Results 1 to 6 of 6

Thread: Show proofs

  1. #1
    Newbie ArchReimann's Avatar
    Joined
    Jul 2007
    From
    Diliman, Manila
    Posts
    4

    Show proofs

    Can anyone help me with these proofs:

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


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

    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; Aug 12th 2007 at 05:18 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by ArchReimann View Post
    2. Prove that for every real number x, there is a real number y such that
    $\displaystyle (x+y = xy) leftrightarrow x neq 1$
    For any $\displaystyle x\not = 1$ choose $\displaystyle y=\frac{x}{x-1}$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie ArchReimann's Avatar
    Joined
    Jul 2007
    From
    Diliman, Manila
    Posts
    4
    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!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    658
    Thanks
    42
    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 $\displaystyle \Im, \ \wp$ respectively. We have that there is some element $\displaystyle \{g_0\}\in \wp$ such that, $\displaystyle (\forall \{f\}\in\Im) \ \{f\}\cap\{g_0\}=\emptyset$. (1)

    Suppose now the sentence "the union of all sets in $\displaystyle \Im$ and the intersection of all sets in $\displaystyle \wp$ are disjoint" is false. This means "the union of all sets in $\displaystyle \Im$ and the intersection of all sets in $\displaystyle \wp$ have a common element". Call this $\displaystyle f$. Certainly, $\displaystyle \{f\}\in \Im$ and $\displaystyle f\in \cap\wp$ (the intersection of this family). Now since $\displaystyle f\in \cap\wp$, we must have that $\displaystyle \{f\}$ intersects all elements of the family: $\displaystyle \{f\}\cap\{g\}\neq\emptyset, \ \forall \{g\}\in\wp$. This and (1) contradict each other.
    Last edited by Rebesques; Aug 12th 2007 at 08:06 AM. Reason: old age.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie ArchReimann's Avatar
    Joined
    Jul 2007
    From
    Diliman, Manila
    Posts
    4
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    658
    Thanks
    42
    Quote Originally Posted by ArchReimann View Post
    Can anyone help me with these proofs:

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

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

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

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

    The other inclusion is up to you



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

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

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

    For the second one. Suppose $\displaystyle x\neq1$. Then choose for $\displaystyle y$what Hacker has proposed. The proof is now complete.
    Last edited by Rebesques; Aug 12th 2007 at 05:49 PM. Reason: myopia.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need help with these proofs
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 23rd 2010, 03:16 AM
  2. how to show show this proof using MAX
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jan 14th 2009, 12:05 PM
  3. need help with 2 proofs
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Mar 30th 2008, 06:10 PM
  4. Set Proofs
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: Mar 5th 2008, 05:31 PM
  5. Replies: 3
    Last Post: Oct 6th 2007, 02:01 PM

Search Tags


/mathhelpforum @mathhelpforum