I'm just starting to learn calculus, but this was the first idea presented: $$0.\overline9=1$$ This would mean that this is true: $$a-0.\overline01=a$$ When I thought about it, then I realised that if the above was true, then this other statement must be considered true: $$((((a-0.\overline01)-0.\overline01)-0.\overline01)...)=(a-0.\overline01)-0.\overline01=a-0.\overline01=a$$ This is confusing because it presents a lot of weird consequences. One such consequence would be a different behavior of open and closed intervals, since numbers that I previously considered to be different, are now considered to be the same. For instance, $2$ is now equal to $1.\overline931415926$.

I can't help but think that I am missing a crucial piece of information.

  • 7
    $\begingroup$ Yes, this is confusing because $0.\bar{0}1$ seems to indicate a decimal with "infinite zeros and then a one at the end." Which, of course, is absurd. $\endgroup$
    – vanPelt2
    May 25, 2020 at 19:31
  • 2
    $\begingroup$ Loosely. you will never get to the $1$ in $0.\bar{0}1$. $\endgroup$
    – copper.hat
    May 25, 2020 at 19:33
  • 3
    $\begingroup$ Does this answer your question? Are infinitesimals equal to zero? $\endgroup$
    – JMoravitz
    May 25, 2020 at 19:36
  • 2
    $\begingroup$ The $0.\overline01$ is not a number, unless you mean $0.\overline0$ (which is really just $0$), or perhaps $0.0\overline1$ (which is just $1/90$) - the periodic repeat overline must be on last digit(s). Also $0.\overline{9}=1$ is quite natural if you already know for example $0.\overline{3}=1/3$ , since $0.\overline{9}=3\cdot 0.\overline{3}=3\cdot \frac{1}{3}=1$ (to get an intuition). $\endgroup$
    – Sil
    May 25, 2020 at 19:36
  • 1
    $\begingroup$ @JMoravitz That post is misisng the point that typically $0.\overline{3}$ is started as a known result from which the $0.\overline{9}=1$ is derived (which is lesser known for some reason). Of course you need to prove the first result first, but again, that one is typically known/accepted already. $\endgroup$
    – Sil
    May 25, 2020 at 19:44

3 Answers 3


If you're just beginning to learn calculus, this will certainly seem a bit confusing! The key is to understand that when we write something such as $0.\overline 9$ we mean something very specific.

In this case, we need a precise definition of what "a decimal expansion with an infinite sequence of nines" even means. One way we can do this is to define it as the infinite series such that: $0.\overline9 = \frac{9}{10} + \frac{9}{100} + ...$

If you are familiar with series, you'll notice this is a geometric series. If you are just starting calculus I'll assume you haven't yet met notions of convergence of sequences and series, but it will suffice to simply state that there is a precise notion of these things. In this sense, the geometric series I described above is equal to $1$, by definition of what it means for an infinite series to converge.

However, consider $0.\overline0 1$. How exactly would we define this? It is not obvious to me, and I don't see a good way to do this. In this sense we cannot apply the usual notions of convergence of series because we're not even sure how to express this 'decimal' as such an infinite series!

The key takeaway is that when we make certain claims in mathematics, they mean very specific things, and they refer to specific notions. In this case we first of all need to define what we mean by $0.\overline 9$, (in this case define it as an infinite series), and then we need to define what it means for an infinite series to converge. For completeness, we say that an infinite series converges if and only if the sequence of partial sums converges. (But note there are also different ways to define convergence in different situations).

With this key takeaway in mind, the very idea of $0.\overline0 1$ is not clear, and hence proving further statements from it is not valid. If you can come up with a precise working definition of what you mean by the symbols $0.\overline0 1$, that's a different story! But as presented, there is no way to define it in the same way as $0.\overline 9$

  • 1
    $\begingroup$ I don't follow this, unless I assume you meant $0.\overline 01$ everywhere you wrote $0.0\overline 1$. $\endgroup$ May 25, 2020 at 19:51
  • 2
    $\begingroup$ you seem to have moved the overline to the wrong digit. It reads currently as the bar is over the $1$ rather than over the $0$. Do instead $0.\overline{0}1$ $\endgroup$
    – JMoravitz
    May 25, 2020 at 19:51
  • $\begingroup$ So sorry, fixed it now $\endgroup$
    – masiewpao
    May 25, 2020 at 19:52
  • $\begingroup$ I don't personally see how this is difficult to work with. If you number the digits with ordinals, those labeled with natural numbers are all 0, and the digit labeled with ω is 1. So this number is 10^-ω. Obviously that doesn't exist in the reals, but is there any reason it should be undefined in some more general field? $\endgroup$ May 25, 2020 at 20:07
  • 2
    $\begingroup$ @SethSchmidt-O'Hainle ... I'm guessing the OP has never heard of "some more general field", so masiewpao answered at the proper level. $\endgroup$
    – GEdgar
    May 25, 2020 at 20:26

In short: The trouble is that you can't do arithmetic on infinite decimal expansions. You can do arithmetic on real numbers, which seems like doing arithmetic on decimal expansions, but it's inherently different. Let me explain, by first trying to build up a world in which we only know about decimals (and not about calculus) and seeing where things go wrong.

Decimal expansions work in a specific way - consider some random decimal expansion: $$812.34512\ldots$$ There's basically only two meaningful kinds of questions we can ask about this decimal expansion:

  1. Which digit is in the $10^n$ place?

  2. If any digit is non-zero, what is the most significant place value occupied?

For instance, it's meaningful to say that the $10^{-1}$ place is occupied by $3$ and the $10^{1}$ place is occupied by $1$, and to say that the most significant place value occupied is the $8$ in the $10^2$ place.

These questions are meaningful without reference to any number system - we can treat these things as just strings of numbers.

Now, it's cumbersome to write infinitely long strings of digits, so we have notation such as $$0.\overline{9}$$ which doesn't involve writing infinitely many digits, but still lets us answer our two questions - the digit in the $10^{n}$ place is $0$ for every $n\geq 0$ and $9$ for every $n < 0$. Similarly, $$0.\overline{123}$$ suffices to tell us that if we want to know the digit in the $10^{-n}$ place, we just repeat the pattern 123123123123 until we reach the desired number of places. Note that this is always a finite process: the place values $10^n$ are indexed by integers $n$, so repeating a pattern over and over always gets us to any desired index. You can also note that $1.\overline{0}$ and $0.\overline{9}$ are definitely different decimal expansions since they, for instance, differ that the $10^{-1}$ place - though remember that we're talking about decimal expansions here, without considering any notion of them representing a number.

The trouble with notation like $$0.\overline{0}1$$ is that it let's us ask the same questions as before: what's in the $10^{-5}$ place of this digits? Well, we just repeat $0$ until we get there - it's a $0$!. What about the $10^{-9}$ spot - oh, that's also a $0$. Oh wait, if I fix any integer $n$, the answer is going to be $0$. So, whatever you might mean by this expression, as a decimal expansion, this is identical to the expansion of all zeros - because the two questions we are allowed to ask of decimal expansions never differ between this expression and $0.\overline{0}$.

Ah, but you ask, if I am so sure of this, what is the result of the following subtraction? $$1.\overline{0} - 0.\overline{9}$$ And here is the trouble that calculus addresses: it is not possible to do arithmetic on decimal expansions in any way resembling what you want them to do.

You might wonder "what about long addition and subtraction?" Well, for integers (or any number where there are only finitely many digits), this is fine: write out the two decimals, one atop the other, then, starting at the rightmost digit, apply your operation and use the rules of carrying/borrowing to proceed to the next one and so on. This algorithm works only through asking (finitely many times) the questions that one may ask of decimal expansions. Try that on an infinite decimal though - you're in trouble at the start, because you can't start from the rightmost digits, because there isn't a rightmost digit to start from! (See that our rules only let us talk about the most significant - i.e. leftmost - digit).

In a concrete example, you can try adding together $0.3333\ldots$ and $0.6666\ldots$. Sure, you could just add each digit and get a valid answer of $0.9999\ldots$, but that's hardly justified by your algorithm - for the first digit you wrote, how do you know that you weren't supposed to have carried into it, without having inspected all the (infinitely many) digits beyond that first one already?

Of course, you could weaken things to ask that your addition is only consistent, in that there is a carry exactly when the previous column summed to at least $10$ and the digit in the answer is the last digit of the sum of the digits and carries in the corresponding column of the addition. You could write $0.333\ldots + 0.666\ldots = 0.999\ldots$ in a long addition table with no carries, or you could write $0.333\ldots + 0.666\ldots = 1.000\ldots$ in a long addition table where you carry on every digit - and either one would look entirely consistent under any finite amount of inspection. Of course, you could ask, where did this carry come from "at the beginning", but this is the sort of question that requires infinitely much inspection all at once, which isn't really allowed in our framework. I think this is what you're getting at with the idea of $0.\overline{0}1$ - the idea of an initial "inciting" carry - but the trouble is that there's no place for a $1$ - indeed, there is no place where we could possibly determine whether to carry or not!

This is really a serious crisis in mathematics - we wished to generalize from integers and fractions of integers (where there are nice rules for computation) to a larger number system encompassing anything that could be written down as a decimal expansion - but the naive way of doing this means we can't do arithmetic!

Instead of trying to make expressions like $0.\overline{0}1$ valid (which is not a bad idea - but I'm not aware of any way to make such an expression work out), the usual solution is to work in the world of calculus instead, where we imagine decimal expansions to represent series such as $$800+10+2+0.3+0.04+0.05+0.001+0.0002+\ldots$$ where the $\ldots$ represents a limit of the partial sums - which is a reasonable object we can define for any decimal expansion solely by asking the questions we may ask of decimal expansions - but this only guarantees that every decimal expansion corresponds to a real number, not vice versa - and, indeed, expressions like $$0.\overline{9} = 1$$ show us that, in the calculus world, decimal expansions are not unique properties of a number: a single number may have multiple expansions.

Indeed, in calculus and its applications, decimal numbers get demoted in importance - usually when we see a number like $0.123\ldots$, we mean something like "this number, rounded to three digits, is $0.123$" or "this number has a decimal representation beginning with $0.123$" (i.e. is in $[0.123,0.124]$) - which are "native" calculus concepts, expressed by distances and intervals - but, for precisely the reasons hinted at by your questions, decimal expansions cease to have the capacity to be foundational building blocks, like they could when working with integers and rationals.


Two nunmbers are different if they have at least one other number between them, in this case there is no number between $0.\bar9$ and $1$, therefore they are the same. When we talk about $0.\bar01$ we are actually talking about $0$ because it is telling us, there are infinetly many $0s$ and the last digit is a 1, but that last digit will never appear since we have infinetly many $0s$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.