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Old 03-07-2002, 12:00 AM   #1
Mayfuck
 
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Talking 2=1

a = b

a2 = ab

a2-b2 = ab-b2

(a+b)(a-b) = b(a-b)

a+b = b

2a = a

2 = 1


[This message has been edited by Mayfuck (edited 03-06-2002).]

 
Old 03-07-2002, 12:01 AM   #2
melancholymystic
 
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Question

wtf?


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I dont' discrimanate, I hate everyone.

 
Old 03-07-2002, 12:02 AM   #3
zekix
 
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..Please tell me that isn't Boolean algebra...

GEORGE BOOLE MUST DIE.

 
Old 03-07-2002, 12:09 AM   #4
bonsor
 
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Quote:
Originally posted by Mayfuck:
a2-b2 = ab-b2
a2-b2 is equal to zero. Factoring that would require dividing. You can't divide numbers into zero.

Owned.


 
Old 03-07-2002, 12:10 AM   #5
Jaggie
 
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Post

Quote:
Originally posted by ******:
a2-b2 is equal to zero. Factoring that would require dividing. You can't divide numbers into zero.

Owned.

Damn, beat me to it.


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i wasted all my years, been chasing all my fears

 
Old 03-07-2002, 12:14 AM   #6
undivinemartyr
 
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Unhappy

Which reminds me, I have to study and I'm not doing it.

 
Old 03-07-2002, 12:14 AM   #7
pale_princess
 
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Red face

u + me = us.

 
Old 03-07-2002, 12:18 AM   #8
Irrelevant
 
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Red face

we hope that you choke.

 
Old 03-07-2002, 12:23 AM   #9
pink_ribbon_scars
 
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Post

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.

 
Old 03-07-2002, 12:25 AM   #10
Mayfuck
 
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Red face

Quote:
Originally posted by pink_ribbon_scars:
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.
It's you!

 
Old 03-07-2002, 12:29 AM   #11
Delta
 
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Post

Quote:
Originally posted by Mayfuck:
a = b
(a+b)(a-b) = b(a-b)
a+b = b
Quote:
Originally posted by pink_ribbon_scars:
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.
dude, going from line 2 to line 3 he divides by zero. im really not big into math at all, but i cant think of any example in any system where division by zero is allowed. i dunno, if it's your major maybe you've seen something i havent, but i honestly cant think of any explanation for that being legit

 
Old 03-07-2002, 12:29 AM   #12
Oblivious
 
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Red face

http://www.aspirin-foundation.com/Headache.jpg

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hello kitten...you'll never know what it's like to be me curled at your feet...

 
Old 03-07-2002, 12:31 AM   #13
lucky_13
 
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liz you're the smartest. http://www.netphoria.org/wwwboard/smile.gif

 
Old 03-07-2002, 01:02 AM   #14
pink_ribbon_scars
 
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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 is a set G, closed under a binary operation *, such that the following axioms are satisfied:
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 03-06-2002).]

 
Old 03-07-2002, 01:11 AM   #15
pink_ribbon_scars
 
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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 (a-b) was a member of a group, than its inverse (a-b)' must be in the group, too.
So (a+b)(a-b)=b(a-b) can be multipled on the right by (a-b)' such that
(a+b)(a-b)(a-b)'=b(a-b)(a-b)'
so
(a+b)e=be
which means
(a+b)=b

take that niggaz

 
Old 03-07-2002, 01:14 AM   #16
zekix
 
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Post

Wow.
Nifty.

 
Old 03-07-2002, 01:15 AM   #17
Delta
 
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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

 
Old 03-07-2002, 01:20 AM   #18
Delta
 
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Post

Quote:
Originally posted by pink_ribbon_scars:
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 (a-b) was a member of a group, than its inverse (a-b)' must be in the group, too.
So (a+b)(a-b)=b(a-b) can be multipled on the right by (a-b)' such that
(a+b)(a-b)(a-b)'=b(a-b)(a-b)'
so
(a+b)e=be
which means
(a+b)=b

take that niggaz
ok, now i dont disagree, you're completely right that that piece is legit. but, wouldnt that mean that a = 0, and therefore b = 0? and if so...

Quote:
Originally posted by Mayfuck:
a+b = b
2a = a
2 = 1
theres now a division of 0 going from the second line to the third line. it should actually read

a+b=b
2a=a
2a - a = 0
a = 0

which makes sense (to me at least)

 
Old 03-07-2002, 01:20 AM   #19
pink_ribbon_scars
 
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Post

Quote:
Originally posted by Delta:
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

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.

 
Old 03-07-2002, 01:26 AM   #20
pink_ribbon_scars
 
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Post

Quote:
Originally posted by Delta:
theres now a division of 0 going from the second line to the third line. it should actually read

a+b=b
2a=a
2a - a = 0
a = 0

which makes sense (to me at least)
That sounds good, I think that original prolem is a proof that if a=b then a must be 0.
I think we nailed it. http://www.netphoria.org/wwwboard/smile.gif

 
Old 03-07-2002, 01:33 AM   #21
Delta
 
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Post

Quote:
Originally posted by pink_ribbon_scars:
That sounds good, I think that original prolem is a proof that if a=b then a must be 0.
I think we nailed it. http://www.netphoria.org/wwwboard/smile.gif
lol, good, so now ill be able to sleep tonite http://www.netphoria.org/wwwboard/wink.gif
this is why i stay away from math. give me a nice qualitative science like biology anyday

 
Old 03-07-2002, 03:39 AM   #22
Hazza
 
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Post

to prove 1 doesn't = 2
if 1=2 then 1=0
proof
1*** = 2
there fore (if 1=2)
1+2=2
1+2-2=2-2
1=0

and
1+j=0
j***-1=0-1
there fore j=0 (because 0=1)
there fore every number = every other number
and it doesn't work.


 
Old 03-07-2002, 04:17 AM   #23
Jaggie
 
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Post

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.

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i wasted all my years, been chasing all my fears

 
Old 03-07-2002, 08:30 AM   #24
Nothing/everything
 
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Talking

***?

 
Old 03-07-2002, 01:25 PM   #25
Brazil.Virex
 
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Post

Quote:
Originally posted by Nothing/everything:
***?

 
Old 03-07-2002, 01:30 PM   #26
Smiley33
 
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Post

numbers is sexy/+

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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

 
Old 03-07-2002, 03:42 PM   #27
Never_Nohen
 
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Post

http://www.netphoria.org/wwwboard/biggrin.gif

 
Old 03-07-2002, 04:02 PM   #28
Mayfuck
 
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Post

Quote:
Originally posted by Jaggie:
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'm not really gonna answer your question other than the existence of numbers and the definite borders between them are there for quantifiable purposes. It's really as simple as that. But the existence of numbers as an autonomous physical embodiment is something to think about. There's an idea that, in a post-Platonic sense of numbers, integers do not exist seperately as individual numbers, but rather a result of counting, because without counting one can conclude that numbers don't have any ontoligical verification. Furthermore, there is a theory that math itself consists only of concepts and theorems and postulates that are only realized through an array of finite steps and procedures, in a limitied space/time, by a finite being. It seems that only through this realization can we associate mathematics with the physical world and thus justifying its existence, because to ask if "1" exists, or if "3.4" exists on its own, ideas begin to get sketchy.


[This message has been edited by Mayfuck (edited 03-07-2002).]

 
Old 03-07-2002, 04:51 PM   #29
Mayonaise
 
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Post

well that made me laugh

 
Old 03-07-2002, 09:01 PM   #30
Irrelevant
 
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Post

Quote:
Originally posted by Jaggie:
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.
have you heard of Zeno's Paradox? here's an example i got off the web:

A runner wants to run a certain distance - let us say 100 meters - in a finite time. But to reach the 100-meter mark, the runner must first reach the 50-meter 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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