
03062002, 11:00 PM  #1 
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2=1
a = b
a^{2} = ab a^{2}b^{2} = abb^{2} (a+b)(ab) = b(ab) a+b = b 2a = a 2 = 1 [This message has been edited by Mayfuck (edited 03062002).] 

03062002, 11:01 PM  #2 
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wtf?
 I dont' discrimanate, I hate everyone. 

03062002, 11:02 PM  #3 
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..Please tell me that isn't Boolean algebra...
GEORGE BOOLE MUST DIE. 

03062002, 11:09 PM  #4  
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Owned. 


03062002, 11:14 PM  #6 
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Which reminds me, I have to study and I'm not doing it.


03062002, 11:14 PM  #7 
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u + me = us.


03062002, 11:18 PM  #8 
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we hope that you choke.


03062002, 11:23 PM  #9 
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I think it's fine. I'm not the official on math, but I am a senior math major. Just because something isn't correct on the integers under usual addition and multiplication doesn't mean it isn't correct, as long as it follows a set of rules. Without defined rules, 1=1 might be bullshit. Not that we had well defined rules for this problem... I don't think I did a good job of explaining that.


03062002, 11:25 PM  #10  
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03062002, 11:29 PM  #11  
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03062002, 11:29 PM  #12 
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http://www.aspirinfoundation.com/Headache.jpg
 hello kitten...you'll never know what it's like to be me curled at your feet... 

03062002, 11:31 PM  #13 
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liz you're the smartest. http://www.netphoria.org/wwwboard/smile.gif


03072002, 12:02 AM  #14 
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thanks julie, you're the coolest. http://www.netphoria.org/wwwboard/wink.gif
er, i'm only doing this to avoid washing dishes, but to end the debate i can tell you about the left and right cancellation laws of abstract algebra. if you've got a few minutes you should read it, it's not as bad as it looks. in abstract algebra, you study groups, rings and fields. here is the definition of a group. Definition (Group) A group G1: Associativity of * For all a,b,c in G, we have (a*b)*c=a*(b*c) G2: Identity e for * e*x=x*e=x G3: Inverse a' of a Corresponding to each element a in G, there is an element a' in G such that a*a'=a'*a=e. So, for example, the integers [...,2,1,0,1,2,...] under addition are a group. G1: We all know that the integers under addition are associative, i.e., (1+2)+3=3+3=6 1+(2+3)=1+5=6 G2: 0 is the additive identity for everything; 2+0=2, 3+0=3 and so on. G3: And for any element, say 2, there is an inverse, 2 in the case, such that 2+(2)=0. So now we can introduce a theorem. Theorem If G is a group with a binary operation *, the the left and right cancellation laws hold in G, that is, a*b=a*c implies b=c, and b*a=c*a implies b=c where a,b,c are elements of G. Proof Suppose a*c=a*c. Then by G3, there exists a', and a'*(a*b)=a'*(a*c). By the associative law, (a'*a)*b=(a'*a)*c. By the definition of a' in G3, a'*a=e, so e*b=e*c. by the definition of e in G2, b=c. Yeah, so the group that mayfuck gave may or may not be a group [I can check if you care but I'm sure no one has read this far anyway!]. But the point is that under the left and right cancellation laws you can "divide by 0" [but mind you you're really not "dividing by 0" since this is a special system...] [This message has been edited by pink_ribbon_scars (edited 03062002).] 

03072002, 12:11 AM  #15 
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i reaize that i'm a huge dork, but i just figured out that you wouldn't even need the left and right cancellation laws in this case. if (ab) was a member of a group, than its inverse (ab)' must be in the group, too.
So (a+b)(ab)=b(ab) can be multipled on the right by (ab)' such that (a+b)(ab)(ab)'=b(ab)(ab)' so (a+b)e=be which means (a+b)=b take that niggaz 

03072002, 12:14 AM  #16 
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Wow.
Nifty. 

03072002, 12:15 AM  #17 
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i kind of get what you're saying, that cancellation in a group isnt really the same thing as dividing both sides by the same term, but i dunno, theres something that still seems off somewhere


03072002, 12:20 AM  #18  
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a+b=b 2a=a 2a  a = 0 a = 0 which makes sense (to me at least) 


03072002, 12:20 AM  #19  
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I'm with you actually, I really doubt that it's a group. I just wanted to show you that it is possible in some cases. 


03072002, 12:26 AM  #20  
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I think we nailed it. http://www.netphoria.org/wwwboard/smile.gif 


03072002, 12:33 AM  #21  
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this is why i stay away from math. give me a nice qualitative science like biology anyday 


03072002, 02:39 AM  #22 
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to prove 1 doesn't = 2
if 1=2 then 1=0 proof 1*** = 2 there fore (if 1=2) 1+2=2 1+22=22 1=0 and 1+j=0 j***1=01 there fore j=0 (because 0=1) there fore every number = every other number and it doesn't work. 

03072002, 03:17 AM  #23 
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I've always pondered something...How can any number truly be defined? How can we say something is 2. There are infinite amount of numbers in between 1 and 2 (1.5, 1.55, 1.555, etc), thus one can never actually make it to 2. Does that make sense? It's kinda like putting a definite number as to the distance to the end of the universe. Assuming the universe is infinite, we can place an arbitrary value (for example, EU can stand for the distance to the end of the universe) but this value will never be defined because there is an infinite amount of numbers in between.
 i wasted all my years, been chasing all my fears 

03072002, 07:30 AM  #24 
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***?


03072002, 12:25 PM  #25  
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03072002, 12:30 PM  #26 
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numbers is sexy/+
 there was a time when you let me know what's really going on below but now you never show that to me do ya 

03072002, 02:42 PM  #27 
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03072002, 03:02 PM  #28  
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Quote:
[This message has been edited by Mayfuck (edited 03072002).] 


03072002, 03:51 PM  #29 
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well that made me laugh


03072002, 08:01 PM  #30  
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Quote:
A runner wants to run a certain distance  let us say 100 meters  in a finite time. But to reach the 100meter mark, the runner must first reach the 50meter mark, and to reach that, the runner must first run 25 meters. But to do that, he or she must first run 12.5 meters. Since space is infinitely divisible, we can repeat these 'requirements' forever. Thus the runner has to reach an infinite number of 'midpoints' in a finite time. This is impossible, so the runner can never reach his goal. In general, anyone who wants to move from one point to another must meet these requirements, and so motion is impossible, and what we perceive as motion is merely an illusion. and calculus deals with these limits. i don't know if it solves this problem, but it does negate its effect. 


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