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Explain why 0.999... = 1 ?

I'm not discussing about the proof, but about why.

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After my research tonight, I can explain my own question. Let's see this.

base \(1\) or unary => \(I.III... = II.\)

base \(2\) or binary => \(0.111... = 01.000...\)

base \(3\) or ternary => \(0.222... = 01.000...\)

base \(4\) or quaternary => \(0.333... = 01.000...\)

base \(5\) or quinary => \(0.444... = 01.000...\)

...

base \(10\) or decimal => \(0.999... = 01.000...\)

...

base \(16\) or hexadecimal => \(0.FFF... = 01.000...\)

and so on.

What's that mean? Now, let's find a job, yes, really job. If you can warranted that you always do the best on your job level, your boss will automatically think that you must get (absolutely get) new level on your job.

so, it's enough to explain why that happen.

"if we have maximum value on a level, it same as we have minimum value on next level." by: YIS (myself)

\(0.999...\) like the border or a limit on a level to level up, \(1.000...\) like the border or a limit on next level to level down.

For your attention in this discussion, I'm very appreciate it. Thank you very much.

another explanation, can be posted freely inside comment.

Note by Yulianto Indra Setiawan
4 years, 3 months ago

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0.999... = 1 because there are no numbers between 0.999... and 1. (Try to name one.) Michael Tang · 4 years, 3 months ago

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@Michael Tang That's a neat way to think about it. Kenneth Chan · 4 years, 3 months ago

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@Michael Tang it can't explain 0.999... = 1

it just explain the meaning of equal (=) symbol.

the (=) symbol actually represent that there are nothing between two things or more. Yulianto Indra Setiawan · 4 years, 3 months ago

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@Michael Tang Why not? :)

Certainly there are no real numbers between 0.999... and 1. That's a part of the definition of real numbers. But you may be surprised to learn that real numbers didn't get a rigorous treatment until as late as near the end of the 19th century.

It is possible to use other numbering systems besides real numbers. We can introduce "infinitesimals" - numbers which are nonzero, but smaller than any real number. And they work consistently. One of the more popular versions is the Hyperreals, http://en.wikipedia.org/wiki/Hyperreal_number . Matt McNabb · 3 years, 12 months ago

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@Michael Tang :)

I have 8 sticks which have same size. I can make 2 squares with them, I also can make 2 squares with only 7 sticks. Yulianto Indra Setiawan · 4 years ago

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@Michael Tang but does that explain why 0.999... = 1 ? it doesn't.. Tunir Saha · 4 years, 3 months ago

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@Tunir Saha If two numbers aren't equal then there are infinitely many real numbers between them. Michael Tang · 4 years, 3 months ago

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@Tunir Saha yes it does... if there aren't any in between it is the same Gaurav Sharma · 4 years, 3 months ago

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@Tunir Saha Tunir S.

yes, it absolutely doesn't explain that. Yulianto Indra Setiawan · 4 years, 3 months ago

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Because there is an infinite number of the digit \(9\), Pay attention to the word "Infinite number". To make it easier, assume there are \(n\) digits of \(9\), using geometric series we get : \[0.\underbrace{99...99}_n =9 \sum_{k=1}^{n} 10^{-k} = 1-10^{-n} .\] Now, Since the we have "an infinite number" of the digit \(9\) take \(n\to \infty\), and you'll get \(1\). Haroun Meghaichi · 4 years, 3 months ago

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Ok, I did a report on this. First of all, algebra. x=.999... 10x=9.99... -x -.999... 9x=9 x=1 .999...=1

Also, a reply I got from a Clay Institute of Mathematics professor when I emailed them about this: The key to resolving questions of this sort in mathematics often lies in being clear over definitions. Before you set about trying to prove your statement, you must first be clear about exactly what it means.

An important point is that .9 recurring is not a number, but a representation of a number as the limit of the infinite sequence

9/10, 99/100, 999/1000, 9999/10000, ...

of fractions. So the (correct) statement that you are making is that the limit of this sequence is 1. The proof has to start with a definition of limit. The conventional definition is that given any choice of 'error' (a positive number), however small, after a certain point the terms of the sequence will all be closer to the limit than the chosen error. The 'certain point', depends on the choice of error: the smaller the error you specify, the further down the sequence you may have to go.

A sequence can have a limit in this sense without any of the terms actually being equal to the limit.

An infinite decimal expansion should be interpreted as the limit of an infinite sequence of fractions with powers of 10 as the denominators. Your example shows that the same number can have different infinite decimal expansions, because

1 = 1.0000 ...

That is, it is also the limit of the sequence

1, 10/10, 100/100, 1000/1000. ...

If you refuse to allow decimal expansions ending in an infinite sequence of 9s, as some people do, then .9 recurring is not allowed, and every number has a unique decimal expansion. Adopting this point of view is neither right nor wrong, but the possibility reinforces the importance of being clear about meaning before you launch into argument. Gaurav Sharma · 4 years, 3 months ago

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@Gaurav Sharma Exactly. \(0.99999...\) is an implied limit. People say that it only "approaches 1" and never "attains " it. But the meaning of \(0.999...\) is to ask what limit it approaches. Shourya Pandey · 1 year, 5 months ago

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@Gaurav Sharma Limits... I agree with you... The idea of converging and diverging series... Saloni Gupta · 4 years, 3 months ago

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x=0.999 10x=9.999 10x-x=9.999-0.999 9x=9 x=1 Sushrut Khanzode · 4 years, 3 months ago

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@Sushrut Khanzode that just a proof, not an explanation.

I already explain that clearly. see the updated context or see my comment in this discussion. Yulianto Indra Setiawan · 4 years, 3 months ago

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@Yulianto Indra Setiawan "Why?, " is a question that requires logic and proof both. So if you ask " why", you can't prevent a proof to come up. Shourya Pandey · 4 years, 3 months ago

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@Shourya Pandey nope. it just require the reason. not the proof / evidence.

proof or evidence or fact is using anything that we have recognized, how to relate to other defined symbol. proof is the result of combining argument or opinion, so that can be the truth for some people that have same opinion or argument. for example, communication. human communication occured because we have same perception or same language. if we have not, we must be missunderstand what another people mean.

but the reason is just require the logical, why that happen. came from analyse and compare something. came from the meaning of something, if we spoke in English, we must comparing to other languange (example: Indonesian Language) to take the same meaning of every phrase. example: the word yes is not absolutely mean iya or ya in Indonesia. so the logical or reason needed the same meaning first before it can be communicated. if we have different perception, we can't talk about the logical.

in this era, proof can be wrong, but the logical absolutely can't be, because the logical must be similarize the meaning first.

proof like we just translate language without similarize the meaning.

logical like we take the meaning of every phrase in a language and then compare or similarize it to other language that we need to communicate for.

is there any error? please give comment, I'm very appreciate it.

if not, is there any explanation about why 0.999... = 1 ? Yulianto Indra Setiawan · 4 years, 3 months ago

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@Yulianto Indra Setiawan A proof is always correct. It is logic, on the other hand, that might fail us. Shourya Pandey · 1 year, 5 months ago

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Comment deleted Feb 28, 2016

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@Papus Chuchotsiri 3=9x doesn't mean x=3. Carefully divide numbers. Shourya Pandey · 1 year, 5 months ago

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@Shourya Pandey oops. I apologize for my mistake. :( Papus Chuchotsiri · 1 year, 5 months ago

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Let x = 0.99999.... Now, 10x = 9.9999.... or, 10x = 9 + 0.99999 or, 10x = 9 + x or, 9x =9 or, x = 1

Therefore, proved that: 0.99999... = 1 Akshit Soota · 4 years, 3 months ago

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@Akshit Soota Wrong! :) 10x =9.999...0 ! So 10x-x = 9.0000...1 so x=1 + 0.000...1 / 9 (and that's insane to write like this, you use wrong tools) Bertrand Delvallee · 4 years, 3 months ago

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@Bertrand Delvallee Actually, 10x=.999...999... since we assumed that there are an infinite number of nines. This is actually a very common proof and you "considere" it equal 1 because it is equal to 1. Pranay N · 4 years, 3 months ago

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I just thought about this, but I'm not sure of its validity. Thoughts?

1/1 = 1

1/1.1 = .909090...

1/1.01 = .99009900...

1/1.001 = .999000999000...

The denominator in the expression, although started at one, if continued to infinity, approaches 1. As the denominator continues to infinity, the right side approaches 1 as well. Even though the number is not technically .9999...., if continued to infinity, what would the difference be if each decimal place becomes 9 at some point. I know this isn't a proof, but it's something that infinity causes oddities within and I found it interesting. Could it still possibly used to help prove .999...=1? Skyler Hill · 4 years, 3 months ago

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We know that 0.333...=1/3 multiply by 3 each side we have 0.999...=1 Arbër Avdullahu · 4 years, 3 months ago

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@Arbër Avdullahu \(0.33333... =1/3 \) cannot be used directly, of course. Shourya Pandey · 1 year, 5 months ago

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Remember , 0.9999........... in this no. the no. of 9s is infinite and between this and 1 there is no possibility of any other no. hence 0.999.... is taken to be equal to 1 Kislay Raj · 4 years, 3 months ago

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@Kislay Raj but does that explain why 0.999... = 1 ? it doesn't.. Tunir Saha · 4 years, 3 months ago

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@Tunir Saha i'm guessing you need the following things to be convinced by Michael and Kislay: "if there are no numbers between a and b, then a = b" and "there are no numbers between 0.999... and 1" Ramon Vicente Marquez · 4 years, 3 months ago

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@Tunir Saha There exist 25+ ways that explain why 0.9999...=1 Arbër Avdullahu · 4 years, 3 months ago

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@Kislay Raj is there no possibility?

how about 0.999...999 ? Yulianto Indra Setiawan · 4 years, 3 months ago

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@Yulianto Indra Setiawan Suppose that 0.999...999 is between 0.999... and 1. Note that 0.999...999... is between 0.999...999 and 1. But 0.999... is the same as 0.999...999... .Therefore 0.999... is between 0.999... and 1, which is a contradiction. Therefore 0.999...999 is not between 0.999... and 1. Ramon Vicente Marquez · 4 years, 3 months ago

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@Ramon Vicente Marquez hahaha. so many vote down ya.

let me explain this.

0.999...999 + 0.000...001 = 1.000...000 => (we know that it will be stop counting or has an ending)

0.999... (the maximum in a level) + 0.000...001 = 1.000...000999... or 1.000...001

0.999... (the maximum in a level) = 1.000... (the minimum of next level) => (because we don't know it will be stop counting / has an ending or not)

so, it will be:

0.999... + 1.000... = 1.999... or 2.000...

0.999... like the border or a limit on a level to level up, 1.000... like the border or a limit on next level to level down.

still don't understand?

is there any error? please give comment, I'm very appreciate it. Yulianto Indra Setiawan · 4 years, 3 months ago

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suppose , x = 0.99999999999...... (1)
\(\Rightarrow\) 10x = 9.999999999...... (2)
subtract the equations (2) - (1)
(10 - 1)x =9
\(\Rightarrow\) x = \(\frac {9}{9}\) =1
so , 9.99999999999...........= 1 Kiriti Mukherjee · 4 years, 3 months ago

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1/9=0,111... is not exactly correct , it is just relative . when you divide 1 into 9 , there will be some part that can't be divided , because if it is divided , there will be other part to divide , permantly !!! however, i aint object to your fact , i call it magic in maths . it like 1+1=2 , of course know it , but you can't explain why . ** sorry if my english is stupid :(( Hung Dinh Thanh · 4 years, 3 months ago

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@Hung Dinh Thanh Your English is quite good, in fact. You have conquered that average American's typing grammar. Justin Wong · 4 years, 3 months ago

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@Justin Wong Hahhahaha :D Noor Muhammad Malik · 4 years, 3 months ago

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1/3 = .33333333... if we add another .3333333333...... we get .66666666666... or 1/3+1/3 = 2/3 if we add another .3333333333..... we get .99999999999... or 2/3+1/3 = 3/3 = 1 implying that 1=.99999999... Djordje Marjanovic · 4 years, 3 months ago

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I mean equal to 9. Oops! Scholastica Okoye · 4 years, 3 months ago

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The last part: 9/9=0.999... = 1 is the key. 9/9 is equal to 1 as you probably might agree and equal to 0.999...

...But if 9/9=0.99999..., then 9x0.9999... should be equal to one, but it's not. It's equal to 8.999..... Scholastica Okoye · 4 years, 3 months ago

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@Scholastica Okoye Why 9×0.999... should equal to 1? Mohit Soni · 4 years, 3 months ago

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Suppose that 0.999... is NOT 1. Then, let x = 1 - 0.999... . Try to think about what x would have to be. If you answer "x has to be infinitely small," then there you go. The only positive number smaller than every other positive number is 0. David Sanchez · 4 years, 3 months ago

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@David Sanchez "The only positive number smaller than ever other positive number is 0" If 0 is a positive, and 0 is smaller that it self ( 0<0 ) you are wrong. Djordje Marjanovic · 4 years, 3 months ago

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@Djordje Marjanovic Oops, you're absolutely right. I should have said "non-negative" instead of "positive." David Sanchez · 4 years, 3 months ago

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Now I read the last row.Can anyone explain me what is need here "I'm not discussing about the proof, but about why."? Arbër Avdullahu · 4 years, 3 months ago

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@Arbër Avdullahu I think it means that he knows that 0.999.... = 1 is true (that it has been proven), but he wants another explanation of intuitively why it is true. Michael Tang · 4 years, 3 months ago

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let \(x = 0.99999....\)

so \(10x = 9.99999....\)

\(10x-x = 9.99999.... - 0.99999....\)

\(9x = 9\)

\(x = 1\)

Looking at the first and last statement, \(0.99999....=1\) Saad Haider · 3 years, 11 months ago

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@Saad Haider the explanation is that since the 9 is recurring, there can be no number greater than this which is lesser than 1. If you got to numberphile, they explain that 0.99999.... is just another way of writing 1 just like 0.33333.... is another way of writing 1/3 and 0.11111.... is another way of writing 1/9 Saad Haider · 3 years, 11 months ago

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0.999.. = x

9.999..=10x -0.999.. = x 9 = 9x x = 1; 0.999..= x x=x 1=0.999.. :) Taze Jared Abubo · 4 years, 2 months ago

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0.999...=0.9+0.09+0.009+....is a geometric series with first term a=0.9 and common ratior=0.1<1. So the infinite series has a partial sum=a/(1-r)=0.9/(1-0.1)=0.9/0.9=1[Here, for higher order of 0.1 decreases rapidly and it becomes negligible for smallness which is very near to zero, we consider limiting value of r^n as zero] Md Abul Kalam Azad · 4 years, 3 months ago

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0.999...=x then 9.99...=10x after that 9+x=10 thus x=1 will this help Kristian Fransali · 4 years, 3 months ago

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Mathematical Induction theory has been used!! Subhrodipto Basu Choudhury · 4 years, 3 months ago

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or, x =0.99.... 10x=9.99... 10x - x = 9.99 - 0.99 9x = 9 x=1 Vaishnav Garg · 4 years, 3 months ago

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@Vaishnav Garg ya..... Even me had same comment.... Vamsi Krishna Appili · 4 years, 3 months ago

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0.99999....=x then 10x=9.999999..... after that subtract the first equation to the second, cancelling the infinite decimal places leaving 9x=9, then x=1. John Errol Obia · 4 years, 3 months ago

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it might be explained using the concept of one sided limit.... Shubhra Aich · 4 years, 3 months ago

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.99999999.....= .9+.09+.009+.0009+........(infinite series) =0.9[1+0.1+.001+.0001+............] =0.9[1/(1-.01)] {infinite g.p series sun formoula =0.9[1/0.9]=1 proved Rohit Aggarwal · 4 years, 3 months ago

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@Rohit Aggarwal When we use Infinite Geometric Series formula, we tend to say that as the series tends to infinity, the (sum of the) series ''converges'' to a particular number... And ''converges'' is the keyword here... Saloni Gupta · 4 years, 3 months ago

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@Saloni Gupta Yes you are right.. The solution never explains the sum of all the numbers that come in the series closer to infinity.. That is the problem.. The exact solution of series that is infinite, is actually, not possible.. Noor Muhammad Malik · 4 years, 3 months ago

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ERROR AND APPROXIMATION Vipin Tiwari · 4 years, 3 months ago

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@Vipin Tiwari please give any explanation according to you about why my explanation may has an error and may like an approximation? Yulianto Indra Setiawan · 4 years, 3 months ago

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After my research tonight, I can explain my own question. Let's see this.

base 1 or unary => I.III... = II

base 2 or binary => 0.111... = 01.000...

base 3 or ternary => 0.222... = 01.000...

base 4 or quaternary => 0.333... = 01.000...

base 5 or quinary => 0.444... = 01.000...

...

base 10 or decimal => 0.999... = 01.000...

...

base 16 or hexadecimal => 0.FFF... = 01.000...

and so on.

What's that mean? Now, let's find a job, yes, really job. If you can warranted that you always do the best on your job level, your boss will automatically think that you must get (absolutely get) new level on your job.

so, it's enough to explain why that happen.

"if we have maximum value on a level, it same as we have minimum value on next level." by: YIS (myself)

so, discussion will be close.

For your attention in this discussion, I'm very appreciate it. Thank you very much.

or, maybe another explanation? Yulianto Indra Setiawan · 4 years, 3 months ago

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@Yulianto Indra Setiawan Let \(x\ngtr { x }^{ 2 }\ngtr { x }^{ 3 }\) and \(y\nless { y }^{ 2 }\nless { y }^{ 3 }\).

Let \(V\) be a string of digits at least one of which is not '0'.

Let \(W\) be a string of infinite digits.

Let \(Z\) be '0'.

Of course, a point always separates the strings in positional notation.

\(x\ngtr { x }^{ 2 }\ngtr { x }^{ 3 }\Leftrightarrow x\) can be represented \(V.W\)

\(y\nless { y }^{ 2 }\nless { y }^{ 3 }\Leftrightarrow y\) can be represented \(Z.W\)

\(A=\left\{ x:x\ngtr { x }^{ 2 }\ngtr { x }^{ 3 } \right\} \Rightarrow \left\{ 1.\overline { 6 } ,2.0 \right\} \subset A\) because \(\left( 2.0 \right) \ngtr { \left( 2.0 \right) }^{ 2 }\ngtr { \left( 2.0 \right) }^{ 3 }\) AND \(\left( 1.\overline { 6 } \right) \ngtr { \left( 1.\overline { 6 } \right) }^{ 2 }\ngtr { \left( 1.\overline { 6 } \right) }^{ 3 }\).

\(B=\left\{ y:y\nless { y }^{ 2 }\nless { y }^{ 3 } \right\} \Rightarrow \left\{ 0.\overline { 3 } ,0.5 \right\} \subset B\) because \(\left( 0.5 \right) \nless { \left( 0.5 \right) }^{ 2 }\nless { \left( 0.5 \right) }^{ 3 }\) AND \(\left( 0.\overline { 3 } \right) \nless { \left( 0.\overline { 3 } \right) }^{ 2 }\nless { \left( 0.\overline { 3 } \right) }^{ 3 }\).

\(1\ngtr { 1 }^{ 2 }\ngtr { 1 }^{ 3 }\) AND \(1\nless { 1 }^{ 2 }\nless { 1 }^{ 3 }\therefore 1\) can be represented \(V.W\) AND \(1\) can be represented \(Z.W\)

In other words, \(1=1.0=0.\overline { 9 } \)

\(1\in A\) AND \(1\in B\).

\(1\) is the least possible \(x\) and the greatest possible \(y\). André Cabatingan · 1 year, 11 months ago

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@Yulianto Indra Setiawan That's not right. Considere 0.9999... = 1 imply that 1.00...01 = 1 too => 0.000...1 equal 0 => 100*0.00...1 equal 0 => 1000..0 * 0.000.1 = 0 => infty / infty = 0 => 1 = 0. Bertrand Delvallee · 4 years, 3 months ago

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@Bertrand Delvallee 0.000... and 0.000...001 are different.

0.000... = 0

0.000...001 = 0.000...001

0.000... + 0.000... = 0.000... = 0

0.999... + 0.000...001 = 1.000...000999...

agree with me? Yulianto Indra Setiawan · 4 years, 3 months ago

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At last, someone absolutely gets why!

This is what I've noticed. If all the digits to the left of the decimal point are \(0\), the absolute value of the real number being represented doesn't get less than the absolute value of it raised to a positive integer. On the other hand, if not all of the digits to the left of the decimal point are \(0\), the absolute value of the real number being represented doesn't get greater than the absolute value of it raised to a positive integer. \(1\) neither gets less nor greater than \(1\) raised to a positive integer, which is why it can be represented in two ways.

Your analogy is great! You better impart that to those who doubt the equality. André Cabatingan · 1 year, 11 months ago

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^^ RIP English Karthik Sridhar · 2 years, 5 months ago

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They aren't the same number in some sense. You could argue that 0.00...1 is their difference, but also it could be argued that that equals zero. I mean, there are infinite zeros, and one 1. So that's just equal to 1 divided by infinity. No matter what you do to infinity, you can't get anything but infinity. The only solution us mathematicians has come up with is zero. Zero times infinity is anything. Including 1. Therefore the distance between 0.99... and 1 is 0. Therefore, they are equal. Finn Hulse · 3 years, 6 months ago

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\(\frac {1}{3} \)\( = 0.333\dots\)

\(\frac {2}{3} \)\( = 0.666\dots\)

\(\frac {3}{3} \)\( = 0.999\dots\)

Since \(\frac {3}{3} \)\( = 1\). Therefore...

\(\frac {3}{3} \)\( = 0.999\dots = 1\)

And that's it. Adam Zaim · 3 years, 9 months ago

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we can do it as let x=0.9999...... 10x=9.99999......... 10x-x=9.9999...- 0.999999 9x=9 which gives x=1 Rishabh Nain · 3 years, 11 months ago

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Because Transforming into Fraction, So We have:

N = 0,999 100N = 99,999...... 100N -N = 99,999.... - 0,999..... 99N=99 N= \frac {99}{99} = 1 Gabriel Merces · 3 years, 11 months ago

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let x=0.999... 10x=9.999... 10x=x+9 9x=9 x=1 since x=0.999... & x=1, 0.999...=1 Ilai Reshef · 4 years ago

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Simple. If you round it to n decimal digits, you are accurate to the nearest \(\frac{1}{10^n}\). You always get 1,for any n. So, let n be infinity. Daniel Leong · 4 years ago

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Aha! By your logic, YIS, there are no celings. Daniel Leong · 4 years ago

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@Daniel Leong Actually, I don't know what its called. Many people say it is logical, but it can be anything languageless.

You're right, there are no floor and no ceiling. They just help us to understand this question, same to another method for proofing.

"If you can see anything unseen, then it means you understand how to use your sense." (YIS)

"If you have found the smallest thing, then you will know how to create everything." (YIS)

Everything that has recursive characteristic, it's not actually recursive, but actually it always become much more detail than before. Yulianto Indra Setiawan · 4 years ago

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IT IS VERY EASY , 1/3=.33333333333333..... MULTIPLYING THE EQUATION BY 3 , WE GET 1=.99999999999999999999....... Shashank Goel · 4 years ago

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@Shashank Goel nope.

using number can't explain this. Yulianto Indra Setiawan · 4 years ago

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@Yulianto Indra Setiawan You want to explain something about numbers without using numbers? Good luck with that… Tim Vermeulen · 4 years ago

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@Tim Vermeulen we need to think before measuring everything. Yulianto Indra Setiawan · 4 years ago

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i think it should be round off the numbers after the decimal and thats why it become one Luy Ii · 4 years, 3 months ago

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hey I know .the answer....

Let x = 0.99999999.... Let 10x - x -------> 1 (10 \times 0.9999....) - 0.9999..... 9.9999........ - 0.9999....... 10x - x = 9 ---------->2 9 x = 9 ----------------------->2

now.... 9 x = 9 x = 1 thus x= 0.99999.... =1 Vamsi Krishna Appili · 4 years, 3 months ago

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coz it rounds up Jess J · 4 years, 3 months ago

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or you can find it in such a manner..as we know 0.999.. is rational so, 0.99...=x (1) we suppose so, 10x = 9.99(2).... then, subtracting 2 frm 1 9x=9 and hence x=1....so here is the prove....and yeah it is because there are no numbers between 0.99.. and 1..so when u write this decimal in the form of p/q..they become 1 Shahan Shaikh · 4 years, 3 months ago

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0.999... is never equal to 1.000... it jest tends to 1.000...so clear your doubts about actuality and approximity. that's all I find best to explain. Ashutosh Pandey · 4 years, 3 months ago

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thats nice way to think Pavneet Bhatia · 4 years, 3 months ago

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thats nice Pavneet Bhatia · 4 years, 3 months ago

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I think the first thing we must mention is the definition of the "equal" symbol. Đức Việt Lê · 4 years, 3 months ago

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it is because of rounding of data. Waqas Tunio · 4 years, 3 months ago

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If \(a\) and \(b\) are equal, then \(a - b = 0\).

Now \(1 - 0.999\cdots = 0.000\cdots = 0\). That can just mean that \(1\) and \(0.999\cdots\) are equal. Parth Kohli · 4 years, 3 months ago

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i think it's just a rounding theory. if you want to write 0.9999... , but i guess you won't do it. so simply rounded to 1 Jembatan Garam · 4 years, 3 months ago

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I always love finding new proofs. Justin Wong · 4 years, 3 months ago

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0.999... is not equal to anything but itself. It's not even a well-formed concept.

You can learn more about the errors in mythmatics here: HTTP://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-672.html I debunk many wrong ideas and concepts at this link. Be sure to visit!

The best proof that 0.999... is not equal to 1 is the rubbish that has been pedaled by academic incompetents and ignoramuses since Euler.

x = 1(object) 10x = 10 (object) 9x = 9(object) x = object

If the object is non-changing which 0.999... is purported to be, the output of the algorithm must equal to the input. There is much more you can learn from me - more than you ever learned in ALL your school years.

Read more about my New Calculus at: HTTP://thenewcalculus.weebly.com John Gabriel · 2 years, 8 months ago

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@John Gabriel Equality and place-value notation are the concepts behind \(0.\overline { 9 } =1\), and they are well-formed concepts. As a matter of fact, it is the concept of infinitesimals that is not even well-formed, and has been replaced by the well-formed concept of limits. As long as infinitesimals are defined to be greater than zero, they will always be proven to be false. Is there a smallest positive real number? André Cabatingan · 1 year, 11 months ago

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So... 0.999...=1.000...? 1.00000000...1-0.99999999...9=0.000...2, but 0.000...2 never reaches 2, so... Yeah. John Muradeli · 3 years, 9 months ago

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.999... does not EQUAL 1 - Equal, by its definition means THE SAME or IDENTICAL - not really really close, it means IDENTICAL. I have ONE apple, not "almost" one apple, not mostly one apple, not .99999... apples. I have ONE. You can throw all the math you want at this, but in the end either you have a whole or you don't. Your logic would mean that .5999... is equal to .6 - Not true! And BTW, the statement above that 9/9 = .999... = 1 is ridiculous. 9/9 = 1.000... Ryan Halpin · 4 years, 3 months ago

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@Ryan Halpin There are multiple ways to represent different numbers, so just because 9/9 = 1 doesn't mean that 9/9 can't be equivalent to something else besides the exact integer "1". 9/9 is also equivalent to e^[(2)(pi)(i)]. I personally believe that there are multiple ways to interpret this and multiple ways to prove this, although not all ones shown in this discussion are correct. Just like some say "1 is the square root of -1" and others say "i squared is 1", these are different and true in different ways and such... I'm personally surprised at the amount of debating on this well-tread path. Justin Wong · 4 years, 3 months ago

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0.99999... = 3 * 0.33333... = 3 * 1/3 = 1 Much easier proof... :) Jubayer Nirjhor · 4 years, 3 months ago

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We can prove this in a simple way

1/3=.333333333333333333..............................................................

1/33=.33333333333333.........3=.9999999999999................... but we know 1/3*3=1

or

1/3=0.333333333333.................. 2/3=0.666666666666..................

1/3+2/3=0.33333333333.................+0.66666666666666666.....=0.9999999999999999..................

But 1/3+2/3=1 ie 0.999999999999999999..........=1 Sreehari Vp · 4 years, 3 months ago

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@Sreehari Vp 1/3 is itself a Recurring number... I don't think it's a good idea to use one of the recurring numbers to prove the other true... :-) Saloni Gupta · 4 years, 3 months ago

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@Saloni Gupta .99999999999.................................is also recurring number. Sreehari Vp · 4 years, 3 months ago

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The last part: 9/9=0.999... = 1 is the key. 9/9 is equal to 1 as you probably might agree and equal to 0.999... Anton Than Trong · 4 years, 3 months ago

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