First, by definition I assume that $0.999...$ actually is defined as: $$\text{lim}_{n\rightarrow\infty}\sum_{i=1}^n 9/10^i$$

Now by geometric series we already know that this equals one. But nonetheless here is an explicit proof. The statement is: $$\forall\epsilon_+\exists\delta\forall n (n>\delta\rightarrow|\sum_{i=1}^n 9/10^i-1|<\epsilon)$$

Which is equivalent to:

$$\forall\epsilon_+\exists\delta\forall n (n>\delta\rightarrow|\sum_{i=0}^n 9/10^i-10|<\epsilon)$$

Let $\epsilon>0$ be a real number. Now note that $\sum_{i=0}^{n} 9/10^i=10-1/10^n$. Choose $\delta=\text{max}(1,\text{ceil(log}(1/\epsilon)))$

As such: $|10-10-1/10^n|=1/10^n<1/10^{\text{ceil(log}(1/\epsilon))}\leq 1/10^{\text{log}(1/\epsilon)}=\epsilon$

Which works out nice...

However, I have learned that $0.999...=1$ doesn't hold in all number systems such as hyperreals and surreals and what not. I am not even sure about rational numbers (although the proof looks like it would work for rationals with small tweaks)... I believe that the statement can't even be formulated in the first order theory of real closed fields as the $n$ is quantified over naturals, so there are number systems which can't even express the fact.

What in the proof goes wrong in the non-standard number systems and what are the primary features of the systems that cause this? Also, a proof of the fact NOT holding in said systems is welcome!

  • $\begingroup$ Usually, the proof is saved by the transfer principle $\endgroup$ Jul 24, 2018 at 17:17
  • $\begingroup$ This number does not exist by the construction of real numbers $\endgroup$ Jul 24, 2018 at 17:23
  • $\begingroup$ @VladislavKharlamov Could you elaborate? How does it not exist? $\endgroup$
    – Jam
    Jul 24, 2018 at 17:38
  • 2
    $\begingroup$ @Jam Many books exclude from decimal expansion the possibility of a tail of $9$'s as an ad hoc assumption in order to get uniqueness of decimal expansions. Vladislav needs to understand that definition and terminology is different from content. While those books don't call $\sum_{n=1}^{\infty}9/10^n$ a decimal expansion, all of them do consider this a valid series and the question of whether the (Cauchy) sum of this series should be $1$ or not is still a valid question in all those texts as well. $\endgroup$
    – user577471
    Jul 24, 2018 at 17:54
  • 2
    $\begingroup$ @ThomasWeller: What do you mean by $0.\bar{0}1$? It certainly doesn't match up any of the usual shorthand for expressing a decimal. In what place is the 1? $\endgroup$
    – user14972
    Jul 25, 2018 at 15:50

1 Answer 1


$0.\bar{9}= 1$ does hold the hyperreals.

The fact you're referring to is something different, and unfortunately that difference is usually not made clear which leads to confusion like you had. What people are trying to say is that in the hyperreals, you can have a terminating decimal that nonetheless has infinitely many $9$'s.

More precisely, you still have decimal notation for the hyperreals, but the places are indexed by hyperintegers rather than ordinary integers. And if you take an infinite hyperinteger $H$, then

$$ 1 - 10^{-H} = \sum_{n=1}^H 9 \cdot 10^{-n} = 0.\underbrace{999\ldots999}_{H\text{ nines}} \neq 1$$

However, $0.\bar{9}$ still refers to the nonterminating decimal that has a $9$ in every fractional place, and $0.\bar{9} = 1$.


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