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Vedamu.org - Vedic Mathematics - Course

INDEX I. Why Vedic Mathematics? II. Vedic Mathematical Formulae Sutras 1. Ekadhikena Purvena 2. Nikhilam navatascaramam Dasatah 3. Urdhva - tiryagbhyam 4. Paravartya Yojayet 5. Sunyam Samya Samuccaye 6. Anurupye - Sunyamanyat 7. Sankalana - Vyavakalanabhyam 8. Puranapuranabhyam 9. Calana - Kalanabhyam 10. Ekanyunena Purvena Upa - Sutras 1. Anurupyena 2. Adyamadyenantya - mantyena 3. Yavadunam Tavadunikrtya Varganca Yojayet 4. Antyayor Dasakepi 5. Antyayoreva 6. Lopana Sthapanabhyam 7. Vilokanam 8. Gunita Samuccayah : Samuccaya Gunitah III Vedic Mathematics - A briefing 1. Terms and Operations 2. Addition and Subtraction 3. Multiplication 4. Division 5. Miscellaneous Items IV Conclusion

file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20maths/www.vedamu.org/Mathematics/course.html12/22/2005 8:49:34 AM

Vedamu.org - Vedic Mathematics - Why Vedic Mathematics

Vedic Mathematics | Sutras EK The Sutra (formula) Ek one”. DHIKENA P jRVE•A

dhikena P

krvena means: “By one more than the previou

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;

Apply the formula to find the squares of the numbers 15, 45, 85, 125, 175 and verify the answers.

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics

Algebraic proof: a) Consider (ax + b)2 Ξ 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) . 10 2 + 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.10 + b2.102 + a . 103 + b 102 + 52 = a2.104 + (2ab + a).103 + (b2+ b)102 +52 file:///C|/My%20Web%20Sites/vedic%20maths/vedic%20mat...g/Mathematics/MathematicalFormulae/Sutras/sutras.html (2 of 12)12/22/2005 8:49:38 AM

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Vedamu.org - Vedic Mathematics - Why Vedic Mathematics

= [ a2.102 + 2ab.10 + a.10 + b2 + b] 102+ 52 = (10a + b) ( 10a+b+1).102 + 25 = P (P+1) 102 + 25, where P = 10a+b. Hence any three digit number whose last...