VEDIC SCIENCE

 

nterest in Vedic maths is growing in the field of education where math’s teachers are looking for a new and better approach to the subject. Even students at IIT (Indian Institute of Technology) are said to be using this ancient technique for quick calculations. No wonder, a recent Convocation speech addressed to the students of IIT, Delhi, Indian Minister for Science & Technology, stressed the significance of Vedic math’s, while pointing out the important contributions of ancient Indian mathematicians, such as Aryabhatta, who laid the foundations of algebra, Baudhayan, the great geometer, and Medhatithi and Madhyatithi, the saint duo, who formulated the basic framework for numerals.

Few of Vedic examples (The Magic of Vedas)

(VEDIC MATH’S FORMULAE):- THERE ARE 16 SUTRA’S


SUTRA MEANING
1 Ekadhikina Purvena By one more than the previous one
COROLLARY: Anurupyena
2 Nikhilam Navatashcaramam Dashatah All from 9 and the last from 10
COROLLARY: Sisyate Sesasamjnah
3 Urdhva-Tiryagbyham Vertically and crosswise
COR: Adyamadyenantyamantyena
4 Paraavartya Yojayet Transpose and adjust
COR: Kevalaih Saptakam Gunyat
5 Shunyam Saamyasamuccaye When the sum is the same that sum is zero.
COROLLARY: Vestanam
6 (Anurupye) Shunyamanyat Sankalana-vyavakalanabhyam
COROLLARY: Yavadunam Tavadunam
7 Sankalana-vyavakalanabhyam By addition and by subtraction
COR: Yavadunam Tavadunikritya Varga Yojayet
8 Puranapuranabyham By the completion or non-completion
COROLLARY: Antyayordashake'pi
9 Chalana-Kalanabyham Differences and Similarities
COROLLARY: Antyayoreva
10 Yaavadunam Whatever the extent of its deficiency
COROLLARY: Samuccayagunitah
11 Vyashtisamanstih Part and Whole
COROLLARY: Lopanasthapanabhyam
12 Shesanyankena Charamena The remainders by the last digit
COROLLARY: Vilokanam
13 Sopaantyadvayamantyam The ultimate and twice the penultimate
COR: Gunitasamuccayah Samuccayagunitah
14 Ekanyunena Purvena By one less than the previous one
COROLLARY: Dhvajanka
15 Gunitasamuchyah The product of the sum is equal to the sum of the product
COROLLARY: Dwandwa Yoga
16 Gunakasamuchyah The factors of the sum is equal to the sum of the factors
COROLLARY: Adyam Antyam Madhyam

 

1 EKADHIKENA PURVENA

The Sutra (formula) Ekadhikena Purvena means: “By one more than the previous one”.
  i) Squares of numbers ending in 5 :

Now we relate the sutra to the ‘squaring of numbers ending in 5’. Consider the example 252.
Here the number is 25. We have to find out the square of the number. For the number 25, the last digit is 5 and the 'previous' digit is 2. Hence, 'one more than the previous one', that is, 2+1=3. The Sutra, in this context, gives the procedure 'to multiply the previous digit 2 by one more than itself, that is, 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.
  Thus: - 252 = 2 X 3 / 25 = 625.
  In the same way,
  352= 3 X (3+1) /25 = 3 X 4/ 25 = 1225;
  652= 6 X 7 / 25 = 4225;
  1052= 10 X 11/25 = 11025;
  1352= 13 X 14/25 = 18225;
  Algebraic proof:
  a) Consider (ax + b)2 ? a2. x2 + 2abx + b2.
  This identity for x = 10 and b = 5 becomes
  (10a + 5)2 = a2 . 102 + 2. 10a . 5 + 52
  = a2 . 102 + a. 102 + 52
  = (a 2+ a ) . 102 + 52
  = a (a + 1) . 102 + 25.
  Clearly 10a + 5 represents two-digit numbers 15, 25, 35, -------,95 for the values a = 1, 2, 3, -------,9 respectively. In such a case the number (10a + 5)2 is of the form whose L.H.S is a (a + 1) and R.H.S is 25, that is, a (a + 1) / 25.
  Thus any such two digit number gives the result in the same fashion.
  Example: 45 = (40 + 5)2, It is of the form (ax+b)2 for a = 4, x=10
and b = 5. giving the answer a (a+1) / 25
that is, 4 (4+1) / 25 + 4 X 5 / 25 = 2025.
  b) Any three digit number is of the form ax2+bx+c for x = 10, a ? 0, a, b, c ? W.
  Now (ax2+bx+ c) 2 = a2 x4 + b2x2 + c2 + 2abx3 + 2bcx + 2cax2
  = a2 x4 +2ab. x3+ (b2 + 2ca)x2+2bc . x+ c2.
  This identity for x = 10, c = 5 becomes (a . 102 + b .10 + 5)2
  = a2.104 + 2.a.b.103 + (b2 + 2.5.a)102 + 2.b.5.10 + 52
  = a2.104 + 2.a.b.103 + (b2 + 10 a)102 + b.102+ 52
  = a2.104 + 2ab.103 + b2.102+ a . 103 + b 102 + 52
  = a2.104+ (2ab + a).103 + (b2+ b)102 +52
  = [ a2.102 + 2ab.10 + a.10 + b2 + b] 102+ 5
  = (10a + b) ( 10a+b+1).102 + 25
  = P (P+1) 102 + 25, where P = 10a+b.
  Hence any three digit number whose last digit is 5 gives the same result as in (a) for P=10a + b, the ‘previous’ of 5.
  Example : 1652 = (1 . 102 + 6 . 10 + 5)2.
  It is of the form (ax2 +bx+c)2 for a = 1, b = 6, c = 5 and x = 10. It gives the answer P(P+1) / 25, where P = 10a + b = 10 X 1 + 6 = 16, the ‘previous’. The answer is 16 (16+1) / 25 = 16 X 17 / 25 = 27225.
  Similar to this other sutra’s has there benefits and mathematic implications to solve the problems.
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