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Boolean Algebra

Started by triele Feb 20, 2007 at 2:10 PM 4 replies 1k views

Original Post

Feb 20, 2007 at 2:10 PM

Ok, this is kicking my arse. I'm looking at the Distributive Law and I understand A(B+C) = AB + AC. What I don't get is: A+BC = (A+B)(A+C). I do some googling and this is how they prove it to me: (a+b)(a+c) = a(a+b) + c(a+b) << WTF right here. How would you even know that you needed to do that in the first place. Everything else, I understand. = aa + ab + ca + cb = a(1 + b + c) + bc = a + bc Any help appreciated. Thanks.

Washu

Feb 20, 2007 at 2:23 PM

Let Q = (A+B), then we have Q(A + C) = QA + QC => (A+B)A + (A+B)C

gunning

Feb 20, 2007 at 2:23 PM

It's just distribution but they do it with the terms switched.

(A+B)(A+C) = (A+C)(A+B) = A(A+B) + C(A+B)

and then the rest of the proof follows

CTar

Feb 20, 2007 at 2:24 PM

EDIT: Too late.

As you know A(B+C) = AB + AC, let A=(a+b), B=a and C=c, then we have:
A(B+C) = (a+b)(a+c) = AB + AC = (a+b)a + (a+b)c. So (a+b)(a+c) is clearly a(a+b) + c(a+b), if you want a reason why this is done, then it's because the expanded equation often makes it easy to spot patterns (at least for equations this small).

triele

Feb 20, 2007 at 2:37 PM

Thankyou guys for the fast response.
Ah, I get it now. I only wished the textbook would explain things clearly like you guys do...but then again, I think its just me.

Thanks again.

thedustbustr

Mar 2, 2007 at 12:01 AM

obv late to discussion but imo its easier to distribute then factor if u need the second factorization

(a+b)(a+c)
= aa + ab + ca + cb
= a(a+b) + c(a+b)

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