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Math Help - Proof with Simple Problem

  1. #1
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    Proof with Simple Problem

    Hey guys. I don't understand how to prove this:

    (a)
    <br />
\text A \cup (null) = A<br />

    (b)
    <br />
\text A \cap U = A<br />
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Foink View Post
    Hey guys. I don't understand how to prove this:

    (a)
    <br />
\text A \cup (null) = A<br />
    one method: Let x \in A \cup \emptyset ......we assume A is non-empty, it's trivial otherwise

    \Rightarrow x \in A or x \in \emptyset

    we cannot have x \in \emptyset, thus, x \in A

    therefore, (A \cup \emptyset ) \subset A

    now assume x \in A. this implies x \in A \cup \emptyset (i'm sure there's some kind of axiom or something that allows us to say this). thus A \subset (A \cup \emptyset )

    since (A \cup \emptyset ) \subset A and A \subset (A \cup \emptyset ), we have A \cup \emptyset = A
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  3. #3
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    Ohhh okay. I guess I was on the right track then. I didn't know you could just say that you could'nt have x \in \emptyset lol Thanks
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Foink View Post
    Ohhh okay. I guess I was on the right track then. I didn't know you could just say that you could'nt have x \in \emptyset lol Thanks
    well, it's the empty set, by definition it has no elements...or it wouldn't be empty, and the name would just be stupid...

    try a similar approach for the second
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  5. #5
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    yea the second one was easier because I didn't have to deal with the emptysets lol thanks
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  6. #6
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    What about these two:

    <br />
\text (a) A - \emptyset = A<br />

    <br />
\text (b)  \emptyset - A = \emptyset <br />

    The damn emptysets
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Foink View Post
    What about these two:

    <br />
\text (a) A - \emptyset = A<br />

    <br />
\text (b)  \emptyset - A = \emptyset <br />

    The damn emptysets
    see where this gets you:

    by definition: A - B = A \cap B' where A, B are sets, and B' means B-compliment
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  8. #8
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    Ahh I see. I got it Thanks!!
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