On Recurring decimals

\[\large{0\red{.}\overline{x_1x_2x_3...x_n}=\dfrac{x_1x_2x_3...x_n}{10^n-1}}\]

Proof of the above statement

Let $l=0\red{.}\overline{x_1x_2x_3...x_n}$ $10^nl=x_1x_2x_3...x_n\red{.}\overline{x_1x_2x_3...x_n}$ $\Rightarrow 10^nl-l={x_1x_2x_3...x_n}$ $(10^n-1)l=x_1x_2x_3...x_n$ ${l=\dfrac{x_1x_2x_3...x_n}{10^n-1}}$ $\boxed{0\red{.}\overline{x_1x_2x_3...x_n}=\dfrac{x_1x_2x_3...x_n}{10^n-1}}$

$\large{a_1a_2a_3...a_p\red{.}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n}=\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_p+b_1b_2b_3...b_q+\dfrac{x_1x_2x_3...x_n}{10^n-1})}$

Proof of the above statement

For any number $a_1a_2a_3...a_p\red{.}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n}$ $a_1a_2a_3...a_p\red{.}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n}=a_1a_2a_3...a_p+0\red{.}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n}$ $=\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_p+b_1b_2b_3...b_q\red{.}\overline{x_1x_2x_3...x_n})$ $=\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_p+b_1b_2b_3...b_q+0\red{.}\overline{x_1x_2x_3...x_n})$ $=\boxed{\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_p+b_1b_2b_3...b_q+\dfrac{x_1x_2x_3...x_n}{10^n-1})}$

Note :

$x_1x_2$ act as number with digits $x_1,x_2$ for example if $x_1=5$ and $x_2=8\Rightarrow x_1x_2=58$ don't confuse ( $x_1x_2\cancel{=}x_1\times x_2$ ), same for $x_1x_2x_3$ and $x_1x_2x_3...x_{n-1}x_n$
$0\red{.}\overline{a}=0\red{.}aaaaa...$ ( $\infty$ times)
Inspired by these note: Recurring Proof Note 1, Recurring Proof Note 2, Recurring Proof Note 3, Recurring Proof Note 4, Recurring Proof Note 5

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On Recurring decimals

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