Simply Boolean expression

Discussion in 'Math' started by ralampur, May 20, 2009.

  1. ralampur

    Thread Starter New Member

    May 20, 2009
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    0
    some one please help with with the solution

    Define the connective * for the two valued variables A, B, and C as follows​
    A *​
    B = AB + A B​

    Let C = A*B, Determine which of the following is valid
    i. A=B*C
    ii. B=A*C​
    iii. A*B*C=1
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Can you confirm what you mean by this bit?

    If you substitute for C in the expressions, you get

    i. A=B*C :::::::: A = B*A*B
    ii. B=A*C ::::::: B = A*A*B
    iii. A*B*C=1 ::: A*B*A*C = 1

    Any thoughts on considering these?
     
  3. ralampur

    Thread Starter New Member

    May 20, 2009
    2
    0
    question i copied was wrong, here is the right question

    Define the connective * for the two valued variables A, B, and C as follows
    A*B=AB+A'B'​

    Let C = A*B, Determine which of the following is valid
    i. A=B*C
    ii. B=A*C​
    iii. A*B*C=1

     
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    So have you had a go at substituting for C ?

    Post your working so far.
     
  5. zgozvrm

    Member

    Oct 24, 2009
    115
    2
    If I understand correctly, what you have is a "custom" operator "*" which takes 2 values, A & B, and performs the following Boolean function on them:
    A*B = AB+A'B'

    If that is the case, then the truth table for this function is:

    A B | A*B
    ----|----
    0 0 | 1
    0 1 | 0
    1 0 | 0
    1 1 | 1

    although it's not necessary to know this in order to solve the problem.

    Assuming you know DeMorgan's Law and the basic Boolean functions (AND, OR, NAND, NOR, NOT, and XOR),
    you are trying to see if:

    1) A = B * C (It is...)
    = B * (A * B)
    = B * (AB + A'B')
    = B(AB + A'B') + B'(AB + A'B')'
    = ABB + A'B'B + B'((AB)'(A'B')')
    = AB + B'(A'+B')(A + B)
    = AB + (A'B' + B'B')(A + B)
    = AB + (A'B' + B')(A + B)
    = AB + B'(A' + 1)(A + B)
    = AB + B'(1)(A + B)
    = AB + B'(A + B)
    = AB + AB' + BB'
    = AB + AB'
    = A(B + B')
    = A(1)
    = A

    2) B = A * C (it is...)
    = A * (AB + A'B')
    = A(AB + A'B') + A'(AB + A'B')'
    = AAB + AA'B' + A'((AB)'(A'B')')
    = AB + A'(A' + B')(A + B)
    = AB + (A'A' +A'B')(A + B)
    = AB + (A' + A'B')(A + B)
    = AB + A'(1 + B')(A + B)
    = AB + A'(A + B)
    = AB + A'A + A'B
    = AB + A'B
    = B(A + A')
    = B

    and

    3) 1 = A * B * C (it is ...)
    = A * (B * C)
    = A * A (as shown in #1 above)
    = AA + A'A'
    = 1 + 1
    = 1

    Granted, I took a lot of shortcuts here (in the interest of saving space), and assumed that you have a pretty good understanding of Boolean algebra. I was guessing that your problem was in visualizing what the actual problem was.

    If this isn't clear enough, I'll be happy to explain in greater detail...
     
  6. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    zgozvrm,

    What is all that heavy thrashing around with Boolean algebra all about? To prove that A=A and B=B? I cannot understand what you are doing. Anyway, that is a common logic function. Specifically, it is a exclusive NOR (XNOR) gate. http://www.tpub.com/content/neets/14185/css/14185_122.htm

    Ratch
     
  7. zgozvrm

    Member

    Oct 24, 2009
    115
    2
    Because he was asked to show that those statements were true. To the uninitiated, it isn't is obvious as it may be to you or me. If he were asked to show if 2x2 was in fact 4, wouldn't you go through some repeated additions to show that it is true?
     
  8. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    zgozvrm

    A truth table or set of truth tables will prove or disprove any logic expression.

    Ratch
     
  9. zgozvrm

    Member

    Oct 24, 2009
    115
    2
    Yes, you're absolutely right. But, if I had shown truth tables, surely someone else would've pointed out that it could've been proven using Boolean algebra.

    You just can't please everyone.
     
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