We now take examples of 1 / a9, where a = 1, 2, -----, 9. In the conversion of such vulgar fractions into recurring decimals, Ekadhika process can be effectively used both in division and multiplication.

a) Division Method  :  Value of 1 / 19.

The numbers of decimal places before repetition is the difference of numerator and denominator, i.e.,, 19 -1=18 places.

For the denominator 19, the purva (previous) is 1.

Step. 1 :  Divide numerator 1 by 20.

       i.e.,, 1 / 20  =  0.1 / 2  = .₁0 ( 0 times, 1 remainder)

Step. 2 :  Divide 10  by  2

       i.e.,,  0.0₀5( 5 times, 0 remainder )

Step. 3 : Divide  5  by  2

       i.e.,,  0.05₁2 ( 2 times, 1 remainder )

 Step. 4 : Divide  ₁2  i.e.,, 12  by  2

       i.e.,,  0.0526 ( 6 times, No remainder )

 Step. 5 :    Divide  6  by  2

       i.e.,,   0.05263 ( 3 times, No remainder )

 Step. 6 :    Divide  3  by  2

        i.e.,,  0.05263₁1(1 time, 1 remainder )

 Step. 7 :    Divide  ₁1   i.e.,, 11  by  2

        i.e.,,  0.052631₁5  (5 times, 1 remainder )

 Step. 8 :    Divide  ₁5  i.e.,,  15  by  2

         i.e.,,  0.0526315₁7 ( 7 times, 1 remainder )

 Step. 9 :    Divide ₁7    i.e.,,   17  by  2

         i.e.,,   0.05263157 ₁8 (8 times, 1 remainder )

 Step. 10 :  Divide  ₁8  i.e.,,  18  by  2

         i.e.,,  0.0526315789 (9 times, No remainder )

 Step. 11 :  Divide  9  by  2

         i.e.,,  0.0526315789 ₁4 (4 times, 1 remainder )

 Step. 12 :  Divide  ₁4   i.e.,,  14  by  2

         i.e.,,  0.052631578947 ( 7 times, No remainder )

 Step. 13 :  Divide  7  by  2

          i.e.,,  0.052631578947₁3 ( 3 times, 1 remainder )

 Step. 14 : Divide  ₁3    i.e.,,   13  by  2

          i.e.,,  0.0526315789473₁6 ( 6 times, 1 remainder )

 Step. 15 : Divide  ₁6     i.e.,,   16  by  2

          i.e.,,   0.052631578947368 (8 times, No remainder )

 Step. 16 : Divide  8  by  2

          i.e.,, 0.0526315789473684 ( 4 times, No remainder )

 Step. 17 : Divide  4  by  2

          i.e.,,  0.05263157894736842 ( 2 times, No remainder )

 Step. 18 : Divide  2  by 2

          i.e.,, 0.052631578947368421 ( 1 time, No remainder )

Now from step 19, i.e.,, dividing 1 by 2, Step 2 to Step. 18 repeats thus giving

                0 __________________        .                        .
        1 / 19 = 0.052631578947368421 or 0.052631578947368421

Note that we have completed the process of division only by using ‘2’. Nowhere the division by 19 occurs.

b) Multiplication Method: Value of 1 / 19

First we recognize the last digit of the denominator of the type 1 / a9. Here the last digit is 9.

For a fraction of the form in whose denominator 9 is the last digit, we take the case of 1 / 19 as follows:

For 1 / 19, 'previous' of 19 is 1. And one more than of it is 1 + 1 = 2.

Therefore 2 is the multiplier for the conversion. We write the last digit in the numerator as 1 and follow the steps leftwards.

Step. 1 :                                        1

Step. 2 :                                      21(multiply 1 by 2, put to left)

Step. 3 :                                   421(multiply 2 by 2, put to left)

Step. 4 :                                 8421(multiply 4 by 2, put to left)

Step. 5 :                              ₁68421 (multiply 8 by 2 =16,                                                                                 1 carried over, 6 put to left)

Step. 6 :                            ₁368421 ( 6 X 2 =12,+1 [carry over]

                                                   = 13, 1 carried over, 3 put to left )

Step. 7 :                          7368421 ( 3 X 2, = 6 +1 [Carryover]

                                                  = 7, put to left)

Step. 8 :                        ₁47368421 (as in the same process)

Step. 9 :                       947368421  ( Do – continue to step 18)

Step. 10 :                   ₁8947368421

Step. 11 :                  ₁78947368421

Step. 12 :                 ₁578947368421

Step. 13 :                ₁1578947368421

Step. 14 :                31578947368421

Step. 15 :               631578947368421

Step. 16 :              ₁2631578947368421

Step. 17 :             52631578947368421

Step. 18 :           1052631578947368421

Now from step 18 onwards the same numbers and order towards left continue.

Thus  1 / 19 = 0.052631578947368421

It is interesting to note that we have

    i)  not at all used division process

    ii) instead of dividing 1 by 19 continuously, just multiplied 1 by 2 and continued to multiply the resultant successively by 2.

Observations :

   a) For any fraction of the form 1 / a9 i.e.,, in whose denominator 9 is the digit in the units place and a is the set of remaining digits, the value of the fraction is in recurring decimal form and the repeating block’s right most digit is 1.

   b) Whatever may be a9, and the numerator, it is enough to follow the said process with (a+1) either in division or in multiplication.

   c) Starting from right most digit and counting from the right, we see ( in the given example 1 / 19)

     Sum of 1^st digit + 10^th digit  =  1 + 8 = 9

     Sum of 2^nd digit + 11^th digit  =  2 + 7 = 9

     - - - - - - - - -- - - - - - - - - - - - - - - - - - -
Sum of 9th digit + 18th digit = 9+ 0 = 9

    From the above observations, we conclude that if we find first 9 digits, further digits can be derived as complements of 9.

     i) Thus at the step 8 in division process we have 0.0526315₁7 and next step. 9 gives 0.052631578

        Now the complements of the numbers
                                0, 5, 2, 6, 3, 1, 5, 7, 8 from 9
                                9, 4, 7, 3, 6, 8, 4, 2, 1 follow the right order

                                        i.e.,, 0.052631578947368421

             Now taking the multiplication process we have

        Step. 8 :         ₁47368421

        Step. 9 :         947368421

    Now the complements of 1,  2, 4,  8,  6,  3, 7,  4, 9 from 9
                               i.e.,, 8, 7, 5, 1, 3, 6, 2, 5, 0 precede in successive steps, giving the answer.

                          0.052631578947368421.

d) When we get (Denominator – Numerator) as the product in the multiplicative process, half the work is done. We stop the multiplication there and mechanically write the remaining  half of the answer by merely taking down complements from 9.

e) Either division or multiplication process of giving the answer can be put in a single line form.