So I recently learned that $0{.}9$ repeating is equal to $1$:

$$ x = 0{.}9\ldots\\ 10x = 9{.}9\ldots\\ 9x = 10x - x = 9{.}9\ldots - 0{.}9\ldots = 9\\ x = 9x/9 = 9/9 = 1\\ x = 1 $$

Or a simpler proof:

$$ x = 1/3 = 0{.}3\ldots\\ 3x = 3/3 = 0{.}9\ldots $$

Which I had no trouble stomaching, this also proves that two different numbers can be equal.

Here's my difficulty though, if the repeating decimal $0{.}9\ldots$ is equal to another, what is $0{.}1\ldots$, $0{.}2\ldots$, $0{.}3\ldots$ etc. up until $0{.}8\ldots$ equal to?

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    $\begingroup$ Conventionally this is not showing "two different numbers can be equal" but that one number can have two different representations. The other repeating decimals you present do not have alternative representations; it's only the ones with repeating 9s. $\endgroup$
    – Joffan
    Jan 9, 2015 at 22:54
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    $\begingroup$ I am guessing that your mother calls you something other than "warspyking". Does that prove that two different people can be equal? $\endgroup$
    – WillO
    Jan 9, 2015 at 22:58
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    $\begingroup$ How did you prove they were different? While they may not look the same, they can represent the same value: $.5=\frac{1}{2}=\frac{50}{100}$ and a bunch of other ways to express that value. $\endgroup$
    – JB King
    Jan 9, 2015 at 23:52

4 Answers 4


The idea is the same, just solve for $x=0{.}\overline{8}$: $$10x=8{.}\overline{8}=8+x\implies 9x=8\implies x=\frac 89$$

Surely you can do the other fractions on your own, or just note that $0{.}\overline{1}=\frac19$ by the same argument, and therefore $0{.}\overline{2}=2\cdot\frac19=\frac29$ and so on.

Note that the "simpler" proof relies on the fact that $\frac13=0{.}\overline{3}$. If you don't understand why this is the case, then the proof is not simpler, since it sweeps the difficulty under rug, rather than dissolving it.

  • $\begingroup$ But wouldn't this be a circular proving the .3… is .3… which is .3… etc? Is there no "rounded" version of them through these calculations? $\endgroup$
    – warspyking
    Jan 9, 2015 at 23:01
  • $\begingroup$ What rounded version? Do you expect $0{.}\overline{3}$ to be equal to $0{.}34$ at some point? $\endgroup$
    – Asaf Karagila
    Jan 9, 2015 at 23:05
  • $\begingroup$ Something like that, why isn't it? $\endgroup$
    – warspyking
    Jan 9, 2015 at 23:10
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    $\begingroup$ Because if those were equal, multiply them by $100$ and you get that $34=33{.}\overline{3}$, which is clearly false. $\endgroup$
    – Asaf Karagila
    Jan 9, 2015 at 23:11
  • $\begingroup$ Ah, I never though of it that way. Thanks. $\endgroup$
    – warspyking
    Jan 9, 2015 at 23:15

this also proves that two different numbers can be equal.

The point is that they are NOT different numbers, just two different representations.

You've already said that $0{,}\overline{3}$ (The overline is common notation for a repeating set of digits) is $\frac{1}{3}$.

To find out what the other repeating representations are equal to, use the same procedre, i.e. multiply by $10$ (or more generally $10^\text{length of period}$), subtract and divide.

You'll find that $0{,}\overline{x} = \frac{x}{9}$.


Your espression $0.3333...x$ is meaningless.

You can imitate the proof you gave like this: $$ x = 0.333..., \\ 10x = 3.333... \\ 10x-x = 3.000... = 3 \\ 9x = 3 \\ x = 3/9 = 1/3 $$

Your other questions can be answered similarly. $0.7777... = 7/9$ and so on.


It's just that $1$ has $2$ different radix representations in base $10$: $\{1, 0.\bar9\}$. In fact, all integers have $2$ different radix representations in base $10$. The integer itself, and the previous integer $+ \space0.\bar 9$. $13$ can be written as either $13$ or $12.\bar 9$. These two representations point to the same value.

$0.\bar8$ is a radix representation for the rational $\dfrac89$.

You can show this:

$$0.\bar 8 = 8 \left(0.1 + 0.01 + 0.001 + \dots \right) = 8\sum_{k=1}^\infty \left(\frac{1}{10}\right)^k = \frac{\frac{8}{10}}{1 - \frac{1}{10}} = \frac89$$


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