1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Boolean algebra: I don't understand how the answer was achieved

  1. Feb 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Factor to obtain a product of sums. (Simplify where possible.)
    BCD + C'D' + BC'D + CD


    2. Relevant equations
    Distributive: X(Y+Z)=XY+XZ and X+YZ=(X+Y)(X+Z)

    Simplification: XY+XY'=X and (X+Y)(X+Y')=X
    X+XY=X and X(X+Y)=X
    (X+Y')Y=XY and XY'+Y=X+Y

    Factoring/multiplying out: (X+Y)(X'+Z)=XZ+X'Y and XY+X'Z=(X+Z)(X'+Y)

    Consensus: XY+YZ+X'Z=XY+X'Z and (X+Y)(Y+Z)(X'+Z)=(X+Y)(X'+Z)


    3. The attempt at a solution
    Here's what I've done so far:

    BCD + C'D' + BC'D + CD Factor out D from BCD and CD

    D(BC + C) + C'D' + B'C'D Factor out C from BC + C

    D[C(B + 1)] + C'D' + B'C'D B + 1 = 1

    DC + C'D' + B'D Factor out D from DC and B'C'D

    D(C + B'C') + C'D' C + B'C' follows XY' + Y = X + Y

    D(B' + C) + C'D' Distribute D

    B'D + CD + C'D'


    Attempt to simplify/factor further

    B'D + CD + C'D' Add consensus of B'D and C'D' = B'C'

    B'D + CD + C'D + B'C' Factor out D from B'D + CD, and C' from C'D + B'C'

    D(B' + C) + C'(D + B') Use associate law to make C' + D

    (C' + D)(B' + C)(D + B') Foil (B' + C)(D + B')

    (C' + D)(B'D' + B' + CD' + B'C) Factor out B' from B'D' + B' + B'C

    (C' + D)[B'(D' + 1 + C) + CD'] D' + 1 = 1

    (C' + D)[B'(1 + C) + CD'] C' + 1 = 1

    (C' + D)[B' + CD'] B' + CD' follows X + YZ = (X + Y)(X + Z)

    (C' + D)(B' + C)(B' + D') Foil (B' + C)(B' + D')

    (C' + D)(B' + B'D' + B'C + D'C) Redundant term B'D' gone due to consensus of B'C + D'C

    (C' + D)(B' + B'C + D'C) Factor out C from (B'C + D'C)

    (C' + D)[C(B' + D') + B'] I stopped here because I felt I was doing it wrong




    The simplified product of sums is: (C' + D)(B' + C + D'). I don't understand how to get that answer from B'D + CD + C'D' . I think I'm using the laws and theorems incorrectly.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Boolean algebra: I don't understand how the answer was achieved
Loading...