Vedic Mathematics: Part II

By Narayana Bhat, PhD

The Sutras or theorems of Vedic Mathematics 

Table 1. The 16 Main Sutras which are explained separately in the following sections.

The sub Sutras or Corollaries

Table 2. The corresponding 14 corollaries for each of the Sutras

Applying the Sutras to Some of the Actual Mathematical calculations

The proposition "by" means the operations this sutra concerns are either multiplication or division. In the case of addition/subtraction proposition "to" or "from" is used.  We cite the examples for only first 2 theorems. This is only to illustrate the power of the theorems.

Ekadhikena Purvena (COROLLARY: Anurupyena) 

"By one more than the one before"

If you want to find the square of 45, you can employ this sutra. The rule says since the first digit is 4 and the second one is 5, you will first have to multiply 4 (4 +1), that is 4 X 5, which is equal to 20 and then multiply 5 with 5, which is 25. Viola! The answer is 2025. Now, you can employ this method to multiply all numbers ending with 5. 

It can also be applied in multiplications when the last digit is not 5 but the sum of the last digits is the base (10) and the previous parts are the same. Consider:       

37 x 33 =  (3x4), 7x3 = 12,21

29 x 21 =  (2x3), 9x1 = 6,09

We illustrate this sutra by its application to conversion of fractions into their equivalent decimal form. Consider fraction 1/19. Using this sutra this can be converted into a decimal form in a single step. This can be done either by applying the sutra for a multiplication operation or for a division operation, thus yielding two methods. 

Method 1: using multiplications

1/19, since 19 is not divisible by 2 or 5, the fractional result is a purely circulating decimal. (If the denominator contains only factors 2 and 5 then it is a purely non-circulating decimal, else it is a mixture of the two.)

So we start with the last digit

            1 Multiply this by "one more", that is, 2 (this is the "key" digit from Ekadhikena) 21 Multiplying 2 by 2, followed by multiplying 4 by 2

            421 => 8421 Now, multiplying 8 by 2, sixteen   68421

              1    <= carry multiplying 6 by 2 is 12 plus 1 carry gives 13

             368421

              1     <= carry

Continuing   7368421 => 47368421 => 947368421

              1

Now we have 9 digits of the answer. There are a total of 18 digits (=denominator-numerator) in the answer computed by complementing the lower half:

             052631578, 947368421

Thus the result is .052631578, 947368421

Method 2: using divisions

The earlier process can also be done using division instead of multiplication. We divide 1 by 2; answer is 0 with remainder 1

            .0     next 10 divided by 2 is five 0.05. Next 5 divided by 2 is 2 with remainder 1

            .052 next 12 (remainder, 2) divided by 2 is 6

            .0526 And so on.

Corollary: A-NURU~PYENA

The upa-Sutra 'anurupyena' means 'proportionality'. This Sutra is highly useful to find products of two numbers when both of them are near the Common bases i.e. powers of base 10. It is very clear that in such cases the expected 'simplicity ' in solving problems is absent.

Example 1.1: 46 X 43

As per the previous methods, if we select 100 as base we get

                    46   -54    This is much more difficult and of no use.

                    43   -57

                   ¯¯¯¯¯¯¯¯

Now by 'anurupyena' we consider a working base in three ways. We can solve the problem. 

Method 1:

 Take the nearest higher multiple of 10. In this case it is 50.

        Treat it as 100 / 2 = 50. Now the steps are as follows: 

1- Choose the working base near to the numbers under consideration.

    i.e., working base is 100 / 2 = 50

2- Write the numbers one below the other  

                               4     6

                               4     3

                           ____________

3- Write the differences of the two numbers respectively from 50 against each number on right side

                    i.e.    46    -04

                             43    -07

                           ¯¯¯¯¯¯¯¯¯

4- Write cross-subtraction or cross- addition as the case may be under the line drawn.

5- Multiply the differences and write the product in the left side of the answer.

                               46    -04

                               43    -07

                           ____________

                            39  / -4 x -7

                                 = 28

6- Since base is 100 / 2 = 50, 39 in the answer represents 39X50.

        Hence divide 39 by 2 because 50 = 100 / 2

Thus 39 ÷ 2 gives 19½ where 19 is quotient and 1 is remainder. This 1 as remainder gives one 50 making the L.H.S of the answer 28 + 50 = 78 (or remainder ½ x 100 + 28)

i.e. R.H.S 19 and L.H.S 78 together give the answer 1978. We represent it as

                    46    -04

                    43    -07

                   ¯¯¯¯¯¯¯¯¯

               2)  39  /   28

                   ¯¯¯¯¯¯¯¯¯                  

                   19½ /  28

                = 19 / 78 = 1978

Example 1.2:  42 X 48.

With 100 / 2 = 50 as working base, the problem is as follows:

                    42    -08

                    48    -02

                   ¯¯¯¯¯¯¯¯¯

                2) 40   /  16

                  ¯¯¯¯¯¯¯¯¯

                    20  /   16

                 42 x 48 = 2016 

Nikhilam Navatashcaramam Dashatah(COROLLARY: Sisyate Sesasamjnah)

"All from nine and the last from ten"

The formula can be very effectively applied in multiplication of numbers, which are nearer to bases like 10, 100, 1000 i.e., to the powers of 10. The procedure of multiplication using the Nikhilaminvolves minimum number of steps, space; it is time saving and involves only mental calculations. The numbers taken can be either less or more than the base considered.

For instance: compute the square of 9. The nearest power of 10 to 9 is 10. Therefore, let us take 10 as our base. Since 9 is 1 less than 10, decrease it still further to 8. This is the left side of our answer. On the right hand side put the square of the deficiency, that is 1^2. Hence the answer is 81. Similarly, 8^2 = 64, 7^2 = 49.

For numbers above 10, instead of looking at the deficit we look at the surplus. For example: 11^2 = 12 1^2 = 121; 12^2 = (12+2) 2^2 = 144 and so on.

If you want to subtract 4679 from 10000, you can easily apply this sutra. Each figure in 4679 is subtracted from 9 and the last figure is subtracted from 10, yielding 5321. Similarly, other sutras lay down such simple rules of calculation.     

Epilogue

Within the Vedas, all disciplines of knowledge transform their identity and get assimilated into the single discipline of organization of knowledge on geometric formats. The knowledge of Mathematics in the Vedas is not confined to what is summarized here.

The Shulba Sutras have preserved only that part of Vedic mathematics which was used for constructing the altars and for computing the calendar to regulate the performance of religious rituals. After the Shulba Sutra period, the main developments in Vedic mathematics arose from needs in the field of astronomy. The Jyotishya, a well known Vedanga, utilizes all branches of mathematics. A close investigation of the Vedic system of mathematics shows that it was much more advanced than the mathematical systems of the civilizations of the Nile or the Euphrates.

By Narayana Bhat, Ph.D.

Narayana Bhat, PhD Mr. Bhat is a member of the Hindu Cultural Center of North Alabama, Huntsville, Alabama. Mr. Bhat will be writing for our Hindu Religion column on regular basis and will choose his topic from readers’ questions. Please send your questions and suggestions to Mr. Bhat at: .(JavaScript must be enabled to view this email address)