Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In a manner similar to how the value $1$ can be represented as $0.(9)$ too, are there any other values that exhibit this property when represented in base 10?

share|cite|improve this question
Every number that has a terminating decimal expansion has a decimal expansion with a tail of $9$s. E.g., $0.5= 0.499999\cdots$. – Arturo Magidin Sep 17 '11 at 22:47
You should try to understand limits and infinite series if you want to know what such representations even mean in the first place. – anon Sep 17 '11 at 22:50
only infinitely many... – user12205 Sep 17 '11 at 23:00
Since only examples between $0$ and $1$ have been posted so far: $2100 = 2099.999\ldots$ – Rahul Sep 17 '11 at 23:04
Jeeze, why edit this century old question? and also another question invovling 0.999999999999999.... – Lost1 Nov 21 '13 at 13:20
up vote 10 down vote accepted

Any number that ends in an infinite series of $9$'s is equal to the number changing all the $9$'s to $0$'s and incrementing the previous place by $1$. So $0.5=0.4999\ldots , 0.1328=0.132799999\ldots$ etc.

share|cite|improve this answer

Every repeating decimal can be represented as a fraction.

Example: Represent $0.\overline{25}$ as a fraction.

First, let $x = 0.\overline{25}$.

Next, multiply both sides of the equation by a power of ten to move the decimal place after the first repeat. In this case, I should choose 100 and get $$100x = 25.\overline{25}.$$

Notice that since there are an infinite number of 25's after the decimal place in $0.\overline{25}$, moving two of them in front of the decimal still leaves an infinite number of them after the decimal. That means we can write that as $$100x = 25 + x.$$

Now we can solve this for $x$.

$$ \begin{align*} 99x &= 25\\ x &= \frac{25}{99} \end{align*} $$

So, $x$ is both equal to $0.\overline{25}$ and $\frac{25}{99}$. That must mean $0.\overline{25} = \frac{25}{99}$.

share|cite|improve this answer
This isn't really what my question was about, but now I know where the rules to convert repeating decimal fractions to ordinary ones come from. :) – Paul Manta Sep 17 '11 at 23:01
@Paul, then I guess my answer isn't really about your question either! Hope it helps, in any case. – The Chaz 2.0 Sep 17 '11 at 23:03
@Paul I interpreted your question as "Are there any other repeating decimals that can be represented as fractions?" Was that not what you intended? – Austin Mohr Sep 17 '11 at 23:06
@Austin I know all rational fractions can be represented as both decimal and ordinary fractions. (I also know that if an ordinary cannot be precisely represented as a decimal, that is a limitation of the number base, not a property of the value.) My question was about numbers that, from an infinitezimal point of view, should not represent the same value, but which in actuality they do. – Paul Manta Sep 17 '11 at 23:13

To generalize Austin's answer (and since I don't know what "this property" is, exactly):

The number $0.\overline{a_1a_2...a_n}$ is equal to $ \frac{ a_1a_2 ... a_n}{99\dots9}$, where there are $n$ nines in the denominator.

So $0. \overline{23} = 23/99$

$0. \overline{123} = 123/999 = 61/333$

share|cite|improve this answer
And I can't, for the life of me, figure out how to fix that LaTex. I intended for there to be 99....9 in the denominator (which could be more precisely represented as nine times a sum of powers of ten...) – The Chaz 2.0 Sep 17 '11 at 23:00
Or a power of ten minus one. – anon Sep 17 '11 at 23:05
Thanks, Brian! And @anon, let's not get into another discussion a lá :) – The Chaz 2.0 Sep 17 '11 at 23:12
The number $0.\overline{a_{p-1}a_{p-2}\ldots a_{1}a_{0}}$ is the sum of a geometric series $$0.\overline{a_{p-1}a_{p-2}\ldots a_{1}a_{0}}=\dfrac{N}{10^{p}}+\dfrac{N}{10^{2p}}+\cdots =\dfrac{N/10^{p}}{1-10^{-p}}=\dfrac{N}{10^{p}-1},$$ where $$N=10^{0}a_{0}+10^{1}a_{1}+\cdots +10^{p-1}a_{p-1}.$$ – Américo Tavares Sep 17 '11 at 23:19

I seem to recall reading somewhere (therefore it's true!! (?)) that Johannes Kepler proposed a base-3 numeral system with three digits: $0$, $1$, and $-1$. In that system, the number $1/2$ can be represented in two different ways: $$ 1.,\ -1,\ -1,\ -1,\ \ldots, $$ and $$ 0.,\ 1,\ 1,\ 1, \ \ldots\ . $$ And similarly for every binary rational number (i.e. rational number whose denominator is a power of $2$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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