PROOF

Let $0.9999.... = x$ . Then $10x = 9.9999....$ .

Now subtracting the first equation from the second, $9x = 9 \Rightarrow x = 1 \Rightarrow \boxed{0.9999.... = 1}$

Alternate way :

$\frac{9}{10} + \ \frac{ 9}{10^2} + \frac{ 9}{10^3} + \frac{ 9}{10^4} + \ldots = \frac{a}{1-r} = \frac{ \frac{9}{10} } { 1 - \frac{1}{10} } = 1$

Instead of getting angry and arguing the point if you got it wrong, read the numerous proofs and sit back and think 'wow I didn't know that and now I do. What an awesome bit of general knowledge I've just learnt.'

$\boxed{1.-}$ If 0.9999999999.... was distinct to 1 then there would be exist a real number x such that 0.9999.... < x < 1 (Impossible).

Theorem.- If x, y are real numbers and x is distintc to y(for example x < y), then there exists z real number, such that x < z < y and z is distinct to x and y.

Proof.- x < $z = \frac{x + y}{2}$ < y . ..

$\boxed{2.-}$ $0.9999.... = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \ldots = \frac{\frac{9}{10}}{1 - \frac{1}{10}} = 1$

$\boxed{3.-}$ $10 \times 0.9999.... - 0.9999999.... = 9.999999... - 0.99999999... = 9 \Rightarrow$ $(10 - 1) \cdot 0.9999.... = 9 \Rightarrow 0.999... = \frac{9}{9} = 1$

$\boxed{4.-}$ Every real number has one unique decimal representation. For example, the decimal representation of $\frac{1}{3} = 0. \hat {3} = 0.3333...$ . Certanly $\frac{1}{3}$ is a rational number and its decimal representation is $0. \hat {3} \Rightarrow 1 = 3 \cdot \frac {1}{3} = 3 \times (0. \hat {3})$ which decimal representation is 1.

Note.- If you don't agree you can view reports and this link $\boxed{5.-}$ There is a last chance, 0,99999.... doesn't exist, then the statement is false too.

1/9 = 0.11111111...
2/9 = 0.22222222...
...
8/9 = 0.88888888...
Therefore
9/9 = 0.99999999...
But also
9/9 = 1

We know that 1/3 = 0.3333... and hence (1/3) 3 = 0.99999.....
But, (1/3)
3 = 1. Therefore, 1= 0.999999999....

What you have to realize is something that is conceptually difficult for many people. The number $0.\overline{9}$ consists of an infinite number of digits after the decimal place. It can not be said the "another nine can always be added" because that would assume the sequence is terminated. There is no number between $0.\overline{9}$ and 1. For an even crazier proof, look up the sum of all real numbers. You will find that $\displaystyle \sum_{i=1}^3 = -1/12$ This is an actual proof and has many physical applications. It is not comprehendable and is extremely counterintuitive, but we know it to be true.

let x=0.99999...

and 10x=9.999999...

If we subtract these two then 9x=9

And then divide by 9: x=1

Therefore 1 = 0.9999... and is not greater than 0.9999...

Forgive my ignorance. Is this a problem that could be simultaneously answered either way? One of the proofs said there had to be a number that you could add to 0.9999.... to make it equal to one, otherwise there was no difference. Wouldn't that number (or expression) be $.1^\infty$ ?

In other words $0.9999... + .1^\infty = 1$ therefor 1 > .9999.... ?

It just seems to me that this is an expression that is neither true nor false. It can exist as both true and false based on the perspective one takes.

as long as time exists, this will be false !!!

This business of repeating decimals is an artifact of the number system. Consider base 10: 1/3 = 0.3333...., 2/3 = 0.666666..... and 3/3 = 0.9999999..... But in base 3, 1/3 = 0.1, 2/3 = 0.2, and 3/3 = 1.0 and the whole problem disappears. No need for fancy proofs once you realize that choosing a different base makes the problem evaporate (though it now makes repeating decimals out of other fractions, but the same argument applies. These are artifacts, nothing more.)

When you subtract a number from itself, the result is zero. For instance, 4 – 4 = 0. So what is the result when you subtract 0.999... from 1? For shorter subtractions, you get:

1.0 - 0.9 = 0.1

1.00 - 0.99 = 0.01

1.000 - 0.999 = 0.001

1.0000 - 0.9999 = 0.0001

1.00000 - 0.99999 = 0.00001

Then what about 1.000... – 0.999...? You'll get an infinite string of zeroes. "But," you ask, "what about that '1' at the end?" Ah, but 0.999... is an infinite decimal; there is no "end", and thus there is no "1 at the end". The zeroes go on forever. And 0.000... = 0.

Then 1 – 0.999... = 0.000... = 0, and 1 = 0.999.

The proof would go: X=1/3 X=.33... 10x=3.33... 9x=3 9*1/3=3 is a true statement. Every algebra text book ever supports this statement.

Honestly it feels like instead of math skills this question is rather "I wonder if you know the little trivia I know :^)"

Most confusion comes from not fully understanding the notation used in the problem. Many people will often automatically think that $0.99999...$ represents an arbitrary term of the sequence $a_{n} = \sum_{i=1}^n \frac{9}{10^{i}}$ , and not the limit $\lim_{n \to \infty} \sum_{i=1}^n \frac{9}{10^{i}}$ . That is, they will consider an arbitrary but only finite number of decimal places. Obviously, every term of the sequence $(a_{n})$ is strictly less than $1$ , but the limit $\lim_{n \to \infty}(a_{n})$ is exactly equal to $1$ . Knowing in advance that the ellipsis means a limit is all that is necessary to correctly solve the problem.

Let x = .99999...

Hence, 10x = 9.99999...

10x - x = 9x = 9.99999... - .99999... = 9

Thus, 9x = 9

And, x = $\frac{9}{9}$ = 1

Hence proven.

1/3 = 0.3333333..... 2/3 = 0.6666666..... 3/3 = 0.9999999.... = 1

Proof: 1/9=0.111111111... 8/9=0.888888888... 1/9+8/9=9/9 9/9=1 0.111111....+0.8888888..... = 0.99999999.... 0.99999999...=1

EVERYONE CALM DOWN! I REPEAT! CALM DOWN!

Now just go here and here and see if you are correct

1 and 0.9999... are just different ways to represent the same thing. Just like 1/3 and 0.3333...

This could be proved in yet another simple way , easy enough to do in some seconds itself.

As $\frac{1}{3}$ = 0.3333333333...................

Multiplying with 3 on the both sides , simply results

1= 0.99999999999999999............

Basically, if 0.9999...... is infinitely going on forever, then at some point it has to reach 1. Like, take a square for example. You half it, and color one section. Then you half the uncolored section, then color one of THOSE sections. Continue. At some point, all of the square's going to be colored. Because I explained this using an enclosed space, it makes more sense right? Therefore, 0.99999.... is equal to 1.

1/3 equals to 0.33333333333333... , 2/3 equals to 0.66666666666666... , then 3/3 in theory should equal to 0.999999999999... but 3 divided by 3 is 1, therefore 0.999999999...= 1

Vihart's video on how .9999999999999999 repeating =1

Though I did get this question right, I must admit it is not very well written, and it really needs the repeating symbol over the decimal on the right and not the ellipses. The use of ellipses is quite disingenuous, as it technically means "an omission." It does not mean that something is being repeated. It could be .999998 for all we know, and not the 3/3 the question attempts to imply.

1/9 = .11111... 2/9 = .22222... 3/9 = .33333...(3/9 = 1/3) 4/9 = .44444... 5/9 = .55555... 6/9 = .66666... (6/9 = 2/3) 7/9 = .77777... 8/9 = .88888... 9/9 = .99999... (9/9 = 3/3 = 1)

Therfore .99999... = 1.

0.999999=x so 9.9999999=10x so 10x-x=9X 9X=9 So x=1

I put 0.99999..... horizontally into my phone calculator and when I flipped it vertically I showed 1.00000...

$\frac{1}{3}$ = 0.33333...

If you multiply both sides by 3 you get:

$\frac{3}{3}$ = 0.99999...

1 = 0.99999...

$1 = \frac{9+1}{10} = \frac{9+\frac{9+\frac{9+\frac{9+⋯}{10}}{10}}{10}}{10} = \frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{9}{10000}⋯= 0.9+0.09+0.009+0.0009+⋯ = 0.9999...$

Your answer is incorrect. Although each 1 of the 10 are each tending(to infinity), to 1, it does NOT mean that each number IS 1. For example - let's just use 3 decimal places then 10x 0.999 (to infinity) is not 10. It is 9.999. What the proposer has done is made 9 of the 'ones' into one each and then made the 'one' left over 0.999 to infinity. It doesn't matter how infinitesimally small it is, you have used two different values for one. Johanna below went on to cite some rubbish about read the proof. This is simple, you accept that 0.999 is not one. Why are you trying to say 0.999to infinity is one? Question, if this was graphed, would the line ever converge with the axes? No. It tends to it but will never touch it.

1 = 1/3 + 1/3 +1/3 = .333... + .333... + .333... = .999...

I think one of the simplest ways to think of this is that x-y=0 only if x=y, and if you subtract 1 - 0,999... the result is a 0 followed by an infinite number of zeros and a 1 in infinity, which is exactly the same as 0. Therefore 0,999... must be equal to 1.

.333333......=1/3

.666666......=2/3

.999999......=3/3=1

There are many different proofs of 0.9999..=1 and here is one of them:

1/3=0.3333... 1=3 * 1/3 = 3 * 0.3333... = 0.99999...

Hence,1=0.9999...

There are many solutions to this problem. The simplest is this:

1/9 = .11111111111111...

9/9 = .999999999999...

Note that 1/9 = 0.11111... So 9/9 = 1 = 0.9999.... Thus they are equal.

The difference between 0.999999... and 1 is 0.00000000...1

but since there are infinite zeros you never get to the 1 at the end

so $0.00000000...1=0$

so they must be the same number

/

suppose $x=0.999999999...$

multiply by 10 $10x=9.99999999999...$

subtract x $9x=9$

divide by 9 $x=1$

Um... it isn't a "Debate" anymore than is 2+2>4? It is a fact... 1/9=.111111, 2/9=.22222,... 8/9=.88888, so 9/9=.99999=1.

Either answer is correct depending on what you want to believe. https://www.youtube.com/watch?v=x-fUDqXlmHM

1/9 = 0,11111...

2/9 = 0,22222...

3/9 = 0,33333...

4/9 = 0,44444...

5/9 = 0,55555...

6/9 = 0,66666...

7/9 = 0,77777...

8/9 = 0,88888...

Then, 9/9 = 0,99999... = 1

May you should reach the answer from your teacher and ask him

Simply explained: Calculate and read the Decimal for 1/9, 2/9, 3/9, 4/9… 8/9 then what is 9/9? We know this is one, but judging by the pattern, it is also .9999…

1-0.999... is 0.000... there are infinite 0's after that decimal point there is no secret 1 at the end. so what's the difference between 0 and 0.0 and 0.000 and 0.0000....

If you allow infinitesimals (which some branches of number theory do have) then yes there is a difference. Generally though infinitesimals are not used and this there is 0 difference between 0.999... and 1

The easiest solution that even a fifth grader could do is as follows 0.99999...=x 10x=9.9999... 10x-x=9.9999...-0.9999... 9x=9 x=1 so without doing a bunch of crazy stuff we can say 1=0.9999... also ... means to infinity

IF 1 can be expressed as 3 x 1/3 or 3 x 0.3333333...

AND 3 x 0.3333333... can also be expressed as 0.99999999...

THEN 1 = 0.9999999...

What do you get by proving this?? 0.9999999.....=1 1.0000000.....=1.00000000...001 1.00000000...001=1.0000000....002 Series goes on.... And on You get 0.999999999......=infinite

It comes down to the axioms of mathematics. Really the basis is the statement: x-y=0 if and only if x=y.

This is a fundamental axiom that is assumed as true mathematically.

So, let's assume 1 $\ne$ $0.\overline{9}$ . That means, there exists a number z such that 0 < z < 1- $0.\overline{9}$

Now your goal is to find z. This would prove, that 1 is not equal to $0.\overline{9}$ .

Unfortunately, no matter how small you choose z to be, because $0.\overline{9}$ extends infinitely (think decimal places), we can never pick z such that 0<z<1- $0.\overline{9}$

1/3=.3333..., so 1=.9999..

1/3 = 0,33333... We multiple both sides by 3, which gives us 1 = 0,99999...

Repeating decimals put into fractions are over 9. For example, 0.333333333.... is 3/9. 9/9 is 1.

If 1/3 is 0.3333..... and 3 times 1/3 is 1 then 3 times 0.3333..... is 1 meaning 0.9999..... is equal to 1

x=0.99999... 10x= 9.99999... 9x = 10x - x = 9.99999... - 0.99999... = 9 9x = 9 x = 1

Its basic mats. 1/3=0.33333333... (infinite number of 3). If u multiply both sides by 3 you get 3/3=0.9999999... 3/3=1 . As simple as that. Prove me wrong.