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

Auxiliary Fractions 1

Auxiliary Fractions 2

Auxiliary Fractions 3

In this lesson, we will revisit the other type of fractions that are amenable to treatment using auxiliary fractions, fractions whose denominators end in a series of 0's followed by a 1. We will see how to modify the method we used for such problems to take into account denominators whose last digit is not a 1, but something close.

To appreciate the method we deal with in this lesson, it is best to review the original methodology for problems such as this which is explained in this previous post. The method we are going to use in this lesson is quite similar to the original method with only slight modifications, so it helps to be familiar with the original method, and to compare this method with it side by side.

The derivation of the auxiliary fraction is actually identical to the method we originally explained. Let us take fractions such as 5/42, 31/103, 641/6002, etc. In each case, we divide the numerator and denominator by the appropriate power of 10 so that the series of 0's (if any) and the number that is close to 1 in the denominator end up after the decimal point. Thus 5/42 becomes 0.5/4.2, 31/103 becomes 0.31/1.03 and 641/6002 becomes 0.641/6.002. Next we reduce the last digit of the numerator by 1, and remove everything after the decimal point in the denominator. Thus the auxiliary fraction of 5/42 is 0.4/4, that of 31/103 is 0.3/1 and that of 641/6002 is 0.640/6.

We can see that the auxiliary fraction is actually the same for both 5/41 as well as 5/42, and similarly it is the same for 31/103 and 31/101. As you may have guessed, it is the same for 641/6002 and 641/6001. If the auxiliary fractions are the same, how do we get different answers for the different fractions? The modifications from the original method obviously lie in the actual division method, so that the answers to the division come out different. The steps are explained in detail below. As always, review the original method so that you are familiar with concepts such as block length, 9's complement, etc.

To perform the actual division, the steps are outlined as below:

- As is the original method explained in this previous lesson, ignore decimal points in the numerator.
- Divide the number in the numerator by the denominator, and write the quotient as a number that has the same number of digits as the block length.
- Write the resulting quotient as the digits of the answer.
- For the next division, take the 9's complement of the quotient (difference from a series of 9's as long as the block-length), append it to the remainder of the previous division, and consider the resulting number to be the next temporary dividend.
- To get the true dividend, find the difference between the last digit of the given fraction and 1. Multiply the quotient by this difference and subtract the result from the temporary dividend to get the true dividend. This is the application of the Anurupyena sutra that we talked about earlier in this lesson.
- Perform the division process once again, writing the quotient as the next digits of the answer, prepending the remainder to the 9's complement of the quotient, subtracting a multiple of the quotient from the resulting dividend to get the true dividend, and performing the division once again and so on.

Ignoring the decimal point in the numerator, our first dividend is 4. Dividing this by 4 gives us a quotient of 1 and a remainder of 0. We put down 1 as the first digit of the answer (we don't have to pad 1 with zeroes because 1 is already 1 digit long, the same length as our block length). The 9's complement of 1 is 8 (the difference from 9), so our next temporary dividend is our remainder (0) prepended in front of the 8, giving us 08.

Our denominator in this case ends in 2 instead of 1. The difference is 1, so we have multiply our quotient by 1 and subtract it from the temporary dividend. This gives us our true dividend, which works out to be 8 - 1*1 = 7.

7 divided by 4 gives us a quotient of 1 (the next digit of our answer), and a remainder of 3. The 9's complement of 1 is 8, so our next temporary dividend becomes 38 (our remainder prepended in front of the 9's complement of the quotient). The true dividend is 38 - 1*1 = 37. 37 divided by 4 gives us a quotient of 9 and a remainder of 1. Our next temporary and permanent dividends therefore become 10 (1 prepended in front of the 9's complement of 9, which is 0), and 10 - 1*9 = 1 and so on. The figure below represents this set of divisions. The top line contains the quotients and constitute the digits of the answer. The second line contains the 9's complements of the quotients. The third line contains the true dividends to be used in the next division. The last line contains the remainders from the divisions. The next dividend in each case is the number on the fourth line prepended in front of the number in the second line - 1*the number on the first line.

7 divided by 4 gives us a quotient of 1 (the next digit of our answer), and a remainder of 3. The 9's complement of 1 is 8, so our next temporary dividend becomes 38 (our remainder prepended in front of the 9's complement of the quotient). The true dividend is 38 - 1*1 = 37. 37 divided by 4 gives us a quotient of 9 and a remainder of 1. Our next temporary and permanent dividends therefore become 10 (1 prepended in front of the 9's complement of 9, which is 0), and 10 - 1*9 = 1 and so on. The figure below represents this set of divisions. The top line contains the quotients and constitute the digits of the answer. The second line contains the 9's complements of the quotients. The third line contains the true dividends to be used in the next division. The last line contains the remainders from the divisions. The next dividend in each case is the number on the fourth line prepended in front of the number in the second line - 1*the number on the first line.

1 1 9 0 4 7 6 1We see that the last two true dividends are actually repeats of the first two dividends, so we can stop at this point, confident that the digits in the answer have started repeating themselves. This then gives us the final answer as:

8 8 0 9 5 2 3 8

07 37 01 19 31 25 07 37

0 3 1 1 3 3 1 3

5/42 = 0.1190476190476190476 . . .

This answer can be verified in a calculator quite easily.

Let us now work out 31/103 in a similar fashion to reinforce our understanding of the technique. We already calculated that the auxiliary fraction of this fraction is 0.30/1. The block length is 2. We also note that since the last digit of the denominator is 3 instead of 1, we have to subtract twice the quotient from the temporary dividend in order to get the true remainder.

Our first division is 30 by 1. This gives us a quotient of 30 and a remainder of 0. The 9's complement of 30 is 69. So, our next temporary dividend is 069, and our next true dividend is 069 - 2*30 = 009. This is divided by 1 to give us the next quotient of 09 (remember that our block length is 2, so the quotient has to be 2 digits long, padded with zeroes in front of it if necessary), and a remainder of 0.

The 9's complement of 09 is 90, which is our next temporary dividend. 90 - 2*9 is 72, which becomes our next true dividend. We then get a quotient of 72 and a remainder of 0 from the division of 72 by 1. The 9's complement of 72 is 27, and our temporary dividend becomes 027. But 027 - 2*72 is a negative number, so we encounter our first complication when using this method.

Our true dividend can not be negative, so we have to adjust the quotient and remainder to make sure that the true dividend becomes positive. Remember that you can always increase the remainder by the divisor every time you decrease the quotient by 1, and the division is still valid. Thus instead of a quotient of 72 and remainder of 0, we find that we have to use a quotient of 70 and remainder of 2 to make the method work (notice that 72*1 + 0 is 72, and 70*1 + 2 is also 72).

Now, our next temporary dividend becomes 229 (the remainder 2, prepended in front of the 9's complement of our quotient, 70). 229 - 2*70 = 89 is our next true dividend. But once again, we find that using a quotient of 89 and a remainder of 0 does not work because our true dividend becomes 10 - 2*89, which is a negative number. We find that to keep the true dividend positive, we need to use a quotient of 87 and a remainder of 2. We repeat the process as long as we want to find more and more digits in the answer. The result is illustrated below:

30 09 70 87 37 86 40 77 66 99 02 91 26 21 35 92 23Our last dividend is 030, which is also the dividend we started with. We can take this as our signal to stop, and conclude that the digits we have found so far repeat in the answer indefinitely. Based on the illustration, we can say that:

69 90 29 12 62 13 59 22 33 00 97 08 73 78 64 07 76

009 072 089 038 088 041 079 068 101 002 093 026 021 036 094 023 030

0 0 2 2 1 2 1 2 2 2 0 2 0 0 1 2 0

31/103 = 0.3009708737864077669902912621359223 . . .

I have used ellipsis (. . .) to indicate that the digits above repeat (in other words, the answer above is complete). In this case, we actually found the full set of repeating digits for this fraction, but in many cases, this may not be possible because the period of the given fraction is too high. In this case, the period is only 34, so it fits on a line without much of a problem!

As you can see, the method we used above is much simpler than doing long division by 103 and finding quotient digits 1 by 1. Note, also, that recognizing when to stop the long division because digit repetition has begun is much harder since individual digits do occur repeatedly in the answer at various places. In our method, all we have to do is keep track of our true dividends and figure out when one of them repeats. That is our signal that the following sets of quotient digits is going to repeat itself. So, you can stop without expending any additional effort.

Auxiliary fractions are a great time-saver in performing divisions with big and seemingly complicated divisors. Once you understand the principles involved in using them, including the modifications introduced in this lesson and the previous one, you can see that you have added several very powerful tools to your arithmetic arsenal. Keep the weapons in your arsenal sharp by practicing with them on a regular basis. Good luck and happy computing!

## 8 comments:

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