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Math Help - Show proofs

  1. #1
    Newbie ArchReimann's Avatar
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    Show proofs

    Can anyone help me with these proofs:

    1. Prove that for any set A and B, the
    P(A \cap B) = P(A) \cap P(B)
    where P(X) = power set of X.


    2. Prove that for every real number x, there is a real number y such that
     (x+y = xy) \leftrightarrow x \neq 1

    3. Suppose F and G are nonempty family of sets and every element of F is disjoint from some element of G. Prove that the union of all sets in F and the intersection of all sets in G are disjoint.

    THANKS! MY MATH 109 GRADE DEPENDS ON THIS!
    Last edited by ArchReimann; August 12th 2007 at 05:18 AM.
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  2. #2
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    Quote Originally Posted by ArchReimann View Post
    2. Prove that for every real number x, there is a real number y such that
     (x+y = xy) leftrightarrow x neq 1
    For any x\not = 1 choose y=\frac{x}{x-1}.
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  3. #3
    Newbie ArchReimann's Avatar
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    Sorry for the messed up statements.

    Um, can you post the step by step proof please? I'm kind of cnfused about the structure of formal proofs. Thanks a lot!
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  4. #4
    Super Member Rebesques's Avatar
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    Suppose F and G are nonempty family of sets and every element of F is disjoint from some element of G. Prove that the union of all sets in F and the intersection of all sets in G are disjoint.

    Let the families be \Im, \ \wp respectively. We have that there is some element \{g_0\}\in \wp such that, (\forall \{f\}\in\Im) \ \{f\}\cap\{g_0\}=\emptyset. (1)

    Suppose now the sentence "the union of all sets in \Im and the intersection of all sets in  \wp are disjoint" is false. This means "the union of all sets in \Im and the intersection of all sets in \wp have a common element". Call this f. Certainly, \{f\}\in \Im and f\in \cap\wp (the intersection of this family). Now since f\in \cap\wp, we must have that \{f\} intersects all elements of the family: \{f\}\cap\{g\}\neq\emptyset, \ \forall \{g\}\in\wp. This and (1) contradict each other.
    Last edited by Rebesques; August 12th 2007 at 08:06 AM. Reason: old age.
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  5. #5
    Newbie ArchReimann's Avatar
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    OMG that's exactly what I wanted. My professor told us that he will give double points on #3 if we use prrof by contradiction. Thanks thanks thanks.

    Now the other two..hmmm.
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  6. #6
    Super Member Rebesques's Avatar
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    Quote Originally Posted by ArchReimann View Post
    Can anyone help me with these proofs:

    1. Prove that for any set A and B, the
    P(A \cap B) = P(A) \cap P(B)
    where P(X) = power set of X.

    Double inclusion: We show P(A \cap B)\subset P(A) \cap P(B) and P(A) \cap P(B)\subset P(A \cap B).

    For the first inclusion, let s\in P(A \cap B). We will prove s\in P(A) \cap P(B).

    We have s\subset A \cap B, so (s\subset A and s\subset B). Then (s\in P(A) and s\in P(B)), which gives s\in P(A) \cap P(B). qed

    The other inclusion is up to you



    2. Prove that for every real number x, there is a real number y such that
     (x+y = xy) \leftrightarrow x \neq 1

    Double implication. We must show  (x+y = xy) \rightarrow x \neq 1 and  (x+y = xy) \leftarrow x \neq 1.

    For the first one. Suppose by contradiction x+y = xy and x=1. Then 1+y = y, a contradiction.

    For the second one. Suppose x\neq1. Then choose for ywhat Hacker has proposed. The proof is now complete.
    Last edited by Rebesques; August 12th 2007 at 05:49 PM. Reason: myopia.
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