# Why does an argument similiar to 0.999...=1 show 999...=-1?

I accept that two numbers can have the same supremum depending on how you generate a decimal representation. So $$2.4999\ldots = 2.5$$ etc.

Can anyone point me to resources that would explain what the below argument that shows $$999\ldots = -1$$ is about?

Here is the most usual proof I see that $$0.999\ldots = 1$$:

$$x=0.999\ldots$$

$$10x=9.999\ldots$$

$$10x - x = 9$$

$$x=1$$

Using this same argument template I can show $$999\ldots=-1$$:

$$x= \ldots9999.0$$

$$0.1x= \ldots9999.9$$

$$0.1x - x = 0.9$$

$$x=-1$$

What might this mean?

Edit from one of the comments:

$$\sum_{k=0}^{\infty}{9 \cdot 10^k}=-1$$

• $999\dots$ is not a real number. And as soon as there is a rightmost $9$, it is a finite, positive real number. Jan 23 '16 at 19:28
• I see you have a Stack Overflow account. Then you may be somewhat familiar with two's complement representation. If you can imagine a two's complement representation with infinitely many bits, you won't be surprised that in that representation the number with all bits $1$ represents $-1$. Analogously, in an infinite ten's complement representation, the number with all digits $9$ represents $-1$. It's not the usual representation of numbers (where an infinite string of nines before the decimal yields an invalid representation), but it makes sense. It's the $10$-adic numbers, as quid mentions. Jan 23 '16 at 20:01
• Closely related: Divergent series and $p$-adics
– MJD
Jan 23 '16 at 20:28
• To those (like me) who read the OP's proof of 0.999... = 1 and thought "Well that's not very convincing..." just realize some steps were elided (full digit manipulation proof can be found on Wikipedia). Jan 23 '16 at 23:48
• You start with $999\ldots$, but seem to continue with $999\ldots9$? I suppose you rather mean $\ldots999$ Jan 24 '16 at 12:41

If you want to understand the mathematics behind these things, it is all based upon the notions of 'convergence' and of 'limits'. If you read any first course in analysis textbook you will find the concept rigorously treated there.

Basically this is the point: Whenever you write 0.999... you are writing down a numeral that represents the 'limit' obtained when an infinite summation $\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...$ is performed. Since we can prove that this sum 'converges' to some real number (namely 1), we are justified in treating the numeral 0.999... as representing some real number.

However, whenever you write down 999... I presume you are writing a numeral to represent the limit obtained when an infinite summation $9+90+900+...$ is performed. Since this limit does not converge to any real number, (it 'diverges'), we are not justified in treating the numeral 999... as any real number. So it does not make sense to divide it by ten, or take it away from itself.

We usually denote such divergent limits by the numeral $\infty$, but this does not denote a real number, and there is no consistent way to define operations such as $\infty - \frac{1}{10}\infty$.

I hope this helps and I hope you are motivated to think more about these things :)

• 1 + 2 + 3 + 4 + ... = -1/12 Jan 24 '16 at 15:25
• @Geinmachi: That is no more true than $0=1$. Jan 25 '16 at 1:01
• @Michael: I can't, he's dead. And just because he said something that is formally incorrect and can be formalized through some particular and nontrivial methods, doesn't mean that everything he says is true. That's a common fallacy known as appeal to authority. Jan 25 '16 at 14:53
• @Michael: Only that a "number" is not a well-defined mathematical notion. It is incorrect to say there is a rational number $x$ such that $x^2=2$. On the other hand, the convergence of a series is a well-defined mathematical notion, and while there are other notions of convergence, there is exactly one notion of convergence where one can omit the type of the convergence. And let me tell you, it ain't Cesaro, Ramanujan or otherwise. And $1+2+3+\ldots$ does not converge in that type of convergence. You want to argue that it does in some other type? Sure. But you need to be explicit. Jan 25 '16 at 17:21
• @Voo: No. I'm saying that $1+2+3+\ldots=-\frac1{12}$ is blatantly false. The statement "The series $1+2+3+\ldots$ converges in the sense of Ramanujan to $-\frac1{12}$" might as well be true. Jan 25 '16 at 19:18

In the $10$-adic numbers it is true that $\dots 9999 = -1$.

More precisely, the series $\sum_{n=0}^{\infty} 9 \cdot 10^n$ converges in $\mathbb{Q}_{10}$ and its limit there is $-1$.

• What do you mean by $\mathbb{Q}_{10}$? I would think it wouldn't be defined because $\mathbb{Z}_{10}$ has zero divisors. Jan 23 '16 at 19:54
• Just as an FYI to the OP: Fernando Gouvêa plays with similar seemingly divergent sums in the introduction of his book, p-adic Numbers. I think you would enjoy reading it. Jan 23 '16 at 19:59
• if $k$ is not prime, $\mathbb{Q}_k$ is a commutative ring instead of a commutative field Jan 23 '16 at 20:00
• @MarkS. The completion of the rational numbers with respect to the $10$-adic distance. It is true it is not a field but I do not see why this would be a problem in this context.
– quid
Jan 23 '16 at 20:03
• Thanks for this gives me a nice thing to look up. @SpamIAm - nice book - bought :) Jan 24 '16 at 10:54

As other users have noted, it doesn't make sense to have infinitely many nines to the left of the decimal point. This is because the sequence $9, 99, 999, \ldots$ doesn't converge to anything, unlike the sequence $0.9, 0.99, 0.999, \ldots$, which converges to $1$.

However, you have touched on something interesting. Are there number systems where this does converge? And if so, does it converge to $-1$? The answer to both is yes.

For an integer $k$, there's a ring called the $k$-adic numbers*. Let's use $k = 10$. It's similar to the regular real numbers in base $k$, but a few things are backwards.

$$\small{\begin{array}{c|c|c} & \textrm{Reals (base }k\textrm{)} & k\textrm{-adic numbers} \\ \hline \textrm{number of digits left of the point} & \textrm{finitely many} & \textrm{infinitely many} \\ \textrm{number of digits right of the point} & \textrm{infinitely many} & \textrm{finitely many} \\ \textrm{two numbers are close if} & \textrm{they match in the leftmost digits} & \textrm{they match in the rightmost digits} \end{array}}$$

That's not really the way that people like to construct the $k$-adics, but it's easier than all that "completion wrt a metric" or "inverse limit" stuff.

Anyways, in this system, $9, 99, 99, \ldots$ does have a limit, and it's $\ldots 999.0$. And if you add $1$ to that, you have the sequence $10, 100, 1000, \ldots$, which converges to $0$. So in that system, $\ldots 999 = -1$.

*Usually people like $k$ to be prime, so they call it $p$. So if you wanna find out more, look up "$p$-adic numbers", not "$k$-adic numbers".

• Thank you. The p-adics sound very interesting Jan 24 '16 at 10:54
• Are you really sure that you don't have everything screwed up in your table? In particular, I have seen reals with infinitely many digits after the point, like pi, but a real number can't have infinitely many digits before the point. (I know nothing about p-adic numbers, so I won't edit) Jan 24 '16 at 12:55
• @wythagoras Yea, he just used before to mean: to the right of, and after to mean: to the left of. Jan 24 '16 at 14:05
• One thing to note: they are called the "$k$-adics" for a reason - unless $k_1$ is a power of $k_2$ or vice versa, the $k_1$-adics and the $k_2$-adics are different rings. So the 10-adics are not the same as the 2-adics. For example, the 10-adics has 0-divisors, while the 2-adics form a field. Jan 26 '16 at 15:47
• Often, when the radix is not (a power of) a prime, they’re called “$g$-adic numbers”, and OP may have success in searching for that term. Additionally, I might point out that $\sum_0^\infty 9\cdot10^n$ is convergent to $-1$ not only $10$-adically, but also $2$-adically and $5$-adically. Jan 26 '16 at 23:13

One thing to think about is how this plays with modular arithmetic, if you're familiar. Basically, arithmetic mod $10^n$ is arithmetic where we only care about the last $n$ digits of a number and is defined by writing $a\equiv b\pmod{c}$ if and only if $a-b$ is a multiple of $c$. So, if $c=10^n$, this is the same as saying the last $n$ digits of $a$ and $b$ coincide. We have a series of equalities expressed as follows: $$9\equiv -1\pmod{10}$$ $$99\equiv -1\pmod{100}$$ $$999\equiv -1\pmod{1000}$$ $$9999\equiv -1\pmod{1000}$$ $$99999\equiv -1\pmod{10000}$$ $$\underbrace{99\ldots 99}_{n\text{ times}}\equiv -1\pmod{10^n}$$ These are all quite easy to prove: clearly, if you add $1$ to $\underbrace{99\ldots 99}_{n\text{ times}}$, you get $1\underbrace{00\ldots 00}_{n\text{ times}}=10^n$, where the last $n$ digits are $0$ meaning, mod $10^n$, this sum is $0$. Since $-1$ is more or less defined as the number whose sum with $1$ is $0$, the equality is proven.

The identity you have is, more or less, what happens when we take all the above identities and send $n$ to infinity. One precise way to do this is to say that two numbers $a$ and $b$ are close to each other whenever $a\equiv b\pmod{10^n}$ for large $n$. This takes us into the $10$-adic numbers, as others have suggested.

Another precise way not involving analysis would be to consider that we can consider a "number" $x$ to be something where we can always ask for the value of $x\pmod {10^n}$ in a consistent way - basically, it is just a string of digits. Then, we can define addition and multiplication of "numbers" in a way consistent with their truncations mod $10^n$. This again gives us that the infinite string of $9$'s equals $-1$, but this time in an algebraic way. (This give us the $10$-adic integers, which is a subset of the $10$-adic numbers. To be precise, the construction one can use for this is called an inverse limit, which is a scary sounding name for a scary looking definition)

It's worth noting that your proof, though not a proof that $\ldots 999$ is a sensible thing to think about, is a proof that if it is defined in any reasonable way (i.e. multiplying by $10$ shifts the digits and subtraction works digitwise when no carrying is at play), it equals $-1$. So, this is going to hold of any "reasonable" notion of summation, as well as in any "reasonable" extension of our algebraic system. For instance, one other answer used a method where we take the sum as a power series $$9x+90x^2+900x^3+9000x^4+\ldots$$ and equated this with $\frac{9}{1-10x}$ near $x=0$, which is a rational function. Without even checking, your proof tells us that this function had better equal $-1$ at $x=1$, since summing by this method allows you to do all the manipulations that you used.

• I like this. It seems similar to @DanielFischer reference to two's complement. Jan 24 '16 at 11:02
• @CommonToad yeah, because two's complement is really just about identifying 2^n with zero, and 2^n - 1 with -1, etc. for some convenient n. The bit about inverting all the bits and adding one to negate is just a convenient bit of sleight-of-hand. The modulus interpretation is the same thing as what's being extended here :) Jan 24 '16 at 20:22
• Actually, this error happens in calculator by this mechanism. Feb 5 '17 at 10:17

This argument is similar to the one $$\sum_{n=1}^\infty n = -1/12$$ which went viral a few years ago. You are actually using methods which were originally designed for manipulating absolutly convergent series on series which are not convergent at all. There has much talk been done and I think our friends of numberphile can explain it the best:

• "our friends of numberphile can explain it the best" Sorry but is this a joke? The N video became famous for the mediocrity of its content. The subsequent piece by the same team (probably the one you call "Clarification") is a dubious self-serving complacent damage control operation made necessary by the justified evisceration of the first piece that happened then. Other videos are doing a much better job hence sending people to this one is a cruel practical joke.
– Did
Oct 20 '17 at 6:38

you can assign a value to $\sum_{k=0}^\infty a_k 10^k$ where each $a_k \in \{0\ldots 9\}$ by saying it is a representation for $\lim_{x \to 1^-} f(x)$ where $f(x) = \sum_{k=0}^\infty a_k 10^k x^k$ for $|x| < 1/10$.

here $a_k = 9$ so $f(x) = 9 \sum_{k=0}^\infty 10^k x^k = \frac{9}{1-10 x}$ hence

$$\ldots999999 = \lim_{x \to 1^-} f(x) = \frac{9}{1-10} = -1$$

• what I'd like to know now is if that way to extend the integers in base $10$ with the Abel summation is related to the $10$-adic numbers Jan 23 '16 at 20:51
• It's worth noting that this is basically achieving the sum by analytic continuation. Jan 24 '16 at 0:15

Sure, $999\dots=1$. It also equals $\pi$, and $\frac00$. At least it would, if it existed (it would be true vacuously).

But decimal notation is defined such that you can only have a finite number of digits to the left of the decimal place (whereas you can have infinite decimals to the right). Therefore, you have a bad assumption: that $999\dots$ existed. See this other answer for a different decimal notation where you can have infinite digits to the left of the decimal point.

• Thank you for this. I will look into the answer you reference Jan 24 '16 at 10:58
• @CommonToad yep, just remember definitions are important, and you can't do anything with a thing that is undefined. Jan 24 '16 at 15:29
• @PyRulez Well, it is equal to $-1$, in the rings where the limit exists. :) Jan 24 '16 at 20:44
• @ThomasAndrews Yeah, I was assuming real numbers (since that's probably what he was familiar with.) Jan 25 '16 at 15:12

The sum of an exponential is well known:

$$\sum_{k=0}^{\infty}{9 \cdot 10^k}=9\sum_{k=0}^{\infty}{10^k}=9\left(\frac{1}{1-10}\right)=-1$$

This is basically, where we have logical problems, since it doesn't appear that the following doesn't make sense for $|a|\ge1$:

$$\sum_{k=0}^{\infty}a^k=\frac{1}{1-a}$$

But it does work for $|a|<1$ and it appears to be continuous, that is we can take a limit to get $|a|=1$, and, if it were continuous, we see that the solution happens to appear as that rational function $\frac1{1-a}$.

And the odd thing is, mathematically, this is nearly sound. Logically, it pretty much isn't.

The problem with such a solution lies in the realization that we are dealing with infinity, and while proving that the sum and rational function are equal, assuming we went with permutation, we found ourselves using the following definition:

$$\infty=\infty+1$$

This means that we face problems when we look at the solution through partial values, ie we try to evaluate the following:

$$\sum_{k=0}^na^k$$

But once we have $n=\infty$, its a whole new ballgame with different rules. You have to understand that simply because our logic says one way and the math says another doesn't mean either is right. Instead, think of $\infty$ as a sensible solution that has application and $-1$ as the other solution obtained mathematically that doesn't appear to mean anything to you or anyone else and probably doesn't have any applications. And, as I think of when I think of String Theory, if it doesn't have any actual effect on the world, it doesn't exist. (quantum foam)

• This is great :) Jan 24 '16 at 10:56

$...999$ is, I assume, an infinite string of $9$s, thus is it infinity. $0.1 \cdot \infty$ is still $\infty$. On line 3, you then say $0.1x - x$, but since both terms are infinity, what you are really saying is $\infty- \infty$, which is an undefined operation, so the rest doesn't hold. If you abuse infinity like it's a number (it's not) you can get all sorts of contradictory results. For example:

$1 + \infty= \infty$

$1 = \infty- \infty= 0$

Thus, $1 = 0$

Which is obviously wrong.

• Compare your comment about $\infty$ with the following. If you abuse zero like it's a number (t's not), you can get all sorts of contradictory results, for example $0 = 0 \cdot (0/0) = (0 \cdot 0)/0 = 0/0 = 1$.
– user14972
May 13 '16 at 6:31

With $x = 0.9999...$, you have an infinite number of 9s following the decimal point; multiplying both sides by 10 you still have an infinite number of 9s following the decimal point, so you can validly subtract one equation from the other to get $9x = 9$.

With $x = 999...9.0$, you have a finite, but arbitrary, number of 9s to the right of the decimal point. (As Henry Swanson points out, $999...9$ does not converge to a value like $0.999...$ does.) Let's try $x = 99999$:

\begin{array}{rcr} x & = & 99999.0 \\ 0.1x & = & 9999.9 \\ \hline 0.9x & = & 89999.1\\ x & = & 99999.0 \end{array}

• I think the OP's intention was to have infinitely many 9's before the decimal point. Jan 24 '16 at 0:16
• So do I, but that's not allowed. With an infinite number of digits on the left, it doesn't matter what those digits; the value goes to infinity Jan 24 '16 at 0:19
• @chepner What's the difference between the fact that the value "goes" to infinity and the fact that it can be mathematically shown to equal $-1$? Think about it, it can depend on the context. Jan 25 '16 at 23:40

It is certainly possible to construct a 'number system' in which $\dots999 = -1$ is true...

Define $\mathbf H = \{n:\mathbb W \to \mathbb Z\}$ to be the set of all infinite sequences of integers and represent the the sequence $n_0, n_1, n_2, \dots$ as $(\dots, n_2, n_1, n_0)$. these sequences will, formally, correspond to the sequences $n_0 + 10n_1 + 100n_2 + \dots$.

Define addition by $(m+n)_i = m_i + n_i$ and multiplication by $(m n)_i = \sum_{k=0}^i m_k n_{i-k}$. Other operation could be define similarly.

Say that $n \in \mathbf H$ is in canonical form if, for every $i \in \mathbb W$, $0 \le n_i \le 9$ or for every $i \in \mathbb W$, $-9 \le n_i \le 0$.

You need to define which sequences are null sequences, that is, sequences that correspond to the integer $0$. For example $(\dots, 0, -1, 10)$ is a null sequence. Say, $n \in \mathbf H$ is a null sequence if, for every $\alpha \in \mathbb W$, there exists a $\beta \in \mathbb W$ such that $\beta > \alpha$ and $\sum_{i=0}^\beta 10^in_i = 0.$ Define two sequences to be congruent if their difference is a null sequence. Prove that congruence is an equivalence relation.

Finally, define $\mathbf H_{10}$ to be the corresponding set of equivalence classes. The set of all sequences in canonical form is a transversal of $\mathbf H_{10}$.

In $\mathbf H_{10}$, it is true that $(\cdots,9,9,9) = (\cdots,0,0,-1).$

The problem is that $\mathbb Z$ is a proper subset of $\mathbf H_{10}$, that is, $\mathbf H_{10}$ is bigger than $\mathbb Z$ and $\dots999$ is in $\mathbf H_{10}$ but it is not in $\mathbb Z$.

$0.99999....$ is a convergence to $1.0$

$...999.9$ using the same template is a "convergence to infinity": contradiction in terms

This line can't be done if you define $x$ as $\infty$

$0.1x − x = 0.9$

since you cannot multiply or subtract infinity as if it were a real number

Assuming "$999...999.0$" could be defined as a real number: $\infty \sum k=0 9⋅10k$ with some integer $k$, using the same invented ellipsis terminology $0.1x - x = -8.999...999.1$

It means that addition is not associative for terms of divergent series. (See conditionally convergent series, https://en.wikipedia.org/wiki/Riemann_series_theorem#Changing_the_sum).

This result is of itself mathematically interesting, but demonstrating it in this absurd way is more likely to yield a "huh" or "rubbish" than an interesting response.

Attempting to "fix" this counter-intuitive mathematical result appears possible, but with unknown repercussions.

• The proof is entirely valid in the 10-adic numbers, which form a quite interesting ring. Addition and multiplication in the 10-adics are associative and commutive. $0$ and $1$ are still identities, and additive inverses still exist. Multiplicative inverses exist if the number is not a 0-divisor, but there are some 0-divisors. This only becomes absurd if someone demands that these be considered real numbers. Jan 26 '16 at 16:33

Folks have put forth the $p$-adic interpretation, but another way of understanding this is to consider it in the context of divergent series. One of the ideas behind the study of divergent series is to view $\sum$ as a linear functional from a particular domain $C(\mathbb{R})$ to $R$, where $C(\mathbb{R})$ is the vector space consisting of all real sequences with convergent series. This operator has many useful properties in and of itself, evidenced by how much we care about series in general.

However, we may also pose the question, Can we extend $\sum$ to an operator, call it $S$, on a vector space $V \subseteq \mathbb{R}^{\mathbb{N}}$ containing $C(\mathbb{R})$? In a very trivial way, the answer is yes, we can. To see this, recall that the axiom of choice implies that every vector space has a basis, so let $\{ e_{i} \in C(\mathbb{R}) : i \in I \} \subseteq \{ e_{j} \in V : j \in J \}$ be bases for $C(\mathbb{R}), V$ respectively. Then we can set $$S (e_{j}) = \begin{cases} \sum (e_{j}) & e_{j} \in C(\mathbb{R}) , \\ 0 & e_{j} \in V \setminus C(\mathbb{R}) . \end{cases}$$ However, this is not very satisfying. Linearity is not the only thing we like about summation. For example, consider the shift operator $\sigma : (a_{k})_{k \in \mathbb{N}} \mapsto (a_{k + 1})_{k \in \mathbb{N}}$. A property of $\sum$ is that $\sum ((a_{k})_{k \in \mathbb{N}}) = a_{1} + \sum(\sigma((a_{k})_{k \in \mathbb{N}}))$. We like this, for example, because it means that when doing convergence tests we know that no matter how $(a_{k})_{k \in \mathbb{N}}$ "starts", if it eventually "looks like" a sequence whose sum converges, then we still have convergence; that is, changing any finite number of terms in a sequence won't change whether or not its series converges. I will call this the Front-End Property (FEP).

It is important to note that the proof you listed in the OP that $0. \overline{9} = 1$, though very common, is as written incomplete. In particular, it assumes (rightfully, of course) that $0. \overline{9}$ is defined. You can show this, for instance, by appealing to the fact the sequence $(0.9, 0.99, 0.999, \ldots )$ is monotonic and bounded by $1$, and thus convergent. From this, we consider that $0. \overline{9} \equiv \sum (9 \cdot 10^{-k})_{k \in \mathbb{N}}$, i.e. $0. \overline{9}$ is defined as the summation of the sequence $(0.9, 0.09, 0.009, 0.0009, \ldots)$. We then use the front-end property. Let $x = \sum (9 \cdot 10^{-k})_{k \in \mathbb{N}}$. Then \begin{align*} 10 x & = \sum(10 \cdot (0.9 \cdot 10^{-k}))_{k \in \mathbb{N}} \\ & = (10 * 0.9) + \sum(10 \cdot (0.9 \cdot 10^{-(k + 1)}))_{k \in \mathbb{N}} \\ & = 9 + \sum(0.9 \cdot 10^{-k})_{k \in \mathbb{N}} \\ & = 9 + x \\ \Rightarrow 9x & = 9 \\ \Rightarrow x = 1 . \end{align*} But line one to line two was just the FEP! What you actually proved in the OP is that $S(10 \cdot (0.9 \cdot 10^{-k}))_{k \in \mathbb{N}} = 1$ for any operator $S$ with the FEP. You're mimicking the same argument to show that $\overline{9} = -1$, i.e. you're showing that IF $S$ has the FEP, THEN $S(9, 90, 900, 9000, \ldots) = - 1$ for a linear map $S : \mathbb{R}^{\mathbb{N}} \to \mathbb{R}$. It just happens that $\sum$ is not defined at $(9 \cdots 10^{k})_{k \in \mathbb{N}}$.

So in essence we're saying, "So we know we have this thing called summation that we can do to sequences that does neat stuff, and its defined in terms of these other sequences of partial sums converging. But we can also talk about this summation as a linear transformation, which is just the same as saying I can multiply by a scalar or add the sequences and get what I expect. But summation has other properties as an operator than linearity. We mentioned the FEP, but it's also what we may call positive, i.e. if $a_{k} \geq 0$ for all $k$, then $\sum (a_{k})_{k \in \mathbb{N}} \geq 0$. So we may ask, Can we extend this to all of $\mathbb{R}^{\mathbb{N}}$ and keep FEP and positivity? As you've shown, no.

Note also that if you apply your standard geometric series formula for when the ratio is $10$, you'll get $\overline{9} = - 1$.