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Boolean Algebra

  1. Sep 10, 2005 #1
    Hi there,

    I've never done this type of Algebra... and it's a bit confusing. I've completed this question, but I don't know if I took the right steps...

    *=AND, +=OR, '=NOT

    If A*B = 0, A+B=1

    Prove: (A+C)*(A'+B)*(B+C) = B*C

    I expanded it a bit and applied some of the rules. Can someone explain it step by step? I got the answer but I'm a bit confused at how I got the answer.

    Thanks!
     
  2. jcsd
  3. Sep 10, 2005 #2

    AKG

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    (A+C)*(A'+B)*(B+C)
    = (A+C)*(B+C)*(A'+B) (commutation)
    = ((A*B)+C)*(A'+B) (distribution)
    = (0+C)*(A'+B) (substitution A*B = 0)
    = C*(A'+B)

    Now I don't know what "rules" you know or how formal you have to be, but you should be able to see that (A'+B) = B given A+B = 1. Clearly, if A'+B = 0, then B = 0. If A'+B = 1, we want B to also necessarily be 1. But suppose it's not, so suppose A'+B=1 but B=0. Then A' = 1, which means A=0. So both A and B would be 0, but we're given A+B = 1, which is a contradiction, so B does indeed have to be 1 if (A'+B) is 1.
     
  4. Sep 16, 2005 #3
    Mhm.. I understood all the steps until the last one...this is what I think...

    Since AB = 0 and A+B=1

    That leaves us with two choices for A or B. A can either be 0 or 1, and B must be the exact compliment.

    Now we're stuck with C(A'+B). *Here's where I'm a bit confused*

    If A is 0, B will be 1, then A' + B = 1
    If A is 1, B will be 0, then A' + B = 0

    From which of these two can we say C*(A'+B) = B*C?

    in otherwords... how does (A'+B) = B?
     
    Last edited: Sep 16, 2005
  5. Sep 16, 2005 #4

    AKG

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    Well I've already told you how (A' + B) = B. However, if you didn't follow the explanation, then consider what you said yourself:

    A can either be 0 or 1, and B must be the exact compliment.

    So B = A', right? So (A' + B) = B + B = B (Idempotence).
     
  6. Sep 16, 2005 #5
    Ah... I think I understand what I just said. A'=B.. then you'd just have to sub it in.

    Man, boolean algebra is mighty weird! I just started a week ago, still getting the hang of it. Thanks for your help!
     
  7. Sep 16, 2005 #6
    how about a proof of A'=B

    given A*B=0
    (or with A')
    A'+A*B=A'
    (and with B)
    A'*B + A*B*B=A'*B
    => (A'+A)*B=A'*B
    => B=A'*B (1)

    given A+B=1
    (and with A') => A'*A + A'*B = A'
    => A'*B = A'

    (substitute from (1) ) => B=A'
     
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