05.09 वैदिक गणित (09) || भाग || A9 वाली संख्या से भाग करना

05.09 वैदिक गणित (09) || भाग || A9 वाली संख्या से भाग करना 

लेखक

ॐ जितेन्द्र सिंह तोमर

M.A., B.Ed., DNYS, MASSCOM

13/5/10/12/2021

05.09 वैदिक गणित (09) || भाग || A9 वाली संख्या से भाग करना 

Dividing by A9

PART 01

Single digit No.÷ 9

0/A9 = ?

1/19 = ?

1/29 = ?

1/39 = ?

1/49 = ?

1/59 = ?

1/69 = ?

1/79 = ?

1/89 = ?

1/99 = ?

PART 01

Dividing by 19

Algebraic proof :

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

 = ___________ [1 - 1/(a+1)x]-1 

      ( a + 1 ) x 

          1 

 = __________ [1 + 1/(a+1)x +1/(a+1)x2+ ----------]

 ( a + 1 ) x 

 = 1/(a+1)x + 1/(a+1)2

x

2

 +1/(a+1)3

x

3

+ ---- ad infinitum

 = 10-1(1/(a+1))+10-2(1/(a+1)2)+10-3(1/(a+1)3) + ---ad infinitum

 This series explains the process of ekadhik. 

 Now consider the problem of 1 / 19. From above we get 

 1 / 19 = 10-1 (1/(1+1)) + 10-2 (1/(1+1)2

) + 10-3 (1/(1+1)3

) + ---- 

 ( since a=1) 

 = 10-1 (1/2) + 10-2 (1/2)2

 + 10-3 (1/3)3

 + ---------- 

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

 = 0.05 + 0.0025 + 0.000125 + 0.00000625+ - - - - - - - 

 = 0.052631

PART 02

Double digit No.÷ 19

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. 

D₁D₁

Hence Ekadhikena purva (one more than the previous) is 1 + 1 = 2 or 1+19=20. 

The sutra is applied in two different context. 

a) by Division Method

b) by Multiplication Method

Now the method of division is as follows: 

A) DIVISION METHOD

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.,, 0.05263157₁8 

(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.,, 0.0526315789₁4 

(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 

1/19 = 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 × 2 =12, + 1 [carry over]  = 13, 1 carried over, 3 put to left ) 

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

 = 7, put to left)

Step. 8 : ₁47368421 

(multiply 7 by 2 =₁4, 1 carried over, 4 put to left) 

Step. 9 : 947368421 

( 4 × 2 =8, + 1 [carry over]  = 9 put to left ) 

Step. 10 : ₁8947368421 

(multiply 9 by 2 =₁8, 1 carried over, 8 put to left) 

Step. 11 : ₁78947368421 

( 8 × 2 = ₁6, + 1 [carry over]  = 16 + 1 = ₁7,  1 carried over, 7 put to left)  

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 1st digit + 10th digit = 1 + 8 = 9

Sum of 2nd digit + 11th 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. 

Example 2: Find 1 / 39 by Ekadhika process. 

 

Now by multiplication method, Ekadhikena purva is 3 + 1 = 4 

1/39 = 1/40

 

1 / 39 = ---------------------1 

 = -------------------------------------41 

 = ----------------------------------₁641 

 = ---------------------------------₂5641 

 = --------------------------------₂25641 

 = -------------------------------₁025641 

 

Here the repeating block happens to be block of 6 digits. 

Now the rule predicting the completion of half of the computation does not hold. The complete block has to be computed by ekadhika process. 

 

Now continue and obtain the result. Find reasons for the non-applicability of 

the said ‘rule’.

Example1 :

 1. Find 1 / 49 by ekadhikena process. 

 Now ‘previous’ is 4. ‘One more than the previous’ is 4 + 1 = 5.

Now by division right ward from the left by ‘5’. 

 1 / 49 = .10 - - - - - - - - - - - -(divide 1 by 50) 

 = .02 - - - - - - - - - (divide 2 by 5, 0 times, 2 remainder ) 

 = .0220 - - - - - - --(divide 20 by 5, 4 times) 

 = .0204 - - - - - - -( divide 4 by 5, 0 times, 4 remainder ) 

 = .020440 -- - -- - ( divide 40 by 5, 8 times ) 

 = .020408 - - - - - (divide 8 by 5, 1 time, 3 remainder ) 

 = .02040831 - - - -(divide 31 by 5, 6 times, 1 remainder ) 

 = .02040811 6 - - - - - - - continue 

 = .0204081613322615306111222244448 - -- - - - - 

 On completing 21 digits, we get 48 

i.e.,,Denominator - Numerator = 49 – 1 = 48 stands. 

i.e, half of the process stops here. The remaining half can be obtained as 

complements from 9. 

 . Thus 1 / 49 = 0.020408163265306122448 

 . 979591836734693877551 

 Now finding 1 / 49 by process of multiplication left ward from right by 5, 

we get 

 1 / 49 = ----------------------------------------------1 

 = ---------------------------------------------51 

 = -------------------------------------------2551 

 = ------------------------------------------27551 

 = ---- 483947294594118333617233446943383727551 

 i.e.,,Denominator – Numerator = 49 – 1 = 48 is obtained as 5X9+3

 Carry over ) = 45 + 3 = 48. Hence half of the process is over. The 

remaining half is automatically obtained as complements of 9. 

 Thus 1 / 49 = ---------------979591836734693877551 

 . = 0.020408163265306122448 

 . 979591836734693877551

Find the recurring decimal form of the fractions 1 / 29, 1 / 59, 

1 / 69, 1 / 79, 1 / 89 using Ekadhika process if possible. Judge whether

the rule of completion of half the computation holds good in such 

cases.