You can find all the previous posts about below:
Introduction to Vedic Mathematics
A Spectacular Illustration of Vedic Mathematics
Multiplication Part 1
Multiplication Part 2
Multiplication Part 3
Multiplication Part 4
Multiplication Part 5
Multiplication Special Case 1
Multiplication Special Case 2
Multiplication Special Case 3
Vertically And Crosswise I
Vertically And Crosswise II
Squaring, Cubing, Etc.
The most obvious problem that confronts one when performing subtraction is the concept of borrowing numbers from the left to perform subtraction. This is particularly confusing for learners, and the procedure is very error-prone since repeated borrowings result in so many numbers struck off and rewritten that it becomes difficult to be sure what was written and what the final number to be subtracted from should be.
This difficulty can be easily illustrated when dealing with a subtraction problem like the following:
Following the classic procedures, we see that the top line becomes a mess of strikeouts and rewrites as we are repeatedly forced to borrow from the left as we try to subtract each digit. It requires a lot of concentration, and it is easy to see how the procedure can be error-prone. As for teaching it to young children, it could become an exercise in frustration both for them and for the teacher.
The second type of problem with subtraction is the problem of multiple subtractions. Consider the problem below:
It is not easy to figure out how to perform this kind of subtraction straight-away. The simplest solution seems to be to add up the numbers to be subtracted separately, then substitute that sum in this problem and do a single subtraction. It is either that or we have to perform 3 separate subtractions. The problem is compounded when multiple subtractions and additions are combined in a single problem. Consider this:
This type of problem is more common than one imagines. Take the simple task of balancing a checkbook for instance. You are immediately confronted with multiple credits and debits interspersed with each other, making the problem very similar to the one above.
Luckily, Vedic Mathematics has a very simple solution that converts all subtraction problems into simple addition problems. It is an application of the Nikhilam Sutra which, if you recall from the lesson on 10's complements, reads Nikhilam Navatascaramam Dasatah, which means All From 9 And The Last From 10. Yes, it involves the use of 10's complements. To illustrate, let us tackle a simple problem: 112 - 98.
To solve the problem, write 112 on one line and the 10's complement of 98 on the second line as below:
Now add them up to get 114. From this, subtract the power of 10 with respect to which the 10's complement was taken. In this case, that is 100, and 114 - 100 is easy to derive mentally to be 14. And that is all there is to it.
The algebraic basis of the method should be obvious to everyone immediately. What we are doing is rewriting a number as the difference between a power of 10 and the number's 10's complement. Thus a - b becomes a - (10^n - d), where d is the 10's complement of b with respect to 10^n. This is obviously the same as a + d - 10^n (^ denotes raising to the power, or exponentiation). Difficult subtraction problems become easy addition problems followed by the subtraction of powers of 10, which is child's play.
Let us now solve the problem we started with as an illustration of the problems with multiple borrowings of numbers on the top row and see how the problem becomes much more pleasant:
+0101011 (10's complement of 9898989 with respect to 10000000)
From this, we subtract out 10000000. This gives us 8801065, which can be verified to be the correct answer.
How do we apply this method to multiple subtractions or combined additions and subtractions? The answer should be obvious from the algebraic explanation of the method: we replace all the numbers to be subtracted by their 10's complements, add them all up, then subtract out all the powers of 10 with respect to which the 10's complements were taken. To illustrate this, we will deal with the two problems we used to illustrate the difficulties encountered with these types of problems:
+5016507 (10's complement of 4983493 with respect fo 10000000)
+8061202 (10's complement of 1938798 with respect fo 10000000)
+3626736 (10's complement of 6373264 with respect fo 10000000)
From this, we subtract out 10000000 thrice, giving us 1687920 right away.
+54554148 (10's complement of 45445852 with respect to 100000000)
+61231152 (10's complement of 38768848 with respect to 100000000)
+62426447 (10's complement of 37573553 with respect to 100000000)
From this, we subtract 100000000 thrice, and get 417138359 immediately and easily.
What happens when we use this method and the final answer is less than the power of 10 to be subtracted from it? To illustrate this case, consider the simple case of 110 - 988. We can easily write that as below:
+012 (10's complement of 988 with respect to 1000)
Now, we need to subtract 1000 from 122, but find immediately, that this is not as simple as it seems because 122 is much smaller than 1000. But the Nikhilam sutra again comes to our rescue. 122 - 1000 is the same as -(1000 - 122). The figure inside the parentheses can immediately be identified as the 10's complement of 122 with respect to 1000. Thus the answer becomes -878, which can be verified to be the correct answer. Thus, the simple rule is that if the final answer of our additions results in a number smaller than the power of 10 to be subtracted to give the final answer, then prepend a "-" sign in front and put down the 10's complement of the number as the rest of the answer. To illustrate this further, let us work out a few more examples:
+652171 (10's complement of 347829 with respect to 1000000)
Since this answer is less than 1000000, the final answer becomes -308354 (negative of the 10's complement of 691646)
+61242 (10's complement of 38758 with respect to 100000)
+65020 (10's complement of 34980 with respect to 100000)
+45196 (10's complement of 54804 with respect to 100000)
This answer is less than the 300000 we need to subtract. So, we first subtract 200000 to get 44931. Then we take the negative of the 10's complement of this number with respect to the third 100000 to get the final answer. It becomes -55069, which can be verified to be correct.
Note that there is nothing special about this method except its antiquity. Computer scientists the world over know that computers perform subtraction by using the 1's complements of numbers (since computers use binary numbers, 1's complement in the world of computers is equivalent to 10's complement in our decimal world) or a slight modification of it, called a 2's complement. But using complements to perform subtraction did not start with the computer age. Instead we see that Vedic sutras have documented this method for millenia before the first electronic computers were invented! As always practice will make one proficient at using this technique to tackle pesky subtraction problems without performing pesky subtractions. Happy computing, and good luck!