In this lesson, we are going to step back from the various arithmetic techniques we have discussed so far. Instead of learning a new arithmetic technique, we are going to deal with a technique that can be used to make those arithmetic techniques easier to perform when large numbers are involved.

You can find all the previous posts about Vedic Mathematics below:Introduction to Vedic Mathematics

A Spectacular Illustration of Vedic Mathematics

10's Complements

Multiplication Part 1

Multiplication Part 2

Multiplication Part 3

Multiplication Part 4

Multiplication Part 5

Multiplication Special Case 1

Multiplication Special Case 2Multiplication Special Case 3Vertically And Crosswise IVertically And Crosswise IISquaring, Cubing, Etc.SubtractionDivision By The Nikhilam Method I

Division By The Nikhilam Method II

Division By The Nikhilam Method IIIDivision By The Paravartya MethodDigital Roots

Straight Division IStraight Division IIWe are all familiar with the place value system of modern mathematics where each digit in a multi-digit number takes on a different place value depending on how far to the left or right of the decimal point it is located. The place value is calculated as the product of the face value of the number with an appropriate power of 10. By the way, this place value notation, which makes numbers much easier to handle (rather than roman numerals, for instance) was

invented in India in ancient times. Positional notation and the

concept of zero were two of the most important contributions of ancient India to modern mathematics.

To put it in more concrete terms, consider a number like 943,875. This number is actually 9 x 100,000 + 4 x 10,000 + 3 x 1,000 + 8 x 100 + 7 x 10 + 5 x 1. In a multi-digit number, the powers of 10 increase steadily from right to left and each power of 10 has a single-digit coefficient associated with it. This single-digit coefficient is almost always a number from 0 to 9.

However, there is nothing inherently wrong with this coefficient being a negative number between 0 and -9. This is usually not very common and may sound very confusing at first, but is actually quite simple to grasp once one becomes familiar with it. We will first explain the concept and then we will deal with some examples of computations that can be simplified significantly by the use of negative coefficients.

The basic concept of using negative coefficients is also referred to as using Vinculums. Vinculum is a latin word that derives from Vincire, to tie. It denotes a bond between two numbers that are linked together by some property.

When we use vinculums in Vedic Mathematics, that bond is usually the 10's complement of individual digits or groups of digits. A number and its 10's complement are tied together because the pair always adds up to a power of 10. This bond is sometimes utilized to rewrite numbers in vinculum form before performing mathematical computations on them.

Let us first take a few examples to see how the concept works. Then we will formalize how to rewrite any number in vinculum form. First, let us take the number 6783. It is easy to see that 6783 = 6 x 1000 + 7 x 100 + 8 x 10 + 3 x 1. It is also easy to see that 6783 = 6 x 1000 + 8 x 100 + (-2 x 10) + 3 x 1. Representing negative numbers by bold, red letters, we can say that 6783 = 68

23.

What we have done above is essentially converted 6783 into one of several vinculum forms that are possible for any given numbers. In this case, we converted one of the digits of the original number (8) to its vinculum form (2). Note that 8 and 2 are 10's complements of each other. As mentioned earlier, the bond referred to in vinculum is the 10's complement of a digit or group of digits.

Now, let us take the same example, 6783, and find a different vinculum form of it. It is easy to verify that 6783 = 7

383. In this case, we have converted the 7 to its vinculum form. One can also verify that 6783 = 7

223. In this case, we have taken the group of digits, 78, and converted the entire group to its vinculum form. Thus, we see that any multi-digit number might have several different vinculum forms, depending on which digit or group of digits is converted to its vinculum form.

But, we are not restricted to converting one digit or a consecutive group of digits to vinculum form to derive one of the many vinculum forms of the whole number. We can convert any number of non-contiguous digits of the original number into their vinculum forms and derive different vinculum forms of the original number also. For instance, 6783 = 7

39

7. In this case, we have taken the vinculum form of the individual digits, 7 and 3 to derive this vinculum form of the original number.

The procedure for deriving a vinculum form of a given number by converting digits and groups of digits into their vinculum form is explained below:

- Identify the digit to the left of the digit or contiguous set of digits that you want to convert to vinculum form. Call this the left digit. If there are no digits to the left of the digit you want to convert to vinculum form, pad the number with zeroes to the left as needed
- Increase this left digit by 1. If this number was already a 9, then it becomes 10, which results in that digit becoming 0 and the digit to its left being incremented (normal carryover from right to left applies)
- Find the 10's complement of the digit or group of digits that you want to convert to vinculum form and substitute them in place of the original digits, but with some indication that they are negative coefficients. Commonly, they are either underlined or a line is drawn on top of them to indicate that they are negative coefficients

That is all there is to it! We will work out a few examples below to illustrate the working of the procedure. In each case, we have colored the digits we are converting to vinculum form in blue and the negative digits in the vinculum form are denoted by red and bold coloring.

83

974 = 84

026 (026 is the 10's complement of 974, and 3 has been changed to 4 as per the procedure above)

7

8934

7 = 8

1135

3 (11 is the 10's complement of 89 and 7 has been changed to 8 as per the procedure above. Similarly, 7 has been changed to its 10's complement and the previous digit, 4, has been incremented to a 5)8

39

74 = 9

60

26 (Note that in this example, the 8 at beginning of the original number was changed to 9 during the conversion of the 3 into its vinculum form. However, instead of 7 (the 10's complement of 3), we have a 6 there because when we convert 74 to its vinculum form, it changes the 9 to a 10, resulting in carryover of a 1 from that digit. This carried over 1 is added to negative 7 to result in the final negative 6)

9

7886 = 10

2114 (9 has been incremented to 10 and 7886 has been changed to its 10's complement, 2114)

Obviously, converting a number to one of its vinculum forms really does not accomplish anything significant, so it is not surprising that the method is quite simple. The question still remains: why do we want to convert a number to any of its vinculum forms?

To understand the answer to that question, one first needs to realize that a number and any of its vinculum forms are equivalent when it comes to any mathematical operations. Addition, subtraction, multiplication and division can all be performed exactly as we are used to doing them after converting one of both numbers involved into any of their vinculum forms. The only caveat is that since the vinculum form has some negative coefficients, one has to be careful about the signs of the products of mathematical operations such as multiplications and divisions. For instance, the multiplication of two negative numbers with each other gives a positive answer while the multiplication of 2 numbers of opposite sign gives a negative answer, and so on.

Most of us also intuitively understand that computations with small numbers are much easier to perform than computations that involve large numbers, even when the numbers are just one digit each. Multiplying or dividing by numbers like 1 and 2 is much easier than multiplying or dividing by numbers like 8 or 9. This combination of the equivalency of a number and its vinculum forms, and the ease of computations using small numbers compared with computations using large numbers, gives the vinculum all its power in vedic mathematics.

To convince ourselves of this, we will tackle a few examples dealing with various techniques we have gone through before, and illustrate how one can make a problem simpler simply by working with an equivalent vinculum form of a number.

Let us start with a multiplication problem that we will try to work out using the

Urdhva-Tiryak sutra (vertically and cross-wise). We will try to calculate 29 x 48. Using the normal method, we work it out as below:

2948---------072520800---------1392Now, let us try it after converting the large numbers, 9 and 8 into their vinculum forms. We get the problem below:

3152-------000201101500-------14121392The answer, as expected is identical, but instead of a vertical product of 9 and 8, we only had to a vertical product of -1 and -2 (which gives us +2). Similarly, instead of 8x2 + 4x9, we had to do -2x3 + -5x1 which gives us -11. The resulting answer is a vinculum number, which can be converted into a normal number simply by following the procedure for producing a vinculum form in reverse.

Let us now try it on 291 x 88. In this case, the normal vertically and cross-wise procedure would be to pad the 88 with a 0 in front and do it as below:

291088---------0000800800088001600000000------------25608Now let us try it as below:

311112--------0000200010004000400030000----------3441225608Notice how the large number of carryovers due to the cross-products of large numbers with each other have been eliminated by the conversion of the large numbers into their much smaller vinculum forms! The sums of the cross-products are also easier to calculate because of the smaller numbers involved in the cross-products.

The same kind of time-savings because of having to deal with smaller numbers can be achieved when this method is applied to

straight division using the Dhvajanka sutra also. To illustrate, let us consider the example of 81356/29.

•9

2•|8 1 3 5|6

••| 4 7 1 5

---------------

••|2 8 0 5|11

Notice the large number of times we had to adjust the intermediate quotients and remainders so that the subtraction of the product of the quotient digit with the flag digit does not result in a negative number. Also, it is easy to see that the products themselves are much harder to calculate because of the large numbers involved in the products.

Let us now try the same problem as below:

•1

3•|8 1 3 5|6

••| 2 2 0 0

--------------

••|2 7 0 5|11

••|••1

---------------

••|2 8 0 5|11

The answer is once again 2805 with a remainder of 11. A few words of explanation may be necessary in this case to explain the procedure though.

First of all, we converted 29 to its vinculum form 31. So, the flag digit becomes -1. Dividing 8 by 3 gives us a quotient digit of 2 and a remainder of 2. The two digit number we get by putting this remainder next to the second numerator digit gives us 21. Now, we need to subtract the product of the flag digit with the first quotient digit from this 21. The product of the quotient digit (2) with the flag digit (-1) is -2. Subtracting -2 from 21 is equivalent to adding 2 to 21, resulting in an intermediate numerator of 23.

This leads to a quotient digit of 7 and a remainder of 2. Putting 2 in front of the next digit of the numerator gives us 23. Subtracting the product of the flag digit (-1) with the latest quotient digit (7) from this gives us 23 - (-7) = 30. When 30 is divided by 3, we get a quotient of 10 (which is a 2-digit number) and remainder of 0. When we get a quotient that has more than 1 digit, normal carryover rules apply, so, we put the 0 down in its own column and the 1 under the column to the left.

However, remember that the previous quotient was 10, so when it comes time to multiply the quotient digit by the flag digit, use 10 as the quotient digit (even though it is technically not a single "digit"). That is how we get our next intermediate numerator of 05 - (-10) = 15. This gives us the next quotient digit of 5 and remainder of 0. The final remainder is calculated as 06 - (-5) = 11.

Now, add up the numbers under each column to make sure the carryover digit is accounted for correctly, and we get our final quotient of 2805 and final remainder of 11. As you can see, the division by 2 or 3 is of approximately the same level of difficulty. But the products of the quotient digits with the flag digit have been greatly simplified because of the use of the vinculum form of 29. Moreover, it was never necessary to adjust the quotient and remainder of any intermediate division to avoid negative numbers.

Another example of taking advantage of the vinculum form is illustrated below:

•98

3••|2 3 1 4|23•••| 2 812 0

----------------

••0|0 5 8 4|-1009•••|0 5 8 1|185•02

4••|2 3 1 4|23

•••| 2 3 3 0

----------------

••0|0 5 7 1|185•••|••••1--------------------

•••|0 5 8 1|185

On some rare occasions, though, using the vinculum form does not provide as much of an advantage as one might expect. In the example below, the denominator is converted to one of its vinculum forms to replace the 8 with the much smaller 2, but the savings in work is not dramatic.

•81

3••|5 3 4 9|89

•••| 2 3 1 0

-----------------

••0|1 4 0 5|-316

••0|1 4 0 4|65

•21

4••|5 3 4 9|89

•••| 1 3 3 2

-----------------

••0|1 3 9 3|446

•••|0 0 1

-----------------

•••|1 4 0 3|446

•••|1 4 0 4|65

Below is another example where converting the denominator to the vinculum form does not provide a huge advantage because the quotient digits become too large, resulting in carryovers (and carryovers imply large quotient "digits" that have to be kept track of carefully during the cross-multiplications). But we do eliminate having to adjust the quotient and remainders to avoid negative results. In general, if the flag digits are all vinculums (negative numbers), one never has to worry about negative results from the subtractions.

•88

1••|9 8 4 7|94•••| 4 6 5 9

---------------••0|5 2 3 8|50

•12

2••|9 8 4 7|94

•••| 1 0 1 0

-----------------••0|4 1 1 5|614•••|1 1 2-----------------

•••|5 2 3 5|614•••|5 2 3 8|50Note that even though we have shown only examples that involve multiplication and division, the use of vinculums is not restricted to these techniques or operations. The vinculum form of a number can be used anywhere where the normal form of the number can be used and vice versa, making the vinculum very powerful and versatile. To demonstrate this, let us work out a simple example as below:

•8

3•|12 4 3 8|9

••|••0 2 1 1

--------------------

••| 4 0 4 5|699••|•1 3 8

--------------------

••| 3 3 4 5|699

••| 3 2 5 5|699

••| 3 2 7 3|15

In this case, the division of 12 by 3 first gives us 4 as the first quotient digit and a remainder of 0. The two-digit number that results from placing this remainder and the next digit of the numerator together is 04. The product of the quotient digit and the flag digit is 32. 4 - 32 = -28. Dividing -28 by 3 gives us a quotient of -10 and a remainder of 2 (another way to say this is that -10 x 3 + 2 = -28). Note that -9 can not be considered the quotient in this case because -9 x 3 = -27 > -28.

We put the -10 down as vinculum digits in the quotient row (with the 1 carried over to the next column as appropriate), and the 2 as a remainder next to the next numerator digit, 3. From 23, we need to subtract the product of the latest quotient digit and the flag digit. This product is -10 x 8 = -80, and 23 - -80 = 23 + 80 = 103. So our next intermediate numerator is 103, and the process is continued as in the figure.

At the end, we add up the columns of numbers, making sure to account for vinculum numbers by considering them to be negative numbers. If any digits in the final answer are vinculum numbers, we convert them to normal form by decrementing the previous digit and substituting the vinculum digit(s) with its 10's complement (the reverse of the procedure for converting a digit to its vinculum form).

In this case, the remainder turns out to be larger than the denominator, so we have to go through the extra step of dividing the remainder by the quotient, adding this quotient to the original quotient and keeping the remainder as the final remainder.

As you can see, this procedure can sometimes be more convenient than the normal procedure we have used for straight division because we no longer have to adjust the quotients and remainders of intermediate divisions to make sure that intermediate numerators do not become negative. In this method, if the intermediate numerator does become negative, we just shrug it off, put down a vinculum digit in the quotient and continue on.Notice that changing the problem so that we make the flag digit a vinculum digit simplifies it much further, as shown below.

•24•|12 4 3 8|9••|••0 2 3 2---------------••| 3 2 6 2|53

••|•••••1

---------------

••| 3 2 7 2|53••| 3 2 7 3|15In general, making the flag digits vinculum digits pays good dividends because it enables us to work on the problem without having to adjust the intermediate quotients and remainders very often to avoid negative intermediate numerators.

Hopefully, this digression into the land of vinculums and vinculum digits gives you one more weapon to add to your arsenal. It is important to note, as always, that not all the tools in your toolbox can be used at all times. One has to consider the problem carefully before making a shrewd guess about which tool might be able to solve the problem with the least difficulty. Practice is the key to making this kind of judgment, so hopefully you will find the time to practice working with vinculums. Good luck, and happy computing!