Talk:0.999.../Arguments

message: I happened to know the interesting equation  through a video on Youtube. I was curious that why people are discussing this for quite a long time since it looks quite obvious that . 
I have to say that I am not an expert on math. To the best of my knowledge, I am giving the following arguments for the interesting debate on  which is intuitively incorrect to me.
However, the correctness of this equality really depends on the definition of .
===  or ? The potential issue of two previous proofs of the equation . ===
Below is a 'proof' of the equation :
The issue of the above 'proof' becomes clear if we write it in another way:
Another well-known 'proof' of the equation  is that: 
::Since  (taught in elementary school), 
::we have  (by algebra), 
::that is  (by algebra). 
However, as it has been widely pointed out, is  correct? It depends on how we define . Nevertheless, we can say: 1 divided by 3 equals 0.3 with a remainder of 0.1, which can be written as: 
::.
Or similarly, we can say:
::,
::,
::.
And generalised as:
::,
Based on the above, if we agree that , it would be clear that .
Also, by algebra, it can be easily derived that:
:: 
which is
:: 
and hence .
=== What is ? ===
As mentioned at the beginning, whether  is true depends on the definition of .
In the wikipedia page of 0.999..., it is said that "0.999... (also written 0.9, among other ways), denotes the repeating decimal consisting of infinitely many 9s after the decimal point (and one 0 before it). "
To me, this explicitly make  as a number (something like  or ), which I believe should be a member of the sequence sequence (0.9, 0.99, 0.999, ...) or the set . In this case,  since we have shown that .
However, after the above definition in the wikipedia page of 0.999..., it is also said that "This repeating decimal represents the smallest number no less than every decimal number in the sequence (0.9, 0.99, 0.999, ...)." which is equivalent to the definition of the notation 0.999... as the limit of the sequence (0.9, 0.99, 0.999, ...). 
There should be no doubt that the limit of the sequence (0.9, 0.99, 0.999, ...) is 1. 
Hence, in this case, since 0.999... is just a notation, there is no problem to say  which is the same as to say something like .
— Preceding unsigned comment added by Snowinnov (talk • contribs) 
----
:Please, reread the definitions: 0. or 0.999... is –within this article–  not defined as , not for any natural number , and additionally, neither n→∞ nor ∞ are numbers in any contexts referred to within this article, so the notations  or  are not covered by the undisputed proposition " for any natural number n". There are no objections to " for any natural number n", however taking the limit "n→∞" takes these notations beyond their capabilities and the rigorous application of formally defined limits must take over. The claim that 0.999... is a member of the sequence (0.(9)n)n∈ℕ is not sustainable, because there is no such  the limit of the sequence is not contained in it. Purgy (talk) 16:53, 19 February 2019 (UTC)
thread_title:  {{math|0.(9)<sub>''n''</sub> &ne; 1}} for any positive integer {{math|''n''}}, but whether {{math|0.999... {{=}} 1}} is true depends on the definition of {{math|0.999...}}  

message: As mentioned at the beginning, whether  is true depends on the definition of .
In the wikipedia page of 0.999..., it is said that "0.999... (also written 0.9, among other ways), denotes the repeating decimal consisting of infinitely many 9s after the decimal point (and one 0 before it). "
To me, this explicitly make  as a number (something like  or ), which I believe should be a member of the sequence sequence (0.9, 0.99, 0.999, ...) or the set . In this case,  since we have shown that .
However, after the above definition in the wikipedia page of 0.999..., it is also said that "This repeating decimal represents the smallest number no less than every decimal number in the sequence (0.9, 0.99, 0.999, ...)." which is equivalent to the definition of the notation 0.999... as the limit of the sequence (0.9, 0.99, 0.999, ...). 
There should be no doubt that the limit of the sequence (0.9, 0.99, 0.999, ...) is 1. 
Hence, in this case, since 0.999... is just a notation, there is no problem to say  which is the same as to say something like .
— Preceding unsigned comment added by Snowinnov (talk • contribs) 
----
:Please, reread the definitions: 0. or 0.999... is –within this article–  not defined as , not for any natural number , and additionally, neither n→∞ nor ∞ are numbers in any contexts referred to within this article, so the notations  or  are not covered by the undisputed proposition " for any natural number n". There are no objections to " for any natural number n", however taking the limit "n→∞" takes these notations beyond their capabilities and the rigorous application of formally defined limits must take over. The claim that 0.999... is a member of the sequence (0.(9)n)n∈ℕ is not sustainable, because there is no such  the limit of the sequence is not contained in it. Purgy (talk) 16:53, 19 February 2019 (UTC)
thread_title:  What is {{math|0.999...}}? 

message: Near the end "This implies that the difference between 1 and x is less than the inverse of any positive integer." This is not true. That symbol is less than or equal, which means that the difference between 1 and x can be equal to the inverse of any positive integer, which includes options that are not 1, so the proof is not complete. It may be correct, I am just a student, but it needs further explanation if so.  — Preceding unsigned comment added by Tyguy338 (talk • contribs) 21:31, 30 September 2021 (UTC)
replies: text: If a quantity is less than or equal to the inverse of any positive integer, then for any positive integer, add one and take the inverse of that: the result will be smaller than the inverse of the original integer. Put another way, if we know that something is less than or equal to a half, and also less than or equal to a third, then it's clearly less than a half. I think this follows so naturally it doesn't need to be spelled out. MartinPoulter (talk) 11:07, 1 October 2021 (UTC)
thread_title:  Small note about the "rigorous proof" 

message: 1. The "Rigorous Proof" section contains no citation. Not one. If this is so "well establish" where are your citations?
2. Without even one credible citation of the rigorous proof, this page is nothing more than propaganda. 
3. It's maintainers blatantly censor Anon contributions, and do not admit a controversy exists, in spite of the very existence of THIS "Arguments" page and widespread strong evidence that every so called rigorous proof is in reality yet another circular argument. 
4. This is also the case for the proof offered in the "Rigorous Proof" section. It is also clearly circular, but who dares to try contradict this overtly false reasoning parading in plain sight.
5. The worst most abusive feature of this "0.999...=0" page is that any attempt to put up a similar page debunking the purported "Rigorous Proof" on this page (eg. "0.999... < 1") would last about 5 minutes. That censorship is the hallmark of book burners. 2003:EB:A714:EC00:1517:3B2E:6E6F:B6C4 (talk) 20:58, 15 July 2022 (UTC)
thread_title:  "Well established" claim is a blatant lie.  

message: If you graph an exponential decay equation where the asymptote is y=1, but you get a y value that equals .999... that can't be incorrect. 96.237.229.98 (talk) 23:04, 4 October 2022 (UTC)
replies: text: 1 and .999... are different ways to write down the same number. If the asymptote of a function is 1, then the asymptote is also .999... . If the function is equal to .999... at some point, then it is equal to 1 at that point. MartinPoulter (talk) 20:12, 5 October 2022 (UTC)
replies: text: Not true. The symbol "1" represents unity. Any other symbol or notation, by it's very nature (not being "1"), represents something else (irrespective of how clever the explaination, it's very existence undermines it's purported purpose). I've been told elsewhere that Wikipedia is a "serious" project only to come across silly articles like this. Iluvlawyering (talk) 06:28, 25 April 2023 (UTC)
replies: text: If you can find a number, other than 1, to which the series ( 0.9, 0.99, 0.999, 0.9999, 0.99999, etc... ) gets, so to speak, "closer and closer without ever reaching it", then you have a point. The symbol "0.999..." is shorthand for "the smallest number to which that series gets closer, whithout ever needing to reach it", and that number is 1. There is nothing to discuss about that. It's in the article, in the last part of section 0.999...#Infinite series and sequences, the only relevant section in the article. - DVdm (talk) 08:26, 25 April 2023 (UTC)

text: "Any other symbol or notation, by it's very nature (not being "1"), represents something else" A truly amazing statement, and hard to treat as serious. There are many other systems of symbols for representing numbers. Look into Eastern Arabic numerals, Suzhou numerals, and the rest. MartinPoulter (talk) 10:13, 25 April 2023 (UTC)
replies: text: No what was said is perfectly valid. You do not ask a man for "0.999... burger" you ask him for 1. The two symbols here represent the same *quantity* but by tbe mere fact of their being different expressions means they are not the same in totality. Same in quantity, different by number (same *in* number). 2600:6C4A:4C7F:D426:B9D1:9261:4B80:6C99 (talk) 13:14, 10 September 2023 (UTC)
replies: text: Your confusion is the standard one between numbers and numerals: "0.999..." and "1" are different numerals, which represent the same number. Just as 壱 does, for example. Imaginatorium (talk) 15:17, 10 September 2023 (UTC)
thread_title:  .999... can't always equal 1