Results 1 to 4 of 4

Math Help - Another sets prove

  1. #1
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261

    Another sets prove

    By using the set algebra , prove that , for any sets A and B
    A u ( A' n B ) = A u B

    Deduce , or using set algebra , show that

    A u ( A' n B ) u [A' n (A u B') n {B u (B' n C}] = A u B u C
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2008
    From
    Singapore
    Posts
    160
    The question test the distributive laws.

    A u ( A' n B ) = A u B
    from L.H.S : A u ( A' n B )
    = ( A U A' ) n ( A u B )
    = S n ( A u B )
    = A u B ( proven )

    You should be able to solve the second part using the law or the procedure listed above

    Quote Originally Posted by mathaddict View Post
    By using the set algebra , prove that , for any sets A and B
    A u ( A' n B ) = A u B

    Deduce , or using set algebra , show that

    A u ( A' n B ) u [A' n (A u B') n {B u (B' n C}] = A u B u C
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by mathaddict View Post
    Deduce , or using set algebra , show that

    A u ( A' n B ) u [A' n (A u B') n {B u (B' n C}] = A u B u C
    I tried this one but i couldn't get the prove . Can someone pls spot my mistake .

    My working :

    (A u B) u (A' n B' ) n ( B u C ) = S n (B u C ) = B u C

    wondering , where is the A .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Proofs using the Laws of Sets

    Hello mathaddict

    Quote Originally Posted by mathaddict View Post
    I tried this one but i couldn't get the prove . Can someone pls spot my mistake .

    My working :

    (A u B) u (A' n B' ) n ( B u C ) = S n (B u C ) = B u C

    wondering , where is the A .
    Here's a proof - there might be a quicker one:

     A \cup (A'\cap B)\cup [A'\cap (A\cup B')\cap (B\cup (B'\cap C)]

    =A\cup B\cup [A'\cap (A\cup B') \cap (B \cup C)], using the first result twice

    =A \cup B \cup[\{(A' \cap A) \cup (A'\cap B')\} \cap (B \cup C)], Distributive Law

     = A \cup B \cup [\{\oslash \cup (A' \cap B') \} \cap (B \cup C)], Complement Law

    = A \cup B \cup [(A' \cap B')  \cap (B \cup C)], Identity Law

    = A \cup B \cup [(A \cup B)'  \cap (B \cup C)], De Morgan's Law

    = [(A \cup B) \cup (A \cup B)']  \cap [(A \cup B) \cup (B \cup C)], Distributive Law

    = \boldsymbol{\text {U}}  \cap [(A \cup B) \cup (B \cup C)], Complement Law

    = A \cup (B \cup B) \cup C, Identity Law, Associative Law

    = A \cup B \cup C, Idempotent Law

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove the following equality of sets
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: July 30th 2010, 09:45 PM
  2. sets prove
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: June 3rd 2009, 02:58 PM
  3. sets prove
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: May 29th 2009, 08:03 AM
  4. sets prove
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: December 15th 2008, 08:55 AM
  5. prove equal sets
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: May 4th 2008, 09:27 AM

Search Tags


/mathhelpforum @mathhelpforum