Boolean Algebra

Discussion in 'Math' started by Sikandar, Nov 7, 2004.

  1. Sikandar

    Thread Starter New Member

    Nov 7, 2004
    2
    0
    I am new to boolean algebra so having these two question to simplify hope you guys will help me,

    Q- Using Demorgan's theorem, show that

    a. (A + B)' (A' + B')' = 0
    b. A + A'B + A'B' = 1
     
  2. arvind

    Member

    Nov 4, 2004
    18
    0
    hi, :D :D :D :p :p :rolleyes: :rolleyes: :rolleyes:
    as far as part b is concerned here is the proof:

    A+ A'B +A'B' = A + A' ( B + B' ) = A + A' = 1

    because by demorgan laws, A + A' =1

    part a is not clear what is that emoticon doing in between? B)
     
  3. Sikandar

    Thread Starter New Member

    Nov 7, 2004
    2
    0
    Thanks for answering b while in a it is B and parenthesis instead of emotion.

     
  4. Perion

    Active Member

    Oct 12, 2004
    43
    1
    DeMorgans theorem says:
    (x AND y)' = x' OR y'
    (x OR y)' = x' AND y'


    For your example a:
    a. (A + B )' (A' + B')' = 0

    This is composed of two expressions ANDed together
    Exp1 is (A + B )'
    Exp2 is (A' + B')'

    Convert each expression using DeMorgans Theorem
    Exp1 becomes A'B'
    Exp2 becomes AB

    So, does ( A'B' )( AB ) = 0 ?

    ( A'B' )( AB ) = A'B'AB -> associative property
    = A'AB'B -> associative
    = (A'A)(B'B') -> associative
    = (0)(0) -> complements identity
    = 0 -> Done

    So, (A + B )' (A' + B')' = 0

    Perion
     
  5. vineethbs

    Well-Known Member

    Nov 14, 2004
    56
    0
    well sikander , one of the best ways to solve boolean algebra probs is to use Venn diagrams , u can really solve problems very quickly .


    |--------|--|------|
    |--------|--|------|
    |----1-- |3|---2--|
    |--------|- |-------|
    1 is A\B
    2 is B\A
    3 is A intersection B

    a+a'b=the total area which is a+b
    now a' wud include all the space including B
    b' all space include all space including A
    their intersection wud include all the space except a and B
    so a+b + all space not including a+b=all space = 1

    well ,the figure looks complicated but this is really easy .
     
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