Algebraic Proof of Finding out the Decimal Expansion of Vulgar Fractions using Ekadhikena Purvena - Suthra

 

Vedic Mathematics

Suthra 1     :        Ekadhikena Purvena

Application : Recurring Decimal Expansion of Vulgar Fractions with denominator ending with ‘9’. (Algebraic Proof)

Meaning    :        By One More than the Previous One

The Algebraic Proof for the same as follows:

Any Vulgar fraction of the form 1/a9

Can be written as

1/a9           =        1/(a+1)x-1) where x=10

                   =                           1              `

                                 (a+1)x[1-1/(a+1)x]

                   =        1/[(a+1)x] [1+1/(a+1)x]^-1

                   =        1/(a+1)x [1+ 1/(a+1)x + 1/(a+1)²x² +…….]

                   =        1/(a+1)x + 1/(a+1)²x²+1/(a+1)³x³+………………….

                   =        10^-1(1/(a+1) + 10^-2(1/(a+1)² + 10^-3 (1/a+1)³ +………

The series explains the process “Ekadhikena….”

Let us consider 1/19

Using the above algebraic proof, we have

1/19                    =        10^-1(1/(1+() + 10^-2(1/(1+1)² + 10^-3(1/(1+1)³+ ………

                   =        10^-1(0.5) + 10^-2 (0.25) + 10^-3 (0.125) +……….

                   =        0.05 + 0.0025 + 0.000125 + 0.00000625 + ……..

                   =        0.052631…….