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LocoDelAssembly
Warning, elementary school undergraduate question is about to begin...
On algebra of real numbers we can have the following equation: Code: X + Y = 0 Now if we want to have X alone on the left side we just substract Y on both sides and we get a sightly more simplified equation: Code: X + Y  Y = 0  Y X = Y Now when I turn into boolean algebra with a very simple equation I found no way to resolve stepbystep, I just get to the end "magically" Code: ; + is OR, * is AND, ~ is NOT A*~B + ~A*B = 1 ; Magic: A = ~B I know that it is valid because it has the same solution set (the one that A XOR B = TRUE has). My problem here is that I jumped to the final expression without any intermediate step, something I found very hateable. So, the question: there is no intermediate step? There is no possible step between the original equation and "the magic equivalent"? The motivation around this is homework, but actually I don't need it at all since I'm not required at all to show in a formal way why I choose one of the conditions of a problem to make up the decision table. If an object can only be red or blue then it is no possible that can't be any of these nor both so using "red" as a condition or "blue" is enough and because I am an hysterical person I want to show mathematically by showing the aforementioned "magic equation" why in the conditions set of a given problem I keep only one of two mutualy exclusive conditions. Thanks for reading PS: And in fact I want this for myself since doing this on the exam can cause me to take the same course the next year for doing wierd nonteached things. PS2: In case you don't know what decision tables are: http://en.wikipedia.org/wiki/Decision_tables 

23 Nov 2007, 01:52 

vid
Sorry, can't help, but... Man, am I happy to be out of university !!!
(no discouragement intended) 

23 Nov 2007, 02:23 

edfed
Code: ;* is and, + is or,  is not, + is nor etc ... 1=0  0=1 A*A=A 1*1=1  1*0=0  0*0=0 1+0=1  0+0=0  1+1=1 (A*B) + (A*B) = 1 A*B=(A*B) ; i didn't put the 1 cause it's a constant, a constant doesn't have condition, refering to differencial computation, Constant'=0 A=A*B*B ;there is no reverse and A*A=B*B ;no div equivalent A=B like and is mul equivalent and can be changed of side without changing with 0 and 1 it's simplest (0*1) + (0*1) = 1 (0*1)=(0*1) 0=0*1*1 0*0=1*1 0=1 ; ohhh, it's the relation of not!^^ please, note that boolean algebra is not algebra! 

23 Nov 2007, 03:22 

Xorpd!
Code: (A*(B))+((A)*B) = 1 ; Given ((A*(B))+((A)*B))*(B) = 1*(B) ; Rightmultiply by (B) ((A*(B))*(B))+(((A)*B)*(B)) = B ; Distributive law and evaluation (A*((B)*(B)))+((A)*(B*(B))) = B ; Associative law (A*(B))+((A)*0) = B ; Evaluation (A*(B))+0 = B ; Evaluation A*(B) = B ; Evaluation [equation 1] (A*(B))+((A)*B) = 1 ; Given A*((A*(B))+((A)*B)) = A*1 ; Leftmultiply by A (A*(A*(B)))+(A*((A)*B)) = A ; Distributive law and evaluation ((A*A)*(B))+((A*(A))*B) = A ; Associative law (A*(B))+(0*B) = A ; Evaluation (A*(B))+0 = A ; Evaluation A*(B) = A ; Evaluation [equation 2] A = B ; Equation 1, equation 2, and transitivity What if we started with (A*(B))+((A)*B) = 0? (A*(B))+((A)*B) = 0 ((A*(B))+((A)*B))+B = 0+B (((A)*B)+(A*(B)))+B = B ((A)*B)+((A*(B))+B) = B ((A)*B)+((A+B)*((B)+B)) = B ((A)*B)+((A+B)*1) = B (B*(A))+(A+B) = B ((B*(A))+A)+B = B ((B+A)*((A)+A))+B = B ((A+B)*1)+B = B (A+B)+B = B A+(B+B) = B A+B = B Then if you OR both sides with A you would get A+B = A So that it follows that A = B I assume that edfed is joking above: A+B = 1 doesn't imply that A = B because they could both be 1. Another possibile approach is to XOR both sides with B and simplify. 

23 Nov 2007, 07:27 

MHajduk
۷ = OR
۸ = AND ↔ = 'iff' i.e. 'if and only if' ~ = NOT Proof: (A ۸ ~B) ۷ (~A ۸ B) = 1 ↔ (A ۸ ~B) = 1 ۷ (~A ۸ B) = 1 ↔ (A = 1 ۸ ~B = 1) ۷ (~A = 1 ۸ B = 1) ↔ (A = 1 ۸ B = 0) ۷ (A = 0 ۸ B = 1) ↔ (B = 0 ۸ A = ~B) ۷ (B = 1 ۸ A = ~B) ↔ (B = 0 ۷ B = 1) ۸ (A = ~B) ↔ 1 ۸ (A = ~B) ↔ A = ~B 

23 Nov 2007, 08:56 

edfed
i don't joke
Code: A+B=1??? is this the joke???? A=(a*b) ? B=(a*b) ? in this condition, A=B A+B=0 A=(a*b) B=(a*b) and then, A=B=0 

23 Nov 2007, 15:15 

LocoDelAssembly
OK, I need to digest all this
I have some doubts still, but I will think some more time before telling a word. Thanks to all 

23 Nov 2007, 16:02 

bitRAKE
edfed wrote: please, note that boolean algebra is not algebra! 

23 Nov 2007, 17:05 

edfed
Quote:
FALSE! Boolean algebra was invented by george boole, a british mathematician. algebra was invented by nobodyknowsbecauseit'stoofar Boolean algebra is only applicated to discretes and quantified algebra binary 0/1 logic proposition between 2 values or more algebra covers the total mathemetic domain. 

23 Nov 2007, 23:13 

bitRAKE
Haha...just because it was discovered afterward doesn't mean it is any less all encompassing. Nothing is more powerful than 0 and 1.


24 Nov 2007, 00:05 

edfed
0/1=infinite
but computer world is not real world computers only understands 0 & 1 Code:  boolean algebra is there
human only understand real values and variables Code:  assembler is there
computers only understand instructions there are many other ways to convert mathematics geometry, optics, analog electrronics, computers, the cheapest and easiest to make is the digital silicium chip. but when quantic processors will appears, we'll need one extra definition of boole algebra. an algebra based on trinary or other.1,0,1 

24 Nov 2007, 00:18 

LocoDelAssembly
Quote:
If you mean quantum computers so far them are not the future for general purpose use, the algorithms also need to be quantic but AFAIK there is no way to translate all the normal algorithms to quantum algorithms. A fragment of the wikipedia: Quote: Consider a problem that has these four properties: http://en.wikipedia.org/wiki/Quantum_computer I felt very dissapointed when a friend, a physicist him, told me that it is not usable for general purpose, I though that all this was just replacing silicon chips by quantum hardware but keeping x86 architecture (or whatever) intact but NO, it is a completely different world 

24 Nov 2007, 01:26 

bitRAKE
I don't deny the limitations, but there is no system without the same limitations. What reality? If we attempt to communicate any part of reality then we have in some way reduced it to binary form, and in doing so lie.
All forms of logic can be emulated with binary. Complex forms are more terse, but it acts merely as a form of compression  no special computablity is granted by brevity. Even fuzzy logic can be represented by binary logic. A choice of percision needs to be made, and the same can be said for any system. 

24 Nov 2007, 01:37 

edfed
the pure math system have no limit.
pure math domain is the meta domain. everything can be explained with maths. 

24 Nov 2007, 18:32 

bitRAKE
Did you run that by Kurt Gödel?
[Edit By loco] Corrected link to use ASCII only chars. It is required for the forum, otherwise the url tags does not work. 

24 Nov 2007, 19:46 

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