Boolean Algebra Help

Discussion in 'Homework Help' started by kurtj, Apr 1, 2013.

  1. kurtj

    Thread Starter New Member

    Apr 1, 2013
    So going through some exercises on understanding Boolean Algebra




    Because P and Q are common factors they can be canceled out and the simplified statement is = P*(Q+S)

    Q2 is



    But this has got me confused as to how to simplify it further.
    I know that Q+Q' = 1 but Q*Q' = 0, but also Q*/+Q = Q.
    But i'm not sure how to start attacking it.

    Truth Table would be
    PQS | Equation
    000 0
    001 0
    010 1
    011 1
    100 1
    101 1
    110 1
    111 1


    All are common factors so,
    R*R= R, S*S'= 0, T*T=T

    Simplified = (R+T)
    It looks wrong to me I think the answer is (R) but I cant tell where I went wrong.

    Is Q1 Correct and is someone able to help step by step to explain what I need to do to work with Q2 and Q3?
  2. tshuck

    Well-Known Member

    Oct 18, 2012
    Q1: it's good, but you should show the step, PQ + PQ +PS , then the identity that allows you to say that PQ + PQ = PQ.

    Q2: Expand out your terms. You know that P = PB + PB', and that you can OR the same term as many times as you'd like, so expand theed terms to be able to cancel the other terms out. Let's see how far you get with that information.

    Q3: Your problem is in your first line, you've split up S*T and haven't done your distribution properly. If you had, say, A*B(B + A*B)), this would come out to, A*B*B + A*B*A*B.

    As an aside, you needn't write the AND operation as A*B, you can just write AB.
  3. WBahn


    Mar 31, 2012
    You've got a terminology problem. P and Q are not "common factors", per se, but rather PQ is a "repeated term" and the duplicate term can be removed. Then P becomes a common factor that can be factored out. So

    PQ + P(Q+S)
    PQ + PQ + PS
    PQ + PS

    Another way to do the dance is to say that P is a common factor, which means you can factor it out, leaving you with Q as a repeated term that can be reduced:

    PQ + P(Q+S)

    What is your basis for pulling in the first two terms and grouping them the way you did?

    Is 4+5+6(7+9) the same as (4+5+6)(7+9)?

    Your truth table is correct. Just be more careful with your boolean manipulations.

    Hint: What does [X + X(anything)] reduce to?

    How did you come up with your second line. Take it one step at a time. You are trying to combine a bunch of steps and getting lost along the way. Distribute things the same way you would in normal algebra.

    You are correct that the final answer is just R.
    spark8217 likes this.
  4. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    Hint 2: This Boolean algebra question from WBahn is a known law. It's critical in solving Q2 and Q3.
    Last edited: Apr 1, 2013
  5. kurtj

    Thread Starter New Member

    Apr 1, 2013
    x1 + xy
    x(1+y) | (1+y=1)
    x1 | (x*1=x)

    No they are different, I suppose I got thinking the wrong way about multiplication with brackets

    P(Q'+Q+S) | Q'+Q=1
    P(Q+S(1)) | S*1=1

    R(S'T+ST) | S'T + ST = 1
    R(1) | R*1=1

    hows does this look? I think its the NOTing of a Value thats messing me up, I know if you break the line change the sign, but because only 1 value is NOTted
  6. WBahn


    Mar 31, 2012

    Nope. You are being sloppy again.

    How did you go from



    P(Q'+Q+S) ?

    There's no PQ term there!

    Again, you're being sloppy.


    Notice that the last two terms in each factor don't contain an R. Yet in your next line you have an R in every term. How does that happen?

    Slow down and do it step by step:


    Distribute the first over the second:

    (R+S*T)*R + (R+S*T)*S'*T

    Now distribute again

    R*R+S*T*R + R*S'*T + S*T*S'*T

    Now remove duplicates and reorder factors

    R + R*S*T + R*S'*T + S*S'*T

    Now factor out the R from the first three terms

    R(1 + S*T + S'*T) + S*S'*T
    R(1) + S*S'*T
    R + S*S'*T

    Now kill off the second term

    R + (S*S')*T
    R + (0)*T

    See? Take it step by step.

    Learning engineering by memorizing "break the line change the sign" and similar sayings is a big part of the problem. If you take the time and effort to learn the fundamentals, then you don't have to memorize cute little slogans that substitute for comprehension. You'll understand WHAT DeMorgan's theorems are and WHAT they mean and HOW they work and it will simply become a part of the way you think.