1 is 0.9999999999999............

"As the number of digits used increases without bound, the difference between 1 and the repeating decimal expansion 0.999... shrinks to be smaller in magnitude than any finite number, thus the magnitude of the difference in the limit of an infinite number of steps must be zero."
ahhh so I learn something... thanks for the info... now that makes sense. [why no non-zero infinitesimals are included typically in this number system]
There is no need for the above proof based solely on this rule..simple: 0.999... = 1 end of story...


edited:
yeah 'twas a stupid question...

thanks for your time...
 
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The most interesting thing about this discussion is not whether 0.999...=1 (I mean really, who cares? What difference does it make?)
It's why our minds rebel from the notion.

Intuition and ambiguous teaching lead students to think of the limit of a sequence as a kind of infinite process rather than a fixed value, since a sequence need not reach its limit. Where students accept the difference between a sequence of numbers and its limit, they might read "0.999..." as meaning the sequence rather than its limit.
- Tall and Schwarzenberger, quoted in Wikipedia

The lower primate in us still resists, saying: .999~ doesn't really represent a number, then, but a process. To find a number we have to halt the process, at which point the .999~ = 1 thing falls apart.
Nonsense.
- Cecil Adams, also quoted in Wikipedia

why do people resist the notion?
If one is made aware of the following :
"As the number of digits used increases without bound, the difference between 1 and the repeating decimal expansion 0.999... shrinks to be smaller in magnitude than any finite number, thus the magnitude of the difference in the limit of an infinite number of steps must be zero."

there would be no problem.. I would think...

but no, as is probably happening in the class room, we end up with proofs like the one I/we have been wasting our time on when in fact no proof is required other than proving the above quoted statement.

Even the wiki article fails to mention the above at the top of the page where it needs to be...
So the teachers need to get or at least reinforce the facts on the table before throwing UNNECESSARY proofs that only confuse and raise suspicion and curiosity unnecessarily.
 
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How do they determine when a number has shrunken sufficiently enough to be smaller in magnitude than any finite number?
That's the question that should have been asked to begin with.. IMO
 
yes, although I can not read the notations [not familiar with the notation] you have made I can see that the limit function may be applied only in one direction...

Could you please explain why the result would end up being asymmetrical if you think this is the case?
Or do you believe the proof can be reversed where 1.000... can be proved to = 0.999...

Basically I am suggesting
that if 0.999... can be proved to = 1
then 1.000.... should be able to be be proved to = 0.999... using the same process from start to finish

and I believe at this stage this is not possible.

btw I do not believe this lack of symmetry, if proven, necessarily invalidates the proof. But it certainly makes it an interesting one to me, for many reasons.

$$\lim_{n \to \infty} (\frac{9}{10^1}+....+\frac{9}{10^n})$$ means adding an INFINITE number of terms in the sum $$S_n=\frac{9}{10^1}+....+\frac{9}{10^n}$$
$$\frac{9}{10^1}+....+\frac{9}{10^n}$$ means adding a FINITE number of terms.
Both concepts are taught in teaching series (of powers), in college. If someone explained this to you earlier in the thread, we would have saved pages and pages of misunderstandings.

Even the wiki article fails to mention the above at the top of the page where it needs to be...
So the teachers need to get or at least reinforce the facts on the table before throwing UNNECESSARY proofs that only confuse and raise suspicion and curiosity unnecessarily.

Wiki is not the best way of learning, taking a class in calculus (in this particular case) is a better option. The chapter on "Limits" and the chapter on "Series"
 
I really don't have time to read through your exceedingly long posts that largely just seem to state the same things eight different ways and somehow manage to come close to (or even crossing the threshold of word salad.

And yet, evidently you have not.


As usual, you've got it arse about face.


Excellent answer. Unfortunately it won't stop him from coming back and serving a word salad even bigger than what he's been serving. He will do so for as long as he's allowed to pollute this forum, this is what he's done under his prior name, this is what he's doing now: pretend to post science while, in reality, posting garbage.
 
$$\lim_{n \to \infty} (\frac{9}{10^1}+....+\frac{9}{10^n})$$ means adding an INFINITE number of terms in the sum $$S_n=\frac{9}{10^1}+....+\frac{9}{10^n}$$
$$\frac{9}{10^1}+....+\frac{9}{10^n}$$ means adding a FINITE number of terms.
Both concepts are taught in teaching series (of powers), in college. If someone explained this to you earlier in the thread, we would have saved pages and pages of misunderstandings.



Wiki is not the best way of learning, taking a class in calculus (in this particular case) is a better option. The chapter on "Limits" and the chapter on "Series"
Thank you for showing this in the previous post... but it took that quote:

"As the number of digits used increases without bound, the difference between 1 and the repeating decimal expansion 0.999... shrinks to be smaller in magnitude than any finite number, thus the magnitude of the difference in the limit of an infinite number of steps must be zero."
to seal the deal....so to speak.

So I ask myself, "Why is there a need to prove something if no proof is necessary?" I tend to think that the teachers maybe trying to allow a bit of irony and humor to play in the classroom. Instead of saying this is a proof maybe saying it is an interesting application of something that doesn't need proving. It certainly hooked me in so I can imagine what it is doing to others.

The question now though is how do we determine when a number's magnitude is too small to be considered finite?

This may not be so easy to resolve except by pointing to the appropriate theorems etc.
But this is what should be in question not the unnecessary proofs or applications. IMO

Again thanks for your time.
 
The question now though is how do we determine when a number's magnitude is too small to be considered finite?

This is precisely what you learn in the chapter called "Limits". The explanation is not trivial, rpenner had an attempt at it but it was overcomplicated and muddled. So, may I suggest once more the Calculus class?

This may not be so easy to resolve except by pointing to the appropriate theorems etc.

Correct, see above.


Again thanks for your time.

You are welcome. take the class, you won't regret it.
 
thanks for your time...
No problem.

How do they determine when a number has shrunken sufficiently enough to be smaller in magnitude than any finite number?
That's the question that should have been asked to begin with.. IMO

Pick a tiny (positive) number, $$\delta$$. Find out if there is a natural number M such for all n > M the difference between the sequence and the purported limit of the sequence is less than $$\delta$$. Prove this is the case for all such tiny positive numbers, typically by writing a relationship between M and $$\delta$$.

For infinite decimals of the form $$x = a_0.a_1a_2a_3a_4a_5a_6... = \sum_{k=0}^{\infty} \frac{a_k}{10^k} = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{a_k}{10^k}$$
such a relationship is $$M = - \lfloor \log_{10} \textrm{min}(1, \delta) \rfloor$$
Which is saying that the difference between (n > 0)$$a_0.a_1$$ (and all expansions with more digits) and $$x$$ is less than 1.
Which is saying that the difference between (n> 1)$$a_0.a_1a_2$$ and $$x$$ is less than $$\frac{1}{10}$$.
Which is saying that the difference between (n>2)$$a_0.a_1a_2a_3$$ and $$x$$ is less than $$\frac{1}{100}$$.

Even if $$\delta$$ is some ridiculously small number like $$\frac{1}{\left( \textrm{1 googol} \right)^{\textrm{1 googol}}}$$ there is a M such that all $$n>M$$ means the finite version digit expansion is closer than $$\delta$$ to x.
So in the limit of an infinite number of terms the digit expansion must equal x.
 
Trippy , I appreciate your post...
however it is also a tautology or self evident that the following is true also

9.999.... - 0.999 = 9 & 9 + 0.999... = 10

same as exampled with 10-6 = 4 & 4 + 6 = 10
We have asymmetry, therefore reversibility to prove the equations.

The proof for 0.999... = 1 can not be reversed and is asymmetrical as far as I can tell..
I believe poster Tach may have something to say about this.

Ummm...

Look again at the proof that uses the decimal expansion of $$\frac{1}{9}$$ for a start off (or the one that used 0.333(3)...

$$\frac{1}{9}$$ = 0.111(1)
Multiply both sides by 9
$$\frac{9}{9}$$ = 0.999(9)
1=0.999(9)
 
Ummm...

Look again at the proof that uses the decimal expansion of $$\frac{1}{9}$$ for a start off (or the one that used 0.333(3)...

$$\frac{1}{9}$$ = 0.111(1)
Multiply both sides by 9
$$\frac{9}{9}$$ = 0.999(9)
1=0.999(9)
That is nice and short, but assuming $$\frac{1}{9}$$ = 0.111(1) seems to be same as assuming 1=0.999(9)
I.e. now we need to prove the starting assumption that:$$\frac{1}{9}$$ = 0.111(1) do we not? {for a valid proof of 1=0.999(9) }
 
Good Morning, Trippy. :)

I really don't have time to read through your exceedingly long posts that largely just seem to state the same things eight different ways and somehow manage to come close to (or even crossing the threshold of) word salad.

Urmmm, Trippy, may I remind you that this is the "Alternative Theories" section, not the "Established Maths" section. Here participants are exploring and discussing alternative views with a view to further refining/replacing established understandings, if possible. Yes?

Moreover, whenever the discussion is moved back to the very Axiomatic/Conventionality 'beginnings', and the reasons/validity/consistency OF those in the 'larger picture' context (and not being just constrained to 'blindly following' of the consequences OF those very same axioms/conventions once 'introduced'), then it is unavoidable to end up with WORDS to arrive at some sort of MUTUAL 'contextual understandings' rather than just being constrained to self-justifying circularity of just repeating the 'blindly following' numerical 'proofs' manipulations which depend on certain 'exceptions' in order to make the axioms/conventionality consistent in a self-serving way rather than exploring the basis for the whole system 'from scratch'.

And that is what I am doing (politely and 'alternatively' exploring/discussing) here certain claims made by you and others IN THE CONTEXT OF EXAMINING whence such claims arose and whether they are self-consistent in both the limited axioms AND in the 'bigger picture' which I have already made clear I am 'coming from' in this discussion.

Given this section, and the nature of the discussion as I point out, it is a bit ungenerous to characterize the INESCAPABLE USE of EXPLORATORY and necessary WORDS and discussion modes which must necessarily precede any 'blindly following' and 'self-serving' REPETITION of 'numerical proofs' and all those very things which this discussion (from where I am coming) are all in question precisely because of the 'from scratch' exploration/discussion points/words presented to treat what went before the axioms/conventions rather than just what follows them (which we all agree DOES blindly follow 'given' those axioms/conventions, but which I now question the 'completeness' of those axioms/conventions THEMSELVES).

So your ungenerous and 'personal' attitude characterizing others use (perforce of the nature of this discussion) of their polite and on-topic descriptive words in fair context, as 'word salad', is not really playing cricket, is it. It brings a certain prejudicial air to your reading in the context, which only makes the common understanding part of a conversation difficult from the very start, and failure to fully understand the other guys' stance AS EXPLAINED in their (so-called by you) 'word salad' indicates laziness and/or reading bias and automatic resort to kneejerking rather than fair and exhaustive conversation on point as presented/explained in context. Yes?

If you do not read this reply like many others) properly and fairly, then you automatically disqualify yourself from, and maybe even distort, the conversation based on your own personal attitudes to 'the source' rather than fairly and considering the points put in context of THIS Alternative Theories section/discussion. Yes? :)

First off, the zeroes are not trivial - that's not what significant means in this context.

With respect, but you seem not to have read or understood my quoted statement (to Pete) recognizing the differences IN CONTEXT.

If you had read/understood properly, then you would have gathered that I said zeros can be trivial in some cases and not trivial in other cases. Context indicates there significance or their non-significance etc. That is what I said. So your attempt at characterizing that I did not recognize/understand the difference is sailing pretty close to a 'strawman'. But perhaps it isn't meant to be a strawman; perhaps you just didn't read/understand that relevant post/comment properly and IN FULL. I will assume so rather than take it as an intentional strawman (since I know you would never do such a thing intentionally). :


Secondly the 'formatting device' is the agreement not to write the non-significant zeroes because it's cumbersome, not the other way around. If you want to deal in fundamentals then you have to realize that it 1.0(0) not 1. The zeroes exist whether you write them or not. Do I need to explain to you what numbers actually mean?

The decimal point is fundamental even though the convention is not to write it unless the zeroes after it. If we follow your logic, then rather than writing "A circle of 565m has an area of 1.00 km[sup]2[/sup]" I should just write "A circle of 565m has an area of 1km[sup]2[/sup]". There's a problem with this - a circle with a radius of 560m also has an area of 1km[sup]2[/sup].


The fundamental fact is that the zeroes before and after the 1 exist whether we choose to write them or not.
The convention is to not write zeroes unless they're significant.

I can't say it any more directly than that.

Ok. Conventions. Let's assume (for the sake of argument) that I agree with your insistence on a "pre-existing zeros" convention; ie:

"0.00000 ->" in an unending series of 'pre-existing' zeros to the right of the decimal point. Where do we get to with that convention?

The example I initially questioned, IN THAT CONTEXT only,was Pete's use of a 'proof' involving the step/operation :

10 x 0.99999... = 9.9999...

I pointed out that fundamentally we added the zero from the 10 to the end of the original 0.9999... unending string, which "forced" the unending string to move the leading "9" one position to the left, past the decimal point, resulting in:

9.9999...0

Pete's 'convention' was to interpret that 'result' as being the (in my view TRIVIAL convention) "action" of "shifting the decimal point to the right".

Whereas I argued the "alternative" view (from observation of the real fundamentals I perceive); ie:

That in fact the REAL action was to add a zero (from the real number 10 factor/operation/entity) to the end of the unending string, and so by that actual action forced the ACTUAL NUMBER leading 9 to move to the left, rather than merely trivially and conventionally making some NON-real "action/operation" trivially described as "moving the decimal point to the right".

If zero is a number; and if that zero IN CONTEXT IS 'significant' and not just trivial, THEN I merely observe that the 'real action involving that adding a zero at the end is MORE fundamental and significant than any mere convention of "moving the decimal point to the right" as per his argument/proof in that case/context.

Oh, and before you come back with the obvious trivial retort that 'the pre-existing zeros' and 'the repeating 9 string never 'ends', you would be trying to have it both ways: ie, such strings are conventionally presumed (via the Limits argument) to END at some limiting 'state' in the context of 'infinity' arguments/conventions; and yet you adjured QQ to "try long division" etc which never ends. So I leave it to the more 'real action' context to tell which 'take' is the real way to understand this aspect being re-examined in this 'alternative theory' section/discussion. :)

Anyhow, that's it, that's where I am coming from in this discussion of alternative exploration 'from scratch' context, mate. No more, no less than that. If it's still all 'word salad' to you, or if you don't bother reading all the arguments in full, then you leave yourself open to creating further UNINTENTIONAL 'strawmen'.

In any case, Trippy, thanks sincerely for your responses regardless; much appreciated all the same. :)
 
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Billy T, $$\frac{1}{9} = 0.111...$$ (or the equivalence any repeating decimal with a particular rational number) follows by either of the division algorithms AND the continuity of the reals. But this topic is typically not taught to anyone other than mathematicians.

But it is easy to see that the only possible repeating decimal for 1/9 is 0.111... because the division algorithm turns into an infinite loop.
Division algorithm 2 for positive rationals
Let $$s'_0 = ( p \; \textrm{mod} \; q ) + q \left[ q|p \; \wedge \; p \neq 0 \right] , \quad a'_0 = \frac{ p - s'_0 }{q}, z'_0 = a'_0$$ then it follows that $$a'_0 \in N_0, \; s'_0 \in N_0 \; s'_0 \leq q, \; z'_0 = \sum_{k=0}^{0} b^{-k} a'_k , \; b^0 p = s'_0 + b^0 q z'_0 $$
Let $$s'_{n+1} = \left( (b s'_n) \; \textrm{mod} \; q \right) + q \left[ q|(b s'_n) \; \wedge \; s'_n \neq 0 \right] , \quad a'_{n+1} = \frac{ b s'_n - s'_{n+1} }{q}$$ then it follows that $$a'_0 \in \mathbb{N}_0, \; s'_0 \in \mathbb{N}_0 \; s'_{n+1} \leq q, \; a'_{n+1} < b, \; z'_{n+1} = \sum_{k=0}^{n+1} b^{-k} a'_k , \; b^{n+1} p = s'_{n+1} + b^{n+1} q z'_{n+1}$$

...

Example 2
Let $$p = 1, q = b-1$$
Then $$a_0 = 0, a_1 = 1, a_2 = 1, a_3 = 1, ..., s_0 = 1, s_1 = 1, s_2 = 1, s_3 = 1, ... , z_0 = 0, z_1 = \frac{b^1 -1}{(b-1)b^1}, z_2 = \frac{b^2 -1}{(b-1)b^2}, z_3 = \frac{b^3 -1}{(b-1)b^3}, ...$$
and $$a'_0 = 0, a'_1 = 1, a'_2 = 1, a'_3 = 1, ..., s'_0 = 1, s'_1 = 1, s'_2 = 1, s'_3 = 1, ... , z'_0 = 0, z'_1 = \frac{b^1 -1}{(b-1)b^1}, z'_2 = \frac{b^2 -1}{(b-1)b^2}, z'_3 = \frac{b^3 -1}{(b-1)b^3}, ...$$
And $$ \lim_{n\to\infty} z_n = \lim_{n\to\infty} z'_n = \frac{1}{b-1} \lim_{n\to\infty} \left( 1 - \frac{1}{b^n} \right) = \frac{1}{b-1}$$

So 0.111... = 0.111... = 1/9.

...

Without infinity and continuity, there is no such concept as the real numbers.

... And with infinity and continuity, 0.999... = 1.

Later I would work through a more interesting example, a decimal that repeats with a period of 3.

Look instead at $$\frac{41}{333} = 0.123123123...$$. How do we know this. By working through the division algorithm with p = 41, q = 333, b=10
then
$$\quad \quad \quad k \quad \quad \quad $$$$s'_k$$$$a'_k$$$$z'_k$$Check
04100$$\frac{41}{333} = 0 + 10^{-0} \frac{41}{333}$$
17710.1$$\frac{41}{333} = 0.1 + 10^{-1} \frac{77}{333}$$
210420.12$$\frac{41}{333} = 0.12 + 10^{-2} \frac{104}{333}$$
34130.123$$\frac{41}{333} = 0.123 + 10^{-3} \frac{41}{333}$$
47710.1231$$\frac{41}{333} = 0.1231 + 10^{-4} \frac{77}{333}$$
510420.12312$$\frac{41}{333} = 0.12312 + 10^{-5} \frac{104}{333}$$
64130.123123$$\frac{41}{333} = 0.123123 + 10^{-6} \frac{41}{333}$$
............because $$\ell = 3$$, we have for any integer n:
3 n + 1771$$z'_{3n} + 1 \times 10^{-3n-1}$$$$\frac{41}{333} = z'_{3n} + \times 10^{-3n-1} + 10^{-3 n - 1} \frac{77}{333}$$
3n + 21042$$z'_{3n} + 12 \times 10^{-3n-2}$$$$\frac{41}{333} = z'_{3n} + 12 \times 10^{-3n-2} + 10^{-3n -2} \frac{104}{333}$$
3n+3413$$z'_{3n} + 123 \times 10^{-3n-3}$$$$\frac{41}{333} = z'_{3n} + 123 \times 10^{-3n-3} + 10^{-3n -3} \frac{41}{333}$$

Thus $$\frac{41}{333} = 0.123123123...$$ which repeats forever.
 
That is nice and short, but assuming $$\frac{1}{9}$$ = 0.111(1) seems to be same as assuming 1=0.999(9)

Correct. This is why all these abbreviated "proofs" lifted off the internet are invalid.

I.e. now we need to prove the starting assumption that:$$\frac{1}{9}$$ = 0.111(1) do we not? {for a valid proof of 1=0.999(9) }

Correct, again. Yet the rigorous proof is trivial. By definition:

$$0.(1)=\lim_{n \to \infty}(\frac{1}{10}+...+\frac{1}{10^n})=\frac{1}{9}\lim_{n \to \infty}(1-\frac{1}{10^n})=\frac{1}{9}$$
 
Excellent answer. Unfortunately it won't stop him from coming back and serving a word salad even bigger than what he's been serving. He will do so for as long as he's allowed to pollute this forum, this is what he's done under his prior name, this is what he's doing now: pretend to post science while, in reality, posting garbage.



Tach, although you have obviously been trying to 'behave' and make polite and useful posts, you have unfortunately 'reverted' to trolling again in this instance. I acknowledge that such posts from you are coming fewer and further between, it is a pity that this post of yours (to Trippy) is once again empty of argument, on-topic or otherwise. It instead just takes a personal opinionated swipe at another member with the obvious intent of further 'coloring' and 'framing' and 'chearleading' purely personal disparagements. You continue to do this sort of thing, trying to 'pre-conclude' an ON-going discussion (in this instance in the Alternative Theories section 'decide' no less!).

Your attempts at labeling (without any shred of supporting argument) a member 'right' or 'wrong' IS YOUR OPINION ONLY at this stage, since the discussion is NOT COMPLETE (as you would see if your ego and prejudices didn't keep getting in the way of polite and on-topic discussion of Alternative views in THIS section specifically created FOR such discussions.

Please learn from your recent bans and avoid making such obvious troll and chearleading and personal intrusions into the contextual flow of on-topic conversation on issues raised. Thanks. :)
 
Tach, this post of yours (to Trippy) is empty of argument, on-topic or otherwise. It instead just takes a personal opinionated swipe at another member with the obvious intent of further 'coloring' and 'framing' and 'chearleading' purely personal disparagements. You continue to do this sort of thing, trying to 'pre-conclude' an ON-going discussion (in this instance in the Alternative Theories section 'decide' no less!).

Your attempts at labeling (without any shred of supporting argument) a member 'right' or 'wrong' IS YOUR OPINION ONLY

Nah, you "forget" that it is also Trippy's and Pete's. Most likely, anyone's who knows science and has seen your posts.


at this stage, since the discussion is NOT COMPLETE (as you would see if your ego and prejudices didn't keep getting in the way of polite and on-topic discussion of Alternative views in THIS section specifically created FOR such discussions.

Please learn from your recent bans and avoid making such obvious troll and chearleading and personal intrusions into the contextual flow of on-topic conversation on issues raised. Thanks. :)

Nah, I simply agreed with Trippy that your posts are devoid of any scientific content (like your fringe denials: "zero is not a number", "0.999 is not a number").
 
Nah, I simply agreed with Trippy that your posts are devoid of any scientific content (like your fringe denials: "zero is not a number", "0.999 is not a number").

Precisely. That is exactly why you get banned. You 'agree' about his OPINION and effectively chearleading personal disparagement, and without any supporting argument on the actual points STILL under discussion. So making you post a 'cheerleading an opinion' post, and thus may result in you getting banned again if you keep up that sort of 'empty posts' activity (for which others also have been banned in the past, so what's good for the goose is good for the gander and the mods would therefore be justified via precedent to ban you just for doing that alone, yes?).

So please, I ask politely for you to resist making such empty posts obviously of a personal opinion nature aimed at framing and coloring the person rather than addressing the issues on-topic in on-going conversation. Thanks.

EDIT: I see you edited your offending 'empty' troll post. And STILL effectively only cheerleading/characterizing OTHERS' premature personal opinions which have also been presented in an ON-going conversation, and hence neither here nor there. You 'piling on' doesn't make it any more right under the rules. Just because "others did it' doesn't mean you have to cheerlead them and emulate them in such ungenerous behaviour. Try to be good for its own sake, and not be forced into it by the rules, hey? Thanks.
 
Precisely. That is exactly why you get banned. You 'agree' about his OPINION and effectively chearleading personal disparagement, and without any supporting argument on the actual points STILL under discussion.

That's not true, I have proved you wrong tens of times. So, did you learn the fact that both 0 and 0.999 ARE indeed numbers?
 
For all numbers, there is an number greater than that number. Because x + 1 > x.
1 is a number.
0.999... is a number.
If 0.999... = 1 then 1 - 0.999... = 0. But this is not what the OP assumes.

Our hypothesis must be 1 > 0.999.. from which the following follows:

If 1 > 0.999... then 1 - 0.999... > 0.
Define x = 1 - 0.999....
Then x is a number. And x > 0.
Then 1/x is a number. Therefore there is a number N > 1/x.

Therefore our hypothesis leads invariably to $$\exists N\quad N \gt \frac{1}{1 - \lim_{n \to \infty } \sum_{k=1}^{n} 9 \times 10^{-k} } = \lim_{n \to \infty } \frac{1}{1 - \sum_{k=1}^{n} 9 \times 10^{-k} } = \lim_{n \to \infty } \frac{1}{ 10^{-n} } = \lim_{n \to \infty } 10^n $$.

But no such N can exists, because there is always a value of M which makes the above untrue for all values of n > M. We can even write:
$$M = 1 + \lceil \log_{10} N \rceil$$ and prove
$$\forall n \geq M \quad N \not\gt 10^M \leq 10^n$$ .


Therefore the hypothesis that $$1 > 0.999...$$ cannot be true because 1 - 0.999... is smaller than any positive number.
Not even if N = $$(\textrm{1 googol})^{\tiny \textrm{1 googol}} = (10^{100})^{(10^{100})} = (10^{10^2})^{(10^{100})} = 10^{\left( 10^2 \times 10^{100} \right)} = 10^{10^{102}}$$ can we avoid this Archimedean property of numbers.
There is always a larger number -- the natural numbers are without bound.
And so $$M = 1 + \lceil \log_{10} 10^{10^{102}} \rceil = \textrm{100 googol} + 1$$

Your proof is invalid.

When dealing with .9(n), you must proceed by some form of induction.

You will note under induction, you can't prove .9(n) is ever anything other than finite.
In other words, see Kunen. When handling finite ordinals, you must make all of your proofs by induction.

Here is a negative proof.

For all n, 1 - .9(n) has an infinite # of digits. Assume that is false. Find the least b such that 1 - .9(b) has an finite # of digits.

Then, .9(b) has an infinite number of digits, which is a contradiction.

So, a limit based on an ordinal sequence never reaches the limit.
 
That's not true, I have proved you wrong tens of times. So, did you learn the fact that both 0 and 0.999 ARE indeed numbers?

And YOU have been proven wrong by others and myself on many occasions. So what? And I recall at least two times when your 'correction' was WRONG. And you later/elsewhere came to the opposite stance which your INCORRECT 'correction' maintained. So it is you who has been both wrong and a hypocrite for changing your tune and not acknowledging that fact to the original interlocutors you 'corrected' so wrongly.

Anyhow, Tach, this is a new discussion, conducted on its own presented arguments in THIS context in an on-going discussion in Alternative Theory section.

Bringing personal 'baggage' and prejudicial attitudes/kneejerks from OTHER past discussions is frowned upon by the rules/management (as advised to me and all other members often enough for you not to be able to pretend you aren't aware of that policy).

So, Tach, unless you are purposely trolling and baiting and bringing personal baggage and prejudices because you are masochistic and tempting banning for good (as mods have warned you would happen if you persisted in your anti-discussion 'bad habits' and disruptions), then I would strongly recommend keeping strictly to the rules and stop such 'noise'. Thanks.
 
Correct. This is why all these abbreviated "proofs" lifted off the internet are invalid.



Correct, again. Yet the rigorous proof is trivial. By definition:

$$0.(1)=\lim_{n \to \infty}(\frac{1}{10}+...+\frac{1}{10^n})=\frac{1}{9}\lim_{n \to \infty}(1-\frac{1}{10^n})=\frac{1}{9}$$

Let's see your proof that your infinite series actually attains the limit.

Remember, you are dealing with finite ordinals and I expect to see a proof based on that fact.
 
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