[FONT=q_serif]Hi guys, I have been asked to solve this question > AB + A'C + BC, the answer to the question is > AB + A'C + BC = AB + A'C + BC(A + A') = AB + A'C + ABC + A'BC = AB + ABC + A'C + A'BC = AB(1 + C) + A'C(1 + B) = AB + A'C .
where did the (A + A') come from?
here is another example (A + C) (A + B + C) so first I will multiply both brackets and I get AA + AB + AC + CA + CB + CC. AA = A and CC = C so now I'm left with A + AB + AC + CA + CB + C, A and AB have A in common so A(B + 1) = A now I'm left with A + AC + CA + CB + C, A and AC both have A in common so A(C + 1) = A. now I'm left with A + CA + CB + C, I will rearrange the variables A + C + CA + CB, C and CA both have C in common so C(A + 1) = C, now I'm left with A + C + CB . C and CB both have C in common so C( B + 1) = C so the final answer is A + C.
I checked a few trusted online boolean algebra calculators and indeed this is the correct answer I also tried this method on many other questions and got the correct answer but how is AB + A'C + BC any different?? how I solved it I did AB + BC + A'C. AB and BC both have B in common so B(A + C + 1) = B. the reason why I did + 1 is because there is an implicit one after AB right? I then got B + AC', I must have gone wrong somwehere? thanks[/FONT]
Properly speaking, you haven't told us the
question. "AB + A'C + BC" is a Boolean expression, not a question. What is it that you were told to
do with it?
What
properties have you been taught? Have you been taught a particular
method to accomplish whatever the goal is here?
Clearly you know you can distribute; there are other
specific properties that can be used; for example, you found that A+AB = A, which is an "absorption law". Your main method was factoring, which is one good tool. On the other hand, you missed the fact that AC = CA, so that AC+CA = AC. But as you saw, you don't need to get to the goal as fast as possible; there are many ways to get there.
But that example didn't have any complements, so you may not be familiar with the properties that you need for the problem you are asking about. Also, you seem to have made a leap in putting in a 1 without justification, which suggests that you are not carefully following a list of rules, which is why I asked about what properties you have learned. That will be important.
I'll try to explain the steps of the solution you found, but how it is explained depends on the particular set of rules you have learned.
AB + A'C + BC
= AB + A'C + BC(A + A') ... because A+A' = 1 (a complementation law), BC = BC1 = BC(A + A'); they probably did this in order to get another A' in there
= AB + A'C + ABC + A'BC ... here they distributed (and commuted)
= AB + ABC + A'C + A'BC ... here they brought the A' terms together
= AB(1 + C) + A'C(1 + B) ... here they factored, in order to use the absorption law (which makes the C and the B redundant)
= AB + A'C
As to what you did, a 1 doesn't just magically appear whenever you need it; when you obtained it correctly, it was the result of factoring. In your wrong work, you said that AB + BC + A'C = B(A + C + 1), which is false: the right hand side is actually AB + BC + B. You can't factor out a B that is not present in all three terms. All you could get is B(A + C) + A'C, which doesn't help.