In India, the culture of maths originated long ago…at the time of the Aryas. Mathematicians of that time were the true pioneers of modern counting. A shocking truth is that despite the absence of calculators, they used to solve massive calculations accurately within moments, some of them not even solvable with modern scientific calculators. Maths at that time was not like today’s research works. It was meant for counting FAST. I emphasize “fast”, because that’s why Vedic Mathematicians own a prestigious position in the history of Mathematics. Unless you have got math-o-phobia, these techniques are bound to interest you.
Computation at the Vedic era was entirely based on two easy definitions, one of them being Base, and other one being Complement.
Bases– The entire number system is built on the numbers 0-9. All of them keep repeating themselves in a specific order, but at some milestones. These “milestones” refer to 10,100,1000….etc. which we call ten’s digit, hundred’s digit….etc. in modern number theory. These numbers(which have “1” as their first digit, followed by zeroes) are termed as Bases.
Complements– Any number from the number line, when subtracted from its nearest base that is greater than it, gives the Complement of the number.
example: (i) Suppose we take 58, it is closest to the base 100, and 100 is greater than it. So its complement will be (100-52)=42.
(ii) Suppose we have taken 42 instead of 58, in that case, 42 is closest to 10 rather than 100. But despite that we take the nearest base to be 100 since 10 is less than 42. Its complement is obviously 58.
# Obtaining Complement:
Like the Vedic Mathematicians defined Complements, they also suggested how to calculate complements of a number. This technique has been referred in a Sutra, which says…
“निखिलं नवतश्चरमम दशतः”
Which means, “All from nine and last from ten“. If we observe carefully, we would notice that it only refers to the basic technique of subtracting higher digit numbers. Let us illustrate it with an example.
Suppose, we are to find the Complement of the number 1234. So, what we have to do is subtract the last digit from ten, i.e. (10-4)=6.
And the rest of the digits from 9, i.e. (9-1)=8, (9-2)=7, (9-3)=6. So the complement becomes 8766. Clearly (1234+8766)=10000.
A point to remember, when the number will end with a zero, i.e. 4810, then the technique is a bit different. We exclude the zero from the number first. i.e. the number becomes 481. Now, we determine its complement following the Sutra, which is found to be 519. Finally we put the zero at the end, and the number that formed is the complement of 4810, that is 5190.
If more than one zeroes are present at the end, then we exclude all of them, determine the complement, and finally put them back at the end.
The proof is even more simple, it is the “borrow” operation that reduces the base from 10 to 9 in each of the subtractions other than the last one. This concept of Complements was also used in the subtraction of two numbers, which we are about to see now…
Let us recall the Sutra I stated a little ago, that is:
“निखिलं नवतश्चरमम दशतः”
Which means, “All from nine and last from ten“. So, we are going to use our pet funda of complements.
I am presenting the process point wise,
1) We take our previous example. (1081-723). In this case, we refer the digits of 1081 to be Upper digits, and the digits of 723 to Lower digits, just for easier representation.
2) In each column of subtraction(by column I refer to unit’s digit,ten’s digit etc.), we take the difference of upper digit and lower digit, i.e. (upper digit ~ lower digit).
3) If in some column, upper digit > lower digit, then the answer becomes (upper digit – lower digit – 1). This extra 1 is subtracted since we are coming out of complements.
4) If in some column, upper digit < lower digit, the answer becomes the complement of (lower digit – upper digit).
Let’s now take our example. I am referring unit’s digit as column 1, ten’s digit as column 2,…etc. So, in column 1, upper digit < lower digit, which means we are to take complement. Answer for column 1 is [10-(3-1)]=8.
For the next column, upper digit > lower digit, so we come out of complements, and the answer is (8-2-1)=5.
For 3rd column, upper digit < lower digit, so we need complement. Thus, the answer is [10-(7-0)]=3.
Finally for the last column, upper digit > lower digit, so the answer becomes, (1-0-1)=0.
So, our final answer is, 0358, i.e. 358. Clearly (358+723)=1081. So, our concept of complement proved to be useful once again.
Vedic Mathematicians intended to solve a lot of problems based on this complement concepts…it is hard to imagine how powerful this technique can actually be for fast calculations!!