When writing recurring decimal to fraction why do we move the decimal place equal to the number of digits in the repeating string? For example, to write $0.10\overline{4357}$ as a fraction, we let $x= 0.10\overline{4357}$, we then multiply both sides by $10^4$ because that's the number of digits in the repeating string ($4357$) to get $10^4x=1043.574357...$, we then subtract $x$ from $10^{4}x$ and $0.1043574357....$ to get $9999x = 1043.47$, we then multiply both sides by $100$ to get $999900x=104347$, then divide to get $x = 104347/999900.$ So, why does this work?

  • $\begingroup$ Simply because using that multiplier lines up one of the decimals to the other so that all decimals beyond some place cancel out on subtraction. Then what's left is a terminating decimal which is easy to make into a rational. $\endgroup$
    – coffeemath
    Sep 27, 2022 at 4:18
  • 2
    $\begingroup$ If it's easier for you, notice that the decimal repeats after 6 digits, so multiply by $10^6$, we get $104,357.4357...$ now you can subtract off $10^2 \cdot 0.104357... = 10.4357...$ and we have $999,900x = 104,347$ $\endgroup$
    – Igor
    Sep 27, 2022 at 5:56
  • $\begingroup$ Why not multpily with $100$ first (the number of decimal digits after the comma and before the period) ? Then , you have $100x=10+\frac{4357}{9999}=\frac{104347}{9999}$ $\endgroup$
    – Peter
    Sep 27, 2022 at 7:21

1 Answer 1


$$0.4357435743574357\cdots = \frac{4357}{9999}. \tag1 $$

Everything in the posted question can be inferred from (1) above. So, the question reduces to determining why (1) above is true.

(1) above may be re-stated as

$$\frac{4357}{10000} \times \left[1 + \frac{1}{(10000)^1} + \frac{1}{(10000)^2} + \frac{1}{(10000)^3} + \cdots \right] $$

$$= \frac{4357}{10000} \times \frac{1}{1 - \frac{1}{10000}} = \frac{4357}{10000} \times \frac{10000}{9999} = \frac{4357}{9999}. \tag2 $$

In order to explain why (2) above is true, you have to understand that it is part of the following general formula for geometric series:

To simplify the situation, if $0 < t < 1$, then

$$1 + t + t^2 + t^3 + \cdots = \frac{1}{1 - t}. \tag3 $$

That is, (2) is an immediate consequence of $(3)$, with $t$ set to the value $~\dfrac{1}{10000}.$

So, the posted question reduces to explaining why (3) is true, for any value of $t$ such that $0 < t < 1.$

This particular question has been beaten to death, many times on MathSE. I presume that you can find pertinent MathSE articles if you search on the string : "geometric series".

Alternatively, see the Wikipedia article on Geometric Series.

The Wikipedia article may be glossing over the fact that for any fixed constant $t$ such that $0 < t < 1,$ you have that

$$\lim_{n\to\infty} t^n = 0. \tag4 $$

This is easily proven by realizing that since $0 < t < 1,$ you must have that $\log(t)$ is some fixed negative number. Therefore, $\displaystyle \log\left[t^n\right] = n \times $ some fixed negative number.

Therefore, the limit as $n \to \infty$ of $\log\left[t^n\right]$ must be $-\infty$.

This demonstrates that $$\lim_{n\to \infty} t^n = 0.$$


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