# Math Myth: ##0.999999999.... =0.\bar{9}= 1##

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• Greg Bernhardt
In summary, the discussion revolves around the misconception that ##0.\bar{9}## is not equal to 1, when in fact it is. This is because the sum of an infinite geometric series is equal to its limit, which in this case is 1. Some confusion may arise when considering nonstandard models of the reals, but this does not change the fact that in the standard model, ##0.\bar{9}## is equal to 1.

#### Greg Bernhardt

From @fresh_42's Insight
https://www.physicsforums.com/insights/10-math-things-we-all-learnt-wrong-at-school/

This isn't actually a problem at school. It is a problem in understanding. Yes, the two numbers are indeed equal. They are only written in a different representation, such as ##\pi = 3.1415926... ## notes pi and everyone knows it is not ending with ##6## or ever. Nevertheless, we write '##=##' and not '##\approx##' because we mean the actual number, not its approximation.

\begin{align*}0.\bar{9}=0.999999999...&=\dfrac{9}{10}+\dfrac{9}{100}+\dfrac{9}{1000}+\ldots\\[10pt] &=\sum_{i=0}^\infty \dfrac{9}{10^{i+1}} =9\cdot\sum_{i=0}^\infty \dfrac{1}{10}\cdot\dfrac{1}{10^{i}}\stackrel{(*)}{=}\dfrac{9}{10}\cdot\lim_{n \to \infty}\sum_{i=0}^n \dfrac{1}{10^{i}}\\[10pt] &=\dfrac{9}{10}\cdot \lim_{n \to \infty}\dfrac{1-\left(\dfrac{1}{10}\right)^{i+1}}{1-\dfrac{1}{10}} \stackrel{(**)}{=} \dfrac{9}{10}\cdot \dfrac{1-0}{1-\dfrac{1}{10}}=\dfrac{9}{10}\cdot \dfrac{10}{9}=1\end{align*}

The crucial points are the limits. While the first one ##(*)## is simply a translation for 'and so on', i.e. the dots in '##0.999999999...##' which shouldn't be controversial, the misconception begins at the second one ##(**)##. A limit isn't a process, it is a number! And ##0.\bar{9}## isn't a process either, it is a number. Number ##1.##

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Something I find interesting about this repeating decimal, it works in other number bases. If you take the highest value digit for that base, and repeat forever then it equals 1.

Take binary for example: 0.111... = (1/2) + (1/2²) + (1/2³) + ... which is a geometric series that most of us are familiar with that equals 1.

Check out https://www.purplemath.com/modules/series5.htm for more on Geometric Series.

I'll note the title of this thread is in direct contradiction with the content of the post and this insight.

PeroK
What's the point?

mathman said:
What's the point?

In a strict sense, this is only true in the standard, countable model for the Reals. If you use a model where infinitesimals are allowed/well-defined, the equalitudoes not hold. Maybe @stevendaryl can elaborate?

WWGD said:
In a strict sense, this is only true in the standard, countable model for the Reals. If you use a model where infinitesimals are allowed/well-defined, the equalitudoes not hold. Maybe @stevendaryl can elaborate?

Which claim is true only in the standard model for the reals?

The nonstandard models of the reals are by definition models of the reals. This means that any theorem (provable statement) about the standard reals is also true of the nonstandard reals. So if ##0.999...## is defined to mean ##\sum_{n=1}^\infty 9 \times 10^{-n}##, that is equal to 1 for both standard and nonstandard models of the real.

What's different about the nonstandard reals is that if ##N## is a nonstandard integer, then you can define a "finite" sum:

##S_N = \sum_{n=1}^N 9 \times 10^{-n}##

Then you can prove
##S_N = 1 - 10^{-N}##

So this number is not equal to 1. It is less than 1. But if ##N## is nonstandard, then ##S_N \approx 1##. It's not equal to 1, but ##S_N - 1## is infinitesimal.

stevendaryl said:
Which claim is true only in the standard model for the reals?

The nonstandard models of the reals are by definition models of the reals. This means that any theorem (provable statement) about the standard reals is also true of the nonstandard reals. So if ##0.999...## is defined to mean ##\sum_{n=1}^\infty 9 \times 10^{-n}##, that is equal to 1 for both standard and nonstandard models of the real.

What's different about the nonstandard reals is that if ##N## is a nonstandard integer, then you can define a "finite" sum:

##S_N = \sum_{n=1}^N 9 \times 10^{-n}##

Then you can prove
##S_N = 1 - 10^{-N}##

So this number is not equal to 1. It is less than 1. But if ##N## is nonstandard, then ##S_N \approx 1##. It's not equal to 1, but ##S_N - 1## is infinitesimal.
Yes, but the models are elementary equivalent but not isomorphic, I understand. Archimedean property is not expressive in 1st order, so is not preserved. My proof of the equality used that 1-0.9999... is indefinitely-small, which follows from archimedean principle. Maybe there is a proof independent of the Archimedean property though.

WWGD said:
Yes, but the models are elementary equivalent but not isomorphic, I understand. Archimedean property is not expressive in 1st order, so is not preserved. My proof of the equality used that 1-0.9999... is indefinitely-small, which follows from archimedean principle. Maybe there is a proof independent of the Archimedean property though.
The infinite summation involves only rational numbers, so it doesn't require the set of all real numbers. It's a sequence of rationals that converges to a rational, namely ##1##.

PeroK said:
The infinite summation involves only rational numbers, so it doesn't require the set of all real numbers. It's a sequence of rationals that converges to a rational, namely ##1##.
Rational numbers are Real numbers; as such, all axioms on Reals apply to them. I believe convergence is a second order property so that notions of convergence in the Standard Reals do not necessarily extend to non-standard models. Besides, even in the Standard Reals, convergent sums of Rationals do not necessarily converge to Rationals, e.g., the decimal expansion of , say , ##\pi##.

WWGD said:
Rational numbers are Real numbers; as such, all axioms on Reals apply to them.
But not the Archimedean property!

Given ##a/b, c/d##, (a/b)*b*c>c/d .

@WWGD
I do not know the what the discussion is about here [e.g. if it is about infinitesimals then I don't have any idea]. If you are talking about models of set theory then, in a sense, you are right. Even though it is a bit beyond my scope, I have some (very imprecise, but still...) idea about what happens in that case. You are kind of right in that case I think.

The thing here is "non-standard natural numbers". The point being that if you take a theory like PA then what we have in mind is the set of natural numbers ##\mathbb{N}## over which its statements are meant to be interpreted/quantified. However, somewhat loosely speaking, PA can't pindown ##\mathbb{N}## exactly. In other words, there are other "structures" which also satisfy the theory PA. These are what are called non-standard natural numbers.

So if ##M_1## is the standard model of PA and ##M_2## is any non-standard model, a number of things happen. For example:
(1) ##1_{M_1}+1_{M_1}=2_{M_1}##
##1_{M_2}+1_{M_2}=2_{M_2}##

So, internally, within the models this identities will hold. In fact, similarly this will hold for all theorems of PA. The above identities are not surprising because "1+1=2" is theorem of PA (note that for a consistent theory, a theorem is true in all its models).

(2) However, you are right, in the sense that if we look at ##M_2## "externally", it seems possible [being cautious in wording here] that an elementary identity of the form: a+a=2a (for some fixed ##a##) is "actually" failing in ##M_2##. The point being:
##a_{M_1}+a_{M_1}=2a_{M_1}##
##a+a=2a## (the ##a## of ##M_1## is the "real/intended" ##a##)

##a_{M_2}+a_{M_2}=2a_{M_2}##

But this ##a_{M_2}## might not have anything to do with the actual ##a##. In fact, ##a_{M_2}## might not even be present in the actual model ##M_1## (since ##M_2## being non-standard has several extra elements).

And further, the identity ##a+a=2a## can potentially fail in ##M_2## (I think) if we are looking at it externally (this is the main qualifier here).

(3) If you look at theory such as "PA+~con(PA)" it is consistent. But it is reasonable to ask how so? Even though ~con(PA) is false for the "actual" natural numbers, "PA+~con(PA)" still have some model. However, this model will not be ##M_1## but some other non-standard model.

(4) The non-standard models in ##PA## have very well-understood order-type. If you search a bit you will find it (or I can link it in case of interest). However, the actual mechanics/details of what happens in these models seems to be complicated (I haven't studied it so I don't know the details about it, besides a couple of things I mentioned above).

(5) So coming to set theory, a similar thing happens in models of it too. However, the issues becomes an order-of-magnitude more complicated because this time we can't just have non-standard numbers but also non-standard ordinals.

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You mean a correction of understanding by posting this thread but not a myth? Because ##0.999...=1## as you showed with the infinite series and the limits.

Another proof i know is ##x=0.999... \Rightarrow 10x=9.999... \Rightarrow## ## 10x-x=9.999...-0.999... \Rightarrow 9x=9 \Rightarrow ## ##x=1 \Rightarrow0.999...=1## and also when dividing 1 with 3 you get 0.333... so it goes to infinity and ##\frac{1}{3}## ##\times 3=0.333...\times 3=1## perhaps this last one does not seem that rigorous.

Greg Bernhardt said:
And ##0.\bar{9}## isn't a process either, it is a number.
No it's not. It's an equivalence class.

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