Help needed with Boolean Algebra and DeMorgan

Discussion in 'Homework Help' started by randb, Mar 27, 2014.

  1. randb

    Thread Starter New Member

    Dec 29, 2012
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    Demorgan problem

    (( AB') (A'C))' One broke bar recd (AB')' + (A'C)'

    then broke the bar again and recd A'+B+A+C' A'+a =1 1+ other variables =1?

    Book states answer is (A+B') ( A'+C)

    Where did I go wrong?
     
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    What's wrong with your title ?
     
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  3. amilton542

    Active Member

    Nov 13, 2010
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    Why are you applying boolean operators to expressions alone?
     
  4. jjw

    Member

    Dec 24, 2013
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    The answer in the book is wrong if the original expression is correct.
     
  5. amilton542

    Active Member

    Nov 13, 2010
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     Y = [(AB')(A'C)]'

     Y = (AB')' + (A'C)'

     Y = (A'+B)(A+C')

    Which is the same as you've got.

    You're operating on expressions alone though, which is O.K, but could you type this "question" in it's entirety because something's missing.
     
    Last edited: Mar 29, 2014
  6. jjw

    Member

    Dec 24, 2013
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    Last expression is missing + should be (A'+B)+(A+C')
     
  7. randb

    Thread Starter New Member

    Dec 29, 2012
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    Applying DeMorgan's theorem and Boolean algebra to the expression [​IMG] results in _____.
     
  8. randb

    Thread Starter New Member

    Dec 29, 2012
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    Thank you for your responses and help
     
  9. amilton542

    Active Member

    Nov 13, 2010
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    Oops, yes you're right. I missed that one somehow.

    In all fairness, I really don't know. Maybe someone else could help you.
     
  10. WBahn

    Moderator

    Mar 31, 2012
    17,737
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    Did you go wrong?

    Let's check it out. You are claiming that the expression

    ((AB') (A'C))' = 1

    This is of the form

    (FG)' = 1

    which is

    FG = 0

    F = AB'
    G = A'C

    If A is False, then F is False and FG is False.
    If A' is False, the G is False and FG is False.

    What does that tell you?
     
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