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  1. #16
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shilz222 View Post
    I was asked to prove: $\displaystyle (A-B) \cap C = \emptyset \Leftrightarrow A \cap C \subseteq B $

    So there are two directions.

    In one direction ($\displaystyle \Leftarrow $) we are given: $\displaystyle A \cap C \subseteq B $ and $\displaystyle (A-B) \cap C \neq \emptyset $ for proof by contradiction.

    In the other direction ($\displaystyle \Rightarrow $) we are given: $\displaystyle (A-B) \cap C = \emptyset $ and then do what you said in your earlier post.
    I apologize if I am misinterpreting something. All of my comments are directed toward your $\displaystyle \implies$ proof by contradition. I have not addressed any comments to the proof in the other direction.

    -Dan
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  2. #17
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    Quote Originally Posted by shilz222 View Post
    I was asked to prove: $\displaystyle (A-B) \cap C = \emptyset \Leftrightarrow A \cap C \subseteq B $ So there are two directions.
    YES, YES, YES! That is correct.

    Quote Originally Posted by shilz222 View Post
    In one direction ($\displaystyle \Leftarrow $) we are given: $\displaystyle A \cap C \subseteq B $ and $\displaystyle (A-B) \cap C \neq \emptyset $
    NO, NO, NO This not correct.

    One direction: Given $\displaystyle \left( {A \cap C} \right) \subseteq B$ prove that $\displaystyle \left( {A - B} \right) \cap C = \emptyset $.

    The other direction: Given $\displaystyle \left( {A - B} \right) \cap C = \emptyset $ prove that $\displaystyle \left( {A \cap C} \right) \subseteq B$.
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  3. #18
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    Why is it not correct?

    We assume that $\displaystyle (A-B) \cap C \neq \emptyset $ for proof by contradiction.

    One direction: Given that $\displaystyle (A \cap C) \subseteq B $ and $\displaystyle (A-B) \cap C \neq \emptyset $ use proof by contradiction to show that $\displaystyle (A-B) \cap C = \emptyset $ (my thinking).

    Your other direction looks good.

    So now is this correct?
    Last edited by shilz222; Aug 13th 2007 at 09:28 AM.
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  4. #19
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shilz222 View Post
    Why is it not correct?

    We assume that $\displaystyle (A-B) \cap C \neq \emptyset $ for proof by contradiction.

    One direction: Given that $\displaystyle (A \cap C) \subseteq B $ and $\displaystyle (A-B) \cap C \neq \emptyset $ use proof by contradiction to show that $\displaystyle (A-B) \cap C = \emptyset $ (my thinking).

    Your other direction looks good.
    Okay, so to be clear. You agree that your proof of the $\displaystyle \implies$ direction is incorrect? And now you are working on the $\displaystyle \leftarrow$ proof. And you have decided to try to prove it by contradiction. Is this all correct?

    -Dan
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  5. #20
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    I asked help for the $\displaystyle \Longrightarrow $ direction.

    Quote Originally Posted by Plato View Post
    This is a very important concept: $\displaystyle X \cap Y = \emptyset \text{ if an only if }X \subseteq Y'$.
    That is: two sets are disjoint iff each is a subset of the complement of the other.

    $\displaystyle \left( {A - B} \right) \cap C = \left( {A \cap B'} \right) \cap C = \left( {A \cap C} \right) \cap B'$, thus if $\displaystyle \left( {A - B} \right) \cap C = \emptyset$ then $\displaystyle \left( {A \cap C} \right) \subseteq B$.

    Now you do the other way.

    Trying to teach oneself this material can be very dangerous if one wants to do higher mathematics.
    I agree with this method that Plato has done.

    For the $\displaystyle \Longleftarrow $ direction I already proved it by contradiction.

    Suppose that $\displaystyle (A-B) \cap C \neq \emptyset $. If $\displaystyle x \in A \cap C $ then $\displaystyle x \in A $ and $\displaystyle x \in C \Rightarrow x \in B $. Then $\displaystyle \exists x \in (A-B) \cap C $ such that $\displaystyle x \in A $ and $\displaystyle x \in C $ but $\displaystyle x \not \in B $. But this is a contradiction and so $\displaystyle (A-B) \cap C = \emptyset $.
    So is this not correct?
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  6. #21
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    Basically I used 2 different techniques to prove the directions.
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  7. #22
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    Here it is.
    Given $\displaystyle \left( {A \cap C} \right) \subseteq B$ that means $\displaystyle \left( {A \cap C} \right) \cap B' = \emptyset $.
    But $\displaystyle \left( {A \cap C} \right) \cap B' = \left( {A \cap B'} \right) \cap C = \left( {A - B} \right) \cap C = \emptyset$
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  8. #23
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    I already know that is correct. But why isnt the proof by contradiction correct?
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  9. #24
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    I guess what I am asking is that is it because my proof is awkward that its not correct (i.e. using direct and proof by contradiction)?
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  10. #25
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    Quote Originally Posted by shilz222 View Post
    I guess what I am asking is that is it because my proof is awkward that its not correct (i.e. using direct and proof by contradiction)?
    Simply because the direct proof is shorter. From what I have seen (I may be wrong) proof by contradiction is usually only used if a direct proof is either lacking or obscenely difficult.

    -Dan
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