1. PrasadArithmetic Operations Revisited1 VEDIC MATHEMATICS : Arithmetic Operations T. K. Prasad http://www.cs.wright.edu/~tkprasad

2. PrasadArithmetic Operations Revisited2 Positional Number System TEN- THOUSANDS THOUSANDSHUNDREDSTENSUNITS 43210 = 4 * 10,000 + 3 * 1,000 + 2 * 100 + 1 * 10 + 0

3. PrasadArithmetic Operations Revisited3 Two Digit Multiplication (above the base) using Vedic Approach 1.Method : Vertically and Crosswise Sutra 2.Correctness and Applicability

4. PrasadArithmetic Operations Revisited4 Method: Multiply 13 * 12 Write the first number to be multiplied and excess over 10 in the first row, and the second number to be multiplied and excess over 10 in the second row. 13 3 12 2

5. PrasadArithmetic Operations Revisited5 13 3 12 2 To determine the 3-digit product: –add crosswise to obtain the left digits (13 + 2) = (12 + 3) = 15 –and –multiply the excess vertically to obtain the right digit. (3 * 2) = 6 13 * 12 = 156

6. PrasadArithmetic Operations Revisited6 Another Example 12 * 14 = 12 2 14 4 16 8 12 * 14 = 168

7. PrasadArithmetic Operations Revisited7 Questions Why do both crosswise additions yield the same result? Why does this method yield the correct answer for this example? Does this method always work for any pair of numbers?

8. PrasadArithmetic Operations Revisited8 Proof Sketch (12 + 4) = (14 + 2) = 16 Why are they same? That is, the sum of first number and excess over 10 of the second number, and …. (12 + (14 – 10)) = (12+14 – 10) = (26 – 10) = 16 (14 + (12 – 10)) = (14+12 – 10) = (26 – 10) = 16

9. PrasadArithmetic Operations Revisited9 Left digits [Crosswise Addition] Left digits [Crosswise Addition] Correctness Argument: Two possibilities 12 = (10 + 2) 14 = (10 + 4) 12 * 14 = (10 + 2) * 14 = 10 * 14 + 2 * 14 = 10 * 14 + 2 * (10 + 4) = 10 * 14 + 2 * 10 + (2 * 4) = 10 * (14+2) + 8 = 10 * 16 + 8 = 168 12 = (10 + 2) 14 = (10 + 4) 12 * 14 = 12 * (10 + 4) = 12 * 10 + 12 * 4 = 12 * 10 + (10 + 2) * 4 = 12 * 10 + 10 * 4 + (2 * 4) = 10 * (12 + 4) + 8 = 10 * 16 + 8 = 168 Right digit [Vertical Product] Right digit [Vertical Product]

10. PrasadArithmetic Operations Revisited10 15 * 12 15 5 12 2 17 10 18 0 Another Example

11. PrasadArithmetic Operations Revisited11 17 * 15 17 7 15 5 22 35 22+3 5 25 5 Need proof to feel comfortable! Yet Another Example

12. PrasadArithmetic Operations Revisited12 Method: Multiply 113 * 106 Write the first number to be multiplied and excess over 100 in the first row, and the second number to be multiplied and excess over 100 in the second row. 113 13 106 6

13. PrasadArithmetic Operations Revisited13 113 13 106 6 To determine the 5-digit product: –add crosswise to obtain the left digits (113 + 6) = (106 + 13) = 119 –and –multiply the excess vertically to obtain the right digits. (13 * 6) = 78 113 * 106 = 11978

14. PrasadArithmetic Operations Revisited14 Questions Why do both crosswise additions yield the same result? Why does this method yield the correct answer for this example? Does this method always work for any pair of 3 digit numbers?

15. PrasadArithmetic Operations Revisited15 Proof Sketch (113 + 6) = (106 + 13) = 119 Why are they same? (113 + (106 – 100)) = (113 + 106 – 100) = 119 (106 + (113 – 100)) = (106 + 113 – 100) = 119

16. PrasadArithmetic Operations Revisited16 Right digits [Vertical Product] Right digits [Vertical Product] Left digits [Crosswise Addition] Left digits [Crosswise Addition] Correctness of Product : Two possibilities 113 = (100 + 13) 106 = (100 + 6) 113 * 106 = 113 * (100 + 6) = 113 * 100 + (100 + 13) * 6 = 113 * 100 + 100 * 6 + (13 * 6) = 100 * (113 + 6) + 78 = 100 * 119 + 78 = 11978 113 = (100 + 13) 106 = (100 + 6) 113 * 106 = (100 + 13) * 106 = 100 * 106 + 13 * (100 + 6) = 100 * 106 + 13 * 100 + (13 * 6) = 100 * (106 + 13) + 78 = 100 * 119 + 78 = 11978

17. PrasadArithmetic Operations Revisited17 160 * 180 160 60 180 80 240 4800 288 00 Note that, the product of the excess over 100 has more than two digits. However, the weight associated with 240 and 48 are both 100, and hence they can be combined. Another Example Breakdown?!

18. PrasadArithmetic Operations Revisited18 190 * 199 190 90 199 99 289 8910 289+89 10 378 10 This approach is valid This approach is valid with suggested modifications! Yet Another Example Breakdown?!

19. More Shortcuts PrasadArithmetic Operations Revisited19

20. Quick squaring of numbers that end in 5 15 * 15 = 225 = (1*2) (5*5) 75 * 75 = 5625 = (7*8) (5*5) 95 * 95 = 9025 = (9*10) (5*5) Proof: Let the two digit number be written as D5. D5 * D5 = (D*10 + 5) * (D*10 + 5) = (D*D*100) + (D*2*50) + 5*5 = (D*(D+1))*100 + 25 PrasadArithmetic Operations Revisited20

21. Quick Multiplication : Special Case Proof: Let two digit numbers be AB and AC. AB * AC = (A*10 + B) * (A*10 + C) = (A*A*100) + (A*10*(B+C)) + B*C = (A*A)*100 + (A)*(B+C)*10 + (B*C) For B+C=10, this reduces to A*(A+1)*100 + B*C For A=12, B=8 and C=2, this reduces to (12)*(13)*100 + 16 = 15616 PrasadArithmetic Operations Revisited21

22. Quicking squaring of numbers that begin with 5 51 * 51 = (5*5+1) *100 + (1*1) = 2601 57 * 57 = (5*5+7) *100 + (7*7) = 3249 59 * 59 = (5*5+9) *100 + (9*9) =3481 Proof: Let the two digit number be written as 5D. 5D * 5D = (50 + D) * (50 + D) = (25 + D)*100 + (D*D) PrasadArithmetic Operations Revisited22

23. Quick squaring of two digit numbers Proof: Let two digit numbers be AB. AB * AB = (A*10 + B) * (A*10 + B) = (A*A)*100 + 2*(A*10)*B + B*B = (A*A)*100 + 20*(A*B) + (B*B) For AB=79, this reduces to 4900+20*63+81 = 4981+1260 =6241 For AB=116, this reduces to 12100+20*66+36 = 12136+1320 =13456 PrasadArithmetic Operations Revisited23

24. Generalized Multplication Using Working Base PrasadArithmetic Operations Revisited24

25. PrasadArithmetic Operations Revisited25 23 +3 24 +4 To determine the product, choose working base as 20: –add crosswise to obtain the left digits with weight 20 (23 + 4) = (24 + 3) = 27 –multiply the excess vertically to obtain the right digits. (3 * 4) = 12 23 * 24 = 27 * 20 + 12 = 540 + 12 23 * 24 = 552

26. 723 +23 724 +24 To determine the product, choose working base as 700: –add crosswise to obtain the left digits with weight 700 (723 + 24) = (724 + 23) = 747 –multiply the excess vertically to obtain the right digits. (23 * 24) = 552 723 * 724 = 747 * 700 + 552 = 522900 + 552 723 * 724 = 523452 PrasadArithmetic Operations Revisited26

27. 783 -17 775 -25 To determine the product, choose working base as 800: –add crosswise to obtain the left digits with weight 800 (783 - 25) = (775 - 17) = 758 –multiply the excess vertically to obtain the right digits. (17 * 25) = 425 783 * 775 = 758 * 800 + 425 = 606400 + 425 783 * 775 = 606825 PrasadArithmetic Operations Revisited27

28. 532 +32 472 -28 To determine the product, choose working base as 1000/2: –add crosswise to obtain the left digits with wt. 1000/2 (532 - 28) = (472 + 32) = 504 –multiply the excess vertically to obtain the right digits. (+32) * (-28) = 896 532 * 472= (504 / 2)*1000 + (104 -1000) = 252000 + 104 - 1000 532 * 472= 251104 PrasadArithmetic Operations Revisited28