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.x1x2x3...xnl=0\red{.}\overline{x_1x_2x_3...x_n} 10nl=x1x2x3...xn.x1x2x3...xn10^nl=x_1x_2x_3...x_n\red{.}\overline{x_1x_2x_3...x_n} 10nll=x1x2x3...xn\Rightarrow 10^nl-l={x_1x_2x_3...x_n} (10n1)l=x1x2x3...xn(10^n-1)l=x_1x_2x_3...x_n l=x1x2x3...xn10n1{l=\dfrac{x_1x_2x_3...x_n}{10^n-1}} 0.x1x2x3...xn=x1x2x3...xn10n1\boxed{0\red{.}\overline{x_1x_2x_3...x_n}=\dfrac{x_1x_2x_3...x_n}{10^n-1}}

a1a2a3...ap.b1b2b3...bqx1x2x3...xn=110q(10q×a1a2a3...ap+b1b2b3...bq+x1x2x3...xn10n1)\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 a1a2a3...ap.b1b2b3...bqx1x2x3...xna_1a_2a_3...a_p\red{.}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n} a1a2a3...ap.b1b2b3...bqx1x2x3...xn=a1a2a3...ap+0.b1b2b3...bqx1x2x3...xna_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} =110q(10q×a1a2a3...ap+b1b2b3...bq.x1x2x3...xn)=\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}) =110q(10q×a1a2a3...ap+b1b2b3...bq+0.x1x2x3...xn)=\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}) =110q(10q×a1a2a3...ap+b1b2b3...bq+x1x2x3...xn10n1)=\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 :

  • x1x2x_1x_2 act as number with digits x1,x2x_1,x_2 for example if x1=5x_1=5 and x2=8x1x2=58x_2=8\Rightarrow x_1x_2=58 dont confuse (x1x2=x1×x2x_1x_2\cancel{=}x_1\times x_2), same for x1x2x3x_1x_2x_3 and x1x2x3...xn1xnx_1x_2x_3...x_{n-1}x_n

  • 0.a=0.aaaaa...0\red{.}\overline{a}=0\red{.}aaaaa...

  • Inspired by these note: Recurring Proof Note 1, Recurring Proof Note 2, Recurring Proof Note 3, Recurring Proof Note 4, Recurring Proof Note 5

Note by Zakir Husain
1 day, 18 hours ago

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@Yajat Shamji

Zakir Husain - 1 day, 17 hours ago

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Well.. I am impressed.

Yajat Shamji - 1 day, 2 hours ago

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✨ brilliant +1

Jeff Giff - 1 day, 2 hours ago

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By the way, I have an interesting #Geometry problem!
Given points A,B,C,DA,B,C,D, find the square PQRS\square PQRS with A on PQ, B on QR, C on RS, D on SP.
I figured out the first part, where we can construct circles with diameters AB, BC, CD, DA respectively, so if a point W is on arc AB, AWB=90.\angle AWB=90^\circ.

Jeff Giff - 1 day, 2 hours ago

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I tried the problem and got an algorithm to construct a rectangle PQRS PQRS with points A,B,CA,B,C and DD on sides PQ,QR,RS,SPPQ,QR,RS,SP respectively. Also there will be infinitely many such rectangles for given points A,B,C,DA,B,C,D

Zakir Husain - 1 day, 1 hour ago

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What about a square?

Jeff Giff - 1 day, 1 hour ago

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I will try it also! and will inform you as I get any results.

Zakir Husain - 1 day, 1 hour ago

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@Zakir Husain Let’s start a discussion! That might help :)

Jeff Giff - 6 hours ago

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square is also a rectangle..

Kriti Kamal - 7 hours ago

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@Kriti Kamal But a rectangle isn’t a square, so I hope to find an algorithm to construct a square (I know it is possible but I don’t know a specific way to do it except for brute-force :P) :)

Jeff Giff - 6 hours ago

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