1. PrasadArithmetic Operations1 VEDIC MATHEMATICS : Arithmetic Operations T. K. Prasad http://www.cs.wright.edu/~tkprasad

2. PrasadArithmetic Operations2 Positional Number System TEN- THOUSANDS THOUSANDSHUNDREDSTENSUNITS 43210 = 4 * 10,000 + 3 * 1,000 + 2 * 100 + 1 * 10 + 0

3. PrasadArithmetic Operations3 Adding Numbers Unary System !!! + !!!!!!!! = !!!!!!!!!!! Roman System III + VIII = XI Too laborious and unreadable! Not uniform, Not incremental!

4. PrasadArithmetic Operations4 Arabic System 45 + 31 = 4 * 10 + 5 + 3 * 10 + 1 = 7 * 10 + 6 = 76

5. PrasadArithmetic Operations5 Arabic System (introducing carry) 45 + 36 = 4 * 10 + 5 + 3 * 10 + 6 = 7 * 10 + 11 = 7 * 10 + 1 * 10 + 1 = 8 * 10 + 1 = 81

6. PrasadArithmetic Operations6 Subtracting Numbers Unary System !!!!!!!!!!! - !!!! = !!!!!!! Roman System XI - IV = VII

7. PrasadArithmetic Operations7 Arabic System 45 – 31 = 4 * 10 + 5 – [3 * 10 + 1] = [4 – 3] *10 + [5 – 1] = 1 * 10 + 4 = 14

8. PrasadArithmetic Operations8 Arabic System (introducing borrow) 65 - 36 = 6 * 10 + 5 – [ 3 * 10 + 6 ] = [6 – 3] * 10 + [5 – 6] = 5 * 10 + 1 * 10 + 5 – [ 3 * 10 + 6 ] = [5 – 3] * 10 + [10 + 5 – 6] = 2 * 10 + [15 – 6] = 29

9. PrasadArithmetic Operations9 Recap: Positional Number System TEN- THOUSANDS THOUSANDSHUNDREDSTENSUNITS 43210 = 4 * 10,000 + 3 * 1,000 + 2 * 100 + 1 * 10 + 0

10. PrasadArithmetic Operations10 Prerequisite / Background Single Digit Addition E.g., 4 + 5 = 9 –With carry E.g., 4 + 8 = 12 Single Digit Subtraction E.g., 9  5 = 4 –With borrow E.g., 12  04 = 08

11. PrasadArithmetic Operations11 Multiplying Single Digit Numbers 1 * 2 = 2 1 x 2 = 2 3 * 4 = 4 * 3 = 12 5 * 5 = 25

12. PrasadArithmetic Operations12 Single Digit Multiplication (of Large Digits in terms of Small Digits) using Vedic Approach 1.Method : Vertically and Crosswise Sutra 2.Correctness and Applicability

13. PrasadArithmetic Operations13 10’s Complement TC(1) = 9 TC(3) = 7 TC(4) = 6 TC(6) = 4 TC(8) = 2 TC(9) = 1 10’s complement of a digit d is (10 – d).

14. PrasadArithmetic Operations14 Method: Multiply 7 * 8 Write the first digit to be multiplied and its 10’s complement in the first row, and the second digit to be multiplied and its 10’s complement in the second row. 7 3 8 2

15. PrasadArithmetic Operations15 7 3 8 2 To determine the 2-digit product: –subtract crosswise to obtain the left digit (7 – 2) = (8 – 3) = 5 –and –multiply the complements vertically to obtain the right digit. (3 * 2) = 6 7 * 8 = 56

16. PrasadArithmetic Operations16 Another Example 8 * 9 = 8 2 9 1 7 2 8 * 9 = 72

17. PrasadArithmetic Operations17 Questions Why do both crosswise subtractions yield the same result? Why does this method yield the correct answer for this example? Does this method always work for any pair of digits?

18. PrasadArithmetic Operations18 Proof Sketch (8 – 3) = (7 – 2) = 5 Why are they same? That is, the difference between first digit and 10’s complement of the second digit. (8 – (10 – 7)) = (8 + 7 – 10) = (15 – 10) = 5 (7 – (10 – 8)) = (7 + 8 – 10) = (15 – 10) = 5

19. PrasadArithmetic Operations19 Left digit [Crosswise Subtraction] Left digit [Crosswise Subtraction] Correctness Argument: Two possibilities 8 = (10 – 2) 7 = (10 – 3) 8 * 7 = (10 – 2) * 7 = 10 * 7 – 2 * 7 = 10 * 7 – 2 * (10 – 3) = 10 * 7 – 2 * 10 + (2 * 3) = 10 * (7 – 2) + 6 = 10 * 5 + 6 = 56 8 = (10 – 2) 7 = (10 – 3) 8 * 7 = 8 * (10 – 3) = 8 * 10 – 8 * 3 = 8 * 10 – (10 – 2) * 3 = 8 * 10 – 10 * 3 + (2 * 3) = 10 * (8 – 3) + 6 = 10 * 5 + 6 = 56 Right digit [Vertical Product] Right digit [Vertical Product]

20. PrasadArithmetic Operations20 5 * 8 5 5 8 2 3 10 4 0 Another Example

21. PrasadArithmetic Operations21 3 * 2 3 7 2 8 – 5 56 – 5+5 6 0 6 Pointless in practice Pointless in practice but need proof to feel comfortable! Yet Another Example