# Writing recurring decimals as a fraction

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?

• 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. Sep 27, 2022 at 4:18
• 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 $999,900x = 104,347$
– Igor
Sep 27, 2022 at 5:56
• Why not multpily with $100$ first (the number of decimal digits after the comma and before the period) ? Then , you have $100x=10+\frac{4357}{9999}=\frac{104347}{9999}$ Sep 27, 2022 at 7:21

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

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