# 0.999… equals one.

But you knew that, right? Check out the Wikipedia article about different proofs that 0.999… equals 1. It’s really quite interesting to see the wildly different methods that can be used to arrive at the same conclusion. While I’m partial to the Cauchy sequence version, the first of the proofs is my favorite, beautiful in its simplicity:

0.333…. = 1/3

3* 0.333… = 3 * (1/3)

0.999… = (3*1)/3 = 3/3 = 1

Q.E.D. =)

## 70 thoughts on “0.999… equals one.”

1. Anonymous says:

But 0.333…. itself does not equal one-third; it is an approximation, just as saying 0.999…. equals one. Technically, 0.3 = 3/10, 0.33 = 33/100, 0.333 = 333/1000, etc. etc. 1/3 = 0.33333…… This is an infinitely repeating decimal. You can’t use this arguement to "prove" 0.999… equals one because saying 0.333…. equals exactly 1/3 is already using an assumption. I don’t know if this makes sense, but it does in my head! :-D

1. Anonymous says:

If you’ve ever taken Calculus, as the number of decimal places increases, 0.999… converges to 1, therefore it equals one.
It is a limit.

2. Anonymous says:

No. .3 is an approximation, .33 or .333 or .333333333333333 are approximations, but .333… (which indicates an infinite number of 3’s) is an exact representation of 1/3. I think the problem is the way we read the number from left to right one digit at a time, we imagine the number getting closer and closer to 1/3 as we read each digit. But in fact, the number is there all at once, and with an infinite number of 3’s it exactly represents 1/3.

1. Anonymous says:

Actually, you can prove .9999…. = 1 matheimatically. (and you can prove .3333…. = 1/3 the same way.

Try this. Let x = .99999 ….. Then 10x = 9.9999…… Now subtract.

10x – x = 9.9999…. – .9999…
9x = 9
x = 1

Since x = .9999…. and x = 1, .9999…. = 1.

Similarly, let x = .3333….; 10x then = 3.3333…..

10x – x = 3.3333…. – .3333…..
9x = 3
x = 3/9 = 1/3
Therefore, .3333…. = 1/3

It’s not an approximation, it’s not a limit – it’s an equality. It’s 7th grade algebra (if I can recall all the way back to 7th grade ;-)

3. Anonymous says:

well it does make sense in my head too! :D

4. Anonymous says:

you’re right. although 0,333 tends to get to infinite (which could stop anywhere if needed), it cannot be 1/3; ever!!!

1. No– "0,333" is a definite number. It does not "tend to" anything.

Reminds me of a joke, though" 2 = 3, for a sufficiently large value of 2.

Windell H. Oskay

2. Anonymous says:

Your proof is wrong! 1/3 is actually not 0.333 but 0.3(3 occuring infinite times!). Simply calculate 1/3 in your scientific calculator and try change it into a fraction.. You will see what I mean.

3. Anonymous says:

The "…" in this case means infinitely continuing. Same as writing .9 with a line over the 9. It’s just impossible to type it that way.

1. Anonymous says:

the problem with the equation is that 3 * 0.333… does NOT = 0.999…

3 * 0.333… = 1

3 * 0.333 (not recurring) DOES = 0.999 (not recurring) and that’s why it looks like it works.

If that made any sense at all!

4. Anonymous says:

it does make sense
.333 equaling 1/3 is just a close estimation, or theory, and not a actual proven fact
Im not saying its right or wrong just that it is possible
‘theory’ and ‘fact’ do not necessarily stand in opposition of each other and the .999… theory equaling 1 is just a theory

1. Anonymous says:

you may have also proven that 3/3 is not equal to 1
because 0.333… times 3 is 0.999…
and 0.999 is not equal to 1

5. Anonymous says:

The proof is wrong – at least in my opinion, here’s why :

0,333333333…. no matter how many 3’s you add to the number it is not equal to 1/3 it is just nearly there(forgive the language, but i’m no mathematician) :D

(0,333…)*3 /= 3*(1/3)
0,333…=333…/1000… right? .Yes
so (333…../1000…..)*3 /= 1 – thus does not equal 3*(1/3)
than : 0,(3)*3 = 0,(9)
thank you and good night :)))) Tankowiec signing off

1. Anonymous says:

lolz. i agree

2. Anonymous says:

Unfortunately, .33333… does equal 1/3
.3333… is the summation of repeating geometric series (3)*(.1)^(n-1).
By doing some simple calculus, we find that the sum of this series is equal to:
lim n=> infinity, the series will add up to equal: .3/(1-1/10)=1/3

likewise, we prove that .9999… is equal to one the same way.
the geometric series for this is 9*(.1)^(n-1)
as n=> infinity, the series will add up to equal .9/(1-1/10)= 1

6. Anonymous says:

This is mathematically proven to be true.

Another proof is:

Let x = 0.9999….

Multiplying both sides by 10, we get:

10x = 9.999….

Subtracting the first equation from the second:

9x = 9

=> x = 1

Therefore, 0.9999…. = 1

<Trumpet call>

End of proof! ;)

1. Anonymous says:

No. Because your second equation contains one fewer digit after the decimal, no matter how many digits of precision you choose. Let’s look at a simple case:

```Let x = 0.99
10x = 9.9
10x-x = 9.9-0.99
9x/9 = 8.91/9
x = 0.99```

Busted! You cannot prove this without trickery, without fooling the reader somewhere along the line. All of these proofs rely on tricking us somewhere, and the more successful proofs hide the falsehood deep in the math where most people won’t catch it.

0.999… does not equal 1 and never will. The best you can say is that the approximation 0.999… approaches one as the decimal precision increases.

1. When we write "0.999…", we mean an infinitely repeated decimal, sometimes written as 0.9 (with a bar over the nine).

By definition, this mathematical object has a non-finite number of decimal places so it is quite meaningless for you to talk about what happens "as the decimal precision increases." The only falsehood I see here is your supposed equivalence between an object with a fixed number of decimal places and an infinitely repeating decimal.

Windell H. Oskay

1. Anonymous says:

No. What you’re suggesting requires "infinity + 1" because for your math to work, you must have LESS precision on the "let x" line and MORE precision on the "10x" line in order for the math to work. If the precision is equal, i.e. infinite, as you say, then the math is broken. This is a trick, plain and simple.

It’s not mathematically useful to discuss decimals with "infinite precision" because any number that cannot be represented with a finite number of decimal places cannot be accurately written in decimal. This only means that decimal is a lousy format for writing certain numbers. It doesn’t mean the numbers are weird. The value 1/3, for example, cannot be accurately written in decimal. Only an approximation can be written, but the true value is finite and well-understood. In base-3, one can write 1/3 accurately.

1. >No. What you’re suggesting requires “infinity + 1” because for your math to work, you must have LESS precision on the “let x” line and MORE precision on the “10x” line in order for the math to work. If the precision is equal, i.e. infinite, as you say, then the math is broken. This is a trick, plain and simple.

No, there is no trick here. If you think that there is, then you simply aren’t fully grasping what it means to say “0.999…”

When we write “0.999…”, we mean an infinite sum:
Summation, from i = 1 to infinity, of 9/(10^i)

This series does not terminate. There *is* no last digit. If there were, then there could be a finite, non-zero difference between 0.999…. and 1.
The term “precision,” loosely speaking, means the number of digits– if you think that the term precision applies, (or in other words, that there is a last digit) then you still aren’t getting it. Your argument is exactly equivalent to stating that infinity is some large but unspecified number– can you see the problem with that argument yet?

You might want to consider reading the original article that I linked to, which discusses a number of other common misconceptions as well. There is also a special FAQ section to deal with basic questions like these.

Windell H. Oskay

1. Anonymous says:

Windell, you seem like a bright guy. I cannot believe that you are STILL arguing that 0.999… = 1. Seriously? What’s next? 2=3? I understand infinity, and I understand asymptotic functions that APPROACH a value but NEVER reach it. That’s what we’re talking about here. They are NOT equal. Period. I rest my case and will stop arguing with you. If you still believe that 0.999… = 1 then I have a bridge in Brookly that I’d like to sell you.

1. >Windell, you seem like a bright guy. I cannot believe that you are STILL arguing that 0.999… = 1. Seriously? What’s next? 2=3?

You can drop the implied slur on my intelligence. If you don’t understand a point in mathematics, turning to ad hominem attacks doesn’t really help your argument much.

> I understand infinity, and I understand asymptotic functions that APPROACH a value but NEVER reach it.

You understand part of what’s going on here, which is a start. You aren’t understanding what it means for something to approach an asymptote. What it means, is that the value *infinitely far away* *IS* equal to that asymptotic value– otherwise it would not be approaching that asymptote, but instead diverging or converging to a different number. From wikipedia (and your math book): "A function f(x) is said to be asymptotic to a function g(x) as x -> infinity" if f(x) &#8722; g(x) -> 0.

(Side note: We have been talking about a number, not a function. If you want to represent the question as a function– you brought up functions here– we can do so, but it gets a little more challenging yet since it’s a function of discrete values. So, let f(m) = sum(1:m)(1/(10^m)). The domain of the function is the set of natural numbers, and the range is the set of natural numbers. Now, for any finite value m, f(m) = 0.999…9. There is a final, terminating 9 in the sequence. In the limit as m goes to infinity, you need to use the limit, which brings us right back to the sum that I had mentioned earlier. )

Windell H. Oskay

1. Anonymous says:

I concede. I read more at the Wikipedia article, and I see many many people are arguing this but the mathematicians all seem to agree with you. It intuitively SEEMS as though 0.999… must be infinitesimally smaller than 1, and it appears as though some of these "proofs" simply try to obfuscate the infinitesimal difference. But there is math far too complicated for me that supposedly proves it. I still don’t understand it, but I’ll take the mathematicians’ words for it, including yours. Please accept my apology for the Brooklyn Bridge comment. :-(

2. Anonymous says:

Um, actually, your argument is wrong, because the entire proof is based on the fact that even if you multiply 0.999… by 10, getting 9.999…, you STILL have infinitely many 9’s after the decimal point.

So my proof is still valid. :)

1. Wow– that was remarkably ambiguous– no way to tell whether you are arguing for a difference between 0.999…. and 1 or not.

If you at least logged in as a user, we might be able to tell which way you were arguing from your name or something.

Windell H. Oskay

2. Anonymous says:

0.999 does not equal 1. When you represent 1/3 as 0.333 you are using a format that approximates the value (less precision). The reason that your calculator outputs 0.333 when you input 1/3 is because 1/3 can not be represented exactly due to the way numbers are stored in a computer. Therefore it uses an approximation. In the actual world however, there are no need for approximations as we are not limited by having to store numbers using a binary format so instead we use fractions. The Calculator provided in Microsoft Windows XP also does this. From the help section:
"Understanding Extended Precision – Extended Precision, a feature of Calculator, means that all operations are accurate to at least 32 digits. Calculator also stores rational numbers as fractions to retain accuracy. For example, 1/3 is stored as 1/3, rather than .333."
0.9999 can never equal one. It will always be less due to it being an approximation.

1. We are talking about "0.999…", not "0.999" There is a very big difference between the two, and it sounds like you’ve missed the point entirely.

Windell H. Oskay

3. Anonymous says:

The idea of any decimal repeating i.e. 0.999….. is simply put is like a 35 year old nerd who lives in his mom’s basement… he’ll just never make it all the way. 0.9999…. itself is an active number, approaching one, but never actually getting there. Just like an asymptote on a graph, the line approaches a number infinitely, but can never and will never equal said number. nuff said.

1. > 0.9999…. itself is an active number, approaching one, but never actually getting there.

There is not any mathematical object that I am aware of called an "active number." 0.999… has an absolute and real value that is completely fixed– it does not vary, and does not approach any other number– it’s just a fixed number like 2, pi, sqrt(2), or so forth. (It just happens to be another way of writing "1".)

> Just like an asymptote on a graph, the line approaches a number infinitely, but can never and will never equal said number.

No. You have a fundamental misunderstanding about what it means to be as asymptote. If a function asymptotically approaches a given value, that means that the limit, infinitely far away, *is* that value. If that were not the case, then the function would be limiting to some other value– and not asymptotically approaching it after all. For any *finite* value of your independent variable, the function can only approach the final value; that analysis certainly does not apply in the limit that you look at an infinite number of digits.

Windell H. Oskay

7. Anonymous says:

0.999 does not equal 1. When you represent 1/3 as 0.333 you are using a format that approximates the value (less precision). The reason that your calculator outputs 0.333 when you input 1/3 is because 1/3 can not be represented exactly due to the way numbers are stored in a computer. Therefore it uses an approximation. In the actual world however, there are no need for approximations as we are not limited by having to store numbers using a binary format so instead we use fractions. The Calculator provided in Microsoft Windows XP also does this. From the help section:
"Understanding Extended Precision – Extended Precision, a feature of Calculator, means that all operations are accurate to at least 32 digits. Calculator also stores rational numbers as fractions to retain accuracy. For example, 1/3 is stored as 1/3, rather than .333."
0.9999 can never equal one. It will always be less due to it being an approximation.

1. Anonymous says:

Very true and you explained it like pro. If i understand corectly though isnt it actually imposible to work this sort of thing exactly?
By definition the reocurring 3 in 1/3 as a decimal figure (3.333…) is never ending so if a computer tried to work it out without limiting itself to a number of decimal places, it could never complete the calculation as it would be repeatedly adding 3’s to the equation literally forever.

2. >0.999 does not equal 1.

Correct you are.
You will never get an argument from me, or anyone else that knows a little math, about that!

Windell H. Oskay

3. Anonymous says:

the number .9999… or .333… is not an approximation. you missed the fact that … means repeats forever. Also its not a good idea to look at the workings of a computer to understand mathematical concepts. In the case of .999…=1 it is not designed to recognize the abstract logic it takes to come to that conclusion which is in fact correct. The only way to get a computer to tell you .999…=1 would be for a person to program it to do so, it will never come to this conclusion on its own. If you know some calculus, there are more complex proofs of .999…=1 that are more satisfying if you think the other proofs are lacking.

8. Anonymous says:

A way to think about this in words and logic instead of equations is that given the fact that .9999… has an infinite number of 9s at the end of it there is no number you can add to .999… to equal 1. Try it if you don’t believe. .999…+.001= 1.000999… etc. Since there is no number between .999… and 1 you can conclude it is the same number. It sounds weird when you first hear it, but then again, no one ever guaranteed that common sense is always right.

1. Anonymous says:

Yup, that’s how I thought of it too when I first heard this proof in a math lec! :)

0.9 + 0.1 = 1
0.99 + 0.01 = 1
0.999 + 0.001 = 1

and so, 0.999…. + 0.000… = 1

where the 0.000… essentially = 0 because the ‘1’ comes after infinite digits, and therefore doesn’t come at all.

I don’t think this is a valid proof though, but I think it kinda helps in ‘understanding’ so to speak.

1. Thanks– that’s an interesting and intuitive approach; might be a good way to approach the topic for youngsters, even.

[With a little limit calculus (or real analysis, if you’re hardcore!) the proof that you are outlining *can* actually be put on a rigorous, mathematically sound basis as well.]

Windell H. Oskay

9. Anonymous says:

Uhh… I’m just a code monkey, not some math genius, but…

1=0.999…
1-0.999…=0
but actually executing that operation gives
1-0.999…=0.000…1
so 0.000…1=0

for f(x)=1/n; f(0.1)=10, f(0.01)=100, f(0.001)=1000, etc. to f(0.000…1)=1000…0.0

1/0.000…1= +inf
1/0=fail
as +inf!=fail, 0.000…1!=0, so 1!=0.999…

Uhh… WTFBBQ???

I’m wondering if mixing open-form ideas like 0.999… and 0.000…1 with closed form ones like equals is fully valid? Or maybe I’ll just grab another banana and go back to my nice, simple, easy chars, ints and longs…

1. If you’re a code monkey, you know about bugs. (And you’ve just added one.)

> 1=0.999…
> 1-0.999…=0
> but actually executing that operation gives
> 1-0.999…=0.000…1

That last line is not valid *unless* the "sequence of 9’s" terminates at some point. (Our "…" here means that it repeats forever; you can’t have something repeat forever and *THEN* have a final number after that!)

Windell H. Oskay

1. Anonymous says:

Bugs. Yes. They have a bad habit of getting into the bananas… yuck. 8-P

What’s the difference between 0.999…999 and 0.999… ? Only 9s trail the decimal point. They both have an infinite number of nines trailing the decimal point. So each 9 in the first should have a corresponding 9 in the second, no? If each digit in one number has a corresponding digit in another, and they are all of the same value at the same place (in this case, 9) should not the two numbers be equal?

So, if 0.999…999 = 0.999… and 0.999…999 does terminate, then 0.999… does also. And if 0.999…999 = 0.999… and 1-0.999…999=0.000…1 then 1-0.999…=0.000…1. Already established is that 1=0.999… so 1-0.999…=0, so 0.000…1 also must equal 0.

Except that by definition, 0.999… does not terminate (so, does 0.999… even exist?). And what is the smallest infinitely small number greater than zero? Would it not be 0.000…1? And if it’s greater than zero, it can’t be equal to zero.

Um… are you sure that using = with … is valid? (In other words, neither statement is true: 1=0.999… and 1!=0.999… ???)

1. Oh boy.

>What’s the difference between 0.999…999 and 0.999… ? Only 9s trail the decimal point. They both have an infinite number of nines trailing the decimal point. […]

No. There is no mathematical object "0.999…9"; it just doesn’t make sense. Something either goes on forever or it does not. Either there is no end and it’s a true repeating decimal (i.e., it goes on forever), or there is a last digit and it does not go on forever.

Let’s rephrase your proposition. Suppose you come across a row of trees and a sign that tells you that it’s an infinitely long row of trees. What that means is that you could keep driving as long as you want– even forever– along this row of trees, and there would never be and end to it. Now suppose you drive along for a few days and come across a point where there’s a last tree in the row. Does that mean that you are now infinitely far away, and that there are infinitely many trees between your present position and where you started? No: It means that the sign was wrong! The row DID NOT go on forever, because it had an end.

>so, does 0.999… even exist?.
Yes, it’s one of the infinitely many ways to write the number "1."

> And what is the smallest infinitely small number greater than zero? Would it not be 0.000…1?

The only infinitely small number is zero.

There is no "smallest number greater than zero." There can’t be. Suppose that my aunt thought she had found the smallest number greater than zero. I would take that number, divide it by two, and produce a smaller yet number that was still greater than zero, proving that her number was not the smallest. Since that can be repeated for any positive number, it shows that there is no such thing.

>Um… are you sure that using = with … is valid?

These symbols can be used in valid ways. I wish you would. ;)

Windell H. Oskay

1. Anonymous says:

Still can’t believe there are human beings trying to devolve our race by claiming things to be true via mathematical manipulation.

1 is not 0.999… I have plenty of arguments as to why it isn’t.

(r–>) = recurring digits in the direction of the arrow

1.000(r–>) – 0.999(r–>) = 0.(<–r)0001

Are you going to argue that 0 = 1? That is exactly what you’re doing when you say 1 = 0.999… There is always going to be that 1 at the end when you subtract 0.999… from 1. Some may argue that if the 9’s don’t end, the 0’s don’t. Fine. They don’t. But how are you going to dictate which direction the recurring digits go? They would go left forever into the decimal point, making the value smaller and smaller but the 1 is still ALWAYS there.

By arguing that 1 = 0.999…, you are saying something infinitely small does not exist. Then we don’t exist. The universe has infinite space, so compared to the universe, we are infinitely small.

1. >1 is not 0.999… I have plenty of arguments as to why it isn’t.

The only trouble is that you don’t have any arguments that are correct. I am truly impressed by the density of mathematical and conceptual errors that you have managed to achieve in such a short comment! You’ve even managed to drag physics errors into this– that’s a new record, even for this discussion.

What’s funny to me is that when I posted the original article, I just thought that all of the different proofs were neat– I didn’t appreciate whatsoever that there would be an significant number of people who found this non-obvious. I remember learning about repeating decimals in about sixth grade. Have so many people really forgotten?

Windell H. Oskay

1. Anonymous says:

Instead of just saying, "You’re wrong," how about you prove me wrong?

1. This whole article was about a list of different proofs that 0.999… = 1.

Each and every one of those proofs *is* a proof that you are wrong. If that’s not enough, Wikipedia even has a whole FAQ section dedicated to the particular error that you are making. (Well, the error that you repeat the most, anyway.)

Now, we’ve been over that particular error (it seems like) twenty times in other comments, and so has Wikipedia, (and your sixth grade teacher) but it comes down to this: If a digit repeats infinitely, it *does not* have a point at which it stops.

That’s what it means for it to repeat infinitely. If it stops at some point, it’s *not* infinite. So long as you keep arguing “this goes on forever and then stops at some point,” you aren’t discussing an infinite situation: you’re discussing some make-believe hybrid between infinite and non-infinite. Mathematical objects have well-defined definitions– your mysterious hybrid number does not.

Windell H. Oskay

2. Anonymous says:

1 is greater than 0. Want a "proof"? 1 + x > 0 + x Period. You are trying to manipulate math by using infinity in an equation. INFINITY IS NOT A NUMBER. How could you possibly think that you could plug it in and expect it to prove anything? That’s like ignoring the properties of 0.

Infinity minus one is still infinity. Meaning:

X – 1 = X

Is that physically possible for any other "number"? No, therefore infinity should have its own set of properties.

3. >1 is greater than 0.

I never argued otherwise.

> INFINITY IS NOT A NUMBER.

I never said it was.

>Infinity minus one is still infinity.

I agree.

> infinity should have its own set of properties.

It does.

Out of curiosity, since we seem to agree on every single point, why are you arguing? ;)

Windell H. Oskay

4. Anonymous says:

I am saying it can’t just be plugged into equations to be considered "proof" because it is considered manipulation – not truth. In calculators, 0.999… might "be" 1 aso to pi "being" 22/7. Also:

1.000000000000…
0.999999999999…

See the first digits in each line of numbers? That is the most important one. Period. 1.000… > 0.999…

Arguing that an infitely small number is equal to zero is saying that we do not exist and seconds do not exist. You can always divide a second into something smaller, but will it ever be equal to no time passing at all? Also, our bodies are infinitely small when compared to the infinite universe. Does that mean we are 0 and that we don’t exist?

Saying 0.999… is 1 is also saying 1/x = 0 where X = infinity. No. That 1 is still a value and no matter how large you make the denominator, the 1 will ALWAYS be there.

5. > See the first digits in each line of numbers? That is the most important one. Period. 1.000… &gt; 0.999…

You can’t just ignore what’s after the decimal place and pretend. Math doesn’t work that way.

>Arguing that an infitely small number is equal to zero is saying that we do not exist and seconds do not exist.

Where the heck do you get that idea? It’s simply not true.

> You can always divide a second into something smaller, but will it ever be equal to no time passing at all?

Wow– bringing Zeno’s paradox into this. Very impressive. I’m sure the ancient greeks would be impressed.

>Also, our bodies are infinitely small when compared to the infinite universe. Does that mean we are 0 and that we don’t exist?

Simply put, not true. It genuinely reflects that you DO NOT understand what it means for something to be infinite. Please stop abusing mathematics– learn the definition before you keep using the terms.

Windell H. Oskay

6. Anonymous says:

(Different Anonymous…)

Windell, I want to congratulate you for your persistence in trying to persuade other Anonymous that you are correct, but I fear you are fighting a losing battle.

I use this proof quite frequently with my (high school) students, and most of them seem to get the idea without too much effort, even though they want to resist the idea at first. Once they get it, you can see the light in their eyes, seeing that maths can actually be interesting.

Makes a change from trying to correct errors like 7/0=0.

10. Anonymous says:

When one constructs the real numbers, either from cauchy sequences or using supremums of sets, you actually create equivalence classes. Just as when one constructs the rational numbers. That is why 1/2 = 2/4, it is because they belong to the same equivalence class. Likewise you see when constructing the reals that 1 = .99999…

P.S.
1/3 = .333…
It is not an approximation.

1. Anonymous says:

Think of it as a geometric sequence.

From [n = 1 to infinity] (n = integers) the sum of 3 * (1/10)^n

For a geometric sequence, the first term divided by (1-[rate of change]) = the sum.

So, the first term = 3 / 10
Rate of change = 1/10

3/10/(1-1/10) = 1/3

2. Anonymous says:

Thank you, sir. Dedekind cuts are indeed the only answer to this question.

11. alphachapmtl says:

A simple proof:

x = 0.999…
10x = 9.999…
10x – x = 9.999… – 0.999…
9x = 9
x=1

1. Anonymous says:

.9999…. can be represented as an infinite geometric series (9/10 + 9/100 +9/1000…) of the form sum(c*X^n) where c=9 X=(1/10) and n goes from 1 to infinity. the sum of a power series is given by (first term in series)/(1-X) if AbsoluteValue(X) is less than one. in this case 1/10 <1. so we can use the equation to solve for the sum. if we plug the above values into this equation, we get (9/10)/(1-(1/10))=(9/10)/(9/10)=1. so .99999…(going on forever) does equal one.

the same steps can be repeated to show that .33333…(going on forever) does equal 1/3 (c=3, X=1/10)

12. Anonymous says:

Jesus..
"Arguing" about this is so silly. People who don’t "believe" it don’t grasp that the … means infinite repeating decimals. That simple. They think they’re somehow clever by noticing that 0.333 isn’t 1/3. Well, golly, really? 0.333… isn’t the same thing as 0.333, and there is no number such as "0.000…1".

EOD

1. Anonymous says:

YES!!!!

There is a concept called Density of Real Numbers. It states between 2 real numbers there is an infinite number of reals between them. There are no numbers between 0.999… and 1, so they’re the same number. Same with the hypothetical 0.000…01 and 0.

1. Anonymous says:

Actually its Dutch. InfoNu does not accept English articles. And I can’t write any good English anyway.

I suggest you try an online-translation programme. They are not perfect, but you should get an impression of what it’s about.

If you have any serious comments on the math, I would be glad to hear it.

Regards,

Bart van Donselaar

1. >Actually its Dutch.

So it is!

>If you have any serious comments on the math, I would be glad to hear it.

I couldn’t translate the whole thing, but it looks like you’re possibly cooking up a non-standard analysis that doesn’t apply to real numbers (and therefore is off-topic to this discussion), or are just blatantly wrong. I’m not sure which. In either case, it certainly is not a valid refutation of the point of our article.

Windell H. Oskay

1. Anonymous says:

It’s clear to everybody who knows the subject that the real numbers 1 and 0.999… are equal. It’s no use discussing the point. The only reason this discussion about 1 and 0.999… is stil going on, is because many people would rather see 1 and 0.999… to be different. They are in fact looking for alternative interpretations of ‘1’ and ‘0.999…’.

Now why not construct a theory or theories in which the signs ‘1’ and ‘0.999…’ have (besides there ordinary meaning) an alternative interpretation which makes them unequal? In that case people who want to delve into the problem can decide for themselves which theory they want to use, and on which occasion.

This would probably – at least – result in interesting recreational mathematics.

The theory I have used in the article is a precursor of Robinsons nonstandard analysis. It is simpler in that it is based on the commutative ring of infinite sequences of real numbers. No free ultrafilters or the like are used.

Two translation-services I have found are:

Regards,

Bart van Donselaar

1. Unfortunately, it isn’t as clear as you and I would both hope– if you read through the comments above, you’ll find a few folks that argue for a difference in the real number values.

I’ll agree that there is a time and a place for advanced and nonstandard analysis, but it’s important that we’re clear to separate that from the more basic issue, which is all that we’re addressing here.

Windell H. Oskay

1. Anonymous says:

I don’t think the Wikipedia-article ‘0.999…’ as a relatively simple explanation can be essentially improved.

If people deny the equality of the real numbers 1 and 0.999… they usually have no idea of what real numbers mathematically are. Real numbers are seen as quantities given by infinite decimal expansions. So they think as long as you are talking in terms of infinite decimal expansions you are in the realm of real numbers. You consequently see a lot of nonsens and sometimes ideas that can be treated rigorously in terms of alternative number systems. I don’t think the battle wil end as long as the denial of the equality of 1 and 0.999… isn’t acknowledged as a theoretical possibility.

Regards,

Bart van Donselaar

13. Anonymous says:

Completely correct

1. Well then, it’s a good thing that mathematics relies on logic and proof, rather than just beliefs!

Windell H. Oskay