Exploring the Convergence of 0.999... and the Concept of Infinity

[russ_watters]

so you consider infinity to be a number governed by its own rules such that infinity is greater than itself (greater than all real numbers)

Looks like we also need to define "real numbers" for you. The word "real" isn't arbitrary as you are using it, it has a specific definition in math as well. http://en.wikipedia.org/wiki/Real_numbers

In mathematics, the real numbers are intuitively defined as numbers that are in one-to-one correspondence with the points on an infinite line—the number line. The term "real number" is a retronym coined in response to "imaginary number".

Notice, while the real number line is infinite, "infinity" is not a point on the line, hence "infinity" is not a real number. Your above objection is based on a misunderstanding of the definition of "real numbers." (you also characterized "infinity" as a "digit" a few posts ago - also incorrect).

Seriously, ram, you have a lot to learn about math. What we're talking about here is largely high school stuff. If you would only accept that you have a lot to learn and decide to learn it, you'd be much better off.

 

[ram2048]

i can goto 5 sites and pull a different definition of real numbers, irrational numbers, and infinity.

and you talk to ME about inconsistency. I can't use your words because your definitions are "Mumbo jumbo"

i say something then you correct me with a different definition that means the same thing.

the point of the matter is, you say the number line is infinitely divisible but then you say when you divide it by infinity "it can't be determined" or "is close enough to zero we'll just call it zero"

i'm trying to establish a notational system that allows for infinite division AND allows for division beyond that.

you guys say the current system works, why mess with it, but totally balk at the idea of replacing it with something that works BETTER.

sentimental value i guess... :|

 

[ram2048]

Implications of your ideas:
a)
Now, you say that a sequence: 0.1, 0.01, 0.001 and so on does not converge to 0,
but to some number 0...1 (with an infinite number of zeroes in between)
Clearly then, a sequence: 0.2, 0.02, 0.002 goes to some number:
0...2 (Right?)

these are not even converging sequences under my notation. you would have .00...02, .00...002, .00...0002, . . .

Have a good day!

i didn't! you cursed me didn't you!

 

[ram2048]

Did you mean last term?

And why would there be a last term? Each term is of the form 1/2^m where m is an integer and there is no last integer...

of course i meant last term. Integral has me talking about digits and then you jump on me when i get a word wrong. okay okay spanish inquisition over here..

and yes there is no last integer by MY notation, but there IS by the current one. such that 1/Infinity = 0 = limit.

it's your system though, you KNOW it better than me why do i have to tell you?

 

[Hurkyl]

i can goto 5 sites and pull a different definition of real numbers, irrational numbers, and infinity.

I wouldn't doubt it. There are often many equivalent ways of defining something.

For instance, (IIRC) the first rigorous definition of the real numbers was that a real number is an equivalence class of cauchy sequences of rational numbers. (In laymen's terms, any real number is identified with the sequences of fractions that approach it)

The usual modern definition preferred would be to define the real numbers as a complete ordered field. (In laymen's terms, +, -, *, /, and < all work "properly", and there are no "holes")

And it can be proved that the first definition satisfies the requirements of the second definition. Conversely, anything that satisfies the second definition is isomorphic (in laymen's terms, the same) to the first definition.


(Incidentally, 5? Really? I can only think of 3 definitions of real numbers you could be reasonably expected to find online, and only one definition of irrational. Different definitions of infinity wouldn't surprise me, because there are a lot of different concepts that are (sloppily) called "infinity")

the point of the matter is, you say the number line is infinitely divisible but then you say when you divide it by infinity "it can't be determined" or "is close enough to zero we'll just call it zero"

Actually, we say "Please be more specific".

"Infinitely divisible", taken entirely literally, means that it can be divided into an infinite number of terms. (Notice I did not say "an infinity of terms") In particular, the real line can be divided into c (= |R|) terms.

c is a thing called a cardinal number. It is not a finite cardinal, so it must be an infinite cardinal. (Some call it a transfinite cardinal, simply because so many laymen get confused when things are called infinite)

Division is not well-defined on cardinals, because multiplication is not very nice. For example, 1 * c = 2 * c. If you could divide by c, you would get 1 = 2, which is bad.

i'm trying to establish a notational system that allows for infinite division AND allows for division beyond that.

Learning math might be helpful to see how to do this.

Here's a simple approach to defining such a system:

Consider all (real) rational functions in x. E.G. things like 6, 7 + x^2, and (1 + 3x + 4x^5)/(4x + 3x^7)

+, -, *, and / can all be defined and function "properly" in this field. You can order this field by decreeing that x is bigger than any real number, and then extending the definition of < to accommodate this decree. So, for example, 7 + x^2 is infinite, because it is bigger than any real number. Proof:

Let r be any real number.
r < x
1 < x
r = 1 * r < 1 * x < x * x < x * x + 7
Thus r < x^2 + 7

Similarly, (1 + 3x + 4x^5)/(4x + 3x^7) is an infinitessimal.

If you don't like x, maybe you could use w (omega), a common symbol for transfinite numbers.


There's another number system (whose technical details are very difficult to follow) called the hyperreals which have transfinites and infinitessimals, but, for the most part, behave exactly like the real numbers. You might find information on this by searching for "non-standard analysis". Last time I went looking, there was actually an undergraduate calculus text in PDF format somewhere on the web that develops calculus using nonstandard analysis (i.e. with infinitessimals and transfinite integers, etc) instead of the usual way.

you guys say the current system works, why mess with it, but totally balk at the idea of replacing it with something that works BETTER.

We're "balking" because you are claiming the current system doesn't work.

of course i meant last term. Integral has me talking about digits and then you jump on me when i get a word wrong. okay okay spanish inquisition over here..

Or, maybe I'm just making sure I knew what you meant.

 

[ram2048]

i never claimed it didn't work, i claimed it was inaccurate ;D

i walk on bridges and fly by airplane. i wouldn't do so if i didn't trust our current set of math at all

in any case, there's little point in arguing as it seems apparent that such things are not going to change overnight.

i thank you for the envigorating discussion, and humbly apologize for getting bent out of shape at you in previous threads, Hurk

 

[matt grime]

yes there is no last integer by MY notation, but there IS by the current one. such that 1/Infinity = 0 = limit.

it's your system though, you KNOW it better than me why do i have to tell you?

There is no last integer in 'our' system, that you think we think there is is your error.

 

[ram2048]

ooooooooooooooooookay

define a number less than 1/infinity but greater than 0 in your system

 

[matt grime]

what do you mean by 1/infinity? infinity isn't a real number, so why should i be able to do that? have you not learned anything from this thread? you aren'ty dealing with the real numbers when you write that kind of thing.


in robinsonian analysis i believe the object you are talking about is labelled epsilon.

 

[Zurtex]

Well I'd like to thank all the people who tried to help ram out on this thread, whether or not he still spouting rubbish you have really made me feel a lot better about numbers . I still can't believe your trying to get the point across and have not just banned him from posting on the maths forums and ignored him here, well done your good people.

 

[Hurkyl]

define a number less than 1/infinity but greater than 0 in your system

This question, in general, doesn't make sense.

It cannot make sense in the context of the real numbers.

In the context of the extended real numbers, this cannot be done, since 1/infinity is 0.

In the context of the hyperreal numbers, if we rephrase your challenge (so that it makes sense) as:
"If w is a positive, infinite number, then define a number less than 1/w but greater than 0"
then an answer to your challenge would be 1/(2w).

 

[ram2048]

i somehow wrote that all wrong :O

what i meant to say is the "last" integer in your system would be a function of infinity such that nint(infinity) = that integer (theoretically if you could USE that function)

but that's still getting off the point.

if you COULD then 1/2^nint(∞) would be the closest step for your consideration.

forgot where i was going with this O_O

meh...

 

[Zurtex]

ram2048 what is a "last integer"?

That makes less sense to me than anything else you have written.

 

[ram2048]

hell if i know, Hurkyl was talking about integers and blah blah no last integer..

but there IS a last integer because you have an upwards limit of infinity.

you can't define that number without it being an expression "of infinity" itself, so it kinda defies itself.

but there was a point i was trying to make such that a sum to infinity, even AT infinity does not equal its limit in such a convergent sieries as Zeno's paradox.

as with everything "Infinity related" it's all theory and you have to apply logic. if every "step" in the process or "term" computed is half the remaining, there will never be a process that is "whole of the remaining" because that breaks the rule set forth in the initial exercise. so even at infinity, or beyond infinity in the case of "hyperreal" blah blah "extended irrationals" or whatever you want the sum STILL doesn't equal the limit.

the same is true of .999~ every digit tacked onto the end brings it closer and closer to 1, but no digit is ever a 10 completing a whole "step"

but it really doesn't matter... whatever :D

 

[chroot]

So there's an integer that is larger than every other integer, eh? Could you do us a favor and show us which integer is the biggest?

- Warren

 

[ram2048]

the one closest to infinity? :O

 

[chroot]

That's the stupidest thing I've ever heard.

- Warren

 

[ram2048]

haha my point is proven

not only can you NOT get to infinity, ever. you can NOT get to the largest integer that is not infinity.

so what hurkyl said about every step being considered is kinda out the window eh?

in any case .999 can never be = to 1 because every step is 9/10's the remaining step towards 1 and "the destination is never in consideration"

there is no step covered that is 10/10ths the remaining distance

sums to infinity are a good approximation

 

[Hurkyl]

What is one plus the "largest integer that is not infinity"?
What is two plus the "largest integer that is not infinity"?

(plus, here, means integer addition)

the same is true of .999~ every digit tacked onto the end brings it closer and closer to 1

Why (and how) are you tacking digits onto .999~?

not only can you NOT get to infinity, ever. you can NOT get to the largest integer that is not infinity.

But you forget, you can "get" to every positive integer. It's a simple proof by induction:

I can get to 1.
If I can get to n, then I can get to n+1 by adding 1 to n.
Thus, by induction, I can get to any positive integer.

If the number system about which you are speaking has numbers to which you cannot "get", then you're not speaking about the integers.

in any case .999 can never be = to 1 because every step is 9/10's the remaining step towards 1 and "the destination is never in consideration"

.999 can never be equal to 1 because it is equal to 999/1000. Of what steps are you speaking?

 

[Hurkyl]

sums to infinity are a good approximation

Approximation of what?

 

[lvlastermind]

ram doesn't understand that you can't have 1/infinity because infinity=0 and you can't divide by zero.

 

[ram2048]

Why (and how) are you tacking digits onto .999~?

doesn't matter why and how, but if you must know it's by process of summation where each "step" resolves into one "digit".

9/10 + 9/100 + 9/1000 ...

.999 can never be equal to 1 because it is equal to 999/1000. Of what steps are you speaking?

you knew what i meant.. :P

if you accept Zeno's conjecture as true, within the confines of the problem set forth you can not reach the destination then you also accept that within mathematics .999~ can never equal 1 simply because they are the same problem with a different curve Zeno's is 1/2 and .999~ is 9/10ths.

 

[lvlastermind]

if you accept Zeno's conjecture as true, within the confines of the problem set forth you can not reach the destination then you also accept that within mathematics .999~ can never equal 1 simply because they are the same problem with a different curve Zeno's is 1/2 and .999~ is 9/10ths.

but you will reach your destination with an infinite number of halfs. This is one example of how infinity doesn't act like a real number. Beacause it doesn't have a value but it still has meaning.

 

[Zurtex]

haha my point is proven

not only can you NOT get to infinity, ever. you can NOT get to the largest integer that is not infinity.

We don't add the things one at a time.

 

[ram2048]

but you will reach your destination with an infinite number of halfs. This is one example of how infinity doesn't act like a real number. Beacause it doesn't have a value but it still has meaning.

no this is an example of how calculus uses "infinity" to approximate.

if something cannot logically EVER be something then infinity and forever it will not be it.

how can you possibly reason that the destination is reached? it means that the last step you took wasn't a half-step but a whole one.

i have described earlier how calculus accepts 1/∞ and 2/∞ as the same number because both would be equal to 0.

indeed if 1/∞ = 0 then it is not actually a step at all, so the runner MUST have reached the destination in step 1/2^nint(∞) where nint defines the nearest integer to infinity.

but then we would have the paradox of being able to halve that distance YET AGAIN. such that 1/2^[nint(∞)+1]

and so on... so somehow the conclusion that the destination CAN be reached has to be wrong. or 1/∞ = 0 is wrong. or both are.

i say both personally ;D

 

[Zurtex]

Your argument is wrong.

1/n > 0

Where n is a positive integer. There is NO nearest integer to infinity, that's just silly. Your problem is that you do not seem to be able to grasp a concept of infinity and that it is not on the real number line.

 

[ram2048]

excuse me, but you guys sum n to infinity ALL THE TIME

that means it's on the same number line, it may not be included in your set of reals, but it's still on the same line.

that means somewhere along the line increasingly greater numbers become infinite. if NOT sum n to infinity has NO meaning. like saying sum n to cow or sum n to vacuum cleaner...

the definition of infinity provides a clear relation of "the concept" to known reals such that even though it's not a number it's a function of numbers so much so that it is possible to use it in calculations.

if you're saying a set of reals can never increase to infinity then you're basically accepting that given the infinite number of steps in Zeno's problem, the man will NEVER reach his destination. clear now?

if you understand that much go back to my other post and read the logical explanation on how even at infinity the destination cannot be reached, so it makes no real difference whether a "largest integer" is real or not. the outcome is still the same

 

[ram2048]

sums to infinity are a good approximation

Approximation of what?

approximation of the actual value that would come out of the calculation if you calculated it out longhand. :|

logically

 

[Zurtex]

excuse me, but you guys sum n to infinity ALL THE TIME

that means it's on the same number line, it may not be included in your set of reals, but it's still on the same line.

that means somewhere along the line increasingly greater numbers become infinite. if NOT sum n to infinity has NO meaning. like saying sum n to cow or sum n to vacuum cleaner...

the definition of infinity provides a clear relation of "the concept" to known reals such that even though it's not a number it's a function of numbers so much so that it is possible to use it in calculations.

if you're saying a set of reals can never increase to infinity then you're basically accepting that given the infinite number of steps in Zeno's problem, the man will NEVER reach his destination. clear now?

if you understand that much go back to my other post and read the logical explanation on how even at infinity the destination cannot be reached, so it makes no real difference whether a "largest integer" is real or not. the outcome is still the same

No, your totally wrong as has been proven and shown on this thread many times.

 

[Hurkyl]

doesn't matter why and how, but if you must know it's by process of summation where each "step" resolves into one "digit".

9/10 + 9/100 + 9/1000 ...

This notation looks like an infinite sum; I don't see any "steps".

(To keep things moving)

I'm presuming by "steps" you are first considering 9/10, then 9/10+9/100, then 9/10+9/100+9/1000, and so on. But, of course, none of these are 9/10 + 9/100 + 9/1000 ... (though the limit of these "steps" is)

if you accept Zeno's conjecture as true, within the confines of the problem set forth you can not reach the destination

I have no problem with that. Zeno only considers the motion up to (but not including) reaching the destination. Thus, it would be silly to think that the destination would be reached in the period he analyzes.

then you also accept that within mathematics .999~ can never equal 1 simply because they are the same problem with a different curve Zeno's is 1/2 and .999~ is 9/10ths.

How do you figure? If I do it with 9/10s, then Zeno considers each of these intervals of position [0, 9/10], [9/10, 99/100], [99/100, 999/1000] ...
Putting all of these intervals together yields the interval [0, 1).

excuse me, but you guys sum n to infinity ALL THE TIME

Right, and sums to infinity are defined by

[tex]
\sum_{i=1}^{\infty} a_i = \lim_{m \rightarrow \infty} \sum_{i=1}^m a_i
[/tex]
which can be further resolved (by applying the definition of limit)
[itex]\sum_{i=1}^{\infty} a_i = L[/itex] if and only if for every positive [itex]\epsilon[/itex] there exists an integer [itex]N[/itex] such that for any integer [itex]m[/itex] greater than [itex]N[/itex], we have
[tex]
|L - \sum_{i=1}^m a_i| < \epsilon
[/itex]

Notice, in particular, that this is not logically equivallent to saying that you keep adding terms one by one until you've reached an infinite number of terms. (Though, IMHO, it's for the most part conceptually equivalent)

that means it's on the same number line, it may not be included in your set of reals, but it's still on the same line.

The number line only has real numbers on it, thus it doesn't have infinity on it.

But, as mentioned before, mathematicians do use an extension of the real numbers which has a positive and negative infinity on each endpoint. But...

that means somewhere along the line increasingly greater numbers become infinite.

No it doesn't. A sequence of increasingly greater numbers can converge to infinity, but none of the individual numbers need be infinite... just like a sequence of numbers can converge to zero, but none of them need to be zero. (e.g. 1, -1, 1/2, -1/2, 1/4, -1/4, 1/8, -1/8, ...)

if NOT sum n to infinity has NO meaning.

I'll say it again, sum to infinity has this meaning:

[tex]
\sum_{i=1}^{\infty} a_i = \lim_{m \rightarrow \infty} \sum_{i=1}^m a_i
[/tex]

if you're saying a set of reals can never increase to infinity

I'm saying no real number may be infinite, an entirely different statement.

you're basically accepting that given the infinite number of steps in Zeno's problem, the man will NEVER reach his destination. clear now?

I accept that; the destination is reached after the "steps" contemplated by Zeno.

approximation of the actual value that would come out of the calculation if you calculated it out longhand. :|

logically

How do you plan to add an infinite number of terms by longhand?

 

[ram2048]

How do you figure? If I do it with 9/10s, then Zeno considers each of these intervals of position [0, 9/10], [9/10, 99/100], [99/100, 999/1000] ...
Putting all of these intervals together yields the interval [0, 1).

how is that different from [0, 1/2], [1/2, 3/4], [3/4, 7/8]...

 

[ram2048]

i'm still not understanding your wording on this

I accept that; the destination is reached after the "steps" contemplated by Zeno

if every step is half the distance (how the problem is defined) then after the "steps" still leaves a whole destination left to cover. if every step is a fraction of 1 whole then somewhere along you believe 1/2 = 1

the only thing i can think of to account for this is your belief that infinity is always equal to itself no matter how you transform it such that 1/2 x ∞ = 1 x ∞.

something like that would make a last step possible but it's not logical at all.

 

[Zurtex]

the only thing i can think of to account for this is your belief that infinity is always equal to itself no matter how you transform it such that 1/2 x ∞ = 1 x ∞.

something like that would make a last step possible but it's not logical at all.

Infinity isn't a real number and doesn't act like that one, so yes that would be correct.

Last possible step? What are you on about?

 

[Hurkyl]

how is that different from [0, 1/2], [1/2, 3/4], [3/4, 7/8]...

Put all of those together and you also get the interval [0, 1).

i'm still not understanding your wording on this

My intent was to state this: the destination is not reached during the sequence of steps considered by Zeno, but that does not imply that the destination cannot be reached at some time that occurs later than the steps considered by Zeno.


However, this reminded me of some of the things I used to point out in Zeno's paradox discussions; the following statement is also true:

At any particular point in time, if it can be said that all of the steps considered by Zeno have occurred, then it is also true that the destination has been reached.

But allow me to emphasize; the destination is not reached during the steps considered by Zeno. In particular, in order to make this statement, I do not make the assumption that one of these steps covers the entire remaining distance. Additionally, I need not use "infinity" anywhere to prove my claim.

the only thing i can think of to account for this is your belief that infinity is always equal to itself no matter how you transform it such that 1/2 x &infin; = 1 x &infin;.

I "believe" it because that is how multiplication by &infin; is defined in the extended real numbers.

something like that would make a last step possible but it's not logical at all.

You mean that it conflicts with your common sense. It is entirely logical because it can be proven rigorously from the definitions and axioms.


But, it shouldn't conflict with common sense. Consider this; when things grow without bound, they approach infinity, right? E.G. the sequence 1, 2, 3, 4, ... approaches infinity, and the sequence 2, 4, 6, 8, ... does as well.

Furthermore, multiplication is a continuous operation. If a sequence approaches a limit, then if I double every number in the sequence, then the limit gets doubled.

Now, 2, 4, 6, 8, ... is the double of the sequence 1, 2, 3, 4, ..., so the limit of 2, 4, 6, 8, ... is double the limit of 1, 2, 3, 4, ..., thus suggesting that &infin; = 2 * &infin; should be correct.

 

[ram2048]

the destination is not reached during the sequence of steps considered by Zeno, but that does not imply that the destination cannot be reached at some time that occurs later than the steps considered by Zeno.

well since zeno considers to infinity, you believe that "beyond infinity" there lies a step such that 1/2 the remaining distance = the whole distance?

what I'm saying is infinity or not you believe you can take 1/2 a distance and it equals the whole distance. basically going back to what Integral said about the smallest thing being a point, what's the 'length" of a point there isn't a length so points cannot be used to fill a "gap" of any distance.

because going backwards in steps you'd also have to have 2x"the length of a point distance left to cover such that you could 'fill' it with your 1x point.

the convergence would look something like...

8 point lengths remaining (move (1/2 x 8) points)
4 point lengths covered

4 point lengths remaining (move (1/2 x 4) points)
2 point lengths covered

2 point lengths remaining (move (1/2 x 2) points)
1 point length covered

1 point length remaining (move (1/2 x 1) point. point is not divisible. move 1 point)
1 point length covered

i don't know if that makes anymore more sense at all, but that last consideration is the one i say wouldn't happen because it would be beyond the confines of the problem.

bear in mind i know a point has no "length" so don't lecture me on that. I'm using it to illustrate a point since it's the smallest conceivable thing.

so using that model, stuff has a limit of divisibility. the end step would break the rules set forth in the problem such that you'd be moving not 1/2 step but a whole step. while we calculate out with infinite steps we still reach that imposed limit caused by the point not being divisible.

but if it WERE divisible it STILL wouldn't be reached because we'd be faced with 1/2 a point 1/4 a point etc etc.