\,\,x\in C-B ...from here the way to get a contradiction is straightforward and thus no such element can exist...

    Tonio
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  3. #3
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    Thats exactly what I did, i'm not sure if I did it correctly though.

    This is what I got :

    (A ∩ 0) ∩ B'

    (0 ∩ B')

    0

    is that correct?
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  4. #4
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    Quote Originally Posted by l flipboi l View Post
    Thats exactly what I did, i'm not sure if I did it correctly though.

    This is what I got :

    (A ∩ 0) ∩ B'

    (0 ∩ B')

    0

    is that correct?

    I've no idea what you had in mind, even assuming that 0=\emptyset...how did you come up with the above? Besides this, X\cap \emptyset =\emptyset for any set X, so from

    the first line you get the empty set...

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    I've no idea what you had in mind, even assuming that 0=\emptyset...how did you come up with the above? Besides this, X\cap \emptyset =\emptyset for any set X, so from

    the first line you get the empty set...

    Tonio
    I excluded alot of the work, but I used the commutative and associative property to get c and c' to cancel out.

    which brought me to my answer
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  6. #6
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    Quote Originally Posted by l flipboi l View Post
    I excluded alot of the work, but I used the commutative and associative property to get c and c' to cancel out.

    which brought me to my answer

    Good, but then why did you say you "did exactly the same" as I did?!!

    Tonio
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  7. #7
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    Thanks for the help! I meant, I set it up like how you did.
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