विनजीत वैदिक गणित सूत्र-उपसूत्र ज्ञान पुस्तक || 3 || प्रथम सूत्र →एकाधिकेन पूर्वेण

विनजीत वैदिक गणित सूत्र-उपसूत्र ज्ञान पुस्तक ||3|| प्रथम सूत्र  →एकाधिकेन पूर्वेण

लेखक

ॐ जितेन्द्र सिंह तोमर

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

(Specialist in Basic and Vedic Maths)

1.    Ekadhikena Purvena 

        (एकाधिकेन पूर्वेण)

पहले से एक अधिक के द्वारा।

‘By one more than the previous one’

Corollary: Anurupyena

Ekadhikena Purvena (एकाधिकेन पूर्वेण)

The Sutra means: “By one more than the previous one”.

पहले से एक अधिक के द्वारा।

Addition of two (or more) numbers using Sūtra 1. ekādhikena pūrveṇa:
As an illustration, let us use this Sūtra for:
28 + 165

	Steps		28 + 165
	1.	Equate the digits of both the numbers by prefixing imaginary Zeroes in front, and start counting upwards.	In this case,   0 2 8 + 1 6 5 --------
	2.	While summing the digits, place a dot on the previous digit for whichever digit caused the count greater than 10, and continue counting by removing the Ten's place.	In this case,   0 2o8 + 1 6 5 --------       3
	3.	While counting consider the dots in the column as an ekādhika, or +1.	In this case,   0 2o8 + 1 6 5 --------   1 9 3, which is the answer!

So, for a practitioner of Vedic Mathematics, for something like:
234 + 403 + 564 + 721

0 2o3o4   0 4 0 3   0o5 6 4 + 0 7 2 1 ----------   1 9 2 2   Thus, 234 + 403 + 564 + 721 = 1922

Again, for something like:
78924 + 27272 + 72684

0o7o8o9 2o4   0 2 7 2o7 2 + 0 7 2 6 8 4 --------------   1 7 8 8 8 0   Thus, 78924 + 27272 + 72684 = 178880

Subtraction of numbers using Sūtra 1. ekādhikena pūrveṇa:
As an illustration, let us use this Sūtra for:
34 - 18

	Steps		34 - 18
	1.	Equate the digits of both the numbers by considering imaginary Zeroes in front, and start counting upwards.	In this case,   3 4 - 1 8 -------
	2.	While subtracting the digits, place a dot on the previous digit for wherever digit being subtracted from a smaller digit, and sum with the digit's pūrak. A 'pūrak' of a digit is its complementary, 10 - x	In this case,   3 4 - 1o8 ------     6   Note: The pūrak of 8 is (10 - 8) = 2 And, 4 + 2 = 6
	3.	Count the number of dots in each digits, and subtract further to get the answer.	In this case,   3 4 - 1o8 ------   1 6, which is the answer!

So, for a practitioner of Vedic Mathematics, for something like:
5124 - 3608

5 1 2 4 - 3o6 0o8 ----------   1 5 1 6   Thus, 5124 - 3608 = 1516

Again, for something like:
78924 - 27272

7 8 9 2 4 - 2 7 2o7 2 ------------   5 1 6 5 2   Thus, 78924 - 27272 = 51652

Although this method can only subtract smaller numbers from bigger numbers, it can also be used in subtracting bigger numbers from smaller ones because x - y = -(y - x), as for something like:
234 - 7381

7 3 8 1 - 0 2 3o4 ----------   7 1 4 7   Thus, 234 - 7381 = -(7381 - 234) = -7147

 
And, this Sūtra obviously works because it is only a variation of the conventional method of subtraction, by using the 'Dot' as a graphic mnemonic device.

Squaring of numbers ending with 5 using Sūtra 1. ekādhikena pūrveṇa:
We have discussed this technique along with this Sūtra, as it is popularly referred. However, we recommend the corollary (Upasūtra 8. antyayordaśake'pi - discussed here ») for a more generalized approach.
 
As an illustration, let us use this Sūtra for:
15²

	Steps		152
	1.	Remember that square of 5 is 25. This is the 2nd part of the answer.	In this case, the 2nd part is 25 - as in all cases.
	2.	Multiply the 1st part (leaving the last digit) with its ekādhika (1st part + 1).	In this case, The 1st part is 1, and 1 × (1+1) = 1 × 2 = 2
	3.	Join both the results to get the answer.	In this case, Joining 2 and 25, we get 225, which is the answer!

Let us take another example, for something like:
35²

The second part is: 25 Product of 1st part, with its ekādhika: 3 × (3+1) = 3 × 4 = 12   Joining both the results, we get 1225, which is the answer!

So, for a practitioner of Vedic Mathematics, for something like:
9995²

The second part is: 25 Product of 1st part, with its ekādhika: 999 × (999+1) = 999 × 1000 = 999000   Joining both the results, we get 99900025   Thus, 99952 = 99,900,025

And, for something like:
10005²

The second part is: 25 Product of 1st part, with its ekādhika: 1000 × (1000+1) = 1000 × 1001 = 1001000   Joining both the results, we get 100100025   Thus, 100052 = 100,100,025

However, this technique may get cumbersome for bigger numbers - which is why it is discussed for 2-digit and 3-digit numbers. As an example, for something like:
7875²

The second part is: 25 Product of 1st part, with its ekādhika: 787 × (787+1) = 787 × 788 = 620156   Joining both the results, we get 62015625   Thus, 78752 = 62,015,625

But, why does it work? For this Sūtra (ekādhikena pūrveṇa), let us consider the following:

Assuming a number (x + 5) Then, (x + 5)2 = (x + 5) × (x + 5) = x2 + 5x + 5x + 25 = x2 + x(5 + 5) + 25 = x2 + 10x + ab = x (x + 10) + 25   This is exactly what this Sūtra makes us do.   Note: Since the results are 'joined', (x + 10) is always adjusted for (x + 100), which is (x + 1).

1. This sutra is useful to the ‘squaring of numbers ending in 5’.

Example: 25².

For the number 25, the last digit is 5 and the 'previous' digit is 2. According to formula 'One more than the previous one', that is, 2+1=3. The Sutra gives the procedure 'to multiply the previous digit 2 (by one more than itself) by 3. It becomes the L.H.S (left hand side) of the result, that is, 2 X 3 = 6. The R.H.S (right hand side) of the result is 5², that is, 25.

25² = 2 X 3 / 25 = 6/25=625.

15² = 1 X (1+1) /25 =225;

95² = 9 X 10/25 = 9025;

135² = 13 X 14/25 = 18225;

 'अन्त्ययोर्दशकेपि' का 'प्रयोग में' अर्थ है: अंतिम वाले का योग 10 हो या 10^n हो. n प्राकृत संख्या है.

अर्थात् 10 हो, या 100,1000, या 10000 या .....  हो.

जैसे- 56x54, 102x108,157x143 651x649 ,1998x1992,1998x1002  इत्यादि में, क्रमशः 6+4=10; 2+8=10; 57+43=100; 51+49=100; 8+2=10; 998+002=1000  है.

अब इनमें एक और बात पर ध्यान दीजिये,  1998x1992 में 'पूर्व' 199 है और 'अंतिम' में 8 और 2 है, जबकि 1998x1002 में 'पूर्व' 1 है और 'अंतिम' में 998 और 002 है. 

13x17 =

'पूर्व' =1, एकाधिक =2.इस प्रकार 1x2/3x7 =2/21. =221.
अतः 13x17 =221.

इसी तरह से 11x19 = 1x2/1x9 =2/09. ----ध्यान दें कि 1x9=9 आता है जबकि हमें दो अंक चाहिए अतः 09 लिखेंगे।
तब 11x19 =209 होगा।

'आधार' में जितने शून्य (0) हो, दाएं तरफ उसका दोगुना अंक चाहिए। इन दोनों में गुणा में आधार 10 था अतः दो अंकों की आवश्यकता थी.

2. Vulgar fractions whose denominators are numbers ending in Nine

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

 

Division Method:  Value of 1 / 19.

The numbers of decimal places before repetition is the difference of numerator and denominator, 19 -1=18 places. For the denominator 19, the purva (previous) is 1. Hence Ekadhikena purva (one more than the previous) is 1 + 1 = 2.

So that 1/19 => 1/20

Now,

1.      0.₁0 

(Divide numerator 1 by 20, 0 times, 1 remainder)

2.      0.0₀5 

(Divide 10 by 2, 5 times, No remainder)

3.      0.05₁2 

(Divide 5 by 2, result is 2 times, 1 remainder)

4.      0.0526 

(Divide ₁2 is read as 12 by 2, 6 times, No remainder)

5.      0.05263 

(Divide 6 by 2, 3 times, No remainder)

6.      0.05263₁1

(Divide 3 by 2, 1 time, 1 remainder)

7.      0.052631₁5 

(Divide ₁1 or 11 by 25 times, 1 remainder)

8.      0.0526315₁7 

(Divide ₁5 or 15 by 2, 7 times, 1 remainder)

9.      0.05263157₁8 

(Divide ₁7 or 17 by 2, 8 times, 1 remainder)

10.       0.0526315789 

(Divide ₁8 or 18 by 2, 9 times, No remainder)

11.       0.0526315789₁4 

(Divide 9 by 2, 4 times, 1 remainder)

12.       0.052631578947 

(Divide ₁4 or 14 by 2, 7 times, No remainder)

13.       0.052631578947₁3 

(Divide 7 by 2, 3 times, 1 remainder) 

14.       0.0526315789473₁6 

(Divide ₁3 or 13 by 2, 6 times, 1 remainder)

15.       0.052631578947368 

(Divide ₁6 or 16 by 2, 8 times, No remainder)

16.       0.0526315789473684 

(Divide 8 by 2, 4 times, No remainder)

17.       0.05263157894736842 

(Divide 4 by 2, 2 times, No remainder)

18.       0.052631578947368421 

(Divide 2 by 2, 1 time, No remainder)

0 

Multiplication Method: Value of 1 / 19

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 (अंश) as 1 and follow the steps leftwards.

1.      1

2.      21

(multiply 1 by 2, put to left)

3.      421

(multiply 2 by 2, put to left)

4.      8421

(multiply 4 by 2, put to left)

5.      ₁68421 

(multiply 8 by 2=16, 1 carried over, 6 put to left)

6.      ₁368421 

(6 X 2 =12, +1 = 13, 1 carried over, 3 put to left)

7.      7368421 

(3 X 2, = 6 +1 = 7, put to left)

8.      ₁47368421 

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

9.      947368421 

(4 X 2, = 8, +1 = 9, put to left)

10.       ₁8947368421

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

11.       ₁78947368421

(8 X 2 =16,+1=17, 1 carried over, 7 put to left)

12.       ₁578947368421

(7 X 2 =14,+1=15, 1 carried over, 5 put to left)

13.       ₁1578947368421

(5 X 2 =10,+1=11,1 carried over, 1 put to left)

14.       31578947368421

(1 X 2 =2,+1=3, 3 put to left)

15.       631578947368421

(3 X 2 =6, 6 put to left)

16.       ₁2631578947368421

(6 X 2 =12, 1 carried over, 2 put to left)

17.       52631578947368421

(2 X 2 =4,+1=5, 5 put to left)

18.       ₁052631578947368421

(5 X 2 =10, 1 carried over, 0 put to left)

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

Thus 1 / 19 = 0.052631578947368421

 

Example: Value of 1 / 7.

1/7 में हर के अंक को 9 बनाने के लिए हर और अंश में 7 से गुणा करते हैं।

(1/7 = 7/49); हर के 49 का पूर्वेण हैं, 4; जिसका एकाधिक 5 हैं। भिन्न के आवृति दशमलव स्वरूप का अन्तिम अंक 7 होगा तथा 7-1 = 6 अंकों के पश्चात् दशमलव अंकों की पुनरावृति होगी।

1.      7

2.      ₃57 (multiply 7 by 5 =35, 3 carried over, 5 put to left)

3.      ₂857 (5 X 5 =25, +3 = 28, 2 carried over, 8 put to left)

4.      ₄2857 (8 X 5 =40, +2 = 42, 4 carried over, 2 put to left)

5.      ₁42857 (2 X 5 =10, +4 = 14, 1 carried over, 4 put to left)

6.      ₂142857 (4 X 5 =20, +1 = 21, 2 carried over, 1 put to left)

अतः

1/7= 142857.....

Method 1

For example, take ${Displaystyle 1/19}$. In the divisor(19), previous one or the number before 9 is 1. By sutra, Ekaadhika or by adding 1 more to the previous one, we get 2. Lets call the previous one+1 (here 2) as "x". In this method,we start from the end. There will be (divisor-1) terms in the answer. Now,

• Assign last number to be 1. Now,multiply it with "x".ie,

2 1
(1*x)|1

• Now go on multiplying with "x" for (divisor-1)/2 times (here,${\displaystyle (19-1)/2=9}$) ,ie,

Result : 9 4 7 3 6 8 4 2 1
Process:(4*x+1)|(7*x)|(3*x+1)|[6*x+1(carry of last multiplication)]|(8*x)|(4*x)|(2*x)|(1*x)

• In the next step,we write the complement of 9 from the last number onwards,(divisor-1)/2 times

Result  : 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1
Process  : (9-i)|(9-h)|(9-g)|(9-f)|(9-e)|(9-d)|(9-c)|(9-b)|(9-a)|i|h|g|f|e|d|c|b|a

• Now, prefix 0.,and this is your final answer, more accurate than a value that your calculator or computer can give.

${displaystyle 1/19}$ = 0.0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1

Following the same steps,${displaystyle 1/29}$,${displaystyle 1/39}$,${displaystyle 1/49}$..can also be fond in seconds,if you practice it.

For example,${\displaystyle 1/29}$ = 0.0 3 4 4 8 2 7 5 8 6 2 0 6 8 9 6 5 5 1 7 2 4 1 3 7 9 3 1 ${displaystyle    29-1=28}$ terms)

Method 2

Using the same method, we can find ${displaystyle 1/7}$, ${displaystyle 1/13}$ etc. also.ie,

${\displaystyle 1/7}$ = ${\displaystyle (7/7)*(1/7)}$ = ${\displaystyle 7*(1/49)}$, where ${\displaystyle 1/49}$ can be found out by above method.

Also, ${\displaystyle 1/13}$ = ${\displaystyle (3/3)*(1/13)}$ = ${\displaystyle 3*(1/39)}$.

Method 3

${\displaystyle 1/7}$=${\displaystyle 7/49}$ previous digit is 4 so multiply 7 by 4+1 (=5,x) Result : 0 . 1 4 2 8 5 7
Process:0.(4*x+1)|(2*x+4)|(8*x+2)|(5*x+3(carry of last multiplication))|(7*x) (${\displaystyle 7-1=6}$ terms)