Below is a conjecture that I can't prove.
It was inspired by a math trick for division when the (base 10) denominator ends in the digit 9.
Can anyone help me out with a proof?
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Here is an excell example,
Below is a conjecture that I can't prove.
It was inspired by a math trick for division when the (base 10) denominator ends in the digit 9.
Can anyone help me out with a proof?
- - - Updated - - -
Here is an excell example,
Last edited by RLBrown; December 18th, 2016 at 00:15.
These numbers are the same that arise in the standard division algorithm (as taught in elementary school), like so:
\begin{tikzpicture}
\draw (0,0) node {$5\,. {}^50{}^{12}0{}^60{}^30{}^{11}0\ldots$} ;\draw (0,-0.5) node{$0\,. {}^{\phantom5}2 {}^{\phantom{12}}6 {}^{\phantom6}3 {}^{\phantom3}1 {}^{\phantom{11}} 5 \ldots$} ;\draw (-2,-0.05) node {$19$} ;\draw (-1.7,0.3) -- (-1.7,-0.3) -- (1.6,-0.3) ;
\end{tikzpicture}
At each stage of that algorithm, you carry forward a remainder $r_{k-1}$, multiply it by 10 and then divide by $10d-1$, getting a quotient $q_{k-1}$ and a remainder $r_k$, so that $$10r_{k-1} = (10d-1)q_{k-1} + r_k.$$ If you write that as $$r_k = 10(r_{k-1}- dq_{k-1}) + q_{k-1},$$ you see that it is exactly the same as the recurrence relation $$b_k = 10(b_{k-1} - da_{k-1}) + a_{k-1}$$ from the Vedic algorithm. (It also has the same initial condition $r_1 = n$.)
So if you believe the standard division algorithm, then you should also believe the Vedic algorithm.