APPLICATIONS OF VEDIC MATHS

1. Straight Squaring:

We have already noticed methods useful to find out squares of numbers. But the methods are useful under some situations and conditions only. Now we go to a more general formula.

The sutra Dwandwa-yoga (Duplex combination process) is used in two different meanings. They are i) by squaring ii) by cross-multiplying.

We use both the meanings of Dwandwa-yoga in the context of finding squares of numbers as follows:

We denote the Duplex of a number by the symbol D. We define for a single digit ‘a’, D =a². and for a two digit number of the form ‘ab’, D=2( a x b ). If it is a 3 digit number like ‘abc’, D =2( a x c ) + b².

For a 4 digit number ‘abcd’, D = 2( a x d ) + 2( b x c ) and so on. i.e. if the digit is single central digit, D represents ‘square’: and for the case of an even number of digits equidistant from the two ends D represent the double of the cross- product.

Consider the examples:

Number		DuplexD
3		32 = 9
6		62 = 36
23		2 (2 x 3) = 12
64		2 (6 x 4) = 48
128		2 (1 x 8) + 22 = 16 + 4 = 20
305		2 (3 x 5) + 02 = 30 + 0 = 30
4231		2 (4 x 1) + 2 (2 x 3) = 8 + 12 = 20
7346		2 (7 x 6) + 2 (3 x 4) = 84 + 24 = 108

Further observe that for a n- digit number, the square of the number contains 2n or 2n-1 digits. Thus in this process, we take extra dots to the left one less than the number of digits in the given numbers.

Examples:1  62²  Since number of digits = 2, we take one extra dot to the left. Thus

                   .62         for 2, D = 2² = 4
                  ____
                   644         for 62, D = 2 x 6 x 2 = 24
                 32           for 62, D = 2(0 x2) + 6^{2
                      _____}                         = 36
                 3844

            62² = 3844.

Examples:2  234²  Number of digits = 3. extradots =2 Thus

               ..234    for 4, D = 4² = 16
              _____
              42546    for 34, D = 2 x 3 x 4 = 24
              1221      for 234, D = 2 x 2 x 4 + 3² = 25
              _____
              54756     for  .234, D = 2.0.4 + 2.2.3 = 12
                           for ..234, D = 2.0.4 + 2.0.3 + 2² = 4

Examples:3  1426². Number of digits = 4, extra dots = 3

               i.e
                      ...1426           6,  D = 36
                    ________
                    1808246           26,  D= 2.2.6 = 24
                      22523           426,  D =2.4.6 + 2² = 52
                   _________
                    2033476        1426,  D = 2.1.6 + 2.4.2= 28
                                        .1426,  D = 2.0.6 + 2.1.2 + 4² = 20
                                       ..1426,  D = 2.0.6 + 2.0.2 + 2.1.4 = 8
                                      ...1426,  D = 1² = 1

                   Thus 1426² = 2033476.

With a little bit of practice the results can be obtained mentally as a single line answer.

Algebraic Proof:

Consider the first example 62²

        Now 62² = (6 x 10 + 2)² =(10a + b)² where a = 6, b = 2
                                            = 100a² + 2.10a.b + b²
                                            = a² (100) + 2ab (10) + b²

i.e. b² in the unit place, 2ab in the 10^th place and a² in the100^th place i.e. 2² = 4 in units place, 2.6.2 = 24 in the10^th place (4 in the 10^th place and with carried over to 100^th place). 6²=36 in the 100^th place and with carried over 2 the 100^th place becomes 36+2=38.

    Thus the answer 3844.

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Find the squares of the numbers  54, 123, 2051, 3146.
       Applying the Vedic sutra Dwanda yoga.

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2.CUBING

Take a two digit number say 14.

i) Find the ratio of the two digits i.e. 1:4

ii) Now write the cube of the first digit of the number i.e. 1³

iii) Now write numbers in a row of 4 terms in such a way that the first one is the cube of the first digit and remaining three are obtained in a geometric progression with common ratio as the ratio of the original two digits (i.e. 1:4) i.e. the row is

                          1    4    16    64.

iv) Write twice the values of 2^nd and 3^rd terms under the terms respectively in second row.

i.e.,
        1    4    16    64
              8    32           ( 2 x 4 = 8, 2 x 16 = 32)

v) Add the numbers column wise and follow carry over process.

        1    4    16    64    Since 16 + 32 + 6 (carryover) = 54
              8    32           4 written and 5 (carryover) + 4 + 8 = 17
        ______________
        2    7     4     4      7 written and 1 (carryover) + 1 = 2.

This 2744 is nothing but the cube of the number 14

Example 1:  Find 18³

                   

Example 2:  Find 33³


                       

Algebraic Proof:

Let a and b be two digits.

Consider the row a³     a²b    ab²     b³
     the first isa³ and the numbers are in the ratio a:b
    since a³:a²b=a²b:b³=a:b

Now twice of a²b, ab² are 2a²b, 2ab²

            a³ +  a²b + ab² + b³
                  2a²b + 2ab^{2
             ________________________________}
            a³ +3a²b + 3ab² + b³ = (a + b)³.

Thus cubes of two digit numbers can be obtained very easily by using the vedic sutra ‘anurupyena’. Now cubing can be done by using the vedic sutra ‘Yavadunam’.

Example 3:   Consider 106³.

i) The base is 100 and excess is 6. In this context we double the excess and then add.

    i.e. 106 + 12 = 118. ( 2 X 6 =12 )

    This becomes the left - hand - most portion of the cube.

    i.e. 106³ = 118 / - - - -

ii) Multiply the new excess by the initial excess

         i.e. 18 x 6 = 108 (excess of 118 is 18)

    Now this forms the middle portion of the product of course 1 is carried over, 08 in the middle. 

         i.e. 106³ = 118 / 08 / - - - - -
                            1

iii) The last portion of the product is cube of the initial excess.

    i.e. 6³ = 216.

    16 in the last portion and 2 carried over.

         i.e. 106³ = 118 / 081 /16 = 1191016
                            1      2

Example 4:  Find 1002³.

i) Base = 1000. Excess = 2. Left-hand-most portion of the cube becomes 1002+(2x2)=1006.

ii) New excess x initial excess = 6 x 2 = 12.

    Thus 012 forms the middle portion of the cube.

iii) Cube of initial excess = 2³ = 8.

    So the last portion is 008.

    Thus 1002³ = 1006 / 012 / 008 = 1006012008.

Example 5:  Find 94³.

i) Base = 100, deficit = -6. Left-hand-most portion of the cube becomes94+(2x-6)=94-12=82.

ii) New deficit x initial deficit = -(100-82)x(-6)=-18x-6=108

    Thus middle potion of the cube = 08 and 1 is carried over.

iii) Cube of initial deficit = (-6)³ = -216
                                         __                 __
            Now 94³ =82 / 08 / 16 = 83 / 06 / 16
                                    _
                             1     2
                                           = 83 / 05 / (100 – 16)
                                           = 830584.

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Find the cubes of the following numbers using Vedic sutras.

            103, 112, 91, 89, 998, 9992, 1014.

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3. Equation of Straight line passing through two given points:

To find the equation of straight line passing through the points (x₁,y₁) and (x₂, y₂) , we generally consider one of the following methods.

1. General equation y = mx + c.

    It is passing through (x₁, y₁) theny₁ = mx₁ + c.

    It is passing through (x₂, y₂) also, theny₂ = mx₂ + c.

    Solving these two simultaneous equations, we get ‘m’ and ‘c’ and so the equation.

2. The formula
                                            (y₂ - y₁)
                               y – y₁ = ________ (x – x₁)   and substitution.
                                             (x₂- x₁) 

Some sequence of steps gives the equation. But the paravartya sutra enables us to arrive at the conclusion in a more easy way and convenient to work mentally.

Example1:  Find the equation of the line passing through the points (9,7) and (5,2).

Step1: Put the difference of the y - coordinates as the x - coefficient and vice - versa.

          i.e. x coefficient = 7- 2 = 5

          y coefficient = 9 - 5 = 4.

          Thus L.H.S of equation is 5x - 4y.

Step 2: The constant term (R.H.S) is obtained by substituting the co-ordinates of either of the given points in

        L.H.S (obtained through step-1)

        i.e. R.H.S of the equation is    

             5(9) -4(7) = 45 - 28 = 17

            or 5(5) - 4(2) = 25 - 8 = 17.

            Thus the equation is 5x - 4y = 17.

Example 2:  Find the equation of the line passing through (2, -3) and (4,-7).

Step 1 : x[-3-(-7)] –y[2-4] = 4x + 2y.

Step 2 : 4(2) + 2(-3) = 8 –6 = 2.

Step 3 : Equation is 4x + 2y =2 or 2x +y = 1.

Example 3 : Equation of the line passing through the points (7,9) and (3,-7).

Step 1 : x[9 - (-7)] – y(7 - 3) = 16x - 4y.

Step 2 : 16(7) - 4(9) = 112 – 36 = 76

Step 3 : 16x- 4y = 76 or 4x – y = 19