Is .999 Repeating Equal to 1?: The Proof Behind the Infamous Debate

[Mark44]

But from the same Wikipedia page where it is stated that 0.9¯=1, it is also mentioned that a geometric series gives a proof of convergence. Proof of convergence is not a proof of equality.

That's true, but if we know what the sequence converges to, then that is a proof of equality.

Or:
$0.\overline{9} = \lim_{n \rightarrow \infty} \frac{\frac{0.9}{0.1}(1-0.1^n)}{\frac{1}{0.1}-1} = \lim_{n \rightarrow \infty} (1-0.1^n) \neq 1$

No, not true. $\lim_{n \to \infty}1−0.1^n=1−\lim_{n \to \infty}0.1^n=1$. Period. Full Stop. End of story. Since you claim that $\lim_{n \to \infty} 1 - 0.9^n \ne 1$, please tell me what is the correct value for this limit, and I will show you why that is wrong.

@WWGD , are you saying that 'converges' is the same as 'equals'? To me converges means tends to a limit, without reaching it.

The limiting expression never reaches the putative limit in any finite number of steps, but the limit is taken as n grows large without bound.

The idea here with limits is that we can make the limiting expression arbitrarily close to the limit value. Mathematically speaking, there is no difference between 0.999... and 1. If you believe there is a difference, please tell me how far away from 1 is 0.999...

$\frac 1 \infty$ cannot be equal to 0. It is close, but not equal. The 1 on the numerator proves it.

Of course $\frac 1 \infty$ is not zero. The division is not defined, so it is incorrect to say that it is close.

 

[jack action]

@Mark44 , although I agree with the concept that the limit is by definition the value that it converges too, it still doesn't reach it until you consider $\infty$, which is not a real number.

If you believe there is a difference, please tell me how far away from 1 is 0.999...

Let's take the example of @Frodo who defines $\frac{1}{3} = 0.333...$ and then goes back to show that $3 \times 0.333... = 1$. But let's look at this some other way:
$$3 \times 0.3 = 0.9$$
$$3 \times 0.33 = 0.99$$
$$3 \times 0.333 = 0.999$$
Therefore, one can see that:
$$3 \times 0.333... = 0.999...$$
So the correct fraction would be $0.333... = \frac{0.999...}{3}$, not $0.333... = \frac{1}{3}$

From this I deduce that:
$$0.9 \lt 0.99 \lt 0.999 \lt 0.999... \lt 1$$
Where $0.999... = 1 - \lim_{n \rightarrow \infty}\frac{1}{n}$. (edit: more precisely $0.999... = 1 - \lim_{n \rightarrow \infty}\frac{1}{10^n}$)

Of course 1∞ it is not zero. The division is not defined, so it is incorrect to say that it is close.

The point is that the only way $\frac{x}{n} = 0$ is if $x = 0$, no matter how large $n$ is. If $x = 1$, then there are no values of $n$ that exist such that $\frac{1}{n} = 0$.

Does that make sense or am I talking through my hat?

 

[Mark44]

@Mark44 , although I agree with the concept that the limit is by definition the value that it converges too, it still doesn't reach it until you consider $\infty$, which is not a real number.

Then you are at odds with everyone in the mathematics community.
We don't care at all about whether a sequence "reaches" the value it converges to; what is of concern to us is that however far away from 1 a particular term in the sequence {0.9, 0.99, 0.999, 0.9999, ...} is, we take a specific term farther out in the sequence, and get closer to the limit value.

The question isn't do we ever get to the limit value, but rather, can we get so close to it, that you can't see daylight between the two things.

Let's take the example of @Frodo who defines $\frac{1}{3} = 0.333...$ and then goes back to show that $3 \times 0.333... = 1$. But let's look at this some other way:
$$3 \times 0.3 = 0.9$$
$$3 \times 0.33 = 0.99$$
$$3 \times 0.333 = 0.999$$
Therefore, one can see that:
$$3 \times 0.333... = 0.999...$$

I agree completely with the first three equations, but the right side of the last equation is 1 by another name.

So the correct fraction would be $0.333... = \frac{0.999...}{3}$, not $0.333... = \frac{1}{3}$

No, both are true statements.

From this I deduce that:
$$0.9 \lt 0.99 \lt 0.999 \lt 0.999... \lt 1$$

0.9 < 0.99 < 0.999, and so on with a finite number of 9 digits,
but 0.999... equals 1.

If you don't believe this, please do this subtraction for me:
1 - 0.999...

Where $0.999... = 1 - \lim_{n \rightarrow \infty}\frac{1}{n}$. (edit: more precisely $0.999... = 1 - \lim_{n \rightarrow \infty}\frac{1}{10^n}$)

First year calculus: $\lim_{n \to \infty} \frac 1 {10^n} = 0$, not "close to 0."
Please do not tell me that the value is 0.00000...1, where there are an infinite number of 0's ahead of the 1.

The point is that the only way $\frac{x}{n} = 0$ is if $x = 0$, no matter how large $n$ is.

Well, of course, but here n is a finite number. In your previous example, you had $\frac 1 \infty$, which isn't defined.

If $x = 1$, then there are no values of $n$ that exist such that $\frac{1}{n} = 0$.

But this limit, $\lim_{n \to \infty} \frac 1 n$ does exist, and is equal to (not close to) 0.

Does that make sense or am I talking through my hat?

When you write things such as the following, yes.
$\lim_{n \to \infty} \frac 1 {10^n} \ne 0$
$\lim_{n \to \infty} \frac 9 {10^n} \ne 1$
$0.333... \ne \frac 1 3$
$0.999... \lt 1$

 

[jbriggs444]

Any person who can divide by 3 should be able see that 1 divided by 3 is exactly equal 0.333333...

That depends on what one accepts the notation 0.333... to mean. If one accepts it as the real-valued limit of the infinite series then sure. But if one accepts that, why not use it to prove that 0.999... is equal to 1 in a more direct fashion?

 

[benorin]

Wow! I've had HW threads I've posted runaway like this with lots of experts arguing the "finer points" of the proof/problem and I did not find it very helpful as the student.

I'm not an expert but it seems to me that the definition of a base 10 number combined with that of a repeating decimal ipso facto implies that infinite series are defined. Therefore, to expedite a helpful discourse, I suggest using the definition

$$0.\bar{9} := \sum_{k=1}^\infty\tfrac{9}{10^k}$$

and from there, there are well known ways to derive the sum of a geometric series, which I leave as an exercise for the OP. Good luck!

 

[Mark44]

I've had HW threads I've posted runaway like this with lots of experts arguing the "finer points" of the proof/problem and I did not find it very helpful as the student.

We've had lots of threads on exactly the same subject: whether 0.999... = 1. The most common misconceptions in threads of this nature are a flawed understanding of what it means mathematically for a sequence of numbers to converge, as well as a misunderstanding about how a limit is defined mathematically.

 

[jack action]

I must admit that all the "proofs" I've seen until now tend to say $0.999... \triangleq 1$ therefore $0.999... = 1$.

But I finally found an explanation that satisfies me:

Some proofs that 0.999... = 1 rely on the Archimedean property of the real numbers: that there are no nonzero infinitesimals. Specifically, the difference 1 − 0.999... must be smaller than any positive rational number, so it must be an infinitesimal; but since the reals do not contain nonzero infinitesimals, the difference is therefore zero, and therefore the two values are the same.

I didn't know about the infinitesimals concept (I guess this is one for the Today I learned thread), but explained this way, it makes sense when there is a distinction between real numbers and infinitesimals, with zero as the only common number in the two sets.

I think this is what the equation in post #33 by @Mark44 states, I'm just not fluent enough in that type of notation to understand it.

 

[Mark44]

From the wiki article:

Specifically, the difference 1 − 0.999... must be smaller than any positive rational number

I asked a question similar to this a couple times: if 0.999... is not equal to 1, then what do you get from the difference 1 - 0.999... ?

I think this is what the equation in post #33 by @Mark44 states, I'm just not fluent enough in that type of notation to understand it.

I didn't say anything about infinitesimals. What I said in that post was that the difference between 1 and $\sum_{i = 1}^N \frac 9 {10^i}$ can be made as small as anyone demands, by choosing an appropriate value for N.
If for a given N the difference isn't as small as you would like, I'll choose a larger value of N.
If for that N the difference isn't as small as you would like, I'll choose an even larger value of N.
If for that N the difference isn't as small as you would like, I'll choose an even larger value of N.
If for that N the difference isn't as small as you would like, I'll choose an even larger value of N.
and so on until you tire of the game.

 

[hmmm27]

I asked a question similar to this a couple times: if 0.999... is not equal to 1, then what do you get from the difference 1 - 0.999... ?

$1-0.\dot9=0.\dot0 1$

 

[jack action]

I didn't say anything about infinitesimals. What I said in that post was that the difference between 1 and $\sum_{i = 1}^N \frac 9 {10^i}$ can be made as small as anyone demands, by choosing an appropriate value for N.
If for a given N the difference isn't as small as you would like, I'll choose a larger value of N.
If for that N the difference isn't as small as you would like, I'll choose an even larger value of N.
If for that N the difference isn't as small as you would like, I'll choose an even larger value of N.
If for that N the difference isn't as small as you would like, I'll choose an even larger value of N.
and so on until you tire of the game.

The problem with that is that there is no end of the game: there is always a difference. A difference necessarily means not equal to zero. This is why you need the concept of infinitesimals, i.e there exist a number that is indivisible. That may put an end to the game.

No matter what, you will have to define something as being true. My previous source has a citation that I find interesting:

Although the real numbers form an extremely useful number system, the decision to interpret the notation "0.999..." as naming a real number is ultimately a convention, and Timothy Gowers argues in Mathematics: A Very Short Introduction that the resulting identity 0.999... = 1 is a convention as well:

However, it is by no means an arbitrary convention, because not adopting it forces one either to invent strange new objects or to abandon some of the familiar rules of arithmetic.

I asked a question similar to this a couple times: if 0.999... is not equal to 1, then what do you get from the difference 1 - 0.999... ?

Let's begin with numbers that are finite. You get:
$$1 - 0.(9)_n = \frac{1}{10^n}$$
Where $n$ is a positive natural number. Even for $n=0$, it works: $1 - 0 = 1$. No matter what is the value of $n$, there is always a difference that is non-zero.

You can even find $\frac{1}{3}$ with something similar:
$$3 \times 0.(3)_n = 0.(9)_n = 1 - \frac{1}{10^n} = 1 \times \left(1 - \frac{1}{10^n}\right)$$
$$\frac{1}{3} = \frac{0.(3)_n}{1 - \frac{1}{10^n}}$$
The difference between $\frac{1}{3}$ and $0.(3)_n$ can be found this way:
$$\frac{1}{3} - 0.(3)_n = \frac{0.(3)_n}{1 - \frac{1}{10^n}} - 0.(3)_n = \frac{1}{3 \times 10^n}$$
And since for each value of $n$ there is a value $n+1$, it is possible that the difference get always smaller.

Now let's go into infinitesimals. If there is always a difference with finite number, then it seems logical that there is also one with infinite number. But it has to be smaller. Let's say $\epsilon$ is indivisible, thus the closest number to zero. One could define the difference as:
$$1 - 0. \bar 9 \triangleq \epsilon$$
If we use that definition to find $\frac{1}{3}$ as we did before, we get:
$$\frac{1}{3} = \frac{0.\bar 3}{1 - \epsilon}$$
Similarly, the difference between $\frac{1}{3}$ and $0.\bar 3$ should be found this way:
$$\frac{1}{3} - 0.\bar 3 = \frac{0.\bar 3}{1 - \epsilon} - 0.\bar 3 = \frac{\epsilon}{3}$$
Now we have a problem, since $\epsilon$ is indivisible by definition.

But if one defines the difference as:
$$1 - 0. \bar 9 \triangleq 0$$
Then repeating everything above yields:
$$\frac{1}{3} = 0.\bar 3$$
And:
$$\frac{1}{3} - 0.\bar 3 = 0$$
That is coherent and it is elegant. So by convention (as stated in my previous quote), accepting that $1 - 0. \bar 9 \triangleq 0$ is OK.

But what if ones defines the difference as:
$$1 - 0. \bar 9 \triangleq \lim_{n \rightarrow \infty} n\epsilon$$
I don't know if it causes other problems, but I know it solves my $\frac{1}{3}$ problem:
$$\frac{1}{3} = \frac{0.\bar 3}{1 - \lim_{n \rightarrow \infty} n\epsilon}$$
and:
$$\frac{1}{3} - 0.\bar 3 = \frac{0.\bar 3}{1 - \lim_{n \rightarrow \infty} n\epsilon} - 0.\bar 3 = \frac{\lim_{n \rightarrow \infty} n\epsilon}{3}$$
Because for a very large value of $n$, there is always a $\frac{n}{3}$ that exists, then it is coherent.

But without looking further, that is already - as stated in my previous quote - a strange new object.

So, it makes sense to define $1 - 0. \bar 9 \triangleq 0$, but I think the $\triangleq$ is important here, it cannot be proven, it is a definition. So the question «What else could it be equals to?» may very well have an answer if we insist that there must be a difference. But - when compared to not having a difference - would it add something without complicating everything? Probably not.

 

[jbriggs444]

$1-0.\dot9=0.\dot0 1$

But that fix is worse than the problem it is intended to solve.

Edit: here we have @Mark44 anticipating you:

Please do not tell me that the value is 0.00000...1, where there are an infinite number of 0's ahead of the 1.

The 9's on the left hand side are presumably indexed by the natural numbers in their standard order. The digits on the right hand side are indexed by some other set [or at least a different order type]. The numbers are incommensurate.

The problem with trying to make an arithmetic complete by adding new numbers presented with new and baroque syntaxes is that you then have to think about how you do addition, subtraction, multiplication and division with the new numbers. Instead of solving problems, one is creating new problems. It is not as easy a game as one might think.

It is also problematic when one throws around notation without defining it first. What exactly is the order type for the digit positions in $0.\dot0 1$ ?

Edit: Backing up and taking a somewhat longer viewpoint...

We ask about the difference between 1.0 and 0.999... because we have an expectation that whenever we write down two numbers, the difference between the two will be another number. This is the principle of closure. We want our number system to be "closed" under the operation of subtraction.

We started with a numbering system where we are allowed to write down numbers with "..." notation at their ends. This numbering system seemed not to be closed under subtraction. So you've invented a notation where numbers can now end with "..." followed by a single digit.

What about the product of 1.000...1 and 1.000...1? Is it going to be 1.000...0002000...1? Where does the madness end?

It turns out that we already have an arithmetic that is closed (the real numbers) and a notation that works (decimal notation). The gotcha is that to make the notation match the arithmetic, we need to accept that both "0.999..." and "1" denote the same number. Now we have saved closure because 1 - 0.999 = 0.

Back when I was in school they used to teach the difference between numbers and numerals. Like everyone else, I did not see much point in the distinction. Why distinguish between a number and the name of a number? "Shut up and add already".

A number of people here now should be lamenting not paying attention to the distinction then.

 

[Mark44]

The problem with that is that there is no end of the game: there is always a difference. A difference necessarily means not equal to zero. This is why you need the concept of infinitesimals

Sure, always a difference, for any finite string of decimal digits. But in the notation 0.999... there are an infinite number of 9 digits to the right of the decimal point.

The game that we both are referring to goes like this:
Loop:

You tell me a small number $\epsilon > 0$ such that $1 - \sum_{k=1}^N \frac 9 {10^k} < \epsilon$.​

I choose a (large) integer N for which the inequality is true.​

If you are satisfied, goto Done​

Else goto Loop​

Done:

End​

As far as infinitesimals being needed, not really, which is why the concept of infinitesimals was superseded by and large by the concept of limits, and only revived by the relatively new studies of surreal numbers, hyperreal numbers, and superreal numbers.

Let's begin with numbers that are finite. You get:
$1 - 0.(9)_n = \frac{1}{10^n}$

No argument there.

But what if ones defines the difference as:
$1 - 0. \bar 9 \triangleq \lim_{n \rightarrow \infty} n\epsilon$
I don't know if it causes other problems,
$\frac{1}{3} = \frac{0.\bar 3}{1 - \lim_{n \rightarrow \infty} n\epsilon}$

The problem here is the limit: $\lim_{n \to \infty} n \epsilon$.
Here we have n growing large without bound that is multiplying a number very close to zero. In ordinary mathematics of real numbers, this is what is called indeterminate.

 

[WWGD]

The problem with that is that there is no end of the game: there is always a difference. A difference necessarily means not equal to zero. This is why you need the concept of infinitesimals, i.e there exist a number that is indivisible. That may put an end to the game.

No matter what, you will have to define something as being true. My previous source has a citation that I find interesting:Let's begin with numbers that are finite. You get:
$$1 - 0.(9)_n = \frac{1}{10^n}$$
Where $n$ is a positive natural number. Even for $n=0$, it works: $1 - 0 = 1$. No matter what is the value of $n$, there is always a difference that is non-zero.

You can even find $\frac{1}{3}$ with something similar:
$$3 \times 0.(3)_n = 0.(9)_n = 1 - \frac{1}{10^n} = 1 \times \left(1 - \frac{1}{10^n}\right)$$
$$\frac{1}{3} = \frac{0.(3)_n}{1 - \frac{1}{10^n}}$$
The difference between $\frac{1}{3}$ and $0.(3)_n$ can be found this way:
$$\frac{1}{3} - 0.(3)_n = \frac{0.(3)_n}{1 - \frac{1}{10^n}} - 0.(3)_n = \frac{1}{3 \times 10^n}$$
And since for each value of $n$ there is a value $n+1$, it is possible that the difference get always smaller.

Now let's go into infinitesimals. If there is always a difference with finite number, then it seems logical that there is also one with infinite number. But it has to be smaller. Let's say $\epsilon$ is indivisible, thus the closest number to zero. One could define the difference as:
$$1 - 0. \bar 9 \triangleq \epsilon$$
If we use that definition to find $\frac{1}{3}$ as we did before, we get:
$$\frac{1}{3} = \frac{0.\bar 3}{1 - \epsilon}$$
Similarly, the difference between $\frac{1}{3}$ and $0.\bar 3$ should be found this way:
$$\frac{1}{3} - 0.\bar 3 = \frac{0.\bar 3}{1 - \epsilon} - 0.\bar 3 = \frac{\epsilon}{3}$$
Now we have a problem, since $\epsilon$ is indivisible by definition.

But if one defines the difference as:
$$1 - 0. \bar 9 \triangleq 0$$
Then repeating everything above yields:
$$\frac{1}{3} = 0.\bar 3$$
And:
$$\frac{1}{3} - 0.\bar 3 = 0$$
That is coherent and it is elegant. So by convention (as stated in my previous quote), accepting that $1 - 0. \bar 9 \triangleq 0$ is OK.

But what if ones defines the difference as:
$$1 - 0. \bar 9 \triangleq \lim_{n \rightarrow \infty} n\epsilon$$
I don't know if it causes other problems, but I know it solves my $\frac{1}{3}$ problem:
$$\frac{1}{3} = \frac{0.\bar 3}{1 - \lim_{n \rightarrow \infty} n\epsilon}$$
and:
$$\frac{1}{3} - 0.\bar 3 = \frac{0.\bar 3}{1 - \lim_{n \rightarrow \infty} n\epsilon} - 0.\bar 3 = \frac{\lim_{n \rightarrow \infty} n\epsilon}{3}$$
Because for a very large value of $n$, there is always a $\frac{n}{3}$ that exists, then it is coherent.

But without looking further, that is already - as stated in my previous quote - a strange new object.

So, it makes sense to define $1 - 0. \bar 9 \triangleq 0$, but I think the $\triangleq$ is important here, it cannot be proven, it is a definition. So the question «What else could it be equals to?» may very well have an answer if we insist that there must be a difference. But - when compared to not having a difference - would it add something without complicating everything? Probably not.

Two("Standard") Reals can't be indefitely close to each other without being equal.

 

[jack action]

The thing I want to emphasize is that a value at infinity doesn't exist. For example, the following graph represents the equation $1/ x$ enlarged at a region close to $x = \infty$ (where $\infty^-$ is the closest point to $\infty$). The important thing to notice is that the function is not defined at $x = \infty$. Intuitively, it always approaches zero. But it is not defined, so nobody can tell the value for sure. Nobody can prove it.

<Quasi profanity edited by mentor> I'm calling it: $\frac{1}{\infty} \triangleq 0$.» There are clever ways with infinitesimals or limits to make it more bearable to define the indefinable, but in the end that's what it is: somebody, somehow, declared: $\frac{1}{\infty} \triangleq 0$. Then it becomes true because this is how it is defined.

But if this is true, then you will end up with $1- 0.\bar 9 = 0$. The bar over the nine represents a form of infinity. But you could have defined $1- 0.\bar 9 \triangleq 0$ and starting with this point, you would end up with $\frac{1}{\infty} = 0$.

This is why when someone asks to be shown the proof that $1 = 0. \bar 9$, it cannot really be done, because it is true because someone defined it as is. I've seen the $\triangleq$ used in a few definitions, and I think it is a very important distinction: $1= 0. \bar 9$ because $1 \triangleq 0. \bar 9$.

This is true:
$$1 - 0.(9)_n = \frac{1}{10^n}$$
This is also true, but it only states that both limits are the same (of the undefined form $\frac{0}{0} = 1$):
$$\lim_{n \rightarrow \infty} (1 - 0.(9)_n) = \lim_{n \rightarrow \infty}\frac{1}{10^n}$$
And the left side is basically a definition for a new notation:
$$0.\bar 9 \triangleq \lim_{n \rightarrow \infty} 0.(9)_n$$
So stating the following is still a comparison of two limits that ends up being an undefined form of $\frac{0}{0}$:
$$1 - 0.\bar 9 = \lim_{n \rightarrow \infty}\frac{1}{10^n}$$
$$0 = 0$$

 

[jbriggs444]

But it is not defined, so nobody can tell the value for sure. Nobody can prove it.

And nobody is trying to. You are arguing against a straw man of your own making.

 

[jack action]

And nobody is trying to. You are arguing against a straw man of your own making.

I think I'm arguing about circular logic.

 

[jbriggs444]

I think I'm arguing about circular logic.

Again, any circularity is in the straw man. You are responsible for any arguments it makes.

 

[PeroK]

I think I'm arguing about circular logic.

Your uncertainty about how real analysis works is really just a "beginner's" mistake. Rather than confront that misunderstanding, you assume that the whole edifice of standard analysis is fundamentally flawed and that the 19th century mathematicians who developed it have duped all the mathematicians of the 20th and 21st centuries into a basic oversight. That's not a credible position.

 

[Mark44]

The thing I want to emphasize is that a value at infinity doesn't exist.

Agreed.

For example, the following graph represents the equation 1/x enlarged at a region close to $x = \infty$ (where $\infty-$ is the closest point to $\infty$).

You have two errors in understanding here: "close to $\infty$" and $\infty-$ being the "closest" point to $\infty$. However "close to" $\infty$ you can name a number, I can come up with a number that is larger. And "closest to" $\infty$ makes no more sense than to say the closest number to 5, just to pick a number out of a hat.

I'm calling it: $\frac{1}{\infty} \triangleq 0$.» There are clever ways with infinitesimals or limits to make it more bearable to define the indefinable, but in the end that's what it is: somebody, somehow, declared: $\frac{1}{\infty} \triangleq 0$. Then it becomes true because this is how it is defined.

No. $\frac 1 \infty$ is not defined. Period.
Can you cite exactly who it was who claimed that $\frac{1}{\infty} = 0$?

But you could have defined $1- 0.\bar 9 \triangleq 0$ and starting with this point, you would end up with $\frac{1}{\infty} = 0$.

Again, $\frac 1 \infty$ is undefined, so it makes no sense to say that it is equal to any number, let alone 0.

 

[fresh_42]

The problem with that is that there is no end of the game: there is always a difference. A difference necessarily means not equal to zero. This is why you need the concept of infinitesimals, i.e there exist a number that is indivisible. That may put an end to the game.

This is wrong. There is no need for infinitesimals and any attempt to derive it is nonsense, since we are perfectly fine without. Your thinking is way too procedural. There is no process, hence no infinitesimal place left.

The quotation of Wikipedia only illustrates that $0.\bar{9}$ isn't properly defined if we do not assume a hidden limit, hence a convention. Namely $0.\bar{9}=\lim_{n \to \infty}9\sum_{k=1}^n (\frac{1}{10})^k$ which can easily be proven to be equal to $1$ without using any infinitesimals. The moment you drop limit and sum is the moment where convention comes into play and this discussion starts.

This thread will be closed. The entire debate depends on notational and therewith semantic arguments, plus we have had it already so many times.

 

[jedishrfu]

Since we have exhausted this thread topic to death and are in the infinitesimal stage of ending the discussion, I feel it's a good time to close it.

Thank you all for participating here.

Jedi