Boolean algebra question: Solving AB + A'C + BC

adam2016

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[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]
 
[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.
 
Thanks doctor,

great answer just a quick follow up why couldn't we just immediately factorise?

in the example we are using

AB + A'C + BC

We can rearrange the terms so it is

AB + BC + A'C

both BC and A'C have C in common so why can't we just factor them?

we will get C(B + A')

but here is the part that gets me,how would we solve this? and why don't we just say C(B + A' + 1) or even two 1's C(B + A' + 1 + 1) ? since we have two variables B and A'?

thanks
 
great answer just a quick follow up...how would we solve this?
Quick question: What were the instructions for the posted expression? What are you supposed to be doing with it? It is not mathematically possible to "solve" an expression, so what do you mean by this language?

Please be specific. Thank you! ;)
 
Thanks doctor,

great answer just a quick follow up why couldn't we just immediately factorise?

in the example we are using

AB + A'C + BC

We can rearrange the terms so it is

AB + BC + A'C

both BC and A'C have C in common so why can't we just factor them?

we will get C(B + A')

but here is the part that gets me,how would we solve this? and why don't we just say C(B + A' + 1) or even two 1's C(B + A' + 1 + 1) ? since we have two variables B and A'?

You still haven't answered my question: What were you told to do with this Boolean expression? There are several things you might want to do; if it is to "simplify", what that means can vary. Similarly, Boolean algebra can be presented in quite different styles. Without knowing the instructions, and what rules you have been taught, I really can't give direct answers.

In general, given an expression, there are all sorts of things you can do. The question is, which will take you toward your goal? You haven't stated the goal, or the system you are being taught for getting there, so I don't even know if there is a reason, in your context, not to do what you are suggesting here. All I can see is that you seem to have driven, following the traffic laws, into a dead end -- it's legal, but not helpful. To do what's helpful, you need to keep the goal in mind. So, what is that goal?

You can't just stick in 1's as you did here, because they change the meaning! It's true that A = 1A, so you can stick as many 1's in there as you want; but it is not true that A = A+1. Do you see that? Adding (or'ing) 1 to B+A' changes it. Here, you didn't just do something unhelpful; you broke the law. Don't do that.
 
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