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Raithed
China7078 Posts
......._ 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? ![[image loading]](http://i32.tinypic.com/3492s1d.jpg)   | ||
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NathanSC
United States620 Posts
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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fight_or_flight
United States3988 Posts
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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Raithed
China7078 Posts
x^2 + 4x + 4 would be (x+2)(x+2) ... Can you explain further, I really don't see it. | ||
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Muirhead
United States556 Posts
Plug in A=1, and you get C(1+B)=C(1)+CB=C+CB | ||
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fight_or_flight
United States3988 Posts
BC + C = C(B + 1) | ||
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Raithed
China7078 Posts
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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Myxomatosis
United States2392 Posts
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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Raithed
China7078 Posts
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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NathanSC
United States620 Posts
On April 21 2008 11:46 Raithed wrote: 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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fight_or_flight
United States3988 Posts
On April 21 2008 11:43 Raithed wrote: 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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NathanSC
United States620 Posts
On April 21 2008 11:53 fight_or_flight wrote: 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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Raithed
China7078 Posts
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fight_or_flight
United States3988 Posts
Same thing if the inputs on an OR have bubbles...it becomes an AND with a bubble on the output. | ||
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Raithed
China7078 Posts
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Raithed
China7078 Posts
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fight_or_flight
United States3988 Posts
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Raithed
China7078 Posts
Bubbled input NAND --- inverter --- bubbles cancel? Only this way? ![[image loading]](http://i28.tinypic.com/11kxzbq.jpg) | ||
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Raithed
China7078 Posts
NAND + inverter = comes out as AB? Instead of (AB)' ? | ||
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fight_or_flight
United States3988 Posts
![[image loading]](http://img225.imageshack.us/img225/8583/drawing1ty7.png) | ||
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Raithed
China7078 Posts
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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fight_or_flight
United States3988 Posts
edit: might be the same thing your saying, I'm getting confused over this whole thread. | ||
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Raithed
China7078 Posts
How is it what I drew not a NAND gate? A NAND gate as a bubble at the output, but I just added two bubbles on the input? | ||
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fight_or_flight
United States3988 Posts
![[image loading]](http://img231.imageshack.us/img231/3885/76077200dc8.png) | ||
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fight_or_flight
United States3988 Posts
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berated-
United States1134 Posts
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azndsh
United States4447 Posts
deMorgan's Law | ||
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sigma_x
Australia285 Posts
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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Raithed
China7078 Posts
X = (A’ + B’) + (B+C) X = 1? I get (A’ + B’) = (AB)' + (B+C) right? What is the next step? | ||
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fight_or_flight
United States3988 Posts
X = A’ + (B’ + B) + C X = A’ + 1 + C X = 1 | ||
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Raithed
China7078 Posts
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