Another Attempt at Rigor

Summary:

Attempt #2 at proving 0.999... is not equal to 1

Another Attempt at Rigor

WillTheSerious 28 Oct 2010 17:32

$0.999... \neq 1$.

Proof:

Let $S=\{x|x \in \mathbb{R}, 0<x<1\}$.
Since $S$ is made up of all $x \in \mathbb{R}$ between 0 and 1, $S$is an infinite set, meaning $S$ is not finite (by the definition of infinite).
$S$has a first point to the right equal to $1$where $1 \notin S$. (This is by the completeness axiom and the definition of an open set).
Let there be a point $p \in S$of the form $p = -{1-10^{n} \over 10^{n}}$where $n \in \mathbb{N}$. (note: It is considered evident that for any number $n$, $p$ then can be written in the form $0.999...9_{n}$ where the decimal expansion is equal in length to $n$).
Let there then be a point $q \in S$ where $q = -{1-10^{n+1} \over 10^{n+1}}$.
It is then demonstrable that $p<q$where $p \in S$ and $q \in S$.
As $S \in \mathbb{R}$ and is an infinite set, by the axiom that between any two points there is always another point in an infinite set, then for any value of $n, p<q<1$.
As $n \to \infty$, the length of $p$'s decimal expantion also $\to \infty$.
Since $q > p$, $q = 0.999...$ as $n \to \infty$.
Since $1$ is the first point to the right of $S$ and since $1 \notin S$ and $q \in S$, $q \neq 1$.
$\therefore 0.999... \neq 1$.

Q.E.D.

Re: Another Attempt at Rigor

Jeffrey Taylor 28 Oct 2010 21:59

First a quibble:

You say

Let there be a point $p \in S of the form p = -{1-10^{n} \over 10^{n}}$

and you use the same form for q. Wouldn't it be simpler to avoid the double negative by writing this as ${10^{n}-1 \over 10^{n}}$??

The problem with your proof, though is the " =" in the next-to-last line. You can truly say that
as $n \to \infty q \to 0.999...$ but q "approaches" only in a non-meaningful way: for any natural number n, there will always be a finite number of 9's in the string, and in that critical sense the number of 9's never comes any closer to infinity than when there was only a single 9. Try not using the verb "approach": as n increases indefinitely, the length of the decimal string of 9's increases indefinitely. It never comes anywhere "near" (whatever that would mean) an infinite length.

BUT it does converge, in the sense we've given that: for any arbitrary open interval you choose around 1, all members of your sequence for n sufficiently large will fall inside the interval. How large n has to be depends on how close to 1 you choose your lower bound, but since this is an open interval, you'll always have to choose something less than 1.

Notice, too, that were your argument correct, we would have an inconvenient incompatibility between our two main notational systems for fractions: 3 x 1/3 would be 3/3, which simplifies to 1, but 3 x 0.333… would be 0.999… and would NOT be equal to the equivalent fractional answer, presumably because 0.333… is somehow "not" 1/3. But it can be demonstrated that you can find no real number to represent the difference between the 0.333… and 1/3, so it would appear there is no reason not to take them as equivalent. In other words, were your position correct, the decimal system (which is forced to SUGGEST (though obviously, never actually to produce) infinite strings of repeating digit patterns to represent any fraction whose denominator cannot be expressed as a product of powers of 2 and 5) would "break" frquently, and we would need to resort to ratio form:

0.5
1/3
0.25
0.2
1/6
1/7
0.125
1/9
0.1

which would needlessly complicate arithmetic. So at the least these indefinite strings of repeating digit patterns have convenience on their side.

Perhaps this model will help. Think of the subtraction 1.000… - 0.999… where the ellipses mean the digits 0 or 9 are repeated indefinitely. If there were some positive number n at which the repetition stopped, we could use an algorithm: Change the 1 to 0, change each of its 0's from 1 to n - 1 by a 9, and replace the nth zero with 10. Then subtract in the nth place. The result would be the difference between the numbers: 0 followed by a decimal followed by n - 1 0's and a 1, = 1/10^n. But the ellipses make a difference when we use the algorithm: the process of using the algorithm just causes us to rewrite the 1 as 0.999… indefinitely, and there is no n for which the new 9 replacing a 0 is not matched by a 9 in the 0.999… sequence. In this case, the attempt to subtract and show a difference turns into a demonstration of the identity of the two numbers. So long as we never stop, replacing 1.000… with 0.999… MAKES NO DIFFERENCE. As there is never an nth place at which we can place a 10 instead of a 9, the two strings are now identical. They represent the same quantity. 1 = 0.999…

Re: Another Attempt at Rigor

WillTheSerious 29 Oct 2010 03:56

If it were accepted that 0.999… does not equal 1, but that the difference is infintitesimal and therefore of no practical consequence for the interchanging of the two forms for convenience sake, I see no problem with continuing to treat the two numbers as identical. I still don't believe they are the same number even when it can be shown that every application of either number leads to the same outcome. After all, $\mathbb{R}$ is a continuum and there has to be a transition between two locations on that continuum where the starting point and the point just before reaching a discrete value look exactly alike but are different. How else can there be any change?

When I am working a math problem and I come across the value 0.999… either on my calculator or
in my calculated algorithms, I WILL simply assume 1. I would not ever expect to have a problem.

Last edited on 29 Oct 2010 16:50 by WillTheSerious + Show more

Re: Another Attempt at Rigor

WillTheSerious 29 Oct 2010 16:42

I agree with your first point about the form of the equation but when I did the simplification I didn't get the right outcome. I assume I made a math error but I didn't work it out to be sure. I just used the form I knew would work.

Last edited on 29 Oct 2010 16:44 by WillTheSerious + Show more

Re: Another Attempt at Rigor

Jeffrey Taylor 29 Oct 2010 16:50

So we really do come back to Zeno: at each instant of its flight, an arrow occupies exactly as much space as the measure of its length, and an has exact position (to simplify without compromising the situation, (x,y) in a 2-D plane defined by its path, denoting the exact position of its tip). At each succeeding instant it will conserve its length, yet the (x,y) will be different, though fixed. So at each instant the arrow is at rest, yet it appears to move, so when/where does the movement take place? Not at any instant of time, nor at any identifiable location.

Zeno, being Greek, was trying to show that "becoming" was not real: that, to put it crudely enough to show the word-game he was trapped in, "becoming," which means change, the not-yet-here-and-now, was incompatible with "being" which to him meant permanence, the "always already completely here-now." Only the eternal could really BE, the phenomenal, changeable world was not only impermanent but irrational. Of course he knew that arrows could move (and potentially kill). That just proved the irrationality of the world, to him.

The question is, what are we trying to show now? You seem to want to equivocate: to concede that there is NO difference that can be stated between the two representations of unity (1 and the unwriteable 0.999…) while insisting that there must be SOME difference. So we seem to reduce the issue to one for which no mathematical argument is appropriate: in what case(s) can two objects of discussion be both the same and different in regard to a single property (here, quantity represented). Both the classic Aristotelian logic of qualifiers, and the conventional 2-valued modern logic of predication have a "law of the excluded middle" as an axiomatic way of excluding such a situation, but these could be replaced with other, perhaps multi-valued, logics that did not include this axiom.

So why would we do that in order to accommodate this case of two representations of unity whose visual-conceptual difference makes you suspect a difference of reference? If it would lead to some better "fit" of the representational system to our measurements of the world, that would be a gain, but you concede that, from a strictly arithmetic point of view we get no advantage (and I hope you'll concede we also suffer some real inconveniences) from the assumption. And philosophically, I don't see any advantage either: both "1" and "0.999…" when looked at as they are "in themselves", are just socially agreed-upon scratchmarks for representing what we mean when we say "unity" or "one" or (in other languages) "une", "eine", "uno". HOW we represent does not alter WHAT we represent. So why worry oneself about the apparent differences of form (the unwriteably indefinite string of 9's suggesting the sum of an endless series of 9/(10^n) + 9/(10^(n+1) … for all n, versus the nice, clean single stroke of a 1) if we use them interchangeably? The point is to recognize the referent of both as the same, and to understand why sometimes one, sometimes the other is used. For example, most scientific calculators doing anything more complicated than the four elementary arithmetic operations will often produce answers of the form "44.999…" to be understood as "45", because of the way they internally represent numbers and operations. So we teach students to RECOGNIZE that "0.999…" is a numerical synonym for "1" so they can interface with their tools more successfully. On the other hand, I cannot enter the string "0.999…" into my calculator, so instead I input "1", confident that the result will be exactly the same.

Re: Another Attempt at Rigor

WillTheSerious 29 Oct 2010 22:46

It is not the number .999… and it equivalence to 1 that I really have any trouble with. I am perfectly capable of accepting either conclusion as fact on faith alone. However, what one assumption says about the nature of Infinity is fundamentally different than what the other assumption says. To say 0.999… is 1 is to consider a final conclusion to the infinite string of 9's and that an infinitesimal value is the same as zero. My concept of Infinity is of the never ending and along with that is the idea that we can not truely know what value we end with because Infinity is an Existential idea that never is. If we are to believe Cantor, then we can not determine which $\aleph_{n}$ we are dealing with when we conclude 0.999… is 1 and Cantor's ideas would imply there is a very very big difference.

When I started this discussion I was of the opinion that Infinity was Infinity and to say there is more then one sized Infinity is paradoxical. I think there are some interesting and fundamental problems with Cantor's Diagonal Arguement in that his table can never represent all permutations of 1's and 0's or any other base series because to do so would never produce a square table from which to take a unique diagonal. Dana challenges me to demonstrate that a table that is infinite in width (the string of 1's and 0's) covers less area then the height (the number of permutations) of the table. But Infinity is not an achievable target. It is not going to be the case that we "build" a table that is complete only at Infinity. When I talk about a number or a string being infinite, I am reminded that it is not going to be the case. The 9's are never going to be infinite. If Cantor is right then the height of the table is a different Infinity then the width and if he is not right then they are the same. If they are the same he proves his case but only because he's wrong. If he can't prove it. It is because he is right. I have therefore, come to the conclusion that Cantor is right, there are different sized Infinities but the Diagonal Arguement doesn't prove it. Cantor did some amazing and very sophisticated mathematical slight-of-hand. Gödel used the same logical idea to prove his incompleteness theory. I think it was perfectly appropriate for Gödel to use it. It was not slight-of-hand the way he did it.

I expect my opinion to be very unpopular but well - like agent Mulder, I want to believe, I just don't yet.

Last edited on 29 Oct 2010 22:51 by WillTheSerious + Show more

Re: Another Attempt at Rigor

Dana Ernst 30 Oct 2010 18:20

When I talk about a number or a string being infinite, I am reminded that it is not going to be the case. The 9's are never going to be infinite.

By the same reasoning, the set of natural numbers is bounded. In this case, this would imply that there is a largest natural number. Really? What number would that be? If you told me what it was, I would just say, "add 1 to it," and we would have a number larger than the one you said was the largest. This can't be.

If Cantor is right then the height of the table is a different Infinity then the width and if he is not right then they are the same.

No, both are countable. In fact, you can easily see how to enumerate them: first column, second column, etc. and first row, second row, etc.

If they are the same he proves his case but only because he's wrong. If he can't prove it. It is because he is right.

Huh?

I have therefore, come to the conclusion that Cantor is right, there are different sized Infinities but the Diagonal Arguement doesn't prove it. Cantor did some amazing and very sophisticated mathematical slight-of-hand. Gödel used the same logical idea to prove his incompleteness theory. I think it was perfectly appropriate for Gödel to use it. It was not slight-of-hand the way he did it.

I am of the opinion that if you have a problem with Cantor's diagonal argument, then you should be trying to burn Gödel at the stake.

Do you take issue with any of the following statements:

1. $\mathbb{N}$ is unbounded on the right and infinite.

2. $\mathbb{R}$ is unbounded and infinite.

3. The sequence $\{1/n\}$ is infinite but converges to 0.

4. The sequence $\{n^2\}$ is infinite, unbounded, and diverges.

5. The expression $\sum_{n=1}^{\infty}\frac{1}{2^n}$ is an infinite series but converges to 1 despite the fact that no partial sum (only finitely many terms) is equal to 1.

6. The expression $.9\overline{99}$ is an infinite string of 9's to the right of the decimal place and is by definition equal to $\sum_{n=1}^{\infty}\frac{9^n}{10^n}$.

7. $\sum_{n=1}^{\infty}\frac{3}{10^n}=.3\overline{33}=\frac{1}{3}$.

8. The open interval $(0,1)$ is bounded but infinite.

I could go on and on.

Re: Another Attempt at Rigor

WillTheSerious 01 Nov 2010 18:32

1. $\mathbb{N}$is unbounded on the right and infinite.

No problem here. There is always another natural number when one needs it.

2. $\mathbb{R}$is unbounded and infinite.

This is not a problem either.

3. The $\{ 1/n\}$ sequence is infinite but converges to 0.

Now we are getting into some areas of doubt. What does $n$$ equal and how do I know when $n$ is big enough to divide $1$ into a small enough part to be called $0$? Can't we always say the $1/n > 0$?

4. The sequence $\{2^{n}\}$ is infinite, unbounded, and diverges.

I agree.

5. The expression $\sum_{n=1}^\infty{} \frac{1}{2^n}$ is an infinite series but converges to 1 despite the fact that no partial sum (only finitely many terms) is equal to 1.

I understand this to be true by the meaning of convergence, but are you saying this series is equal to 1? Or, are you saying that as $n$ grows without limits the difference between $\sum_{n=1}^\infty{} \frac{1}{2^n}$ and 1 becomes infinitesimal? Is the infinitesimal an inverse expression of the infinite? Does this series stop when $n$ gets to Infinity or are we saying that $n$ can grow and grow and grow without interuption, forever, so even though it can't actually get to a number that could ever be big enough to be called infinite in size, it would get there if it were at all possible? If Infinity were achievable for $n$ might it then stop? Or is there another infinite marker to go towards beyond there. Maybe it wouldn't make any difference by that point. Just because a number is considered infinite doesn't mean that's it. There are different infinities and a countable infinity may infact continue on to an uncountable infinity farther down the same road. I don't know, I'm asking. It doesn't seem like a number is done just because it is infinite but there isn't much more that can be said about it until we can talk about whether we are keeping to a limited (countable) Infinity as an invariable property of one size fits all. Meaning that this series is the same size as that series no matter what as long as it is understood that they are both infinite. In the chart of allowable operations with Infinity; on Infinity can not be subtracted from another Infinity but adding them is okay. We just don't go anywhere when we do that. Maybe we should consider the possibility that we could.

Dana says that my statement about the $0.999...$ string never concluding to Infinity means that $\mathbb{N}$ is bonded.

By the same reasoning, the set of natural numbers is bounded.

That is exactly what I am trying to say is not the case. $\mathbb{N}$ is not bounded but is there a number $n \in \mathbb{N}$ that is infinite? I have been taught 'no'. I can understand that idea too, there is always another number to add and no $n$ can be infinite for the reason that it is achievable and surpassible. So what of the $9$'s? that are $n$ digits long?

Last edited on 02 Nov 2010 17:09 by WillTheSerious + Show more