Talk:0.999...

Can anyone make a nice pretty picture to go along with this stroke of genious?

.999 + .111 does not equal 1, as it says. Instead, .999 + .000...1 equals 1. Even in a content-less encyclopedia like this, you gotta have your math right. EDIT: Nevermind. I read to the end....

Heh, look at that, I look at the article and say "wait a second, doesn't 0.111... + 0.999... = 1.111...? Shouldn't that be 0.999... + 0.00...1 = 1?". Oh well, beaten to it. » Brig Sir Dawg | t | v | c » 23:33, 28 October 2006 (UTC)

.999... equals exactly 1 --Phantom-Lord 22:59, 11 April 2008 (UTC)

Numbers are about as funny as a square horse

Take that, school :/

         Math is in fact funny. This article, however, is not.

0.999 is equal to 1! The people that wrote this article need to realize that 0.999... does not have an ending 9 it goes on for infinity, it is just that humans don't have the ability to understand a number like that. Also, 0.000...1 is controlled by term "N" (determined the position of 1) or if the 0 manage to go on infinitely, the 1 will never come.

Here is how to prove 0.999... = 1 by the definition of 1 upic.me/i/js/0.999...1.jpg

NEVER-MIND I JUST LEARNED THAT PI GOES ON FOREVER BUT THE NEXT NUMBER ISN'T NECESSARILY THE SAME OF AS THE LAST ONE.

Some proofs that should be in the article and this is a long section title.

1. $.9999999999999999=.9sigma₁^inf.1ⁿ$

If anyone has an objection to me putting this in, or has a solution, reply here.~ 21:30, 21 December 2008 (UTC)

I decided to censor my proof because it will likely asplode uncyclopedian readers. Here is a better way of expressing this:"0.999... is 0.9 times the so-called geometric series with ratio r=0.1 and starting term a=0.9. Now, using the geonetric formula, a/(1-r), we get that 0.999...=1. Now, this is bullshit because the geometric formula predicts that 1+2+4+16...=-1 which is obviously wrong.– Preceding unsigned comment added by Citation Needed (talk • contribs)

Aparrently, censoring doesn't work!~ 22:50, 21 December 2008 (UTC)

Nor does strike-outs!~ 22:51, 21 December 2008 (UTC)

the proof is invalid, as it uses itself to proof it (a is true, because a is true.)

【1 paradox】Why 0.999... is not euqal to 1?

The current mathematic theory tells us, 1>0.9, 1>0.99, 1>0.999, ..., but at last it says 1=0.999..., a negation of itself. So it is totally a paradox, name it as 【1 paradox】. You see this is a mathematic problem at first, actually it is a philosophic problem. Then we can resolve it. Because math is a incomplete theory, only philosophy could be a complete one. The answer is that 0.999... is not euqal to 1. Because of these reasons:

1. The infinite world and finite world.

We live in one world but made up of two parts: the infinite part and the finite part. But we develop our mathematic system based on the finite part.

0.999... is a number in the infinite world, but 1 is a number in the finite world. For example, 1 represents an apple. But then 0.999...? We don't know. That is to say, we can't use a number in the infinite world to plus a number in the finite world. For example, an apple plus an apple, we say it is 1+1=2, we get two apples, but if it is an apple plus a banana, we only can say we get two fruits. The key problem is we don't know what is 0.999..., we can get nothing. So we can't say 9+0.999...=9.999... or 1, etc.

We can use "infinite world" and "finite world" to resolve some of zeno's paradox, too.

2. lim0.999...=1, not 0.999...=1.

3.The indeterminate principle.

Because of the indeterminate principle, 1/9 is not equal to 0.111....

For example, cut an apple into nine equal parts, then every part of it is 1/9. But if you use different measure tools to measure the volume of every part, it is indeterminate. That is to say, you may find the volume could not exactly be 0.111..., but it would be 0.123, 0.1142, or 0.11425, etc.

Now we end a biggest mathematical crisis. But most important is this standpoint tells us, our world is only a sample from a sample space. When you realized this, and that the current probability theory is wrong, when you find the Meta-sample-space, you would be able to create a real AI-system. It will indicate that there must be one God-system in the system, which is the controller. Look our world, there must be one God, as for us, only some robots. Maybe we are in a God's game, WHO KNOWS?

More info, three other download points(written in Chinese):

yourfilelink.com/get.php?fid=780934

d01.megashares.com/dl/0LZix2o/the end of the world.rar

localhostr.com/file/3LtuSLb/the%20end%20of%20the%20world.rar

Is the math tag broken?

(or should I have said, "Is LaTeX broken?")

It just seems to be showing "Invalid tag extension name: math" in every place it is used here. --Pentium5dot1 (semi-retired) t~^_^~c 17:35, July 4, 2012 (UTC)

Me too. What is this? Cat the Colourful (Feed me!) Zzz 17:40, 4 July, 2012 (UTC)

(EC) Something changed in the last minute or so ... now the raw wikicode is showing, at least when I looked at P-1MP My Matrix. --Pentium5dot1 (semi-retired) t~^_^~c 17:41, July 4, 2012 (UTC)

Um, the symptoms are still the same on *this* article. Sorry I was lazy and let that edit conflict happen without my rechecking that. --Pentium5dot1 (semi-retired) t~^_^~c 17:42, July 4, 2012 (UTC)

I also have this problem, using Firefox. It either shows the "invalid" red font message or just the raw text, including the <math> brackets in plain view. It was like this a few hours ago. Other Wikia sites don't seem to have this problem with the code, and the "invalid" message is extremely obvious on the main page. --Alpha Quintesson (talk) 18:06, July 4, 2012 (UTC)

It seems to be working now, but when I tried to check my preferences (to see if there is an option for MathJax, which Wikipedia has), it said "Internal error." --Pentium5dot1 (semi-retired) t~^_^~c 19:26, July 10, 2012 (UTC)

It appears this is a byproduct of the current MediaWiki update process. Two days ago, we were at MediaWiki 1.15 or some such, now we are at 1.19.1. As of right now, every script on the site is broken, so there's that. We can only assume that all this will be sorted, since the math tags started working again. ~ BB ~ (T) ~ Wed, Jul 11 '12 11:42 (UTC)

The Uncomfortable subject of Remainder and incompatible counting systems

First and most importantly, remainders should be written as fractions, this is to represent them in the form that started the infinite string, this is useful when we are trying to reassemble the remainder to get a restitution to the original number or when we are mixing different forms of infinity.

${\begin{aligned}1/3&=0.3\ldots +{\tfrac {1}{3}}r\\\end{aligned}}$

This reminder of a third is tacked onto the singular unit, it could be discarded but when all the numbers are added, without the remainder, it will only achieve 0.999... and won't return to the integer value You technically do not need to write the number string out if you know the remainder.

${\begin{aligned}1/3&=1-{\tfrac {2}{3}}r\\\end{aligned}}$

Why two thirds? Well you have to remember that you are subtracting from the whole and that - 2 thirds from 1 whole is 1 third.

Now this basic premise which can demonstrated in any long divisional equation if written manually is observed in notation form, lets get onto the problem at hand.

Wiki would have it that...

${\begin{aligned}x&=0.999\ldots \\10x&=9.999\ldots \\10x-x&=9.999\ldots -0.999\ldots \\9x&=9\\x&=1\end{aligned}}$

...however, a five year old learns about our good friend remainder, just as described above, it is the property which gives infinite strings of any rational number their infinity property from a division.

${\begin{aligned}x&=0.999\ldots \\10x&=9.999\ldots -{\tfrac {9}{1}}r\\10x-x&=9.999\ldots -{\tfrac {9}{1}}r-0.999\ldots \\9x&=9-{\tfrac {9}{1}}r\\x&=1-{\tfrac {1}{1}}r\\1-{\tfrac {1}{1}}r&=0.999\ldots \\x&=0.999\ldots \\\end{aligned}}$

A little bit odd how the wiki equation manages to pull a sudden change in the numbers out of thin air whilst the second one has restitution and CAN BE REVERSED as seen below.

${\begin{aligned}x&=0.999\ldots \\0.999...&=1-{\tfrac {9}{1}}r\\x&=1-{\tfrac {1}{1}}r\\9x&=9-{\tfrac {9}{1}}r\\9x+x&=9.999-{\tfrac {9}{1}}r\ldots &=10-{\tfrac {10}{1}}r\\(10-{\tfrac {10}{1}}r)/10&=1-{\tfrac {1}{1}}r\\x&=1-{\tfrac {1}{1}}r\\1-{\tfrac {1}{1}}r&=0.999\ldots \\x&=0.999\ldots \\\end{aligned}}$

Whilst it suits some people to ignore remainders in equations, it is because of the remainders that you have the string in the first place and so ends the lesson.

There is a much easier way that is sound and reliable:

  x = 1(0.999...)
10x = 10(0.999...)
 9x = 9(0.999...)
  x = 0.999...

The false assertion that 0.999... = 1 rests on Euler's ill-formed definition of an infinite sum. See article on 0.999... at http://johngabrie1.wix.com/newcalculus. 173.228.7.80 00:35, 7 March 2013 (UTC)

Eww... math.  SIR Peasewhizz de NY Biblio HOS Awards 00:44 3.07.13