Is $.\overline{99}$ equal to 1?

Discussions (sometimes quite colorful) regarding the equation $.\overline{99}=1$ seem to pop up from time to time in certain mathematics newsgroups, including alt.algebra.help. These are some of the more common questions and answers.

(Q1) No matter how many decimal places one cares to expand it, it's always going to be less than 1. You tell me how many digits to carry it out, and I'll tell you what the difference is between that and 1. Doesn't this imply they are not equal?

(A1) If I tell you any particular decimal place to carry out the expansion (no matter how many digits that may be) then true, the result will always be less than 1. However, we wouldn't be talking about $.\overline{99}$ anymore, but rather something else entirely that has the form $.9999\ldots 9$ , where ...could be any finite number of nines (regardless of how many that may be,) followed by a final terminating 9. This is not equivalent to $.\overline{99}$ , which has no such terminating position. The pattern repeats indefinitely, which is the very meaning of the overbar in the expression. The decimal expansion never ends. Realize, if there is to be a ``difference'' between $.\overline{99}$ and 1 then we could equate $.\overline{99}$ to $.9999\ldots 9$ by specifying an appropriate number of digits in $\ldots$ . In other words, it would be a terminating decimal. This is a contradiction, since the expression $.\overline{99}$ by definition (of the overbar) is a repeating decimal.

(Q2) Could it be that a real number between $.\overline{99}$ and 1 does in fact exist, but we just don't know how to properly describe it with decimal notation?

(A2) If such a number exists, it could certainly be described to some extent as a decimal, since all real numbers have decimal expansions that either:

• terminate, e.g. $\frac{1}{2}=.5$ , $\frac{12}{100}=.12$
• do not terminate but contain some pattern that repeats indefinitely, e.g. $\frac{7}{9}=.\overline{77}$ , $\frac{9}{11}=.\overline{81}$
• neither terminate nor contain any repeating pattern, e.g. $\sqrt{2}=1.4142135\ldots$ , $\pi=3.1415926\ldots$ .

Such a number (between $.\overline{99}$ and 1) would clearly have a decimal expansion of the form $.999\ldots(something)$ . First, if it terminated anywhere it would certainly be less than 1 (as desired) but it would also necessarily be less than $.\overline{99}$ , since $.\overline{99}$ is greater than anything of the form $.999\ldots9$ for any number of digits in $\ldots$ , as was explained in (A1). Therefore, such a number would not be between $.\overline{99}$ and 1. Second, if such a number did not terminate but repeated, then it would be of the form $.999\overline{999}$ which is, well, equal to $.\overline{99}$ , hence not between $.\overline{99}$ and 1. Finally, if such a number neither terminated nor repeated, then some of the digits in the decimal expansion must be less than 9, meaning this number would be less than $.\overline{99}$ , hence not between $.\overline{99}$ and 1. Since all three cases have been ruled out, we conclude that no such real number exists that is between $.\overline{99}$ and 1, therefore they must be the same number.

(Q3) Isn't the difference between $.\overline{99}$ and 1 something like $.000\overline{000}0001$ ?

(A3) That is not a valid expression for a real number. The repeating pattern $\overline{000}$ means you would never get to the terminating 1, so if this expression was any number at all--and I'm not saying it is--it would have to be 0 since $.000\overline{000}=0$ . Furthermore, if the difference between two real numbers is 0, then they are in fact the same number $(a=b \iff a-b=b-a=0.)$

(Q4) I have seen some rather simple algebraic proofs that show $.\overline{99}=1$ . Are they valid, and if not where do they break down?

(A4) Although not formal proofs, they are indeed valid demonstrations. The assumptions they rely on can be proven, with appropriate methods, to be valid (including the assumptions made in A1-A3.)

$$x = .\overline{99}$$

$$10x = 10( .\overline{99})$$

$$10x = 9 .\overline{99}$$

$$10x - x = 9 .\overline{99}- .\overline{99}$$

$$9x = 9$$

$$x = 1$$

One assumption made here is that it is indeed legal to move the decimal point over one position when multiplying such a repeating decimal by 10, just like you would a terminating decimal.

Another simple demonstration relies on the assumption that $\frac{1}{3}$ is equal to $.\overline{33}$ . Many who are at first skeptical of $.\overline{99}=1$ seem to have no problem understanding that $.\overline{33}=\frac{1}{3}$ , perhaps because it is easily verified with long division:

So we have $.\overline{33}=\frac{1}{3}$ . Multiplying both sides of this equation by 3 yields $.\overline{99}=1$ . But for that matter, we can simply divide 1 by itself and draw the same conclusion...

Don't be too quick to assume a logical fallacy here. Although it is certainly true that 1 goes into 1.0 1.0 times, it is also true that 1 goes into 1.0 .9 times with a remainder of .1. Then, 1 goes into .10 .09 times with a remainder of .01, and so forth. Of course this is not the standard way long division is usually performed, but it is valid nonetheless.

(Q5) What's all this I hear about series? What does a series have to do with $.\overline{99}$ ?

(A5) Using an infinite sum (a.k.a. series) is a more appropriate method of proving $.\overline{99}=1$ . A series is basically a never ending addition problem, so $.\overline{99}$ can be represented by the series:

$$.9 + .09 + .009 + .0009 + \cdots$$

which is the same as ...

$$\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{9}{10000}+\cdots$$

This expression is a direct result of the very meaning of the base 10 number system that we all learned in school (think place value.) Examining this expression, we can easily see a pattern. Each numerator is 9 and each denominator is a successive power of 10 ( $10^1, 10^2$ , etc.). Using Sigma notation we have:

$$\sum^\infty_{n=1} \frac{9}{10^n}$$

Not all series actually have a sum. Those that do are said to converge while those that do not are said to diverge. Convergent and divergent series are defined as follows:

For the infinite series $\Sigma a_n$ , the $\mathbf{n^{th}}$ partial sum is given by:

$$S_n = a_1 + a_2 + \cdots + a_n$$

If the sequence of partial sums $\lbrace S_n \rbrace$ converges to L, then the series $\Sigma a_n$ converges and L is called the sum of the series. If $\lbrace S_n \rbrace$ diverges, then the series diverges.

In other words, the sum of an infinite series is equal to the limit of its $\mathbf{n^{th}}$ partial sum, if this limit exists (if it does not exist, the series has no sum.) Also, an infinite sequence converges to L if the limit of its $\mathbf{n^{th}}$ term is L. Furthermore, the sum of the series is also equal to L. Using the definition, we can directly prove $.\overline{99}=1$ . Let's take some partial sums:

$$S_1 = \frac{9}{10}$$

$$S_2 = \left(\frac{9}{10}+\frac{9}{100}\right) = \frac{99}{100}$$

$$S_3 = \left(\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}\right) = \frac{999}{1000}$$

$$S_4 = \left(\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{9}{10000}\right) = \frac{9999}{10000}$$

A pattern is easily recognizable that lets us generalize the $\mathbf{n^{th}}$ partial sum. The denominators of the partial sums are successive powers of 10. The numerators are all 1 less than the denominator, so we can write the $\mathbf{n^{th}}$ partial sum as:

$$S_n = \frac{10^n-1}{10^n}$$

According to the definition, the series $\sum^\infty_{n=1}\frac{9}{10^n}$ will converge to $\lim_{n \to \infty} \frac{10^n-1}{10^n}$ , if the limit exists. The limit indeed exists (it's 1, proof omitted) and we conclude that:

$$.\overline{99}= \sum^\infty_{n=1} \frac{9}{10^n} = \lim_{n \to \infty} \frac{10^n-1}{10^n} = 1$$

At this point, it is inevitable that someone will argue something to the effect of, ``just because that limit is 1 doesn't necessarily mean the sum of the series is 1 because it is well known that things don't have to equal their limit value.'' But this line of reasoning is invalid, since the very definition of convergent series tells us this limit is the sum of the series.

Of course, we don't necessarily have to use the definition each and every time we need to find the sum of a series. There are various formulas that we can use to find the sums of certain types of series. In our case, you may recognize $\sum^\infty_{n=1}\frac{9}{10^n}$ as a geometric series with a first term $a$ of $\frac{9}{10}$ and a ratio $r$ of $\frac{1}{10}$ . If $0<\left\lvert r \right\rvert <1$ , a geometric series will converge. Furthermore, it will converge to the sum:

$$\sum^\infty_{n=0}ar^n=\frac{a}{1-r}\qquad 0<\left\lvert r \right\rvert <1$$

Letting $a=\frac{9}{10}, r=\frac{1}{10}$ we have:

$$.\overline{99}=\sum^\infty_{n=0}\bigg(\frac{9}{10}\bigg)\bigg(\frac{1}{10}\bigg)^n=\frac{\frac{9}{10}}{1-\frac{1}{10}}=\frac{9}{10-1}=\frac{9}{9}=1$$

Some Final Thoughts.

Since all repeating decimals are rational numbers, $.\overline{99}$ itself is a rational number. But which rational number is it? Is it $\frac{999}{1000}$ ? Is it $\frac{999999}{1000000}$ ? Is it $\frac{9999999999999}{10000000000000}$ ? It can be none of these, even if you carry the pattern out as far as you like, because they all have decimal representations that terminate. So, how can we easily convert not only $.\overline{99}$ , but any repeating decimal into its fractional form? One commonly used method to accomplish this is to divide the part that repeats (the period) by the same number of nines as there are digits in the period. This will become clear with a few examples, that are easily verifiable with long division:

• $.\overline{33}$ has a period consisting of one digit. Dividing this one digit by one nine yields $\frac{3}{9}$ which simplifies to $\frac{1}{3}$ .
• $.\overline{77}$ has a period consisting of one digit. Dividing this one digit by one nine yields $\frac{7}{9}$ .
• $.\overline{81}$ has a period consisting of two digits. Dividing these two digits by two nines yields $\frac{81}{99}$ which simplifies to $\frac{9}{11}$ .
• $.\overline{857142}$ has a period consisting of 6 digits. Dividing these 6 digits by 6 nines gives $\frac{857142}{999999}$ which simplifies to $\frac{6}{7}$ .

And finally, $.\overline{99}$ has a period consisting of one digit. Dividing this one digit by one nine yields $\frac{9}{9}$ which simplifies to 1.