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Saturday, January 1, 2011

Vedic Mathematics Lesson 55: Auxiliary Fractions 4

A very happy new year to one and all of you! Hope you had a wonderful 2010 and are looking forward to an even more fantastic 2011!!

It has been a while since I have updated this series of posts on Vedic Mathematics. Various other things have come up in the meantime preventing me from posting more thoughts on these vedic techniques for arithmetic. But there are still a few topics to cover, so I have decided it is time to create some time for this purpose.

In the last lesson, I hinted that the method we have used for deriving auxiliary fractions and how we used them to perform complicated divisions can be extended even to problems where the denominator does not conform to the patterns we have used so far. In particular, the patterns we have used are denominators that end in a series of 9's or that end in a series of 0's followed by a single 1. The procedure for using the auxiliary fraction technique for solving such division problems was explained in the previous three posts on Vedic Mathematics. In this post, I will talk about a simple extension of this technique that will enhance the usability of this method and make it more versatile.

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 extend the auxiliary fraction technique to fractions in which the denominator does not end in a series of 9's, but instead ends in a series of 9's followed by a number close to but not equal to 9. For instance, this technique can be used to find the solution to problems such as 51/898, 336/19997, and so on. The extension we are talking about is made possible by the application of a vedic sutra that we are already quite familiar with: The Anurupena sutra, which literally means proportionally. How do we use this in this context?

Let us first review the technique we used to perform calculations where the denominator ended in a series of 9's. That technique is explained in detail in this previous post. In particular, for the extension we are proposing in this lesson, the method for deriving the auxiliary fraction is the same. Thus:

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 and the ending digit (which is not a 9) after the decimal point
  • Now, modify the denominator by removing the series of 9's along with the ending digit 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 see this in action for a few fractions.

Let us take 5/28, for instance. Here the denominator contains a series of just o 9's and the ending digit is 8, which is one less than 9. We therefore have to divide the numerator and denominator by 10 to get that ending digit behind the decimal point. This gives us 0.5/2.8 first. Next, we modify the denominator by removing all the 9's and the ending digit (in this case, there are no 9's to remove, just the ending digit), and increasing the number before the decimal point by 1. This then gives us the auxiliary fraction 0.5/3.

Similarly, for 23/6997, we get the auxiliary fraction 0.023/7. For 432/13998, we get 0.432/14, and for 7446/16996, we get 7.446/17, and so on.

Obviously, since the auxiliary fractions for 5/28 as well as 5/29 is just 0.5/3 (similarly, the auxiliary fractions for 432/13998 as well as 432/13999 are both 0.432/14, and so on), the division method must be slightly different to account for the different answers we have to get for these similar-looking but different division problems. So, let us see what the difference is.

The difference comes from the use of the Anurupyena sutra. To understand the difference, it may be best to look at the original technique (which is explained in this previous post) and the modified technique side by side.

We will first deal with the fraction 5/28. As we saw earlier, its auxiliary fraction is 0.5/3. Now we removed just one digit from the denominator (the number of digits removed from the denominator is the number of 9's plus the ending digit). This means that we need to consider the numerator in blocks of one digit (if we had removed two or three digits 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 digit 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 digits the denominator loses during the conversion from the given fraction to the auxiliary fraction). Now, divide 05 by 3 (the denominator of the auxiliary fraction). We get a quotient of 1 and a remainder of 2. Put the quotient, 1, as the first digit of the answer.

Now, prepend the remainder, 2 before the quotient to get 21 as our next temporary dividend. Before we divide again by the denominator though, we apply the Anurupyena sutra at this point. Since the original denominator, 28, is 1 less than the denominator we have previously worked with in auxiliary fractions (29, which ends in a series of 9's), we modify the temporary dividend by adding the product of 1 and our quotient to it. This gives us the correct dividend of 21 + 1*1 = 22.

We then divide this by the 3 (the denominator of the auxiliary fraction). We get a quotient of 7 and a remainder of 1. The quotient, 7, becomes the second digit of the answer. Once again, prepend 0 in front of the quotient 7 to get 17, which becomes our temporary dividend. Our actual dividend is obtained by adding the product of 1 and our quotient to the temporary dividend. Thus, our next dividend becomes 17 + 1*7 = 24. We then divide once again by 3. We get a quotient of 8 (which becomes the third digit of the answer), and a remainder of 0. Prepend it in front of 8 (to get 08) and add 1*8 to it get the next dividend of 16. Once again, divide this by 3 to get the next quotient digit, 5, and a remainder of 1.

So, the main difference from the original technique is the addition of a product of our quotient to the temporary dividend to get the true dividend. The number to multiply by is proportional to the difference between 9 and the last digit of our denominator, hence the application of the Anurupyena sutra. In this example, the difference is 1, so we add the product of 1 and our quotient to each temporary dividend to get our true dividend for the next division.

Let us carry this method forward a few more times. I want to do this to illustrate a complication that may arise in this method, and show you how to get around it. Our last quotient and remainder were 5 and 1. This gives us a temporary dividend of 15 and a true dividend of 15 + 1*5 = 20. Our next quotient then becomes 6 and our next remainder is 2. Our next temporary dividend is therefore 26 and our next true dividend is 26 + 1*6 = 32.

Now, you can see what the complication is: 32 is more than 10 times our denominator, 3. Thus, our next quotient becomes a two digit number. This certainly would seem to throw a wrench into the works here! After all, in the original method, we always got dividends such that the next quotient was the appropriate number of digits. We did not have to deal with the problem we are facing now.

The solution to this problem is not that difficult though. Put down 10 as the next dividend and 2 as the next remainder. The reason we prepend the remainder to the quotient to get our next temporary dividend is so that we have a 2-digit dividend. Here, since the quotient is already 2 digits long, instead of prepending the remainder to it, simply add the remainder to the first digit of the quotient. Thus, our next temporary dividend comes out as 30.

Again, we follow the same procedure for finding the true dividend by adding the product of 1 and the quotient to this temporary dividend. Thus, our true dividend is 30 + 1*10 = 40. Once again, this results in a 2-digit quotient (13), and a remainder of 1. Following the procedure we just explained, our temporary dividend becomes 23, and our true dividend becomes 23 + 1*13 = 36. The next quotient is 12 with a remainder of 0. This gives us a temporary dividend of 12 and a true dividend of 12 + 1*12 = 24. This then gives us our next quotient and remainder of 8 and 0 respectively.

All this will become clearer from the illustration below. The first line of the illustration contains the quotients of our divisions, and thus constitutes the answer we are calculating. The third line contains the remainders from the successive divisions. Thus, our temporary dividends are formed by either prepending the digit in the third line to the digit in the first line (if the number in the first line is one digit long), or by adding the digit in the third line to the first digit of the number in the first line (if the number in the first line is more than one digit long). The second line contains the true dividends obtained by adding the product of 1 and the quotient to the temporary dividend):
 1  7  8  5  6 10 13 12  8  5  6 10 13 12  8
22 24 16 20 32 40 36 24 16 20 32 40 36 24 16
2 1 0 1 2 2 1 0 0 1 2 2 1 0 0
The last quotient in the series is 8 and the remainder is 0. So, our next division would be of 16 by 3, which we have already done twice in this series before. So, there is nothing to be gained by repeating the process any more. In fact, we could have stopped when we encountered our first repetition 6 steps back, when our quotient was once again 8 and our remainder was 0.

This then seems to give us the answer as 0.17856101312856101312 . . .. But anyone with a calculator can tell you that this is not the correct answer! What happened here? Does the technique even work at all? Well, it does work. But we do need one more modification to our original technique to account for the 2-digit quotients in our answer line. What we need to do is carry over the first digit of the 2-digit quotients to the previous quotients. Thus, our true answer becomes:

0.1785(6+1)(0+1)(3+1)285(6+1)(0+1)(3+1)285 . . .

Which can be written in more conventional form as:

0.1785714285714857142 . . .

Now, the above number can be verified to be the correct answer to the original problem, 5/28. So, the technique does work, it just requires a few extra steps to take care of the complications we encountered!

We will now illustrate this method with one more problem (23/6997) to make sure we understand the principles fully. In this case, we already derived an auxiliary fraction of 0.023/7. Our block size is 3 since we removed 3 digits from the denominator. As always, we first ignore the decimal points in the numerator and just take the numbers, 23. We then pad the numbers with zeroes in front so that the total number of digits is one more than the block size of 3. Thus, our first division is of 0023 by 7.

This gives us a quotient of 3 and a remainder of 2. Since our block size is 3, we pad the quotient, 3, with zeroes to the front to make it 3 digits long. Thus, our quotient (and the first three digits of our answer) become 003. Prepending the remainder in front of it gives us 2003. Now, since our denominator did not end with a series of 9's but rather with a series of 9's followed by a 7, we need to make the appropriate adjustments based on the application of the Anurupyena sutra. Since the difference between 7 and 9 is 2, we add 2*the quotient to the temporary dividend at each step to get the true dividend.

Thus, our first true dividend becomes 2003 + 2*3 = 2009. Dividing 2009 by 7 gives us a quotient of 287, and a remainder of 0. 287 is already 3 digits long, so there is no need to pad it with zeroes in front. Our next temporary dividend therefore becomes 0287, and our next true dividend is 0287 + 2*287 = 0861. We can then repeat the procedure as many times as we want to add digits to our answer. See the illustration below to see how this is done:
 003  287  123  052  736  886 1236  1386 1451  764 1041 1160
2009 0861 0369 5156 6208 8658 9708 10158 5353 7292 8123 6480
2 0 0 5 4 6 6 6 1 5 5 3
As you can see in the illustration above, we have several dividends that are greater than 7000, and several blocks of quotient digits that are more than 3 digits long. We have to be careful to make sure we apply the modifications we came up with in this lesson correctly, otherwise, we will end up with the wrong answers. But ultimately, from the illustration above, all we have to do is perform the carryovers of the answer digits, and we are left with the answer below:

23/6997 = 0.003287123052736887237387451765042160 . . .

In this case, I have used ellipsis (. . .) to indicate that the answer is incomplete, not to indicate that the digits I have put down actually repeat. In fact, it would be a little tedious to continue this calculation to that point since the true answer has a period of 1749! You can verify that the answer above is correct, but it is a little tricky without doing division by hand since most calculators don't show this many digits. But, as you can see, we have computed so many digits in the answer with very little effort compared to actually doing long division by 6997 and finding the answer digits one by one.

We already knew that auxiliary fractions are extremely useful in reducing computational complexity in division problems with some types of large denominators. In this lesson, we have seen how the types of large denominators can be expanded to include more numbers so that the method has wider applicability. The techniques in this lesson may appear confusing at first, but practice will make the techniques second nature. In the next lesson, we will look at one more enhancement of the auxiliary fraction technique. In the meantime, good luck, and happy computing!

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