You are not logged in.
Pages: 1
I know A + 1 = 1
but does A (negated) + 1 = 1 too?
or in any case.. would AB + 1 = 1 too?
and would AC(negated) * A(negated)C = 0?
Offline
I'll use "~A" to mean A negated.
We know that A + 1 = 1.
The OR operation needs *just one* of its operands to be 1 in order to evaluate to 1... so no matter what expression is "ORed" with 1 will give 1.
Therefore, ~A + 1 = 1, and AB + 1 = 1.
As for the second part:
Boolean And is commutative, so we can write:
A(~C) * (~A)(C) = A(~A) * C(~C)
And in general, for any boolean X: X * ~X = 0.
(ie, something can't be true AND not true at the same time!)
Therefore, A(~A) * C(~C) = 0 * 0 = 0, as you said.
Last edited by polylog (2006-10-17 12:57:22)
Offline
Wow, thanks a lot!!
Now, I'm having a bit of trouble with these three problems. I have the answers, but I can't seem to get there.
q = RST (all negated at the same time) ( R+S+T) (all negated at the same time)
the answer is supposed to be ~R~S~T ..but I just can't get there. Spent hours on this one.
The other one is:
y = ~(C+D) + ~AC~D + A~B~C + ~A~BCD + AC~D
and the answer is: ~D + A~B~C + ~A~BC .. this one I got kinda close to.. but not quite.
and the last one (and hardest) is
x = ~(M+N+Q) + ~(M+~N+Q) + ~(~M+N+Q)
and ALL of that negated
answer, apparently, is simply MN + Q.. I'm frustrated
Offline
For the 1st one:
You need De Morgan's Law, the two forms of it :
q = ~(RST) * ~(R + S + T)
= ~(RST) * (~R~S~T)
= (~R + ~S + ~T) * (~R~S~T)
now distribute:
= ~R(~R~S~T) + ~S(~R~S~T) + ~T(~R~S~T)
now notice that (~X)(~X) = ~X for any boolean X, and apply that:
= (~R~S~T) + (~R~S~T) + (~R~S~T)
now notice that ORing X with itself any number times is just X: X + X + X + .... = X in boolean (weird but true)
Thus
= (~R~S~T)
Offline
thanks..!!!!!
I'll keep trying on that last one..it's tough.
Offline
For the 2nd, are you sure the answer is supposed to be:
~D + A~B~C + ~A~BC
I got ~D + A~B~C + ~A~BCD
ie, with a D at the end...
It's a pain to type though... I just applied De Morgen to ~(C+D), and then factored out ~D on every term possible, then what was left over simplified to 1, which left ~D.
Offline
For the 2nd, are you sure the answer is supposed to be:
~D + A~B~C + ~A~BC
I got ~D + A~B~C + ~A~BCD
ie, with a D at the end...
It's a pain to type though... I just applied De Morgen to ~(C+D), and then factored out ~D on every term possible, then what was left over simplified to 1, which left ~D.
that's exactly what I got! But apparently it's not correct..the book has the answer I told you. Maybe it's a mistake in the printing? (lol I always tell myself that).
Offline
It's quite possible that it's a typo in the book.
I can't see how to get that answer!
I got the third one too now, I'll need to learn the latex code for boolean to type it nicely, hold on a few minutes..
Offline
This question is based on applying De Morgen ad nauseum, like 15 times:
Offline
Wow. You are truly a kind man. Thanks a lot, you made it all very clear for me.
Offline
I figured out another way to get a smaller answer for #2..
= ~C~D + ~AC~D + A~B~C + C (~A~BD + A~D)
= ~D [(~C+~AC) + AC] + C (~B + A~D)
= ~D [~C + (~A + AC)] + ~BC + AC~D
= ~D[ (C+~C) + A] + ~BC + AC~D
= ~D (1+A) + ~BC + AC~D
= ~D (1+AC) + ~BC
y = ~D + ~BC
don't know if it's right or not
Offline
You're welcome it was fun !
In your simplification, can you tell me how you got the 2nd line? I can't quite follow that step.
Offline
Yeah..I think I made a mistake there because I'm figuring out a way to explain it, and can't think of what I did. Sorry, hehe.
Offline
Pages: 1