You can find all my 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 2

Multiplication Special Case 3

Vertically And Crosswise I

Vertically And Crosswise II

Squaring, Cubing, Etc.

Subtraction

Division By The Nikhilam Method I

Division By The Nikhilam Method II

Division By The Nikhilam Method III

Division By The Paravartya Method

Digital Roots

Straight Division I

Straight Division II

Vinculums

Divisibility Rules

Simple Osculation

Multiplex Osculation

Solving Equations 1

Solving Equations 2

Solving Equations 3

Solving Equations 4

Mergers 1

Mergers 2

Mergers 3

Multiple Mergers

Complex Mergers

Simultaneous Equations 1

Simultaneous Equations 2

Quadratic Equations 1

Quadratic Equations 2

Quadratic Equations 3

Quadratic Equations 4

Cubic Equations

Quartic Equations

Polynomial Division 1

Polynomial Division 2

Polynomial Division 3

Square Roots 1

Square Roots 2

Square Roots 3

Recurring Decimals 1

Recurring Decimals 2

Recurring Decimals 3

The basic idea behind the use of auxiliary fractions is to convert the given division problem into a slightly different one and then use a modified division technique to solve the problem. The slightly different fraction is called the auxiliary fraction of the given fraction which we are trying to evaluate.

There are several types of auxiliary fractions, depending on the nature of the given fraction. We will deal with the first of these types in this lesson. Those are fractions whose denominator ends with one of more 9's. Examples of such fractions include 5/19, 43/899, 22/199, 632/1299, 8765/19999, etc.

To derive the auxiliary fraction for these types of fractions, follow these steps:

- Divide the numerator and denominator by the appropriate power of 10 so that the denominator has the entire series of 9's after the decimal point
- Now, modify the denominator by removing the series of 9's after the decimal point and adding one to the part before the decimal point (this is the application of the Ekadhikena Purvena sutra)
- The auxiliary fraction is now the new numerator divided by the new denominator

Let us take 5/19, for instance. Since the denominator contains a series of just 1 9, we have to divide the numerator and denominator by 10 to get that 9 behind the decimal point. This gives us 0.5/1.9 first. Next, we modify the denominator by removing all the 9's and increasing the number before the decimal point by 1. This then gives us the auxiliary fraction 0.5/2.

Similarly, for 43/899, we get the auxiliary fraction 0.43/9. For 632/1299, we get 6.32/13, and for 1456/13999, we get 1.456/14, and so on.

What do we do with the auxiliary fraction and how do they help us derive the value of the given fraction? The procedure for that is best explained with a couple of examples. So, we will first deal with the fraction 5/19. As we saw earlier, its auxiliary fraction is 0.5/2. Now we removed just one 9 from the denominator. This means that we need to consider the numerator in blocks of one digit (if we had removed two or three 9's from the denominator, the block size goes up correspondingly).

Ignore the decimal point in the numerator, and just look at the number. This will give us our first dividend, which is 5. Since we removed one 9 from the denominator when we converted the original fraction to the auxiliary fraction, our block length is 1. Therefore, pad the numerator with a zero to the left to get a 2-digit dividend, 05. Our first dividend is always going to have one more digit than the block-length (and the block-length is always going to be the number of 9's the denominator loses during the conversion from the given fraction to the auxiliary fraction). Now, divide 05 by 2 (the denominator of the auxiliary fraction). We get a quotient of 2 and a remainder of 1. Put the quotient, 2, as the first digit of the answer.

Now, prepend the remainder, 1 before the quotient to get 12 as our next dividend, and divide again by the denominator, 2. We get a quotient of 6 and a remainder of 0. The quotient, 6, becomes the second digit of the answer. Once again, prepend 0 in front of the quotient 6 to get 06, and divide once again by 2. We get a quotient of 3 (which becomes the third digit of the answer), and a remainder of 0. Prepend it in front of 3 (to get 03) and divide by 2 to get the next quotient digit, 1, and a remainder of 1.

Repeat the procedure of prepending the remainder to the quotient and dividing by the denominator of the auxiliary fraction repeatedly to get the following series of answer digits (the remainders are written on the next line so that you can follow the actual divisions that were performed by 2):

2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1 0 5The last quotient in the series is 5 and the remainder is 0. So, our next division would be of 05 by 2, which is the first division we performed to start the process. That is the signal that we have to stop at this point. This then gives us the answer 5/19 = 0.263157894736842105 . . ..

1 0 0 1 1 1 1 0 1 0 1 1 0 0 0 0 1 0

As you can see, you can work out this series of digits in your mind with some practice. Obviously, this will work for finding the reciprocal of 19 if you choose also (using 1 as the numerator instead of 5 as in this case), and you will notice that this in fact the exact same method we used in our very first lesson about the reciprocals of numbers ending in 9. But the method is applicable to multiples of these reciprocals also.

What if our fraction contained more 9's in a series rather than just 1? Let us now take the case of 43/899 next. We saw that the auxiliary fraction in this case is 0.43/9. In this case, we removed two 9's from the denominator during the derivation of the auxiliary fraction, so we need to consider the numerator in blocks of 2 digits (in other words, our block-length is 2). Let us see how that works.

First ignore the decimal point in the numerator and just look at the number, 43. Since our block length is 2 digits long, pad the numerator with zeroes to the left to make the first dividend three digits long. This gives us 043 as our first dividend. Divide 043 by 9, and express the quotient as a 2-digit number (since our block size is now 2).

This will give us a quotient of 04 and a remainder of 7. 04 becomes the first two digits of the answer. Prepend the 7 in front of the quotient to get our next dividend, 704. Once again divide by 9 to get a quotient of 78 and a remainder of 2. Since 78 is already a 2-digit number, there is no need to pad it with zeroes to the left. We just write 78 as the next two digits of the answer. Now prepend the remainder, 2, in front of 78 to get 278 as our next dividend. Divide by 9 to get 30 as the next two digits of the answer and 8 as the remainder. This will then lead to 830 as our next dividend, and so on.

Continuing this process a few more times, we get the following series of digits in our answer. The remainders in each case are written in the next line so that we know what the dividend was for each division by 9:

04 78 30 92 32 48 05 33 92 65 85 09 45 49 49 94 . . .I have not completed the full answer because in this case, the recurring decimal actually has a period of 420! But you can see how easy it is to actually add digits to the answer using a simple process of division by 9 rather than trying to do long division by 899. That is the power of auxiliary fractions. Note also, how I have written some quotients as two digit numbers on the top line by padding them to the left with zeroes. In fact, we had to perform such padding in the very first division we performed. But in this case, we also got 05 and 09 as quotients in subsequent steps by padding out the single-digit quotients, 5 and 9, with zeroes to the left. This is important to remember, otherwise your answer will not be correct.

7••2••8••2••4••0••3••8••5••7••0••4••4••4••8••3•• . . .

In this case, our answer actually becomes 43/899 = 0.04783092324805339265850945494994 . . .. In this case, I am using ellipsis (. . .) to indicate that the answer is not complete rather than to indicate that the previous set of digits repeats (because the actual answer all the way out to the point where it starts recurring would occupy too much space). Note that your signal to stop (if you choose to follow the process and derive all 420 digits of the recurring decimal representation of 43/899) would come when your next dividend becomes 043, which was our first dividend.

We will stop this lesson with a couple more examples, so that we can be sure we fully comprehend the process and its steps. For this let us take the example of 632/1299. The conversion of this fraction to its auxiliary form gives us 6.32/13. In this case, the block-length is 2, and our first dividend should be 3 digits long (one longer than the block-length). This gives us 632 as our first dividend.

Dividing 632 by 13, we get the two-digit quotient 48, and a remainder of 8. This makes 848 our next dividend, giving us the next two digits of our answer as 65, and a remainder of 3. This makes our next dividend 365. The series continues as below:

48 65 28••09 85 37 33 64 12 . . .Note how our fourth remainder was not a single digit number but a 2-digit one. This does not in any way affect the procedure at all. We just prepend it to the quotient just as if it were a single-digit remainder and proceed on. After the fourth step, our next dividend becomes 1109, and we just continue on as usual. When the divisor is 2 digits long (as in this case), sometimes the remainders can be 2 digits long too. I chose this example to illustrate that this does not change the procedure.

8••3••1••11••4••4••8••1••8•• . . .

So, 632/1299 = 0.486528098537336412 . . .. Once again I use ellipsis to indicate that the answer is not complete since the period of 632/1299 is 432!

Let us take 631/999 as our last example. In this case, we see that the denominator of the given fraction consists of nothing but 9's. How are we supposed to handle this? There are actually two equivalent ways of handling this. The first is to consider the denominator to be 0999. We then divide the numerator and denominator by 1000 to get 0.631/0.999, which then gives us the auxiliary fraction of 0.631/1. Otherwise, we can divide by 100, and get 6.31/9.99, which gives us an auxiliary fraction of 6.31/10.

Let us work out the answer to the given fraction using both auxiliary fractions. In the first case, our first dividend is 0631 (notice how the block-length in this case is 3, and therefore the dividend has to be 4 digits long) and our divisor is 1. This gives us a quotient of 631 and a remainder of 0. Our next dividend is therefore 0631, which is precisely what our first dividend was. Therefore, this is our signal to stop, which gives us the answer 0.631 . . . (in this case, the ellipsis is being used to signal recurrence rather than incompleteness).

In the case of the fraction 6.31/10, our first dividend is 631 (we don't pad it with a zero because our block length is only 2 and therefore our first dividend needs to be only 3 digits long). This then gives us our first quotient of 63 and a remainder of 1. This makes our next dividend 163. The next quotient is then 16, with a remainder of 3. This gives us our next dividend of 316. This finally leads to a quotient of 31 and a remainder of 6, which makes the next dividend 631. We stop at this point since this is the same as our first dividend. The series of digits derived using this auxiliary fraction is given below:

63 16 31We see that our answer now becomes 0.631631 . . .. Actually this is exactly the same as the previous answer, but the recurring part is repeated twice in this answer rather than just once. The difference arises because the period of the given fraction is 3. Therefore when we find the answer using a method whose block-length is 3, we find the answer with no repetition. But when we use a method that has a block-length of 2, it is forced to extend the answer out to 6 digits which is the lowest common multiple (LCM) of our block-length (2) and period (3).

1••3••6

This shows you how your choice of auxiliary fraction can sometimes make the answer appear different even though it eventually works out to the same. Since you don't normally have a choice of multiple auxiliary fractions to work with, this should not be a big concern, but it is good to keep in mind if you do encounter such a case.

Dealing with denominators which end with a series of 9's is just one application of auxiliary fractions. There are auxiliary fractions for dealing with fractions whose denominators have other patters also. Those will have to wait till later, though.

Auxiliary fractions are very useful in reducing the computational burden significantly for divisions in which the denominators are very large, and long division is cumbersome and error-prone. We will expand on our understanding of auxiliary fractions in subsequent lessons and show how it is possible to tackle several different division problems using the concept of auxiliary fractions. In the meantime, good luck, and happy computing!

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