View Full Version : 2=1


Mayfuck
03-06-2002, 10:00 PM
a = b

a<sup>2</sup> = ab

a<sup>2</sup>-b<sup>2</sup> = ab-b<sup>2</sup>

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

a+b = b

2a = a

2 = 1
<font color=black size=1>

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

melancholymystic
03-06-2002, 10:01 PM
wtf?


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I dont' discrimanate, I hate everyone.

zekix
03-06-2002, 10:02 PM
..Please tell me that isn't Boolean algebra...

GEORGE BOOLE MUST DIE.

bonsor
03-06-2002, 10:09 PM
<font face="Arial, Verdana" size="2">Originally posted by Mayfuck:
a<sup>2</sup>-b<sup>2</sup> = ab-b<sup>2</sup></font>

a<sup>2</sup>-b<sup>2</sup> is equal to zero. Factoring that would require dividing. You can't divide numbers into zero.

Owned.

Jaggie
03-06-2002, 10:10 PM
<font face="Arial, Verdana" size="2">Originally posted by ******:
a<sup>2</sup>-b<sup>2</sup> is equal to zero. Factoring that would require dividing. You can't divide numbers into zero.

Owned.

</font>

Damn, beat me to it.


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i wasted (http://www.jaggie.net) all my years, been chasing all my fears

undivinemartyr
03-06-2002, 10:14 PM
Which reminds me, I have to study and I'm not doing it.

pale_princess
03-06-2002, 10:14 PM
u + me = us.

Irrelevant
03-06-2002, 10:18 PM
we hope that you choke.

pink_ribbon_scars
03-06-2002, 10:23 PM
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.

Mayfuck
03-06-2002, 10:25 PM
<font face="Arial, Verdana" size="2">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.</font>

It's you!

Delta
03-06-2002, 10:29 PM
<font face="Arial, Verdana" size="2">Originally posted by Mayfuck:
a = b
(a+b)(a-b) = b(a-b)
a+b = b
</font>
<font face="Arial, Verdana" size="2">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.</font>

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

Oblivious
03-06-2002, 10:29 PM
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...

lucky_13
03-06-2002, 10:31 PM
liz you're the smartest. http://www.netphoria.org/wwwboard/smile.gif

pink_ribbon_scars
03-06-2002, 11:02 PM
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 <G,*> 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).]

pink_ribbon_scars
03-06-2002, 11:11 PM
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

zekix
03-06-2002, 11:14 PM
Wow.
Nifty.

Delta
03-06-2002, 11:15 PM
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

Delta
03-06-2002, 11:20 PM
<font face="Arial, Verdana" size="2">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</font>

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...

<font face="Arial, Verdana" size="2">Originally posted by Mayfuck:
a+b = b
2a = a
2 = 1</font>

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)

pink_ribbon_scars
03-06-2002, 11:20 PM
<font face="Arial, Verdana" size="2">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</font>


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.

pink_ribbon_scars
03-06-2002, 11:26 PM
<font face="Arial, Verdana" size="2">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)</font>

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

Delta
03-06-2002, 11:33 PM
<font face="Arial, Verdana" size="2">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</font>

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

Hazza
03-07-2002, 01:39 AM
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.

Jaggie
03-07-2002, 02:17 AM
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 (http://www.jaggie.net) all my years, been chasing all my fears

Nothing/everything
03-07-2002, 06:30 AM
***?

Brazil.Virex
03-07-2002, 11:25 AM
<font face="Arial, Verdana" size="2">Originally posted by Nothing/everything:
***?</font>

Smiley33
03-07-2002, 11:30 AM
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

Never_Nohen
03-07-2002, 01:42 PM
http://www.netphoria.org/wwwboard/biggrin.gif

Mayfuck
03-07-2002, 02:02 PM
<font face="Arial, Verdana" size="2">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.

</font>

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.<font color=black size=1>


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

Mayonaise
03-07-2002, 02:51 PM
well that made me laugh

Irrelevant
03-07-2002, 07:01 PM
<font face="Arial, Verdana" size="2">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.</font>

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.

bonsor
03-07-2002, 07:23 PM
<font face="Arial, Verdana" size="2">Originally posted by Irrelevant:
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.</font>

The only reason space is allegedly infinitely divisible is because we apply the abstract concept of mathematics to it.

For instance, the quantam leap, the phenomenon on electrons traveling from different electron shells of an atom. When they do that, they don't (appear to) exist anywhere between the two electron shells, which in a way helps prove that space is definitely divisible.

twice
03-07-2002, 07:43 PM
*head implodes*

jenny4ever
03-07-2002, 09:35 PM
holy shit lawson was just telling me about this like a week ago.

Nate the Grate
03-07-2002, 09:45 PM
Let's talk about PI!

*awaits discussion*

raindrops + sunshowers
03-07-2002, 09:45 PM
Dorks.

Jaggie
03-07-2002, 10:16 PM
I agree with both of you, Irrelevent and ******. I don't know if there is an answer really. Numbers are arbitrary.

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i wasted (http://www.jaggie.net) all my years, been chasing all my fears

Mayfuck
03-07-2002, 10:18 PM
There's a lengthy answer to Zeno's Paradox, but it has to do with the collapse of waves and quantum mechanics, and other stuff that goes over my head.

SPFreak37
03-07-2002, 10:44 PM
<font face="Arial, Verdana" size="2">Originally posted by pale_princess:
u + me = us.</font>
that's what i'm talking about!!


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dying of boredom...i'll try it all

Ammy
03-07-2002, 10:45 PM
this thread makes me all warm inside.

scouse_dave
03-08-2002, 12:31 PM
<font face="Arial, Verdana" size="2">Originally posted by Irrelevant:
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.</font>

well, this would certainly be true if 'infinity', a much misunderstood word, was a definitive number. in reality, it is agreed that there are various types of infinities, some 'larger' than others, or diverging more rapidly than others.

in this paradox, you're essentially summing an infinite series of increasingly small distances which sum to 100m (50+25***2.5+etc). corresponding to this series is the time taken to traverse these distances. each of these specific times is finite, but there is not a hard and fast rule with regards to summing an infinite series of finite numbers. in some instances, the total diverges to infinity and in others it converges to a finite answer, as in this example.

Assuming that the runner maintains a constant pace throughout the race, and it takes him X seconds to reach the 50m mark, then the total time for the race is:

T = X + X/2 + X/4 + X/8 + X/16 + ... (TO INFINITY)
= sum(X/(2^(n-1)) ; where n corresponds to the nth number in the series

This can be proven to sum to 2X (i'm sure as hell not gonna go into that now). in essence, the subsequent terms added on to the total each time are not 'strong' enough to break free from a finite asymptotic constraint, and hence this may be considered a 'weak' infinity. (otherwise it would sum to infiity)

hence the time it takes the runner to reach the 50m mark is exactly half the total race time. this is a result which you'd expect.

yeah, it's weird, but then so is infinity. deal with it. technically, it's not a paradox either...it's just phrased in a manner to deliberately confuse people. a paradox holds up to scrutiny. and yes, i'm a bit bored.

Jaggie
03-08-2002, 12:59 PM
<font face="Arial, Verdana" size="2">Originally posted by scouse_dave:
well, this would certainly be true if 'infinity', a much misunderstood word, was a definitive number. in reality, it is agreed that there are various types of infinities, some 'larger' than others, or diverging more rapidly than others.

in this paradox, you're essentially summing an infinite series of increasingly small distances which sum to 100m (50+25***2.5+etc). corresponding to this series is the time taken to traverse these distances. each of these specific times is finite, but there is not a hard and fast rule with regards to summing an infinite series of finite numbers. in some instances, the total diverges to infinity and in others it converges to a finite answer, as in this example.

Assuming that the runner maintains a constant pace throughout the race, and it takes him X seconds to reach the 50m mark, then the total time for the race is:

T = X + X/2 + X/4 + X/8 + X/16 + ... (TO INFINITY)
= sum(X/(2^(n-1)) ; where n corresponds to the nth number in the series

This can be proven to sum to 2X (i'm sure as hell not gonna go into that now). in essence, the subsequent terms added on to the total each time are not 'strong' enough to break free from a finite asymptotic constraint, and hence this may be considered a 'weak' infinity. (otherwise it would sum to infiity)

hence the time it takes the runner to reach the 50m mark is exactly half the total race time. this is a result which you'd expect.

yeah, it's weird, but then so is infinity. deal with it. technically, it's not a paradox either...it's just phrased in a manner to deliberately confuse people. a paradox holds up to scrutiny. and yes, i'm a bit bored.

</font>

OK, so what's your take on my earlier question?

scouse_dave
03-08-2002, 01:42 PM
<font face="Arial, Verdana" size="2">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?</font>

numbers are totally man-made. before mathematicians, there wasn't a '2' - it didn't exist. in fact there's a case to say that it doesn't exist now. http://www.netphoria.org/wwwboard/tongue.gif
Even if you accept that numbers exist, it's hard to argue against them being so fundamentally basic that even giving a common-sense definition to a friend is impossible. How do you explain to someone 'two' without holding up two bananas and saying "look!!! TWO bananas!!!" or whatever?? we're all so comfortable with numbers cos we use them all the time, not cos we understand what they are.

It's true that by starting from 1 and slowly increasing in increments, it may be impossible to reach 2, but i don't see why that's a problem. again, we've invented an increasingly complex number system because various modern-day practices require more accuracy; engineering etc. greeks knew that pi was roughly 22/7, yet they were quite happy approximating it to 3 most of the time.

so i don't even know what you're trying to say. if we were being picky we couldn't count up to 2, that's true; but we invented 2 first http://www.netphoria.org/wwwboard/smile.gif it's just not a problem.

<font face="Arial, Verdana" size="2">Originally posted by Jaggie:
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.
</font>

well as far as i know, the universe is still expanding, so this number EU couldn't ever be defined anyway. my maths is better than my physics, so i'm probably not the best person to ask on this. i don't see why the infinite amount of numbers inbetween is a problem tho. you could divide your shoelaces into infinitely small segments; but that doesn't stop you saying that they're 12 inches long or whatever.

maths is natural. the way in which we define maths is completely artificial

Jaggie
03-08-2002, 03:44 PM
But that's my whole point, numbers are not only artificial, but completely undefined. If something is undefined, how can it be used with any validity. The EU example was the same idea. You said that the universe is expanding and that the true distance to the end is undefined, thus how can we use that number in a formula?

So very sad about me
03-08-2002, 06:18 PM
<font face="Arial, Verdana" size="2">Originally posted by Jaggie:
The EU example was the same idea. You said that the universe is expanding and that the true distance to the end is undefined, thus how can we use that number in a formula?</font>

cos even tho it's changing, you're still able to calculate it?

i dunno, i'm not sure about any of this.

i think the universe discussion should be entirely separate from the numbers one...

pink_ribbon_scars
03-08-2002, 08:31 PM
<font face="Arial, Verdana" size="2">Originally posted by Jaggie:
But that's my whole point, numbers are not only artificial, but completely undefined. If something is undefined, how can it be used with any validity. The EU example was the same idea. You said that the universe is expanding and that the true distance to the end is undefined, thus how can we use that number in a formula?</font>


numbers aren't completely undefined, are they? they ARE defined, and that's what makes them what they are. (?)

Jaggie
03-08-2002, 09:53 PM
<font face="Arial, Verdana" size="2">Originally posted by pink_ribbon_scars:

numbers aren't completely undefined, are they? they ARE defined, and that's what makes them what they are. (?)

</font>

But what we're saying is that because there are infinite amounts of numbers in between a set of numbers, nothing except for 0 can be defined.

bonsor
03-08-2002, 10:17 PM
<font face="Arial, Verdana" size="2">Originally posted by Jaggie:
But what we're saying is that because there are infinite amounts of numbers in between a set of numbers, nothing except for 0 can be defined.</font>
Yes it can.

1 is 1. 2 is 2. Numbers are defined. That is why mathematics exists.

Jaggie
03-08-2002, 11:49 PM
<font face="Arial, Verdana" size="2">Originally posted by ******:
Originally posted by Jaggie:
But what we're saying is that because there are infinite amounts of numbers in between a set of numbers, nothing except for 0 can be defined.</font>
Yes it can.

1 is 1. 2 is 2. Numbers are defined. That is why mathematics exists.

What is 2 defined as? It cannot be defined because all it is is 1 + 1. But there are infinite amount of numbers between 1 and 2 (also between 0 and 1). Thus we can substitute infinity for any number other than 0. So then the formula becomes infinity plus infinity = 2? No, can't be, you cannot add an undefinable (infinity) number to anything. Infinity plus 1 still equals infinity.

Mustard
03-09-2002, 02:02 AM
i like inner toobs! http://www.netphoria.org/wwwboard/smile.gif

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If God is inside of me like everybody says he is, then I hope he likes enchiladas, cause that's what he's getting.

the_boy
03-09-2002, 02:17 AM
<font face="Arial, Verdana" size="2">Originally posted by scouse_dave:

Even if you accept that numbers exist, it's hard to argue against them being so fundamentally basic that even giving a common-sense definition to a friend is impossible. How do you explain to someone 'two' without holding up two bananas and saying "look!!! TWO bananas!!!" or whatever?? we're all so comfortable with numbers cos we use them all the time, not cos we understand what they are. </font>
Ya, i think you hit the nail on the head with this here. This is exactly like an discussion we had in one of my Philosophy Of Language classes. Basically I think that understanding exactly what numbers are and what they mean is much less important than understanding what they represent . So long as we have this knowledge, we can proceed with mathematics and our daily lives without having to worry about all this "Zeno's Paradox" bullshit which will probably never really have an answer. And thats good enough for me. http://www.netphoria.org/wwwboard/smile.gif