matha namo namaha

 
ADDITION AND SUBTRACTION
ADDITION:
    In the convention process we perform the process as follows.
234 + 403 + 564 + 721
write as  234
                  403
                  564
                  721
Step (i): 4 + 3 + 4 + 1 = 12     2 retained and 1 is carried over to left.
Step (ii): 3 + 0 + 6 + 2 = 11    the carried ‘1’ is added
i.e., Now 2 retained as digit in the second place (from right to left) of the answer and 1 is carried over to left.
step (iii): 2 + 4 + 5 + 7 = 18    carried over ‘1’ is added
i.e., 18 + 1 = 19. As the addition process ends, the same 19 is retained in the left had most part of the answer.
thus    234
               403
               564
             +721
             _____
              1922    is the answer
we follow sudhikaran process Recall ‘sudha’ i.e., dot (.) is taken as an upa-sutra (No: 15)

consider the same example




             
i) Carry out the addition column by column in the usual fashion, moving from bottom to top.
(a) 1 + 4 = 5, 5 + 3 = 8, 8 + 4 = 12 The final result is more than 9. The tenth place ‘1’ is dropped once number in the unit place i.e., 2 retained. We say at this stage sudha and a dot is above the top 4. Thus column (1) of addition (right to left)
         .
         4
         3
         4
         1
        __
         2

b) Before coming to column (2) addition, the number of dots are to be counted, This shall be added to the bottom number of column (2) and we proceed as above.
Thus second column becomes
            .
            3        dot=1,    1 + 2 = 3
            0                      3 + 6 = 9
            6                      9 + 0 = 9
            2                      9 + 3 = 12
           __
            2
2 retained and ‘.’ is placed on top number 3

c) proceed as above for column (3)

 2      i) dot = 1          ii) 1 + 7 = 8
 4      iii) 8 + 5 = 13    iv) Sudha is said.
 .
 5      A dot is placed on 5 and proceed
 7      with retained unit place 3.
__
 9      v) 3+4=7,7+2=9 Retain 9 in 3rd digit i.e.,in 100th place.

d) Now the number of dots is counted. Here it is 1 only and the number is carried out left side ie. 1000th place
                 ..
    Thus    234
               403
               .
               564
             +721
             _____
              1922    is the answer.
Though it appears to follow the conventional procedure, a careful observation and practice gives its special use.

eg (1):
                 .
               437
               .  .
               624
                 .
               586
             +162
            ______
              1809

Steps 1:
i) 2 + 6 = 8, 8 + 4 = 12 so a dot on 4 and 2 + 7 = 9 the answer retained under column (i)
ii) One dot from column (i) treated as 1, is carried over to column (ii),
thus 1 + 6 = 7, 7 + 8 = 15 A' dot’; is placed on 8 for the 1 in 15 and the 5 in 15 is added to 2 above.
5 + 2 = 7, 7 + 3 = 10 i.e. 0 is written under column (ii) and a dot for the carried over 1 of 10 is placed on the top of 3.
(iii) The number of dots counted in column (iii) are 2.
Hence the number 2 is carried over to column (ii) Now in column (iii)
2 + 1 = 3, 3 + 5 = 8, 8 + 6 = 14 A dot for 1 on the number 6 and 4 is retained to be added 4 above to give 8. Thus 8 is placed under column (iii).
iv) Finally the number of dots in column (iii) are counted. It is ‘1’ only. So it carried over to 1000th place. As there is no fourth column 1 is the answer for 4th column. Thus the answer is 1809.

Example 3:

                     



Check the result verify these steps with the procedure mentioned above.
The process of addition can also be done in the down-ward direction i.e., addition of numbers column wise from top to bottom

Example 1:

                   



Step 1:    6 + 4 = 10, 1 dot ; 0 + 8 = 8; 8 + 4 = 12;
            1 dot and 2 answer under first column - total 2 dots.
Step 2:  2+2 ( 2 dots) = 4; 4+9 = 13: 1 dot and 3+0 = 3; 3+8 = 11;
     1 dot and 1 answer under second column - total 2 dots.
Step 3:  3+2 ( 2 dots ) = 5; 5+6 = 11:1 dot and 1+7 = 8; 8+7 = 15;
     1 dot and 5 under third column as answer - total 2 dots.
Step 4:  4 + 2 ( 2 dots ) = 6; 6 + 5 =11:
     1 dot and 1+3 = 4; 4+2 = 6. - total 1 dot in the fourth 6 column as answer.
Step 5:  1 dot in the fourth column carried over to 5th column (No digits in it) as 1
     Thus answer is from Step5 to Step1; 16512

Example 2:

                   



Steps
    (i):  8 + 9 = 17; 7 + 4 = 11; 1 + 1 = (2) (2dots)
    (ii): 7 + 2 = 9; 9 + 1 = 10; 0 + 8 = 8, 8 + 9 = 17, (7) (2dots)
    (iii): 2 + 2 = 4; 4 + 6 = 10; 0 + 0 = 0; 0 + 7 = (7) (1 dot)
    (iv): 3 + 1 = 4; 4 + 4 = 8; 8 + 3 = 11; 1 + 1 = (2) (1 dot)
    (v): 1
        Thus answer is 12772.
Add the following numbers use ‘Sudhikaran’ whereever applicable.

        1.                          2.                         3.
              486                      5432                     968763
              395                      3691                     476509
              721                      4808                   +584376
            +609                   +6787                   ¯¯¯¯¯¯¯¯
           ¯¯¯¯¯                  ¯¯¯¯¯¯                  ¯¯¯¯¯¯¯¯
           ¯¯¯¯¯                  ¯¯¯¯¯¯
Check up whether ‘Sudhkaran’ is done correctly. If not write the correct process. In either case find the sums.
           
           
SUBTRACTION:
The ‘Sudha’ Sutra is applicable where the larger digit is to be subtracted from the smaller digit. Let us go to the process through the examples.
Procedure:
i) If the digit to be subtracted is larger, a dot ( sudha ) is given to its left.
ii) The purak of this lower digit is added to the upper digit or purak-rekhank of this lower digit is subtracted.
Example (i): 34 - 18

                      34
                      .
                    -18
                   _____
                                                                           .
     Steps: (i): Since 8>4, a dot is put on its left i.e. 1
          (ii) Purak of 8 i.e. 2 is added to the upper digit i.e. 4
                                                                         _
                    2 + 4 = 6. or Purak-rekhank of 8 i.e. 2 is
                                                       _
                    Subtracted from i.e. 4 - 2 =6.
Now at the tens place a dot (means1) makes the ‘1’ in the number into 1+1=2.This has to be subtracted from above digit. i.e. 3 - 2 = 1. Thus

                      34
                      .
                    -18
                   _____
                      16
Example 2:

                      63
                      .
                    -37
                   _____
                                                                     .
     Steps: (i) 7>3. Hence a dot on left of 7 i.e., 3
          (ii) Purak of 7 i.e. 3 is added to upper digit 3 i.e. 3+3 = 6.

                     This is unit place of the answer.
                Thus answer is 26.
Example (3) :

                      3274
                      ..
                     -1892
                   _______
Steps:
    (i)  2 < 4. No sudha . 4-2 = 2 first digit (form right to left)
                                                                                       .
         (ii) 9 > 7 sudha required. Hence a dot on left of 9 i.e.  8

         (iii) purak of 9 i.e. 1, added to upper 7 gives 1 + 7 = 8 second digit
                                 .
         (iv) Now means 8 + 1 = 9.
                                                                                                  .
         (v) As 9 > 2, once again the same process: dot on left of i.e., 1

         (vi) purak of 9 i.e. 1, added to upper 2 gives 1 + 2 = 3, the third digit.
                        .
         (vii) Now 1 means 1+1 = 2

         (viii) As 2 < 3, we have 3-2 = 1, the fourth digit
Thus answer is 1382
Vedic Check :
Eg (i) in addition :  437 + 624 + 586 + 162 = 1809.
By beejank method, the Beejanks are

      437 4 + 3 + 7 14 1 + 4 5

      624 6 + 2 + 4 12 1 + 2 3

      586 5 + 8 + 6 19 1 + 9 10 1 + 0 1

      162 1 + 6 + 2 9
Now

      437 + 624 + 586 + 162 5 + 3 + 1 + 9 18 1 + 8 9

      Beejank of 1809 1 + 8 + 0 + 9 18 1 + 8 9 verified
Eg.(3) in subtraction :

                    3274 – 1892 = 1382

    now beejanks

        3274 3 + 2 + 7 + 4 3 + 4 7

        1892 1 + 8 + 9 + 2 2

        3292-1892 7-2 5

        1382 1 + 3 + 8 + 2 5  Hence verified.
Mixed addition and subtraction using Rekhanks:
Example 1 : 423 - 654 + 847 - 126 + 204.
In the conventional method we first add all the +ve terms

                        423 + 847 + 204 = 1474
Next we add all negative terms
                 - 654 - 126 = -780
At the end their difference is taken
                1474 - 780 = 694
Thus in 3 steps we complete the problem
But in Vedic method using Rekhank we write and directly find the answer.

                4 2 3
                _ _ _
                6 5 4

                8 4 7
                _ _ _
                1 2 6

                2 0 4
                _____
                   _
                7 1 4     This gives (7 -1) / (10 - 1) / 4 = 694.
Example (2):

             6371 – 2647 + 8096 – 7381 + 1234
                         ____               ____
          = 6371 + 2647 + 8096 + 7381 + 1234
                   _      _           _      _            _      _            _      _
          = (6+2+8+7+1)/(3+6+0+3+2)/(7+4+9+8+3)/(1+7+6+1+4)
                  _
          = 6 / 4 / 7 / 3

          = (6 – 1) / (10 – 4) / 73

          = 5673
* Find the results in the following cases using Vedic methods.

        1)  57 -39                3)   384 -127 + 696 -549 +150

        2) 1286 -968           4)   7084 +1232 - 6907 - 3852 + 4286
* Apply Vedic check for the above four problems and verify the results.

Publised by sri  

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