# 1=0.999... infinities and box of chocolates..Phliosophy of Math...

Discussion in 'General Philosophy' started by Quantum Quack, Nov 2, 2013.

1. Thanks for your response, rpenner.

Have a bad cold and going back to bed, so can't engage in detailed discussions, even if I wanted to. back in a few days, I hope. But while I am here, I will again give some indication of where I am coming from and going to wards on this and other non-conventional maths perspectives of mine arising from my from scratch review of current maths as it is (from which you are observing, and thanks for that, but I am not in the same construct' as you are). Briefly then...

The essence and intents of my further-seeing approach to this review is encapsulated by the comment made by some mathematician (expounding on Fractals or Fracta calculus or some such topic) whom I can't recall at present, to the effect:

If you will recall way back when, I made the observation that my perspective takes into account the surrounding dynamical environment of the otherwise static and sterile number line construct/points treatment. That is why I spoke then of a new CONTEXTUAL MATHS system with re-jigged axioms which will cater for all cases, not just notional infinite/dimensionless 'points/geometries, but also for the real meanings of the maths system elements per se which will re-tool all the current conventional treatments/definitions (which currently output 'undefined/undetermined' etc etc) to better prepare for the needs of the final complete reality based ToE physics modeling.

Anyhow, sorry but I must go. That's all I am at liberty to say at this stage. Back whenever. Thanks again, rpenner, for your time and trouble and polite response. Cheers and stay well as can be, everyone!

3. ### hansdaValued Senior Member

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$T_n \ne 0$ is also true, as "$n$ approaches infinity($n \to \infty$)".

So we can say, $T_n \ne 0$ for any value of $n$.

Now considering the equation $S_n = 1 - \frac{1}{9} \times {T_n}$, we can say that $S_n \ne 1$ for any value of $n$ though $\lim_{n \to \infty}{S_n} = S = 1$.

As $n \to \infty$, the value of $S_n = 0.999...$.

So we can say that $1 \ne 0.999...$ ( as $S_n \ne 1$ for any value of $n$ ).

5. ### rpennerFully WiredValued Senior Member

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You are inventing new notation that has no definition. Therefore you have left the world of reasonable dialogue. Specifically, the term $T_n$ is only defined for the counting numbers 1,2,3,4,.... So the only way infinity is used in a meaningful way is in a formal notation for the unique limit of a sequence, i.e. $T = \lim_{n\to\infty} T_n = 0$. This has a meaning and you know it has a meaning because the references you bring to this discussion define it.

Since $T_n$ is only defined for counting numbers it is possible to use set notation to create a set that has all values of $T_n$:
$\mathbb{T} = \left{ T_n | n \in \mathbb{N}^+ \right}$​
and in that case we simply write $0 \not \in \mathbb{T}$ because there is NO COUNTING NUMBER that makes $T_n = 0$ true. Whatever you are trying to say in the first line here isn't meaningfully distinguished from the formal notation that I have used, so you are dropping the ball on your side of the dialogue to explain what you mean.

Only with the understanding that we have limited the universe of possibilities of n to be counting numbers.

With the same warnings as before. Also, for symmetry in argument, let us define:
$\mathbb{S} = \left{ S_n | n \in \mathbb{N}^+ \right}$​
just to be clear that $0.999... \not\in \mathbb{S}$ because all counting numbers are finite, all values of $S_n$ have a decimal representation with just a finite number of 9's and the 9's in 0.999... go on without end. That's why in a concrete sense we can write $0.999... = \sum \mathbb{T} = \textrm{sup} \mathbb{S} = \lim_{n\to\infty} S_n$ and as proven from analysis, $\lim_{n\to\infty} S_n = 1$.

This doesn't follow if in the same (undefined!) notation you write “$T_n \ne 0$ is also true, as "$n$ approaches infinity($n \to \infty$)".”

As you lack either definitions or demonstrations that these statements are true, these are empty assertions unconnected to the definitions of $T_n$ and $S_n$ and you are not communicating why someone should agree with your assertions.

That argument clearly doesn't hold water because $S_n \ne 0.999...$ for any value of $n$. That's why I thought it was important that you clearly communicate what you meant in line 1 of your post.

7. ### Quantum QuackLife's a tease...Valued Senior Member

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An interesting outcome to the refutation of my contention could be expressed as follows:
Under the current Minkowski /Einstein space time paradigm we see two light cones, past and future converging on a point in between them labelled the Hyper Surface of the Present.

On the basis of what has been stated that point between the past and future cone, must be zero dimensional.

This poses a problem unfortunately as far as I can tell. It means that the present moment doesn't actually exist at all....
For at t=0 [ present moment the time span is zero duration - no absolute rest is allowed] distance in all co-ordinates must also equal zero. Therefore the present moment is non-existent.

If as I contended zero was a 3 dimensional volume held with in a diameter of 1/infinity this problem goes away IMO...but as this has been clearly stated to not be the case the universe appears to be non-existent at any given t=0.

I believe the time span between past and future could be infinitesimal without invoking the problem of absolute rest and allow the universe to exist at t= infinitesimal duration.
as described with this diagram. As discussed earlier the Planck unit is a arbitrary length and can not be used as this invokes the issue of absolute rest.

Last edited: Mar 28, 2014
8. ### hansdaValued Senior Member

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I think you want clarifications for my quoted statement below.

Here consider the arrow symbol "$\to$" as "is approaching". So, "$x \to b$" means " $x$ is approaching $b$"; where $x$ is a variable and $b$ is constant. $x \to 5$ means "$x$ is approaching $5$". So, "$n \to \infty$" means " $n$ is approaching $\infty$".

So, my quoted statement above means: $T_n \ne 0$ is also true, as $n$ is approaching $\infty$.

Here $T_n = \frac {9}{10^n}$ and $n > 0$ where $n$ is a counting number.

$T_n$ is maximum, when $n = 1$. As $n$ increases, $T_n$ decreases.

As $n$ is increasing towards infinity, $T_n$ will decrease towards $0$.

Here $T_n$ is a member of the infinite geometric series, whoose constant ratio is $\frac {1}{10}$.

So, $\frac {T_2}{T_1} = \frac {T_{n+1}}{T_n} = \lim_{n \to \infty} {\frac {T_{n+1}}{T_n}} = \frac {1}{10}$

So, as $n$ is approaching infinity, $T_n$ will take a value such that the constant ratio is maintained.

As $n$ is approaching infinity, $T_n$ also approaches $0$.

So, as $n$ is approaching infinity, $T_n$ has to take a non-zero value in the neighbourhood of $0$; to maintain the constant ratio.

Thus, $T_n \ne 0$ is also true, as $n$ is aproaching $\infty$.

9. ### rpennerFully WiredValued Senior Member

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How can n approach infinity? There is no end to the number line so there is no finite distance from any finite number to infinity. In what sense can a variable approach something that it can't in any sense get closer to? What you want to use is the formal concept of the limit of an unending series.

You write "As $n$ is increasing towards infinity, $T_n$ will decrease towards 0." which is not a precise as $\lim_{n\to\infty} T_n = 0$ which says whenever you give me a positive number $\epsilon$, I can give you a number N such that $n > N$ will always mean $\left| T_n - 0\right| < \epsilon$. There is no infinity in the expansion of $\lim_{n\to\infty} T_n = 0$ because saying $N \lt n \lt \infty$ is redundant since n is limited to being a counting number.

Likewise saying $\forall n \in \mathbb{N}^+ \quad \quad T_n \neq 0$ is stronger than saying "$T \neq 0$ is also true, as $n$ is approaching [sic] $\infty$." because 1) no one knows what you mean when you write " as $n$ is approaching [sic] $\infty$" as I explained above, and 2) it doesn't matter as I pointed this out in post #915 that the expression $\frac{9}{10^n}$ is positive number divided by positive number and thus is a positive number which means greater than zero which means not equal to zero regardless.

10. ### arfa branecall me arfValued Senior Member

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I was taught that the notation $n \to k,\, k = constant$ means n gets 'arbitrarily' close to k. That's fine, because n and k are finite numbers, so when we write $n \to \infty$, that's strictly an abuse of notation.

Logically it says n gets "close to" a number which n is always a non-finite distance from, which is a contradiction. I think I can recall a lecturer saying something like: "Think of n as being arbitrarily large, such that 1/n is arbitrarily small, which goes to being arbitrarily 'close' in the limit".

11. ### rpennerFully WiredValued Senior Member

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I agree. "As n gets arbitrarily large" is a good translation of "$n\to\infty$" but the epsilon-N definition is even more explicit.

12. ### hansdaValued Senior Member

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$n$ can approach infinity as in the Limits to Infinity.

See this wiki quote on Limit:
You are trying to define the limit of $T_n$ at infinity($n \gt N$), whereas I am trying to find the actual value of $T_n$ at infinity($n \gt N$), whether it is zero or non-zero.

When you write $\lim_{n \to \infty} T_n = 0$, in the limit itself you are using the infinity symbol $\infty$.

Here $n$ approaching $\infty$ means $n \gt N$.

So, $T_n = \frac{9}{10^n} \ne 0$ as $n \gt N$.

13. ### hansdaValued Senior Member

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When the notation $n \to \infty$ is used in the limit, for example as $\lim_{n\to \infty} T_n = 0$; do you think, this is an abuse of notation?

14. ### rpennerFully WiredValued Senior Member

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This answers my question "how can n approach infinity" and you answered "Only in limit notation from analysis." Thus your source supports my claim that your objections to post #915 are nonsense.
Yes, since that's the whole point of using real numbers and analysis.
As I said before, that only makes sense in a mathematical system than allows you to do arithmetic with the symbol $\infty$, and it specifically excluded from my discussion because I restrict n to be a counting number. So there is no "at infinity" in my discussion.
But not in the expansion of that shorthand into the epsilon-N description of what it means.
Your own sources tell you different. The value of N is contingent on specifying what positive number epsilon is, but no matter what positive number epsilon is, there is a counting number N that makes this statement true: $n \in \mathbb{N}^+ \wedge n \gt N \quad \rightarrow \quad \left| T_n - 0 \right| \lt \epsilon$. And that's the definition of the limit of a sequence:
$\left( \lim_{n\to\infty} T_n = 0 \right) \quad \leftrightarrow \quad \left( \forall \epsilon \in \mathbb{R}^+ \; \exists N \in \mathbb{N}^+ \; \forall n \in \mathbb{N}^+ \; \left( n \gt N \quad \rightarrow \quad \left| T_n - 0 \right| \lt \epsilon \right) \right)$​
You can't just write $n \gt N$ without the contingency on epsilon and you can't just write $n\to\infty$ without using the concept of limits and real numbers.

Alternately you can think of the contingency of N on epsilon like this: $N = f(\epsilon)$ which is certainly the easiest way to prove the right side it true, but f() is not unique because if $\forall \epsilon \in \mathbb{R}^+ \; \forall n \in \mathbb{N}^+ \; \left( n \gt f(\epsilon) \quad \rightarrow \quad \left| T_n - 0 \right| \lt \epsilon \right)$ is true, then $\forall \epsilon \in \mathbb{R}^+ \; \forall n \in \mathbb{N}^+ \; \forall m \in \mathbb{N}^+ \; \left( n \gt m + f(\epsilon) \quad \rightarrow \quad \left| T_n - 0 \right| \lt \epsilon \right)$ is also true.

Gibberish repeated is still gibberish. The fact that no member of the unending sequence $(T_n)$ is zero was in post #915 which you were expected to read by both myself and James R. The phrase "$n \gt N$" cannot add to that statement. And $N$ is meaningless absence the contingency on epsilon as part of the formal limit definition.

I do not. I think trying to break apart the defined notation for the limit of a sequence into pieces which are not defined is an abuse of notation. If a mod gave you a warning for recommending "Cialis" to another poster, you don't get to complain that it must be OK to endorse "Cialis" because another poster urged the first poster to see a "Specialist". You can't just break apart words willy-nilly and assign meaning to their individual letters. This is the same problem I have with your use of "$n \to \infty$" outside the context of limits.

Likewise you don't get to break "$n \gt N$" outside of the context that defines what N is.

Last edited: Apr 2, 2014
15. ### Quantum QuackLife's a tease...Valued Senior Member

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Just out of curiosity, how would you show a value reducing infinitely?

from say 1 to 0?
[infinitely smaller as distinct from infinitely larger]

16. ### rpennerFully WiredValued Senior Member

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It depends on WHICH sequence you are considering.
• $\lim_{n\to\infty} \frac{1}{n} = 0$ because for every positive number $\epsilon$ we can assign $N = f(\epsilon) = \left\lfloor \frac{1}{\epsilon} \right\rfloor$ such that for all counting numbers n, if $n>N$ is true then we are guaranteed that $\left| \frac{1}{n} - 0 \right| \lt \epsilon$.
• $\lim_{n\to\infty} \frac{1}{n^2} = 0$ because for every positive number $\epsilon$ we can assign $N = f(\epsilon) = \left\lfloor \epsilon^{\tiny - \frac{1}{2}} \right\rfloor$ such that for all counting numbers n, if $n>N$ is true then we are guaranteed that $\left| \frac{1}{n^2} - 0 \right| \lt \epsilon$.
• For any real constant k, $\lim_{n\to\infty} (2 + k^2)^{-n} = 0$ because for every positive number $\epsilon$ we can assign $N = f(\epsilon) = \textrm{max} \left(0, \left\lfloor - \log_2 \epsilon \right\rfloor \right)$ such that for all counting numbers n, if $n>N$ is true then we are guaranteed that $\left| (2 + k^2)^{-n} - 0 \right| \lt \epsilon$.
• For any real constant k, $\lim_{n\to\infty} (2 + n^k)^{-n} = 0$ because for every positive number $\epsilon$ we can assign $N = f(\epsilon) = \textrm{max} \left(0, \left\lfloor - \log_2 \epsilon \right\rfloor \right)$ such that for all counting numbers n, if $n>N$ is true then we are guaranteed that $\left| (2 + n^k)^{-n} - 0 \right| \lt \epsilon$.
• ...

Related graphs demonstrating the value of in the absolute signs is less than the value epsilon:

Last edited: Apr 3, 2014
17. ### Quantum QuackLife's a tease...Valued Senior Member

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Thanks rpenner, I guess one of the reasons for asking was that when considering infinity being applied to largeness we have no actual upper boundary or limit[ excuse my language] where as when applying infinity in reductions we have a "natural limit" [ for want of any better words ] and that being zero.
I thought that by comparing the two situations, infinitely large and infinitely small it may [but probably may not] help you guys resolve the issues you appear to be discussing.

18. ### Quantum QuackLife's a tease...Valued Senior Member

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One of the interesting things that have come out of this thread is that we have conclusively proved that the zero point between past and future is non-existent. Which ultimately means that at that zero point distance is also non-existent.

So I do not know how this can be reconciled with the fact that we appear to exist, even though that moment of our apparent existence is zero dimensional.
any ideas?

19. ### rpennerFully WiredValued Senior Member

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That's a ridiculous and unsupported claim.

// Edit: After careful consideration of both [post=3181475]post #1157[/post] and [post=3175768]post #1144[/post], where no argument connects the assertion that something is zero-dimensional and in some specific philosophical sense "doesn't exist" (that's what I meant by "unsupported"), my position remains unchanged to the point that no further post from me is required. It is "ridiculous" to nakedly assert controversial claims about reality, physics or philosophy without supporting them. Asserting that "here and now" doesn't exist is clearly controversial, in that relativity goes out of its way to point out that neither "here" nor "now" can exist in any absolute sense honored by the laws of physics but nevertheless is happy to work with events of which the zero-dimensional "here and now" is just the closest example.

Last edited: Apr 10, 2014
20. ### Quantum QuackLife's a tease...Valued Senior Member

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please refer to post #1144 for a summary of my premises.
You only have to ask yourself,
"If the duration of time at the zero point between past and future is zero then....?"
and work it out for yourself....

Last edited: Apr 10, 2014
21. ### Quantum QuackLife's a tease...Valued Senior Member

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rpenner , thank you for your serious and careful consideration,

I am NOT asserting a claim.
The math itself is..
The light cone diagram often used is.

If the time duration between past and future, light cones, is zero then what mathematics can be used to describe the reality of the universe [ or any relevant observer frame of reference ] at that point in time?
This is a question for mathematics not philosophy.
It is not I that is being controversial, it is the simple axiomatic logical outcome of a scientific position that is being controversial.
I am merely the messenger not the actor.

If we are satisfied that a zero dimensional universe [ or SRT reference frame] as described by our theoretic's at any given t=0 duration point, can somehow exist then so be it. But that is not Science nor is it philosophy. That is something else entirely.

Again I repeat:
I am merely the messenger not the actor.
I did not create this situation nor am I going to be the one to fix it... [if it is found to be in need of fixing]

IMO, The issue transcends contemporary science and all the light cones do is highlight the issue for us all to observe.

Last edited: Apr 14, 2014
22. ### hansdaValued Senior Member

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From the examples you can see that the "actual values" and the "limit values" of $f(n)$ are not same, as $n$ approaches infinity.
...

A limit also can be explained in terms of arrows. For example "$\lim_{n \to \infty}{T_n} = 0$", can also be explained as:"$T_n \to 0$ as $n \to \infty$".

Do you think here usage of "$n \to \infty$" is an abuse of notation?

23. ### rpennerFully WiredValued Senior Member

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That's a pretty generic properties of the limits of most every non-constant sequence and its limit. This should be as no surprise as I said this in [post=3164396]#915[/post]. Your source and your examples support my post #915.

That's not "explaining" a limit. That's just alternate notation for expressing a limit.

I said "$n \to \infty$" IS an abuse of notation if you attempt to extract it away from the full statement of the limit of a sequence where is a mere part of a complete expression whose meaning doesn't actual involve manipulation of any larger-than-finite number.