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Wednesday, June 16, 2010

Vedic Mathematics Lesson 53: Auxiliary Fractions 2

In addition to numbers ending in 9 or a series of 9's (as we saw in the previous lesson), auxiliary fractions are available for numbers ending in 1 also. In this lesson, we will learn how to use this second type of auxiliary fraction to solve division problems that can appear quite complicated at first sight because of the large size of the denominator.

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

As mentioned earlier, the second type of division for which auxiliary fractions are available involve division by numbers that end with 1 or end with a series of 0's followed by 1. Examples of such fractions include 25/81, 39/101, 14/121, 632/3001, 30/91, etc.

The derivation of the auxiliary fraction for such fractions is also quite simple. Divide both the numerator and denominator by the appropriate power of 10 so that the 1 or the series of 0's followed by 1 in the denominator is after the decimal point. Then, reduce the last digit of the numerator by 1 (this last digit could be beyond the decimal point), and remove all the numbers beyond the decimal point in the denominator. The astute reader will have deduced by now that this is an application of the Ekanyunena Purvena sutra rather than the previous type of auxiliary fraction which was an application of the Ekadhikena Purvena sutra.

Let us see how this works in some examples. Take 25/81, for instance. The denominator ends in 1, so we have to derive the second type of auxiliary fraction in this case. Since the denominator contains just a 1 (with no series of 0's in front of it), we divide both the numerator and denominator by 10 to get 2.5/8.1 first. Next, we reduce the last digit of the numerator (5) by 1, and remove the digits beyond the decimal point in the denominator. This gives us the auxiliary fraction 2.4/8.

Similarly, for 39/101, we first divide by 100 (because the denominator consists of a 0 followed by a 1) to get 0.39/1.01, and then from there, 0.38/1. 14/121 becomes 1.3/12. 632/3001 becomes 0.631/3.

Now consider the case of 30/91. Based on the structure of the denominator, we divide the numerator and denominator by 10 to initially get 3.0/9.1. But now, how do we reduce the last digit of the numerator by 1? It is already 0. It turns out, normal borrowing rules apply. To reduce the last digit by 1, we simply borrow from the digit to the left, and then do the subtraction of 1 from it, so we get 2.9 in the numerator. The auxiliary fraction then becomes 2.9/9.

Now that we have derived the auxiliary fraction, how do we use it to perform the division and get the value of the given fraction? The procedure is slightly different from the one we used to perform the division in the case of the first type of auxiliary fraction.

For this type of auxiliary fraction, first of all, the block size is the number of digits removed from the denominator. Thus the block size after conversion of 25/81 to 2.4/8 is 1. The block size after conversion of 39/101 to 0.38/1 is 2. The block size after conversion of 632/3001 to 0.631/3 is 3, and so on.

To perform the actual division, the steps are outlined as below:
  • As is the case with the previous type of auxiliary fraction, 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 dividend.
  • 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 and performing the division once again and so on.
  • When you reach a point where your next dividend is the exactly same as the first dividend, that is your signal to stop.
Let us see how we apply this methodology by working out a few examples. Let us take 25/81 first. As we saw earlier, the auxiliary fraction for this given fraction is 2.4/8, and the block length is 1.

Ignoring the decimal point in the numerator, our first dividend is 24. Dividing this by 8 gives us a quotient of 3 and a remainder of 0. We put down 3 as the first digit of the answer (we don't have to pad 3 with zeroes because 3 is already 1 digit long, the same length as our block length). The 9's complement of 3 is 6 (the difference from 9), so our next dividend is our remainder (0) prepended in front of the 6, giving us 06.

6 divided by 8 gives us a quotient of 0 (the next digit of our answer), and a remainder of 6. The 9's complement of 0 is 9, so our next dividend becomes 69 (our remainder prepended in front of the 9's complement of the quotient). 69 divided by 8 gives us a quotient of 8 and a remainder of 5. Our next dividend therefore becomes 51 (5 prepended in front of the 9's complement of 8, which is 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 remainders from the divisions. The next dividend in each case is the number on the third line prepended in front of the number in the middle line.
•3 0 8 6 4 1 9 7 5
•6 9 1 3 5 8 0 2 4
0 6 5 3 1 7 6 4 2
We see that the next dividend in this case would be 24, which is the same dividend we started our divisions with. This is our signal that the set of recurring digits has been fully enumerated. We can now say that our answer is 25/81 = 0.308641975 . . .. As you can see, this can all be worked out quite a bit more easily than by trying to perform long division by 81. All we are doing is performing divisions by the much smaller 8, and taking differences from 9!

Let us now tackle 14/121 next. We saw earlier that the auxiliary fraction for this fraction is 1.3/12. Moreover, the block length is 1. The first division in this case is of 13 by 12. The quotient is 1 and the remainder is 1. The 9's complement of the quotient, 1, is 8, so we prepend our remainder 1 to this 8 to get our next dividend, 18. The process continues just as in the case of 25/81, with the resulting figure shown below:
•1 1 5 7 0 2 4 •7 9 3 •3 •8 8 4 •2 9 7
•8 8 4 2 9 7 5 •2 0 6 •6 •1 1 5 •7 0 2
1 6 8 0 2 5 9 11 4 4 10 10 5 3 11 9 6
As you can see, because our divisor is larger than 10, we get a remainder greater than 10 in several of the steps in the division process above. This does not affect the procedure in any significant way. We just prepend whatever remainder we get (whether it is 1 digit or 2 digits long) in front of the 9's complement of the quotient and continue our division without any changes or modifications.

In the case of the above division, we have not completed the process because we have not yet encountered a dividend equal to 14, which was our very first dividend. I stopped the process because I don't want to take up the space required to list out all 22 digits in the answer we are looking for (yes, 14/121 has a period of 22). We have stopped by saying that 14/121 is approximately 0.11570247933884297 . . . (with the ellipsis signifying incompleteness rather than repetition in this case).

Let us now work out 39/101 as our next example. In this case, our auxiliary fraction is 0.38/1. The block length is not 1 in this case though. Since we eliminated the 1 and the preceding 0 from the denominator, our block length is 2. The only change this creates in the process is the requirement that our quotient be in blocks of 2 digits length, and we have to take the complements of this 2-digit quotient from 99 (series of 9's of the same length as the block length).

Thus, our first division gives us a quotient of 38 and a remainder of 0 (in fact in this case, our remainder is always going to be 0 since we are dividing in each step by 1). The 99's complement of 38 is 61. So our next dividend is 061, which gives us a quotient of 61 and a remainder of 0. Our next dividend then becomes 038 once again, which is what we started the process with, so we stop at this point, with an answer of 0.3861 . . ..

We could, of course, have derived an auxiliary fraction of 3.8/10 for this problem, with a block length of 1. This would have led to the same final result finally, but through a slightly lengthier division process as illustrated below:
•3 8 6 1
•6 1 3 8
8 6 1 3
At this point, we are left with a dividend of 38 once again, which is what we started the process with, so we stop, and deduce that the answer, once again, is 39/101 = 0.3861 . . ..

We will conclude this lesson by dealing with one final problem, 632/3001. This fraction yields an auxiliary fraction of 0.631/3, with a block length of 3. You could consider the auxiliary fraction to be 6.31/30 with a block length of 2 or 63.1/300 with a block length of 1. But in general, you will see that the smaller the denominator of the auxiliary fraction (the actual divisor in the series of divisions you will be performing), the easier the divisions are and the faster it is to derive the final answer. So, we will consider 0.631/3 as the auxiliary fraction with a block length of 3.

Dividing 631 by 3 gives us a quotient of 210 and a remainder of 1. Since the quotient is already 3 digits long, we don't have to pad it with 0's to make its length equal to the block length in this instance, but if our quotient was ever less than 3 digits long, we would have to do that. Also, since our block length is 3, we have to take complements from 999 rather than 9 or 99. Our next dividend then becomes 1789. The series of divisions is illustrated below:
•210 596 467 844 051 982 672
•789 403 532 155 948 017 327
1•••1•••2•••0•••2•••2•••1
Notice how we have padded the quotient in the first step (17) with a zero in front to make its length equal to the block length (3). This is important to remember so that the final answer is derived correctly. Our final answer is 632/3001 = 0.210596467844051982672 . . . (with the ellipsis indicating incompleteness in this case. The period of the given fraction is 1500, so it would take several more steps to complete the fraction fully!).

Hope this second type of auxiliary fraction has added another weapon in your arsenal for dealing with recurring fractions. Used correctly, auxiliary fractions have the potential to make division problems significantly easier to work out than solving them using long division. Familiarizing yourself with how to convert fractions to their auxiliary form and then how to perform the actual division will make it easy for you to apply it when the need arises. Good luck, and happy computing!

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