Is it true that $0.999999999\ldots=1$?

Question

I'm told by smart people that $$0.999999999\ldots=1$$ and I believe them, but is there a proof that explains why this is?

Answer 1

What does it mean when you refer to $.99999\ldots$? Symbols don't mean anything in particular until you've defined what you mean by them.

In this case the definition is that you are taking the limit of $.9$, $.99$, $.999$, $.9999$, etc. What does it mean to say that limit is $1$? Well, it means that no matter how small a number $x$ you pick, I can show you a point in that sequence such that all further numbers in the sequence are within distance $x$ of $1$. But certainly whatever number you choose your number is bigger than $10^{-k}$ for some $k$. So I can just pick my point to be the $k$th spot in the sequence.

A more intuitive way of explaining the above argument is that the reason $.99999\ldots = 1$ is that their difference is zero. So let's subtract $1.0000\ldots -.99999\ldots = .00000\ldots = 0$. That is,

$1.0 -.9 = .1$

$1.00-.99 = .01$

$1.000-.999=.001$,

$\ldots$

$1.000\ldots -.99999\ldots = .000\ldots = 0$

Answer 2

Suppose this was not the case, i.e. $0.9999... \neq 1$. Then $0.9999... < 1$ (I hope we agree on that). But between two distinct real numbers, there's always another one in between, say $x=\frac{0.9999... +1}{2}$, hence $0.9999... < x < 1$.

The decimal representation of $x$ must have a digit somewhere that is not $9$ (otherwise $x = 0.9999...$). But that means it's actually smaller, $x < 0.9999...$, contradicting the definition of $x$.

Thus, the assumption that there's a number between $0.9999...$ and $1$ is false, hence they're equal.

Answer 3

What I really don't like about all the above answers, is the underlying assumption that $1/3=0.3333\ldots$ How do you know that? It seems to me like assuming the something which is already known.

A proof I really like is:

$$\begin{align} 0.9999\ldots × 10 &= 9.9999\ldots\\ 0.9999\ldots × (9+1) &= 9.9999\ldots\\ \text{by distribution rule: }\Space{15ex}{0ex}{0ex} \\ 0.9999\ldots × 9 + 0.9999\ldots × 1 &= 9.9999\ldots\\ 0.9999\ldots × 9 &= 9.9999\dots-0.9999\ldots\\ 0.9999\ldots × 9 &= 9\\ 0.9999\ldots &= 1 \end{align}$$

The only things I need to assume is, that $9.999\ldots - 0.999\ldots = 9$ and that $0.999\ldots × 10 = 9.999\ldots$ These seems to me intuitive enough to take for granted.

The proof is from an old high school level math book of the Open University in Israel.

Answer 4

Assuming:

1. infinite decimals are series where the terms are the digits divided by the proper power of the base
2. the infinite geometric series $a + a \cdot r + a \cdot r^2 + a \cdot r^3 + \cdots$ has sum $\dfrac{a}{1 - r}$ as long as $|r|<1$

---

$$0.99999\ldots = \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} + \cdots$$

This is the infinite geometric series with first term $a = \frac{9}{10}$ and common ratio $r = \frac{1}{10}$, so it has sum $$\frac{\frac{9}{10}}{1 - \frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1.$$

Answer 5

\begin{align} x &= 0.999... \\ 10x &= 9.999... \\ &= 9 + 0.999... \\ &= 9 + x \\ 10x - x &= (9 + x) - x \\ (10 - 1)x &= 9 + (x - x) \\ 9x &= 9 \\ x &= 1 \end{align}

Answer 6

There are genuine conceptual difficulties implicit in this question. The transition from the rational numbers to the real numbers is a difficult one, and it took a long time and a lot of thought to make it truly rigorous. It has been pointed out in other answers that the notation $0.999999\ldots$ is just a shorthand notation for the infinite geometric series $\sum\limits_{n=1}^{\infty} 9\left( \frac{1}{10} \right)^{n},$ which has sum $1.$ This is factually correct, but still sweeps some of the conceptual questions under the carpet. There are questions to be addressed about what we mean when we write down (or pretend to) an infinite decimal, or an infinite series. Either of those devices is just a shorthand notation which mathematicians agree will represent some numbers, given a set of ground rules. Let me try to present an argument to suggest that if the notation $0.99999\ldots$ is to meaningfully represent any real number, then that number could be nothing other than the real number $1$, if we can agree that some truths are "self-evident".

Surely we can agree that the real number it represents can't be strictly greater than $1$, if it does indeed represent a real number. Let's now convince ourselves that it can't be a real number strictly less than $1,$ if it makes any sense at all. Well, if it was a real number $r < 1,$ that real number would be greater than or equal to $\sum\limits_{n=1}^{k} 9\left( \frac{1}{10} \right)^{n}$ for any finite integer $k.$ This last number is the decimal $0.99 \ldots 9 $ which terminates after $k$ occurrences of $9,$ and differs from $1$ by $\frac{1}{10^{k}}.$ Since $0 < r <1,$ there is a value of $k$ such that $\frac{1}{10^{k}} < 1-r,$ so $1 - \frac{1}{10^{k}} >r.$ Hence $\sum\limits_{n=1}^{k} 9\left( \frac{1}{10} \right)^{n} > r.$ But this can't be, because we agreed that $r$ should be greater than or equal to each of those truncated sums.

Have I proved that the recurring decimal is equal to $1$? Not really- what I have proved is that if we allow that recurring decimal to meaningfully represent any real number, that real number has to be $1,$ since it can't be strictly less than $1$ and can't be strictly greater than $1$. At this point, it becomes a matter of convention to agree that the real number $1$ can be represented in that form, and that convention will be consistent with our usual operations with real numbers and ordering of the real numbers, and equating the expression with any other real number would not maintain that consistency.

Answer 7

Okay, I burned a lot of reputation points (at least for me) on MathOverflow to gain clarity on how to give some intuition into this problem, so hopefully, this answer will be at least be somewhat illuminating.

To gain a deeper understanding of what is going on, first we need to answer the question, "What is a number?"

There are a lot of ways to define numbers, but in general numbers are thought of as symbols that represent sets.

This is easy for things like natural numbers. So 10 would correspond to the set with ten things -- like a bag of ten stones. Pretty straightforward.

The tricky part is that when we consider ten a subset of the real numbers, we actually redefine it. This is not emphasized even in higher mathematics classes, like real analysis; it just happens when we define real numbers.

So what is 10 when constructed in the real numbers? Well, at least with the Dedekind cut version of the real numbers, all real numbers correspond to a set with an infinite amount of elements. This makes 10 under the hood look drastically different, although in practice it operates exactly the same.

So let's return to the question: Why is 10 the same as 9.99999? Because the real numbers have this completely surprising quality, where there is no next real number. So when you have two real numbers that are as close together as possible, they are the same. I can't think of any physical object that has this quality, but it's how the real numbers work (makes "real" seem ironic).

With integers (bag of stones version) this is not the same. When you have two integers as close to each other as possible they are still different, and they are distant one apart.

Put another way, 10 bags of stones are not the same as 9.9999999 but 10 the natural number, where natural numbers are a subset of the real numbers is.

The bottom line is that the real numbers have these tricky edge cases that are hard to understand intuitively. Don't worry, your intuition is not really failing you. :)

I didn't feel confident answering until I got this Terence Tao link:

Wayback Machine
PDF,page 12

Answer 8

One argument against this is that 0.99999999... is "somewhat" less than 1. How much exactly?

      1 - 0.999999... = ε              (0)

If the above is true, the following also must be true:

9 × (1 - 0.999999...) = ε × 9

Let's calculate:

0.999... ×
9        =
───────────
8.1
  81
   81
     .
      .
       .

───────────
8.999...

Thus:

     9 - 8.999999... = 9ε              (1)

But:

         8.999999... = 8 + 0.99999...  (2)

Indeed:

8.00000000... +
0.99999999... =
────────────────
8.99999999...

Now let's see what we can deduce from (0), (1) and (2).

9 - 8.999999... = 9ε                      because of (2)
9 - 8.999999... = 9 - (8 + 0.99999...) =  because of (1)
                = 9 -  8 - (1 - ε)        because of (0)
                =   1    -  1 + ε         
                =               ε.

Thus:

9ε = ε

8ε = 0

ε = 0

1 - 0.999999... = ε = 0

Quod erat demonstrandum. Pardon my unicode.

Answer 9

If you take two real numbers x and y then there per definition of the real number z for which x < z < y or x > z > y is true.

For x = 0.99999... and y = 1 you can't find a z and therefore 0.99999... = 1.

Answer 10

.999... = 1 because .999... is a concise symbolic representation of "the limit of some variable as it approaches one." Therefore, .999... = 1 for the same reason the limit of x as x approaches 1 equals 1.

Answer 11

Given (by long division):
$\frac{1}{3} = 0.\bar{3}$

Multiply by 3:
$3\times \left( \frac{1}{3} \right) = \left( 0.\bar{3} \right) \times 3$

Therefore:
$\frac{3}{3} = 0.\bar{9}$

QED.

Answer 12

The problem isn't proving that $0.9999... = 1$. There are many proofs and they all are easy.

The problem is being convinced that every argument you are making actually is valid and makes sense, and not having a sinking feeling you aren't just falling for some parlor trick.

$0.99...9;$ (with $n$ 9s) is $\sum_{i= 1}^n \frac 9 {10^i}$ so "obviously" $0.999....$ (with an infinite number of 9s) is $\sum_{i = 1}^{\infty} \frac 9{10^i}$.

The obvious objection is: does it even make sense to talk about adding an infinite number of terms? How can we talk about taking and adding an infinite number of terms?

And it's a legitimate objection.

So when we learn math in elementary school we are told: Every real number can be written as a decimal expansion (maybe infinite) and every possible decimal expansion is a real number. And this is true. But we are not told why and we are expected to take it on faith, and we usually do.

IF we take this on faith then a proof is very easy:

$0.9999.... = \sum_{i = 1}^{\infty} \frac 9{10^i}$

$10*(0.9999....) = 10*\sum_{i = 1}^{\infty} \frac 9{10^i}= \sum_{i = 1}^{\infty} \frac {90}{10^i}=$

$\sum_{i = 1}^{\infty} \frac 9{10^{i-1}} = 9/10^0 + \sum_{i = 2}^{\infty} \frac 9{10^{i-1}}= 9 + \sum_{i = 1}^{\infty} \frac 9{10^i}$ (Look at the indexes!)

So...

$10*(0.999...) - (0.9999...) = (10 - 1)*0.9999.... = 9*0.99999.... = $

$9 + \sum_{i = 1}^{\infty} \frac 9{10^i} - \sum_{i = 1}^{\infty} \frac 9{10^i} = 9$.

So...

$0.9999.... = 9/9 = 1$.

Easy! !!!!!!!IF!!!!!!! we take it on faith that: Every real number can be written as a decimal expansion (maybe infinite) and every possible decimal expansion is a real number.

So why can we take that on faith? That's the issue: why is that true and what does it mean?

So....

We've got the Integers. We use them to count discrete measurements. We can use an integer to divide a unit 1 into $m$ sub-units to measure measurements of $1/m$. As the $m$ can be as large as we want the $1/m$ can be as precise as we want and the system of all possible $n/m; m \ne 0$ can measure any possible quantity with arbitrary and infinite precision.

We hope. We call these $n/m$ numbers the Rationals and everything is fine until we discover that we can't actually measure measurements such as the square root of two or pi.

But the Rationals still have infinite precision. We can get within 1/10 away from pi. We can get within 1/100 away from pi. Within $1/10^n$ for any possible power of 10.

At this point, we hope we can say "we can't measure it with any finite power of 10 but we can always go one more significant measure, so if we go through infinite powers of 10 we will measure it to precision" and we hope that explanation will be convincing.

But it isn't really. We have these "missing numbers" and we can get infinitely close the them, but what are they really?

Well, we decide to become math majors and in our senior year of college we take a Real Analysis course and we find out.

We can view numbers as sets of rational numbers. We can split the rational numbers at any point into two sets. We can split the rational numbers so that all the rational numbers less than 1/2 are in set A and all the rational numbers greater than or equal to 1/2 are in set B (which we ignore; we are only interested in set A.)

These "cuts" can occur at any point but they must follow the following rules:

--the set A of all the smaller rational numbers is not empty. Nor does it contain every rational number. Some rational number is not in it.

--if any rational number (call it q) is in A, then every rational number smaller than q is also in A. (This means that if r is a rational not in A, then every rational bigger than r is also not in A.)

-- A does not have a single largest element. (So it can be all the elements less than 1/2 but it can't be all the elements less than or equal to 1/2).

And we let $\overline R$ be the collection of all possible ways to "cut" the rational numbers in half that way.

Notice sometimes the cut will occur at a rational number (all the rationals less than 1/2), but sometimes it will occur at points "between" the rational numbers. (All the rationals whose squares are less than 2). So the collection $\overline R$ is a larger set than the set of Rational numbers.

It turns out we can define the Real numbers as the points of $\overline R$ where we can cut the rationals in two.

We need to do a bit or work to show that this is actually a number system. We say $x, y \in \overline R; x < y$ if the "Set A made by cutting at x" $\subset$ "Set A made by cutting at y". And we say $x + y = $ the point where we need to cut so that the set A created contains all the sums of the two other sets created by cutting at x and y. And we have to prove math works on $\overline R$. But we can do it. And we do.

But as a consequence we see that every real number is the least upper bound limit of a sequence of rational numbers. That's pretty much the definition of what a "cut point" is; the point that separate all the rationals less than it from all the other rationals.

I like to say (somewhat trivially) that: the real number $x$ is the least upper bound of all the rational numbers that are smaller than $x$. And it's true!

In the real numbers, every real number is the limit of some sequence of rational numbers. And every bounded sequence of rational numbers will have a real number least upper bound limit.

...

Let that sink in for a minute.

=====

Okay, so given a sequence {3, 3.1, 3.14, 3.141,....} = {finite decimals that are less than pi} is a bounded sequence of rational numbers so $\pi = $ the limit of the sequence which is also the limit of the infinite sequence 3.1415926....

It now makes sense to talk of $0.9999.... = \sum_{i=1}^{\infty}9/10^i = \lim\{\sum_{i=1}^n9/10^i\}$ = a precise and real number.

And from there we can say with confidence that that number is $1$. (By any of these proofs.)

Answer 13

Indeed this is true. The underlying reason is that decimal numbers are not unique representations of the reals. (Technically, there does exist a bijection between the set of all decimal numbers and the reals, but it is not the natural/obvious one.)

Here's a very simple proof:

$$\begin{align} \frac13&=0.333\ldots&\hbox{(by long division)}\\ \implies0.333\ldots\times3&=0.999\ldots&\hbox{(multiplying each digit by $3$)} \end{align}$$

Then we already know $0.333\ldots\times3=1$ therefore $0.999\ldots=1$.

Answer 14

There are some situations in which something like $0.99999\ldots < 1$ indeed holds. Here is one coming from social choice theory.

Let $w_1>w_2>\ldots$ be an infinite sequence of positive numbers, and let $T$ be a number in the range $(0,\sum_i w_i)$. Pick an index $i$. Choose a random permutation $\pi$ of the positive integers, and consider the running totals $$ w_{\pi(1)}, w_{\pi(1)} + w_{\pi(2)}, w_{\pi(1)} + w_{\pi(2)} + w_{\pi(3)}, \cdots $$ The Shapley value $\varphi_i(T)$ is the probability that the first time that the running total exceeds $T$ is when $w_i$ is added.

We will particularly be interested in the case in which the sequence $w_i$ is super-increasing: for each $i$, $w_i \geq \sum_{j=i+1}^\infty w_j$. The simplest case is $w_i = 2^{-i}$. Every number $T \in (0,1)$ can be written in the form $$ T = 2^{-a_0} + 2^{-a_1} + \cdots, \qquad a_0 < a_1 < \cdots. $$ In this case we can give an explicit formula for $\varphi_i(T)$: $$ \varphi_i(T) = \begin{cases} \sum_{t\colon a_t>i} \frac{1}{a_t \binom{a_t-1}{t}} & \text{if } i \notin \{a_0,a_1,\ldots\}, \\ \frac{1}{a_s \binom{a_s-1}{s}} - \sum_{t\colon a_t>i} \frac{1}{a_t \binom{a_t-1}{t-1}} & \text{if } i = a_s. \end{cases} $$

The first two functions are plotted here:

What happens for different sets of weights? The same formula applies, for $$ T = w_{a_0} + w_{a_1} + \cdots, \qquad a_0 < a_1 < \cdots. $$ In general not all $T$ will be of this form; for $T$ not of this form, we take the lowest upper bound which is of this form. What we get for $w_i = 3^{-i}$ is:

Notice all the horizontal parts, for example the blue line at $y=1$ at $x \in (1/6,1/3)$. Where does this stem from? Note that $1/3 = 3^{-1} = w_1$, whereas $1/6 = \sum_{i=2}^\infty 3^{-i} = \sum_{i=2}^\infty w_i$. If we substitute $w_i = 2^{-i}$, then $1/3$ corresponds to $0.1$ (in binary), whereas $1/6$ corresponds to $0.011111\ldots$. So in this case there is a (visible) gap between $0.011111\ldots$ and $0.1$!

For more, take a look at this question and this manuscript.

Answer 15

You can visualise it by thinking about it in infinitesimals. The more $9's$ you have on the end of $0.999$, the closer you get to $1$. When you add an infinite number of $9's$ to the decimal expansion, you are infinitely close to $1$ (or an infinitesimal distance away).

And this isn't a rigorous proof, just an aid to visualisation of the result.

Answer 16

Another approach is the following: $$\begin{align} 0.\overline9 &=\lim_{n \to \infty} 0.\underbrace{99\dots 9}_{n\text{ times}} \\ &= \lim_{n \to \infty} \sum\limits_{k=1}^n \frac{9}{10^k} \\ &=\lim_{n \to \infty} 1-\frac{1}{10^n} \\ &=1-\lim_{n \to \infty} \frac{1}{10^n} \\ &=1. \end{align}$$

Answer 17

Often times people who ask this question are not very convinced by a proof. Since they may not be particularly math inclined, they may feel that a proof is a sort of sleight-of-hand trick, and I find the following intuitive argument (read "don't down-vote me for lack of rigor, lack of rigor is the point") a bit more convincing:

STEP 1) If $.99...\neq1$, everyone agrees that it must be less than $1$. Let $\alpha$ denote $.99...$, this mysterious number less than $1$.

STEP 2) Using a number line, you can convince them that since $\alpha<1$, there must be another number $\beta$ such that $\alpha<\beta<1$.

STEP 3) Since $\alpha<\beta$, one of the digits of $\beta$ must be bigger than the corresponding digit of $\alpha$.

STEP 4) However it is usually intitively clear that you cannot make any digit of $.99...$ bigger without making the resulting number (ie $\beta$) bigger than $1$.

STEP 5) Thus no such $\beta$ can exist, and thus $.99...$ cannot be less than $1$.

Answer 18

The real number system is defined as an extension of the rationals with the property that any sequence with an upper bound has a LEAST upper bound. The expression " 0.9-repeated" is defined to be the least real-number upper bound of the sequence 0.9. 0.99, 0.999,..... , which is 1. The rationals (and the reals) can also be extended to an arithmetic system (an ordered field) in which there are positive values which are less than every positive rational. In such systems the expression ".9-repeated" has no meaning.

Answer 19

Here's my favorite reason why $.999\ldots$ should equal $1$:$^{*}$ \begin{align*} .999\ldots + .999\ldots &= (.9 + .09 + .009 + \cdots) + (.9 + .09 + .009 + \cdots) \\ &= (.9 + .9) + (.09 + .09) + (.009 + .009) + \cdots \\ &= 1.8 + .18 + .018 + .0018 + \cdots \\ &= (1 + .8) + (.1 + .08) + (.01 + .008) + (.001 + .0008) + \cdots \\ &= 1 + (.8 + .1) + (.08 + .01) + (.008 + .001) + \cdots \\ &= 1 + .9 + .09 + .009 + \cdots \\ &= 1 + .999\ldots \end{align*} It follows subtracting $.999\ldots$ from both sides that $.999\ldots = 1$.

The reason I like this explanation best is that addition of (positive) infinite decimal expansions (defined in a particular way) is both commutative and associative even if you insist that $.999\ldots$ and $1$ are different numbers. That is, it forms a commutative monoid. But the cancellation property fails: if $a + b = a + c$, then we can't necessarily conclude $b = c$. The example of this is above, and the most fundamental reason why $.999\ldots = 1$ is arguably so that the cancellation property can hold.

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$^{*}$The calculation given here (using rearrangmenet and regrouping of terms) is informal, and not intended to be a proof, but rather to give some idea of how you can add infinite decimal expansions in the monoid where $.999\ldots \ne 1$. It does end up being true that $.999\ldots + .999\ldots = 1 + .999\ldots$ in this monoid.

Answer 20

One cool way I learned to prove this is that, assuming by $0.99999...$ you mean $0.\bar{9}$. Well we can say that $$0.\bar{9}=\sum_{n=1}^{\infty}9\cdot 10^{-n}=9\sum_{n=1}^{\infty}\frac{1}{10^n}$$ Which we know converges by fact that this is a geometric series with the ratio between terms being less than $1$. So we know that $$9\sum_{n=1}^{\infty}\frac{1}{10^n}=9\left(\frac{1}{1-\frac{1}{10}}-1\right)=10-9=1$$ Note that we subtract off the $1$ in the parentheses because we started indexing at $1$ rather than at $0$, so we have to subtract of the value of the sequence at $n=0$ which is $1$.

Answer 21

Use the Squeeze Theorem:

$$0<1-0.999...=0.1+0.9-0.999...=0.1-0.0999...<0.1=0.1^1$$ $$0<0.1-0.0999...=0.01+0.09-0.0999...=0.01-0.00999...<0.01=0.1^2$$ $$...$$ $$0<1-0.999...<0.1^n$$ $$0\le 1-0.999... \le \lim\limits_{n\to\infty}0.1^n=0.$$

Answer 22

Rather than giving an intuitive explanation as most people have done, let me give a first principles formal proof of this fact. If $\epsilon>0$ and $N=\max\left(\lceil{{log}_{10}\frac{1}{\epsilon}}\rceil+1,1\right)$, then for all natural numbers $n\geq N$,

$$\left |\Sigma_{i=1}^{n}\frac{9}{{10}^i}-1\right|=\frac{1}{{10}^n}\leq\frac{1}{{10}^N}=\frac{1}{{10}^{\max\left(\lceil{{log}_{10}\frac{1}{\epsilon}}\rceil+1,1\right)}}\leq\frac{1}{{10}^{\lceil{{log}_{10}\frac{1}{\epsilon}}\rceil+1}}<\frac{1}{{10}^{\lceil{{log}_{10}\frac{1}{\epsilon}}\rceil}}\leq\frac{1}{{10}^{{log}_{10}\frac{1}{\epsilon}}}=\frac{1}{\left (\frac{1}{\epsilon}\right )}=\epsilon$$

and thus $\left |\Sigma_{i=1}^{n}\frac{9}{{10}^i}-1\right|<\epsilon$. Since $\epsilon$ was arbitrary, it follows that

$$.999\ldots=\Sigma_{i=1}^{\infty}\frac{9}{{10}^i}=\lim_{n\rightarrow\infty}\Sigma_{i=1}^{n}\frac{9}{{10}^i}=1$$

Clear as mud, but this is how you’d prove it in, say, $ZFC$, or the second-order theory of real numbers, from first principles.

Answer 23

If you allow a "decimal representation" of a number to end finally with period $9$ as in $0.\bar9$ or $1.123\bar9$ this "decimal representation "of a number would not be unique.

We know by definition that $0.\bar9=\sum_{n=1}^{\infty}(9/10)^n=1$, but for the sake of the uniquenes of the decimal representation $0.\bar9$ is not a decimal representation of any number.

Answer 24

If we take a version of decimal notation with full complement it is indeed so.

In this system, instead of allowing $0.2$ and $0.1999...$ we restrict decimal notation to use only infinite version $0.19999...$ shortly written as $0.1\overline{9}$

Simply, we do not allow an infinite trail of zeros.

In this system, there is no $0$ written as $0.000...$ instead it is $...999.999...$ or with our succinct notation $\overline{9}.\overline{9}$

Negative numbers are written in complement notation. For example, $...998.999...=\overline{9}8.\overline{9}=-1$

All rules of multiplication addition subtraction are totally valid.

In this system, it is indeed $0.99999...=0.\overline{9}=1$ because we cannot represent $1$ as $1.0000...$ .

Answer 25

[Note: this is my original answer, but completely rewritten to clarify its purpose.]

This answer takes up Trevor Richards' point that people asking this question often aren't convinced by rigorous mathematical proofs and instead feel tricked by them. In this situation one thing that might help is a convincing visible demonstration that $0.999999 . . . =1$ has some chance of being true.

The usual demonstration consists of getting someone to agree that $\frac13=0.33333 . . . $ and then multiply it by $3$ to get $0.99999 . . . $. At this point they might be convinced, but might equally feel puzzled or duped.

This, I think, is where more examples come in. We have to see that $\frac13$ isn't some sort of special case that can be used to trick us.

When I first encountered $0.999999. . .$, I found looking at multiples of $\frac19$ helpful. Once you've convinced yourself that this can be represented by an infinite string of $1$'s, it's easy to see that repeatedly adding it gives $0.222222. . .$, $0.333333. . .$, $0.444444. . .$ all the way up to $0.999999. . .$

There's a complete inevitability about this process, especially if you write it out on paper. But . . . maybe it's still just a trick with a repeating digit?

OK then: let's try multiples of $\frac17=0.142857 . . . $. This one's fun because of the way the cycle of digits behaves:

$\frac17=0.142857 . . .$
$\frac27=0.285714 . . .$
$\frac37=0.428571 . . .$

and the pattern continues nicely, and soon it's "obvious" that the digits will just keep rotating round. But then, all of a sudden, they don't:

$\frac67=0.857142 . . .$
$\frac77=0. 999999 . . .$

—There it is again!

We can try with other fractions too, like $\frac{1}{13}$ and $\frac{1}{37}$, that recur after a manageable number of digits. Always we end up at $0.999999 . . .$.

At this stage, it should seem clear (but not formally proved) that accepting the idea of infinitely recurring decimals entails accepting that $0.999999. . . =1$.

The remaining issue, of course, is the acceptance of infinitely recurring decimals. That's addressed in other answers.

Answer 26

Let $R$ be any ring containing an element $x$ such that $1-10x=0$. Suppose further that $R$ contains the formal power series $\sum_{i=1}^\infty x^i$.

Formally this means there is a ring homomorphism $f\colon S\to R$, from a subring $S\subseteq\mathbb{Z}[[t]]$ containing $\sum_{i=1}^\infty t^i$, such that $f(t)=x$.

Then: $$1=\sum_{i=1}^\infty 9x^i.$$

Proof: We have $$0=(1-10x)\left(1+\sum_{i=1}^\infty x^i\right)=1-\sum_{i=1}^\infty 9x^i.$$

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Apology: I looked through the previous $31$ answers and did not see any that made clear that this result can be proved algebraically (independently of any topology, analysis or order structure). I think this is a legitimate contribution to the discussion, as once someone has understood why the identity follows from the definitions, they tend to wonder if they could not have defined the reals "better", to avoid it (or perhaps that is what they were wondering in the first place, without realising).

Answer 27

We can do this by computing the square root of $0.99...$ by long division method.

Now, we will firstly get a zero and decimal point in the quotient as $0<0.99...\leq 1$. Then $9$ as the divisor for the first pair of $9$'s, then $189$ for the next pair of $9$'s, $1989$ further and so on($19989$, $199989$, ...). The quotient is appended with a $9$ at each step and we can see a pattern in the dividends, divisors and the remainders so as to assure that the quotient will have only $9$'s after the decimal point. Though I'm not sure how to prove that there will be no digit except $9$, maybe observing the pattern is enough. (Inputs on this proof will be appreciated.)

Further we get $\sqrt{0.999...}=0.999...$, moreover $0<0.99...\leq 1$.

Now, if $x\in\mathbb{R},\ x=\sqrt{x} \implies x =0\ \text{or}\ x=1$

Thus we get $0.999...=1$

Answer 28

Both proposed number expressions represent the same Dedekind cut of the set of rational numbers, i.e. the same real number.

The answer is "true".

Answer 29

One missing link in other answers.

If $0.9999....$ and $1$ are to represent real numbers then they have to follow the properties of the set of real numbers.

One of them is saying that real numbers are densely ordered, meaning there is always another real number between two different real numbers.

If $0.9999...$ and $1$ are different, there would have to be another real number between the two. However you cannot change any of the digits of either of the two to record this number.

Because of that, $0.9999...$ and $1$ must be considered the same element of the set of real numbers.

Answer 30

The more 9's there are in the sequence, the smaller the margin between 1 and that number becomes, and after every 9, the margin become 10 time smaller. For example, for the number 0.9, it is ${1 \over \mathrm{10}^{1}}$ away from reaching 1, 0.99 is ${1 \over \mathrm{10}^{2}}$ away from reaching 1, and 0.999 is ${1 \over \mathrm{10}^{3}}$ away from reaching 1. So if $0.99999999 \dots$ has an $n$ number of 9's in it, the 'spacing' between that number and 1 is ${1 \over \mathrm{10}^{n}}$. Now, since $0.99999999 \dots$ has an infinite number of 9's, it makes sense that the 'space' between $0.9999999 \dots$ and 1 is ${1 \over \mathrm{10}^{\infty}}$, or $\mathrm{10}^{-\infty}$, which is 0. Now, to prove that $\mathrm{10}^{-\infty}$ is 0. Simply take the log base 10 of both sides, and you'll find that $ \log_{10} 0 = -\infty$. This means that the 'spacing' between $ 0.9999\dots $ and 1 is 0, meaning that $0.99999\dots + 0 = 1$.

Therefore, $0.9999\dots = 1$