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1. When people write B' does it mean B(not)? Like B with a bar on top of it? ......._ B' = B
Also...
2. X = B(A+C) + C X = BA + BC + C X = BA + C(B+1) <-- this part confuses me, can someone explain how the BC + C turn into C(B+1)? X = BA + C*1 (B + 1 = 1 this is from rules) x = BA + C (C * 1 = C this is from rules)
And 3. This NAND gate, with the inverter bubble there, would it be A' B' into A+B? Since the inverters cancel?
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As far as BA + BC + C = BA + C(B + 1), that's really simple factoring.
Look at it as 1BC + 1C. Factor out the C.
C(1B + 1).
And clearly the coefficient 1 before the B is extraneous, so it's not written.
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1) Yes
2) See above post, just factoring.
3) So this is an AND gate with inversions on all inputs and outputs?
Basically, an AND gate with inverted inputs is the same as a NOR by demorgan's theorem A'B' = (A+B)'
Then you have a NOR with an inverted output which is the same as OR.
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I really don't see how BA + BC + C = BA + C(B + 1) as factoring, I mean..
x^2 + 4x + 4 would be (x+2)(x+2) ...
Can you explain further, I really don't see it.
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In general C(A+B)=CA+CB for any three numbers A,B,C. This is called the distributive law of multiplication.
Plug in A=1, and you get C(1+B)=C(1)+CB=C+CB
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Forget the BA for a moment.
BC + C = C(B + 1)
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On April 21 2008 11:38 fight_or_flight wrote: 1) Yes
2) See above post, just factoring.
3) So this is an AND gate with inversions on all inputs and outputs?
Basically, an AND gate with inverted inputs is the same as a NOR by demorgan's theorem A'B' = (A+B)'
Then you have a NOR with an inverted output which is the same as OR. For number 3, so it's still (AB)'? Because there's an inverter in the input which makes it:
A' B' = AB? It doesn't work that way?
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On April 21 2008 11:38 Raithed wrote: I really don't see how BA + BC + C = BA + C(B + 1) as factoring, I mean..
x^2 + 4x + 4 would be (x+2)(x+2) ...
Can you explain further, I really don't see it. x^2 + 4x + 4 = (x+2)(x+2) x^3 + x^2 + 4x + 4 = x^3+(x+2)(x+2)
BC + C = C(B+1) BA + BC + C = BA + C(B+1)
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On April 21 2008 11:40 Muirhead wrote: In general C(A+B)=CA+CB for any three numbers A,B,C. This is called the distributive law of multiplication.
Plug in A=1, and you get C(1+B)=C(1)+CB=C+CB Oh, wow, nice explanation. But what if you don't know the answer, would you always plug A = 1?
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On April 21 2008 11:46 Raithed wrote:Show nested quote +On April 21 2008 11:40 Muirhead wrote: In general C(A+B)=CA+CB for any three numbers A,B,C. This is called the distributive law of multiplication.
Plug in A=1, and you get C(1+B)=C(1)+CB=C+CB Oh, wow, nice explanation. But what if you don't know the answer, would you always plug A = 1? Lol, you kinda missed the point.
He was using an example to demonstrate the factorization of BC + C = C(B + 1).
He pretended that the 1 in the above example was an A, which would have been written as:
BC + CA = C(B + A).
Exact same problem, just with another variable.
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On April 21 2008 11:43 Raithed wrote:Show nested quote +On April 21 2008 11:38 fight_or_flight wrote: 1) Yes
2) See above post, just factoring.
3) So this is an AND gate with inversions on all inputs and outputs?
Basically, an AND gate with inverted inputs is the same as a NOR by demorgan's theorem A'B' = (A+B)'
Then you have a NOR with an inverted output which is the same as OR. For number 3, so it's still (AB)'? Because there's an inverter in the input which makes it: A' B' = AB? It doesn't work that way? Does not work that way. Look at the truth tables:
out = A'B'
A B out 0 0 1 0 1 0 1 0 0 1 1 0
----------------------------
out = AB
A B out 0 0 0 0 1 0 1 0 0 1 1 1
------------------------------
Therefore you conclude that A'B' = NOR. (look up demorgan's theorem)
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On April 21 2008 11:53 fight_or_flight wrote:Show nested quote +On April 21 2008 11:43 Raithed wrote:On April 21 2008 11:38 fight_or_flight wrote: 1) Yes
2) See above post, just factoring.
3) So this is an AND gate with inversions on all inputs and outputs?
Basically, an AND gate with inverted inputs is the same as a NOR by demorgan's theorem A'B' = (A+B)'
Then you have a NOR with an inverted output which is the same as OR. For number 3, so it's still (AB)'? Because there's an inverter in the input which makes it: A' B' = AB? It doesn't work that way? Does not work that way. Look at the truth tables: out = A'B' A B out 0 0 1 0 1 0 1 0 0 1 1 0 ---------------------------- out = AB A B out 0 0 0 0 1 0 1 0 0 1 1 1 ------------------------------ Therefore you conclude that A'B' = NOR. (look up demorgan's theorem) Yeah, this is exactly accurate. Not A AND Not B evaluates as true in very different circumstances than A AND B evaluates as true.
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Without looking at the truth table, I thought not from input, with a not in output would cancel, this just confirms that it's NOR. I didn't mean exactly as (AB)' == AB, I just meant that the ' cancels.
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No, they don't cancel. If you have 2 bubbles on the input of an AND, you can make it into an OR with a bubble on the output.
Same thing if the inputs on an OR have bubbles...it becomes an AND with a bubble on the output.
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So basically, the bubbles DON'T cancel, whatsoever?
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But wait, the gate is a NAND gate...
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So if it's
Bubbled input NAND --- inverter --- bubbles cancel? Only this way?
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Which is basically
NAND + inverter = comes out as AB? Instead of (AB)' ?
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I think you're mixing that with something else fof, I didn't mean if they're the equivalence of anything but such as that:
bubble NAND gate....
NAND gates are basically (AB)' from the beginning, from that diagram alone, but if the input already gives A' and B' the output is already to be (AB)' wouldn't the NOTs cancel? Because the diagrams you provided have bubbles from input to bubbles of output of equivalent gates.
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Thats not a NAND gate though. What you do is imagine the output bubble is far away. Then when it changes from that pic you have 2 bubbles on the output which cancel.
edit: might be the same thing your saying, I'm getting confused over this whole thread.
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Thats how you solve it using "bubble logic". You can also do it algebraically with boolean algebra.
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thats some pretty drawing, boolean algebra is so much easier though
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A' NAND B' = (A' AND B')' = A OR B deMorgan's Law
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"break the line change the sign" as they say. btw: BC+B=BC+B*1 (identity law) BC+B*1=B(C+1) (distributive law)
p.s. For those using examples from the real numbers to show this, why are you trying to explain properties of boolean algebra/rings using field axioms?
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Alright, and guys, how does...
X = (A’ + B’) + (B+C) X = 1?
I get (A’ + B’) = (AB)' + (B+C) right? What is the next step?
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X = (A’ + B’) + (B+C)
X = A’ + (B’ + B) + C
X = A’ + 1 + C
X = 1
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Thank you fof, I'd like to thank everyone for staying up with me while I continue to do this silly stuff. It's 2AM EST. I appreciate it, thanks again fof.
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