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Tuesday, 26 March 2013

Vedic Mathematical Formulae part 1

1. Ekadhikena Purvena
     
EKĀDHIKENA PŪRVEŅA

The Sutra (formula) Ekādhikena Pūrvena means: “By one more than the previous one”.
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 is52, 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.


Algebraic proof: 

 a)  Consider (ax + b)2 Ξ a2. x2+  2abx + b2.

Thisidentity for  x = 10   and b = 5 becomes

(10a + 5) 2 =  a. 102  + 2. 10a . 5 + 52

                 =  a. 102+ a.102+ 52

                 =  (a2+ a ) . 10+52

                 =  a (a + 1) . 10+ 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)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 Є W.

    Now (ax2+bx+ c) 2 =  ax+ b2x+ c+ 2abx3 + 2bcx + 2cax2

                                = ax4+2ab.x3+(b2+ 2ca)x2+2bc . x+ c2.

    This identity for x = 10, c = 5becomes  (a . 10+ b .10 + 5) 2

                        = a2.104+ 2.a.b.10+ (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.103+ b2.102+ a . 103 + b 102+ 52

                        = a2.10+ (2ab + a).10+ (b2+ b)10+52

                        = [ a2.10+2ab.10 + a.10 + b+ b] 102+ 52

                        = (10a + b) ( 10a+b+1).10+ 25

                         = P (P+1) 102+ 25,  where P = 10a+b.

Hence any three digit number whose last digit is 5 gives the same result as in (a) for P=10a + b, the ‘previous’ of  5.

Example :    1652= (1 . 102 + 6 . 10 + 5) 2.

It is of the form (ax2+bx+c)2for a = 1, b = 6, c = 5 and x = 10. It gives the answer P(P+1) / 25, where P = 10a + b = 10 X 1 + 6 = 16, the ‘previous’. The answer is 16 (16+1) / 25 = 16 X 17 / 25 = 27225.



        Apply Ekadhikena purvena to find the squares of  the numbers 95, 225, 375, 635, 745, 915, 1105, 2545.


ii) Vulgar fractions whose denominators are numbers ending in NINE :

    We now take examples of 1 / a9, where a = 1, 2, -----, 9. In the conversion of such vulgar fractions into recurring decimals, Ekadhika process can be effectively used both in division and multiplication.

a) Division Method  :  Value of 1 / 19.

The numbers of decimal places before repetition is the difference of numerator and denominator, i.e.,, 19 -1=18 places.

For the denominator 19, the purva (previous) is 1.

Hence Ekadhikena purva (one more than the previous) is 1 + 1 = 2.

The sutra is applied in a different context.  Now the method of division is as follows:

Step. 1 :  Divide numerator 1 by 20.

       i.e.,, 1 / 20  =  0.1 / 2  = .10 ( 0 times, 1 remainder)

Step. 2 :  Divide 10  by  2

       i.e.,,  0.005( 5 times, 0 remainder )

Step. 3 : Divide  5  by  2

       i.e.,,  0.0512 ( 2 times, 1 remainder )

 Step. 4 : Divide  12  i.e.,, 12  by  2

       i.e.,,  0.0526 ( 6 times, No remainder )

 Step. 5 :    Divide  6  by  2

       i.e.,,   0.05263 ( 3 times, No remainder )

 Step. 6 :    Divide  3  by  2

        i.e.,,  0.0526311(1 time, 1 remainder )

 Step. 7 :    Divide  11   i.e.,, 11  by  2

        i.e.,,  0.05263115  (5 times, 1 remainder )

 Step. 8 :    Divide  15  i.e.,,  15  by  2

         i.e.,,  0.052631517 ( 7 times, 1 remainder )

 Step. 9 :    Divide 17    i.e.,,   17  by  2

         i.e.,,   0.05263157 18 (8 times, 1 remainder )

 Step. 10 :  Divide  18  i.e.,,  18  by  2

         i.e.,,  0.0526315789 (9 times, No remainder )

 Step. 11 :  Divide  9  by  2

         i.e.,,  0.0526315789 14 (4 times, 1 remainder )

 Step. 12 :  Divide  14   i.e.,,  14  by  2

         i.e.,,  0.052631578947 ( 7 times, No remainder )

 Step. 13 :  Divide  7  by  2

          i.e.,,  0.05263157894713 ( 3 times, 1 remainder )

 Step. 14 : Divide  13    i.e.,,   13  by  2

          i.e.,,  0.052631578947316 ( 6 times, 1 remainder )

 Step. 15 : Divide  16     i.e.,,   16  by  2

          i.e.,,   0.052631578947368 (8 times, No remainder )

 Step. 16 : Divide  8  by  2

          i.e.,, 0.0526315789473684 ( 4 times, No remainder )

 Step. 17 : Divide  4  by  2

          i.e.,,  0.05263157894736842 ( 2 times, No remainder )

 Step. 18 : Divide  2  by 2

          i.e.,, 0.052631578947368421 ( 1 time, No remainder )

Now from step 19, i.e.,, dividing 1 by 2, Step 2 to Step. 18 repeats thus giving

               0 __________________        .                       .
        1 / 19 =0.052631578947368421 or 0.052631578947368421

Note that we have completed the process of division only by using ‘2’. Nowhere the division by 19 occurs.

b) Multiplication Method: Value of 1 / 19

First we recognize the last digit of the denominator of the type 1 / a9. Here the last digit is 9.

For a fraction of the form in whose denominator 9 is the last digit, we take the case of 1 / 19 as follows:

For 1 / 19, 'previous' of 19 is 1. And one more than of it is 1 + 1 = 2.

Therefore 2 is the multiplier for the conversion. We write the last digit in the numerator as 1 and follow the steps leftwards.

Step. 1 :                                        1

Step. 2 :                                      21(multiply 1 by 2, put to left)

Step. 3 :                                   421(multiply 2 by 2, put to left)

Step. 4 :                                8421(multiply 4 by 2, put to left)

Step. 5 :                              168421 (multiply 8 by 2 =16,   1 carried over, 6 put to left)

Step. 6 :                            1368421 ( 6 X 2 =12,+1 [carry over]

                                                   = 13, 1 carried over, 3 put to left )

Step. 7 :                          7368421 ( 3 X 2, = 6 +1 [Carryover]

                                                  = 7, put to left)

Step. 8 :                        147368421 (as in the same process)

Step. 9 :                       947368421  ( Do – continue to step 18)

Step. 10 :                   18947368421

Step. 11 :                  178947368421

Step. 12 :                1578947368421

Step. 13 :               11578947368421

Step. 14 :                31578947368421

Step. 15 :               631578947368421

Step. 16 :              12631578947368421

Step. 17 :             52631578947368421

Step. 18 :           1052631578947368421

Now from step 18 onwards the same numbers and order towards left continue.

Thus  1 / 19 = 0.052631578947368421

It is interesting to note that we have

    i)  not at all used division process

    ii) instead of dividing 1 by 19 continuously, just multiplied 1 by 2 and continued to multiply the resultant successively by 2.

Observations :

   a) For any fraction of the form 1 / a9 i.e.,, in whose denominator 9 is the digit in the units place and a is the set of remaining digits, the value of the fraction is in recurring decimal form and the repeating block’s right most digit is 1.

   b) Whatever may be a9, and the numerator, it is enough to follow the said process with (a+1) either in division or in multiplication.

   c) Starting from right most digit and counting from the right, we see ( in the given example 1 / 19)

     Sum of 1st digit + 10th digit  =  1 + 8 = 9

    Sum of 2nd digit + 11th digit  =  2 + 7 = 9

     - - - - - - - - -- - - - - - - - - - - - - - - - - - -
Sum of 9th digit + 18th digit = 9+ 0 = 9

    From the above observations, we conclude that if we find first 9 digits, further digits can be derived as complements of 9.

     i) Thus at the step 8 in division process we have 0.052631517 and next step. 9 gives 0.052631578

        Now the complements of the numbers
                                0, 5, 2, 6, 3, 1, 5, 7, 8 from 9
                                9, 4, 7, 3, 6, 8, 4, 2, 1 follow the right order

                                        i.e.,, 0.052631578947368421

             Now taking the multiplication process we have

        Step. 8 :         147368421

       Step. 9 :         947368421

    Now the complements of 1,  2, 4,  8,  6, 3, 7,  4, 9 from 9
                              i.e.,, 8, 7, 5, 1, 3, 6, 2, 5, 0 precede in successive steps, giving the answer.

                          0.052631578947368421.
d) When we get (Denominator – Numerator) as the product in the multiplicative process, half the work is done. We stop the multiplication there and mechanically write the remaining  half of the answer by merely taking down complements from 9.

e) Either division or multiplication process of giving the answer can be put in a single line form.

Algebraic proof :

Any vulgar fraction of the form 1 / a9 can be written as

 1 / a9 =  1 / ( (a + 1 ) x - 1 ) where x = 10

                                    1
              =   ________________________               
                 ( a + 1 )  x [1 - 1/(a+1)x ]

                    1
                =    ___________ 
[1 - 1/(a+1)x]-1
                       
( a + 1 ) x


                      1
            = __________ [1 + 1/(a+1)x + 1/(a+1)x2+ ----------]
               ( a + 1 ) x 
     =  1/(a+1)x + 1/(a+1)2x2 +1/(a+1)3x3+ ----ad infinitum

      = 10-1(1/(a+1))+10-2(1/(a+1)2)+10-3(1/(a+1)3) + ---ad infinitum

    This series explains the process of ekadhik.

     Now consider the problem of 1 / 19. From above we get

    1 / 19  =  10-1 (1/(1+1)) + 10-2 (1/(1+1)2) + 10-3 (1/(1+1)3) + ----
                                                                                 ( since a=1)

               =  10-1 (1/2) + 10-2 (1/2)2 + 10-3 (1/3)3 + ----------

               =  10-1 (0.5) + 10-2 (0.25) + 10-3 (0.125)+ ----------

               = 0.05 + 0.0025 + 0.000125 + 0.00000625+ - - - - - - -

               = 0.052631 - - - - - - -

    Example1 :

    1. Find 1 / 49 by ekadhikena process.

    Now ‘previous’ is 4. ‘One more than the previous’ is 4 + 1 = 5.

    Now by division right ward from the left by ‘5’.

    1 / 49  =  .10 - - - - - - - - - - - -(divide 1 by 50)

     = .02 - - - - - - - - - (divide 2 by 5, 0 times, 2 remainder )

     = .0220 - - - - - - --(divide 20 by 5, 4 times)

     = .0204 - - - - - - -( divide 4 by 5, 0 times, 4 remainder )

     = .020440 -- - -- - ( divide 40 by 5, 8 times )

     = .020408 - - - - - (divide 8 by 5, 1 time, 3 remainder )

     = .02040831 - - - -(divide 31 by 5, 6 times, 1 remainder )

     = .02040811 6 - - - - - - -    continue

               = .0204081613322615306111222244448 - -- - - - -

   On completing 21 digits, we get 48

i.e.,,Denominator - Numerator = 49 – 1 = 48 stands.

i.e, half of the process stops here. The remaining half can be obtained as complements from 9.
                                 .
         Thus 1 / 49 = 0.020408163265306122448
                                                                     .
                                 979591836734693877551

   Now finding 1 / 49 by process of multiplication left ward from right by 5, we get

        1 / 49 =  ----------------------------------------------1

                  =  ---------------------------------------------51

                  =  -------------------------------------------2551

                  =  ------------------------------------------27551

                  =  ---- 483947294594118333617233446943383727551

     i.e.,,Denominator – Numerator = 49 – 1 = 48 is obtained as 5X9+3

  ( Carry over ) = 45 + 3 = 48. Hence half of the process is over. The remaining half is    automatically obtained as complements of 9.

    Thus 1 / 49 = ---------------979591836734693877551
                               .
                         = 0.020408163265306122448
                                                                   .
                               979591836734693877551

   Example 2:  Find 1 / 39 by Ekadhika process.

   Now by multiplication method, Ekadhikena purva is 3 + 1 = 4

       1 / 39 =  -------------------------------------1

                = -------------------------------------41

                = ----------------------------------1641

               = ---------------------------------25641

               = --------------------------------225641

               = -------------------------------1025641

 Here the repeating block happens to be block of 6 digits. Now the rule predicting the completion of half of the computation does not hold. The complete block has to be computed by ekadhika process.

     Now continue and obtain the result. Find reasons for the non–applicability of the said ‘rule’.


 Find the recurring decimal form of the fractions 1 / 29, 1 / 59,
1 / 69, 1 / 79, 1 / 89 using Ekadhika process if possible. Judge whether the rule of completion of half the computation   holds good in such cases.



Note : The Ekadhikena Purvena sutra can also be used for conversion of vulgar fractions ending in 1, 3, 7 such as 1 / 11, 1 / 21, 1 / 31 - - -- ,1 / 13, 1 / 23, - - - -, 1 / 7, 1 / 17, - - - - - by writing them in the following way and solving them.
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