How big is infinity?

Page 2 - Seeking answers? Join the AnandTech community: where nearly half-a-million members share solutions and discuss the latest tech.

Hazer

Member
Feb 16, 2003
104
0
0
Originally posted by: dejitaru
1+1-1+1-1...=1+1-1+1-1...
so
1+(1-1)+(1-1)...=1+1+(-1+1)+(-1+1)...
simplified
1+(0)+(0)...=2+(0)+(0)...
therefore
1=2...

In line 1: 1+1-1+1-1...=1+1-1+1-1...
In line 2: 1+(1-1)+(1-1)...=1+1+(-1+1)+(-1+1)...
Here's what you missed though

Line3: 1 + Z(1-1) +1 = 1 + 1 + Z(-1+1)
You started your parenthesis early in the first half of your equation, and hence did NOT add the last +1 in it. Thats where your confusing everyone. You cannot simply manipulate your equation using infinite to bound a larger part of the equation on one side, and bounding a smaller portion on the other. This would negate equality.


Its not that you added an extra +1 on the right side of the =, its that you left out the last +1 on the left side.
 

RaynorWolfcastle

Diamond Member
Feb 8, 2001
8,968
16
81
Here is a very simple and intuitive way to demonstrate the different "sizes" of infinity:

take the function ln(x)/x, now take the limit as x -> infinity
we now have

lim (x->??) ln(x)/x = ??/?? = undefined
but L'Hopital's rule (which basically says there are different sizes of infinity, that is you can approach infinity at different "speeds")
states that the limit for a fraction where lim (x->k) p/q = ??/?? we can say that lim (x->k) (dp/dx)/(dq/dx)

so now we have lim (x->??) (1/x)/1 = 0

so the infinity associated to ln(x) as x -> ?? is smaller than the infinity of x as x-> ?? :)
You can check this with a graph if you like, or by plugging this into your calculator and using a very large number for x

Of course this is far from formal, but it can be easily verified.
 

dejitaru

Banned
Sep 29, 2002
627
0
0
You started your parenthesis early in the first half of your equation, and hence did NOT add the last +1 in it. Thats where your confusing everyone. You cannot simply manipulate your equation using infinite to bound a larger part of the equation on one side, and bounding a smaller portion on the other. This would negate equality.
No, I didn't. Look at it again.
There is no "last", as it is infinite.
The fact that it is a nonterminating equation is what negates equality.
 

Hazer

Member
Feb 16, 2003
104
0
0
If it negates equality, then you cannot conclude that 1=2 then. The equality has been compromised from line 2 of your proof. But regardless, I stand fast on my line 3. The infinite series you bounded with parenthesis is represented by Z(1-1), and as so, you must include the last +1 in the equation. If you dont, you are misrepresenting the equality of the equation. You are manipulating the equation for your own purpose, and does not follow the rules for proving in mathematics.
 

Belegost

Golden Member
Feb 20, 2001
1,807
19
81
Dejitaru is absolutely correct, this is a commonly known property of certain alternating infinite series, they're called conditionally convergent.

To explain his method:

we start with an obviously true statement, that one thing equals itself:

1+1-1+1-... = 1+1-1+1-...

Then, we realize that in R we can group terms of addition in any order we please:

1+(1-1)+(1-1)+... = 1+(1-1)+(1-1)+...

This is the part where you have to understand the meaning of infinity to realize why we can state the next line:

1+(1-1)+(1-1)+... = 1+1+(1-1)+(1-1)...

The thing here is to dispel any notion of infinity-1 being less than infinity, both are equally infinite. Therefore, I have an infinite number of -1's with which I can cancel down with an infinite number of +1's. This means, that even though it appears I haven't accounted for all +1 and -1 terms on the left, I HAVE, because there is an infinite number of each term, therefore I can always find another term, there's no end to them. So I could in fact write this in any form I wish, for instance:

(1-1)+(1-1)+... = 1+1+1+1+(1-1)+(1-1)...
=> 0 = 4

And this remains true, because there are an infinite number or terms on either side, and I can add in any order. So while it looks like I should have 4 extra -1's somewhere on the right side, I never do, because I can always find a +1 to cancel any -1 value with.

This is a common problem with solutions to certain series, especially the harmonic series, they converge to a logical value only for certain conditions: hence, conditionally convergent.

<edit> I finally remembered who proved this, it was Riemann, and you can find information about it under the topic of Reimann's Series Theorem, or Riemann's Rearrangement Theorem.</edit>
 

syberscott

Senior member
Feb 20, 2003
372
0
0
Perhaps infinity does not even really exist. It was stated that 1/3=0.3333... is an infinite series. But let us calculate 1/3 in a base 3 number system. We get 0.1. Hardly infinite.
 

Hazer

Member
Feb 16, 2003
104
0
0
Just read it. How nice to prove my point ;)

Reimann theory of devergence states: that, by a suitable rearrangement of terms, a conditionally convergent series may be made to converge to any desired value, or to diverge.

Such that: 1-1+1-1+1-1+... = (1-1) + (1-1) + (1-1) + ... = 0

OR 1-1+1-1+1-1+...= 1-(1-1)-(1-1)-(1-1)-... = 1

BUT, dejitaru claims that two DIFFERENT Riemanns divergent series equal each other to form 1 = 0. Once you delve into Riemanns proof, you are incorrect in stating (1-1) + (1-1) + (1-1) + ... =1-(1-1)-(1-1)-(1-1)-...

In order for each side to remain equal, the series must be manipulated in the same exact way on each side. You have to read the rules for the conditionally convergent series to get this straight.
 

yodayoda

Platinum Member
Jan 8, 2001
2,958
0
86
the set of integers is said to be "countably infinite" and has a cardinality of aleph null (aleph is the first letter of the hebrew alphabet, looks like a bent N).

the set of real numbers is said to be "uncountably infinite" and has a cardinality of 2 to the aleph null (2 ^ aleph null).

hence, while the integers are inifinite, the reals are infinite and larger than the integers. if you want proof for this, forget it. i'm not going to explain the pigeonhole principle--i'm not in grad school anymore.
 

Kuroyama

Member
Nov 22, 2001
175
0
0
2 = limit_{x->0} 2 = limit_{x->0} 2x/x = 0/0 = 1 (since any number divided by itself is 1)

Digitaru is just doing the same thing. We are both using something that is not well defined as an intermediate term, and therefore can "prove" something that is false. As said a few times, 1+1-1+1-1+... can be made to equal anything, and therefore the sum is not well defined. Both are just meant to be fun tricks.
 

sao123

Lifer
May 27, 2002
12,653
205
106
Regarding the infinite series discussion...

1 + infinity = 2 + infinity = 0 + infinity is a true statement....
However this identity property of infinity cannot be used as a foundation statement to prove 1=2=0 this is just mathematically incorrect.

I could say the same thing along this...

0/0 = 1/0 = 2/0 = infinity...
This also cannot be used as a proof that 0=1=2, it has no mathematical basis.
As most prople try to prove here...

x = y
x^2 = x Y (mult both sides by x)
x^2 - y^2 = x y - y^2 (subtract y^2)
(x + y) (x - y) = y (x - y) (Factor both sides)
(x + y) = y (Divide both sides by x-y)
y + y = y (Subst y for x since x = y)
2y = y
2 = 1 (Divide both sides by y)

Math involving integers cannot be properly mapped on a one to one basis compared mathematically with Infinite math.
 

gururu

Platinum Member
Jul 16, 2002
2,402
0
0
There cannot be a larger sum between two classes of numbers if both extend to infinity. This is because there are no sums. No limits with which to calculate a sum. Without two sums, one can not calculate which is larger (primes or integers).
Infinity is like zero. Neither are numbers. They are abstract concepts that are only primitively defined by mathematics. These concepts define all or nothing. Asking which two infinite classes of numbers is larger is like asking which two classes of nothing is smaller.
 

dejitaru

Banned
Sep 29, 2002
627
0
0
If it negates equality, then you cannot conclude that 1=2 then.
That's exactly what I was trying to prove by my post.
Perhaps infinity does not even really exist. It was stated that 1/3=0.3333... is an infinite series. But let us calculate 1/3 in a base 3 number system. We get 0.1. Hardly infinite.
But it is infinite in a base-10 system. Pick a perspective and stick with it.
in a base-pi (perhaps as unary) system, wacky decimals such as 1 (0.3183) will be nonterminating.
x = y
x^2 = x Y (mult both sides by x)
x^2 - y^2 = x y - y^2 (subtract y^2)
(x + y) (x - y) = y (x - y) (Factor both sides)
(x + y) = y (Divide both sides by x-y)
y + y = y (Subst y for x since x = y)
2y = y
2 = 1 (Divide both sides by y)
x^2 - y^2 = x y - y^2 (subtract y^2)
(x + y) (x - y) = y (x - y) (Factor both sides)

You can't factor x y - y^2, that's divide by zero.

Asking which two infinite classes of numbers is larger is like asking which two classes of nothing is smaller.
Exactly.
 

Cuda1447

Lifer
Jul 26, 2002
11,757
0
71
To me its like this.


Infinity = Infinity


Kinda like .99999...=1

Remember that thread? In one explanation in that thread someone used a little algorithm that said something along the lines of 10.9.... minus .9.... = 10 or something to that affect, why Im getting at is they were both infinity and thus both equal.



 

sash1

Diamond Member
Jul 20, 2001
8,896
1
0
The way I percieve it is that infinity is a term used to describe something.

You can never count infinity because thats just what it is. It's infinite; not finite.

You can count the natural set of numbers out as far as you'd like, as well you can count the real number set as far as you'd like, and they'd both be infinite. You'll never reach the end; that's the concept of infinity. While from what it may seem there are more real numbers than natural numbers, they both equal infinity. And infinity is infinity.

~Aunix
 

Jeff7

Lifer
Jan 4, 2001
41,596
20
81
Originally posted by: AunixM3
The way I percieve it is that infinity is a term used to describe something.

You can never count infinity because thats just what it is. It's infinite; not finite.

You can count the natural set of numbers out as far as you'd like, as well you can count the real number set as far as you'd like, and they'd both be infinite. You'll never reach the end; that's the concept of infinity. While from what it may seem there are more real numbers than natural numbers, they both equal infinity. And infinity is infinity.

~Aunix

These things that show 1 equaling 2 seem to prove what I had said, that standard laws of math simply break down with infinity. They have to, since you can't measure infinity; going on what AunixM3 said, infinity is probably more of a concept, not something that can really be measured. To really store data on something infinite, the storage media itself must be infinite in capacity in the first place, and the viewer would likely need infinite perceptions. We're just a bit limited on those two aspects.
 

drag

Elite Member
Jul 4, 2002
8,708
0
0
Hey 42, pretty "deep thought" on your part, earthling. But I wouldn't want to tell the Vogons that. That sort of thing gets em into a intergalatic road building sort of mood, but I think their chiefs would be hip to it. Almost as neat as digital watches that idea was, all though I doubt the little bits of green paper you hold so dear would scarcy care at all! But around here people would cotton to that idea about as much as I did when this hot chick I was hitting on ran off with a guy with a birdcage on his shoulder, just because he claimed to have a second head!! BTW have you seen my ford prefect around? I seem to have forgot were I parked it. hrmf. How am I suppose to finish this play I have been working on for the past 3 years, if I can't even find my car? Oh back to subject matter, I actually think the only people would be interested in that sort of thing would be the pan-galatic sort, if they were not so busy pretending to chase for cheese down mazes and dropping dead at unexpected moments. Oh, and do you know were the dolphins went too? Thinking this way kinda makes be feel like a small dog that swallowed a invading fleet of angry space aliens. I kinda feel wozy now, I need to eat my peanuts, for protein replacement ya, know. I think I'll suck on the corner of my towel for a while, wish I had some tea... sigh.

Hey if that didn't blow your mind, there are as many numbers between 1 and 1.01 then there are in the entire universe!!!!!!! just think we could of easily had a hundred billion toes, if God wasn't so carefull were he put the decible! I would like any one of you PROVE me wrong! haha

So long and thanks for all the fish!
 

grant2

Golden Member
May 23, 2001
1,165
23
81
Originally posted by: dejitaru
x = y
x^2 = x Y (mult both sides by x)
x^2 - y^2 = x y - y^2 (subtract y^2)
(x + y) (x - y) = y (x - y) (Factor both sides)
(x + y) = y (Divide both sides by x-y)
y + y = y (Subst y for x since x = y)
2y = y
2 = 1 (Divide both sides by y)

x^2 - y^2 = x y - y^2 (subtract y^2)
(x + y) (x - y) = y (x - y) (Factor both sides)
You can't factor x y - y^2, that's divide by zero.

I think it's the "(Divide both sides by x-y)" that is the division by zero.
 

sao123

Lifer
May 27, 2002
12,653
205
106
I know where the divide by zero is, I invented that illegal proof. dividing by (x-y)
my point was...
using infinity or undefined numbers in an equations does not mathematically prove that 1 = 2 or any other mathematical falsehood.