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JigPu
10-15-03, 07:40 PM
I was bored in the library today, so started working on some of my somewhat-abstract math, and ended up writing this out. I've been looking over this, and I can't find any problems with it. My abstract math theories have been somewhat hinting at this being true (which is probably why I ended up writing it) so I'm really wondering what you guys thing of it.

1 - 0.9 = 1x10^-1
1 - 0.99 = 1x10^-2
1 - 0.999 = 1x10^-3
1 - 0.[n number of 9s] = 1x10^-n

1 - 0.999... = 1x10^-infinity
0.999... = 1
1 - 1 = 0
1 - 0.999... = 0
1x10^-infinity = 0

JigPu

Burning Phoenix
10-15-03, 08:24 PM
I don't understand how you came up with this:

0.999... = 1

JigPu
10-15-03, 08:32 PM
That's already been proven, and this is how it was explained to me:

1 / 9 = 0.11111...
2 / 9 = 0.22222...
3 / 9 = 0.33333...
.
.
.
8 / 9 = 0.88888...
9 / 9 = 0.99999... = 1

There's a much more involved proof for it (I'll have to search for it) if you really want it though.

JigPu

DaveSauce
10-15-03, 08:42 PM
hmmm yea, that's what my TI-89 says.....

Burning...... 0.99999 = 1, for Lim(x->0) of 1-x. It's not technically 1, but as X gets smaller and smaller, 1-x is so close to 1 that is is considered 1. The fact is, you will get .99999 for as many decimals as the instrument can handle. At a certain point, it is rediculous to keep calling it less than 1, so it is in fact 1. Consider this:

1-0.000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000001

Now, theres absolutly no point in writing the answer as 0.999999999999999999999999999999999999999999999999 99999999999999999999999999999999999999999999999999 99999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999

because if you even breathe on it the wrong way it will be 1, lol.

Anyway, jig, there's nothing at all wrong with that. My TI-89 and the program Mathematica (a very powerful math program) give me 0 for the answer. I'm not sure if the proof given will work, though. I believe that you will have to use limits in order to get an acceptable proof....

but then again, I'm not much of a math person, lol. just wait until Frodo or Nihili come by, they can show you what's up, hehe.

edit:

Forgot to mention this part.

Infinity comes in for the power thingy in the limit....10^-infinity is really 1/(10^infinity). Since 10^infinity is infinately large, then 1/(10^infinity) is infinately small. It's impossible to "prove" due to the nature of infinity, but it is generally accepted as true...

JigPu
10-15-03, 08:55 PM
Yeah, if only my teachers would be so kind as to actually teach me to write real proofs :rolleyes: I agree that limits are probably needed to get an acceptable proof (I'm pretty sure even the 0.999...=1 proof has them), which means I'll have to bug some of the Calculus people to help me o.O.

JigPu

macklin01
10-15-03, 09:09 PM
I'd try to prove that .999999999.... = 1 by using convergence of the following series:

For each n >= 1, define

S_n = sum( 9*10^(-i), i=1..n);

e.g., S_1 = .9, S_2 = .99, etc.

To prove convergence to 1, let epsilon > 0 be arbitrary. Find an N such that if n > N, then

abs( S_n - 1.0 ) < epsilon.

As epsilon is arbitrary, the limit of S_n as n goes to infinity is 1.0.

Here, abs is absolute value.

That's the sort of thing that you'll want to do to prove it rigorously. -- Paul

FunkDaMonkMan
10-15-03, 09:16 PM
I can do limits. They are no biggie, and actually I had done them in my head before they "showed" me in calculus. I'm sure you do too. Any good math thinker does them.

Want to know what a limit is?

what is 1/0?
Algebra 2 student would say "undefined"
Calculus student says ... well... bad example

ok... what is 4/sqrt(x) for x=0?

Algebra 2 student says undefined
calculus student says "lets find the limit as x APPROACHES 0"
you find the value for x=.1 then for x=.01 then x=.001 By now you should see a pattern. The answer is positive infinity.

Your answer is correct, and here is how you write it. lim x -> 0 of x is 0

You should have no problem with calculus.

JigPu
10-15-03, 09:21 PM
Originally posted by FunkDaMonkMan
what is 1/0?
Algebra 2 student would say "undefined"
Calculus student says ... well... bad example
LOL, that's probably the WORST example you could give me :D (since I'm so determined to define it ;) ..... but that's another thread :cool: )

If it's right, then me = MUCHO happy :) Any other math gurus care to help verify for me?

JigPu

FunkDaMonkMan
10-15-03, 09:21 PM
Originally posted by macklin01
I'd try to prove that .999999999.... = 1 by using convergence of the following series:

For each n >= 1, define

S_n = sum( 9*10^(-i), i=1..n);

e.g., S_1 = .9, S_2 = .99, etc.

To prove convergence to 1, let epsilon > 0 be arbitrary. Find an N such that if n > N, then

abs( S_n - 1.0 ) < epsilon.

As epsilon is arbitrary, the limit of S_n as n goes to infinity is 1.0.

Here, abs is absolute value.

That's the sort of thing that you'll want to do to prove it rigorously. -- Paul

Can't this be proved just the same by using a simple limit?

I don't remember a whole lot about the part you are talking about, so it could be required to prove it, I'm not sure. My calc teacher asks me for help :(.

JigPu
10-15-03, 09:24 PM
Originally posted by FunkDaMonkMan

Can't this be proved just the same by using a simple limit?

I don't remember a whole lot about the part you are talking about, so it could be required to prove it, I'm not sure. My calc teacher asks me for help :(.
I'll see if I can dig up the proof I found since you sound interested :)

JigPu

JigPu
10-15-03, 09:51 PM
I can't keep what little text formatting there is here, so this is a cropped picture of the page I found it at:

JigPu

Cjwinnit
10-16-03, 05:32 AM

ok, *^0 = 1 (* means anything)

so 1^0 = 1, 43876^0 = 1, etc......

and 0^* = 0 (so 0^2 = 0 x 0 = 0)

and there's only one answer for a real power.

so what's 0^0 ?

Cjwinnit
10-16-03, 05:36 AM
Oh and another way of writing 0.9999..

0.9999.... = 0.9 + 0.09 + 0.009 + 0.0009.....
= a + ar + ar^2 + ar^3............

a = 0.9, r = 0.1

with an infinite geometric series the summation is

...a...
1 - r

= ....0.9....
....1 - 0.1

= 1

macklin01
10-16-03, 10:39 AM
That's the ideal formula to use.

JigPu, Cjwinnit, and I used the same series:
$\sum_{i=1}^\infty{9 (.1)^{i}}.$

To prove the convergence of a series, you prove the convergence of the sequence of partial sums.

The formula "works" because it has already been proven a priori that any series of the form $\sum_{i=1}^\infty{a (r)^i }$ with
$\abs{r} < 1$ converges.

Do note that in any proof you do with limits, it's incorrect to manipulate the limit as a symbol (such as in JigPu's proof) until you have proven (either by your own proof or by citation of a proper theorem) that the limit indeed exists, i.e., that the series converges.

So, the approach is first to prove that limit exists, and then determine what the limit is. In Cjwinnit's proof, he implicitly used a theorem that states that any such series with acceptable $r$ has a limit for part 1, then used the known result for part 2.

Or, if you know what the limit is in advance, prove it directly. (By whatever means you can with the definition of a limit of a series.) ;)

BTW, by convention, 0^0 is generally (but not always) taken to be 1. A similar case is 0!, which is also taken (by convention) to be 1.

-- Paul

Frodo Baggins
10-16-03, 03:10 PM
Originally posted by Cjwinnit

ok, *^0 = 1 (* means anything)

so 1^0 = 1, 43876^0 = 1, etc......

and 0^* = 0 (so 0^2 = 0 x 0 = 0)

and there's only one answer for a real power.

so what's 0^0 ?

That's not entirely correct,

Infinity^0 is indeterminate

0^0 is one of those things that nobody really knows about. I think Euler wrote a paper on it once. I think the options are either to consider it as indeterminate, non-existent, or 1.

Notice that x^0 as x --> 0, the limit is 1
But 0^x as x --> 0, the limit is 0.

So the graph of x^y is discontinuous there, so it doesn't exist

Blah

Someone mentioned to me that you can get a pretty good answer by looking at the proof of L'Hopital's Rule (Which deals with Cauchy's variation of the Mean Value Theorem), but I skipped through that proof (and I'm too lazy to look at it atm)

macklin01
10-16-03, 03:14 PM
Frodo, it's good to see you! Please remind me -- I have an important PM to work on! -- Paul

Frodo Baggins
10-16-03, 03:36 PM
Originally posted by FunkDaMonkMan
I can do limits. They are no biggie, and actually I had done them in my head before they "showed" me in calculus. I'm sure you do too. Any good math thinker does them.

Want to know what a limit is?

what is 1/0?
Algebra 2 student would say "undefined"
Calculus student says ... well... bad example

ok... what is 4/sqrt(x) for x=0?

Algebra 2 student says undefined
calculus student says "lets find the limit as x APPROACHES 0"
you find the value for x=.1 then for x=.01 then x=.001 By now you should see a pattern. The answer is positive infinity.

Your answer is correct, and here is how you write it. lim x -> 0 of x is 0

You should have no problem with calculus.

.....1/0 is undefined, why should there be a discrepancy between what a cal student and what an alg student says?

lim(x --> 0) 1/x from the right (or left) is positive or negative infinity though

FunkDaMonkMan
10-16-03, 04:45 PM
Originally posted by Frodo Baggins

.....1/0 is undefined, why should there be a discrepancy between what a cal student and what an alg student says?

lim(x --> 0) 1/x from the right (or left) is positive or negative infinity though

That was what I was going to get at, but the postive and negative infinity thing might have been a bit confusing for a simple limit explination.

JigPu
10-16-03, 05:04 PM
Hold on. If we've just determined 0 to be 1x10^-infinity, then why on earth is division by zero undefined???
(by the way... PLEASE don't say it's because you can't add 0 to infinity, because that is probably the easiest to prove of all things dealing with infinity)

1 / 0 =
1x10^0 / 1x10^-infinity =
(1/1)x10^(0+infinity) =
1x10^infinity

:confused: 1x10^-infinity is the limit of the sequence, and as I (probably don't) understand, the limit is what it equals.

Case in point being 0.9999... Since it limits at one, it is equal to one, and math operations can even be perfomed on it with the same results. You can do whatever you wish to 0.9999... and the result will be the same as if you did it to 1 because the error when comparing the two results will have an error less than the epsilon (note: this is what I've read elsewhere... not anything I have real experience with).

Sorry, this is turning into another one of my infinity discussions...
JigPu

Cjwinnit
10-16-03, 05:14 PM
Originally posted by Frodo Baggins
That's not entirely correct,

Infinity^0 is indeterminate

0^0 is one of those things that nobody really knows about. I think Euler wrote a paper on it once. I think the options are either to consider it as indeterminate, non-existent, or 1.

Notice that x^0 as x --> 0, the limit is 1
But 0^x as x --> 0, the limit is 0.

So the graph of x^y is discontinuous there, so it doesn't exist

Perhaps I should have said that * should be anything that exists (Including real, imaginary and complex numbers) but It was sorta implied when I thought about it, doh :-/

I forgot there are other maths ninjas with GD access :)

Cjwinnit
10-16-03, 05:23 PM
Originally posted by JigPu
Hold on. If we've just determined 0 to be 1x10^-infinity, then why on earth is division by zero undefined???

JigPu

ok...

let's substitute 0 for 1x10^-infinity.

Now if we have y = 1 / x, and we want to rewrite it as x = 0, then x also = 1x10^-infinity.

so y = 1 / x = 1 / 0 = 1 / 1x10^-infinity

y = 1 / 1x10^-infinity.

1x10^-infinity = A * B, A = 1 and B = 10^-infinity

so y = 1 / (A * B) = [1 / A * 1 / B] = 1 / B

y = 1/10^-infinity

then

y = 10^infinity

remember what an index is, it's a number times itself a defined number of times.

so 3^5 = 3*3*3*3*3

then 10^infinity = 10*10*10*10*10..................

it's neverending, hence undefined. It doesn't exist.

so y is undefined, hence 1/0 is undefined.

JigPu
10-16-03, 06:31 PM
Originally posted by Cjwinnit
ok...

LOTS OF MATH STUFF I DON'T CARE TO QUOTE :D

it's neverending, hence undefined. It doesn't exist.

so y is undefined, hence 1/0 is undefined.
But in that case, anything involving you doing something an infinite number of times would be undefined.

0.9999... for example would be undefined since you cannot possibly add enough 9s to get an infinite number of them so you can define it. However, it is easily seen that the limit of adding additional 9s is 1, and once you manage to get an infinite number of nines (which only happens with the number 0.9999...) it will be equal to one.

0.9999... (you already know the repeating operation) limits to 1, and so must be 1.
0.1^0.1^0.1^0.1^... limits to 1, and so must be 1.
0.8*0.8*0.8*0.8^... limits to 0, and so must be 0.
PI/4 = 1/1 - 1/3 + 1/5 - 1/7 + ... limits at pi/4, and so must be pi/4.
10*10*10*... limits to infinity, and thus is infinity.

If you can define something that's the result of an infinite number of operations, then you can most certainly define this.

JigPu

modenaf1
10-16-03, 06:36 PM
OMG! :eek: i have left the 9x9 domain! :eek:

/me runs out of this thread!

really though, you guys are math geniouses or something!

JigPu
10-16-03, 06:40 PM
Originally posted by modenaf1
OMG! :eek: i have left the 9x9 domain! :eek:

/me runs out of this thread!

really though, you guys are math geniouses or something!
Well I'm certainly not :D I'm inventive, stubborn, and abhore things that are illogical :D

Math genious though... :rolleyes:

JigPu

nihili
10-16-03, 06:49 PM
*cough* Abraham Robinson *cough*

JigPu
10-16-03, 07:28 PM
Want some cough suppressant Nihili? ;)

JigPu

Frodo Baggins
10-16-03, 08:10 PM
really though, you guys are math geniouses or something!

Not really...

All math ninjas though :sn:

JigPu
10-16-03, 08:24 PM
OK, other than Nihili's reference to Abraham Robinson, does anybody else have suggestions on what I should look in the library for tomorrow? I can only find so much on infintessimals/infinites on the net, so how about it?

BTW... My new favorite shape because of infinity :)
http://www.math.duke.edu/education/ccp/materials/mvcalc/parasurfs/torus.gif

JigPu

nihili
10-16-03, 08:39 PM
Look for information on Non-Standard analysis and the Hyperreal numbers. Robinson's book is supposed to be a good introduction, though I haven't gotten to it yet. I'm working my way through some chapters of Goldblatt's Lectures on the Hyperreals right now. It's pretty good but will be very thick going if you're not used to reading math. I understand there are some calculus books that use non-standard analysis, but I haven't tracked any down yet.

If you need some help on some of the concepts (like filters and such) just give a holler here.

OH, and doesn't a torus have more to do with topology and knot theory than infinity? For infinity I would have though you would want a klein bottle or mobius strip.

One more thing. Check out Moore's book on the ifinite. It's not real mathematical, but it does a good job of laying out the history and a lot of the problems.

JigPu
10-16-03, 09:02 PM
I'm sure I will since I doubt I'm going to be able to read them very easily. I can handle a certian level of depth to books, but most math books seem to get REALLY deep.

Actually, the reason I like the torus is because that's how the coordnite plane would have to be "shaped" (as far as I can tell) if you made a graph from -infinity -> +infinity on both the X and Y axis (since +infinity=-infinity [or is an infintessimal distance away from it] if 0=10^-infinity).

Though I think I'm falling in love with that klein bottle... :drool:
JigPu

nihili
10-16-03, 09:06 PM
You only get a Torus if you take -infinity=+infinity.

Erg, I see you already said that. Ok, try this one. Explain to me which axis encircles the hole of the donut and why.

Also, since the circumference of the torus is infinite, wouldn't it have 0 curvature?

JigPu
10-17-03, 12:26 AM
Originally posted by nihili
You only get a Torus if you take -infinity=+infinity.

Erg, I see you already said that. Ok, try this one. Explain to me which axis encircles the hole of the donut and why.

Also, since the circumference of the torus is infinite, wouldn't it have 0 curvature?
I'm not totaly sure which axis you were refering to, so I'll explain them both.

Imagine a standard coordnate plane. Going up/down on it increases/decreases y, and going right/left on it increases/decreases x. Both y and x have a range from +infinity to -infinity. Since they're equal, they must both curve into a cyndrilical shape. Curving the y-axis gives an infinatly long cylinder that is laying down. Curving the x-axis now brings it into a torus.

Going "up" or "down" the y-axis moves you to the very inside of the torus, stopping at the +-infinity line at the middle of the inside.

Going "right" or "left" along the x-axis moves you around the circumfrence of the torus increasing or decreasing in value as you move. When you've traveled all the way around to the "back" of the torus, x is equal to +-infinity.

Yes, the torus would have no curvature at all which is why standard coordnate planes work so well. The only time the curve can be seen is when you mess with the infinities which require the normal plane to be bent so that +infinity and -infinity on each axis meet.

Somewhat related to this is the z-axis. Visulising it is extremely interesting. Going into and out of the torus have to eventually hit the same point, which would require a 4-dimensional representation of the figure. Since two dimensions (x and y axis) require a third dimension to represent, this would be expected. So long as the z-axis remains finite however, it is sufficient to represent it in only 3 dimensions (just like how if the two x and y dimensions remain finite, it can be sufficiently expressed in only 2 dimensions [with 2 dimensions being sufficient for 1 infinite axis]).

JigPu

nihili
10-17-03, 07:32 AM
Originally posted by JigPu

Going "up" or "down" the y-axis moves you to the very inside of the torus, stopping at the +-infinity line at the middle of the inside.

Going "right" or "left" along the x-axis moves you around the circumfrence of the torus increasing or decreasing in value as you move. When you've traveled all the way around to the "back" of the torus, x is equal to +-infinity.

But why does the x-axis move around the circumference while the y moves you inside the torus? Why not the other way around? Are you presuming that the y-axis gets connected "before" the x-axis?

The plane at least appears to have some symmetry under rotation that the torus lacks. Why is that?

JigPu
10-17-03, 11:25 AM
It could be the other way I guess, but that would be like asking why the x-axis is not vertical on the coordnite plane, and the y-axis is :) In the end, I don't think it really matters which axis is assiciated with which direction, with the only difference being a visual mirror around the x=y line.

Even if the x-axis was connected before the y-axis, the torus shape would still result. You get a cylinder that is standing up which you have to bend in on itself vertically so the y-axis meets up.

Originally posted by nihili
The plane at least appears to have some symmetry under rotation that the torus lacks. Why is that?
Can you explain what you mean by this?

I'm goning to hazard a guess at your meaning the plane can be rotated 90, 180, and 270 degrees with symetrical mirroring resulting. This happens on the torus as well. When rotating the plane, all you are really doing is changing the position and direction of the each axis. Rotating 90 degrees swaps the x and y-axis and changes the direction of the y-axis (bigger y values to the left, smaller to the right). Finding the corresponding point involves only swaping the X and Y values (like the axis), and then multiplying X by -1 (because it has changed direction).

If you change the x and y-axis on the torus in the same way (since that is what rotation is) the same results happen. To find a corresponding point on the torus, you follow the same steps when rotating the plane. The results will be symetrical around the torus. (Unfortunatly I have nothing I can really use to show this, only my fingers plotting in air the function tan(x) for no rotation and 90 degree rotation.)

JigPu