## Wednesday, May 19, 2010

### Vedic Mathematics Lesson 50: Recurring Decimals 2

In the previous lesson, we saw how it is possible to calculate the reciprocals of numbers that don't have 2 or 5 as factors, using a simple method that utilizes the Anurupyena sutra and the Sesanyankena Caramena sutra. In this lesson, we will extend that methodology beyond finding reciprocals. We will use a slight modification of that method to find the values of fractions like 2/3, 4/7, etc.

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
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

Note that fractions like 2/3 and 4/7 still can not be expressed in terminating decimal form. Since the denominators of these fractions have factors other than 2 and 5, their decimal representation will still be a recurring decimal. In the case of a fraction such as 2/3, the decimal representation is well known (as 0.66666 . . ., usually rounded off to 0.6666667), while in other cases, such as 4/7, the answer is not well-known. Hopefully, this lesson will enable us to find the answer to a problem like 4/7 without having to do long division with its attendant difficulties.

The method we are going to use in this lesson is a direct derivation of the method we used in the previous lesson. In fact, the steps are almost identical. To save time and effort, I have therefore reproduced the steps from the previous lesson below. For the steps that require modification, I have mentioned the necessary modification in italics. That way, if you are familiar and thorough with the previous methodology through practice, the new method will require very little adjustment to get used to.
1. Pretend that you are going to use long division to find the fractional form of the given number whose denominator ends in 1, 3, 7, or 9. As such, you would put a decimal point on the quotient line, then add a zero to the dividend and then attempt to perform a division. This will result in a quotient and a remainder
2. We are not interested in the quotient though. We are interested in the remainder. More precisely, we are interested in the ratio of the remainder to the numerator. Call this ratio our multiplier. In general, the multiplier need not be a whole number (in the previous lesson the multiplier was always a whole number since we were taking ratios with respect to 1)
3. Write down the first remainder
4. Now, multiply the first remainder by the multiplier, and cast out the divisor (keep subtracting the divisor from it until the number left is less than the divisor). This becomes the next remainder. Write it next to the first remainder, separated from it by a comma. If our multiplier is not a whole number, and the first remainder can not be multiplied by the multiplier to arrive at a whole number answer (for instance 7 multiplied by 1.5 will result in 10.5, which is not a whole number), add the divisor to the remainder and then try the multiplication. In most cases, adding the divisor once to the remainder will give us a number that can be multiplied by the multiplier to give us a whole number, but sometimes, we may have to add the divisor more than once to the remainder
5. Now multiply the next remainder by the multiplier, and again cast out the divisor. Once again write this remainder next to the previous ones as before. In each of these steps, whenever multiplication of the remainder by the multiplier will result in a non-whole number, remember to add the divisor to the remainder and try the multiplication once again
6. Repeat step 5 until at some point, the remainder becomes the numerator. Add that last remainder to the list of remainders and stop repeating step 5
7. Now, multiply the last digit of each remainder by the last digit of the recurring decimal appropriate for the denominator, the sesanyanka (remember that the sesanyanka of a divisor that ends in 1 is 9, that for a divisor that ends in 3 is 3, that for a divisor that ends in 7 is 7 and that for a divisor that ends in 9 is 1), and put the last digits of the products down in order. This step is what is meant by the vedic sutra, Sesanyankena Caramena, the remainders by the last digit
8. At the end of the process, you have a set of digits that represents the recurring decimal representation of the given fraction!
As you can see, the overall method is very similar to the method we used in the previous lesson. The only difference is that the Anurupyena sutra can be a little harder to apply in this case because our multiplier need not be a whole number. Because the multiplier is not a whole number, the product of the multiplier with the remainder need not be a whole number either. The main adjustment we have to make to the method in the previous lesson is how to take this fact into account. We get over it by allowing the remainder to be adjusted by the denominator until the product with the multiplier comes out to be a whole number.

We will apply this methodology to a simple example to illustrate this. Let us try it on 4/7 and see how the method and the adjustments within it perform.
1. We pretend to perform long division. We add a zero to the end of 4 to get 40. This gives us a quotient of 5 and a remainder of 5
2. We don't care about the quotient. But our remainder being 5 tells us that our multiplier is 5/4 or 1.25
3. We write down the remainder, 5
4. Multiplying 5 by our multiplier, 1.25, gives us 6.25. This is not a whole number. So, we perform the adjustment explained in the method above: we add our denominator, 7, to the remainder, 5 to get 12. The product of 12 and 1.25 is 15, which is a whole number. Now, we cast out our divisor, 7, from 15, giving us 1. This becomes our next remainder. Write it next to the 5, giving us 5,1
5. Once again, multiplying 1 by our multiplier 1.25, results in a non-whole number. Therefore, we add 7 to 1, and then multiply by the multiplier, giving us a whole number, 10. Casting out 7 from 10 leaves us with 3. This gives us 5,1,3
6. Multiplying 3 by 1.25 does not give us a whole number. In fact, even after we add 7 to 3, the product with 1.25 is not a whole number. We see that we need to add the divisor, 7, 3 times to the remainder to get 24, whose product with 1.25 is a whole number (30). Casting out 7's from 30 gives us the next digit to add to our series, 2 (the series now reads 5,1,3,2). Our next remainder, 6, comes from multiplying 16 (we have to add 7 twice to our previous reminder, 2, to get a whole number when multiplied by our multiplier) by 1.25 and casting out 7's (the series now becomes 5,1,3,2,6). When we multiply 20 by 1.25 (once again, we have to add 7 twice to the last remainder, 6, to get a number whose product with 1.25 is a whole number), and cast out 7's, our remainder becomes 4. Since 4 is our numerator, this is our signal to stop. Our list of remainders now reads 5, 1, 3, 2, 6, 4
7. The last digit of the product of 5 and 7 (remember that the sesanyanka of 7 is 7 itself) is 5. The last digit of the product of 1 and 7 is 7, and so on.
8. We ultimately get 571428, and we say that 0.571428. . . is the recurring decimal form of 4/7. We can verify that 7x0.571428571428 . . . is very nearly 4 (as we increase the number of times we include 571428 in the product, the number of 9's in the product increases, and when the number of 571428's is infinite, you get 4 as the product).
As you can see, once you are familiar with the method as explained in the previous lesson, the modifications required to apply that method to numerators other than 1 are quite straight-forward and simple. Obviously, as the multiplier becomes more and more complicated, the adjustment can become more cumbersome, but overall, it should still be possible to apply the method reasonably quickly and accurately.

Let us now apply it to 2/13 as another illustration. In this case, when we perform step 1 by adding a zero to 2 and dividing by 13, we get a remainder of 7. We write down 7, and we know that our multiplier is 7/2. Since the product of 7 and 7/2 is not a whole number, we add 13 to the 7, making it 20 before we multiply by 7/2, giving us a product of 70. This then leads to the next remainder in the series, 5. Following the same procedure over and over again, we can fill out the rest of the series until we hit 2 as the remainder (at which point, we stop). This gives us the following series of remainders: 7, 5, 11, 6, 8, 2.

The sesanyanka of 13 is 3 (since the last digit of 13 is 3), so, we multiply the last digits of these remainders by 3 and put down the last digits as the series of digits in the decimal form of 2/13. This gives us 2/13 = 0.153846 . . .. It can be verified that this is indeed the correct answer we are looking for.

Even though our first two examples have shown that our multiplier is a fractional value, this need not always be the case even though the numerator is not 1. Take 2/31 for instance. In this case, our first remainder is 20 (since 20 is less than 31). The multiplier then becomes 20/2, which is 10. This is a whole number, allowing us to calculate 2/31 with no adjustments to the method we put forth in the previous lesson.

Note that you can still use the method in this lesson, which is a more general version of the method of the previous lesson, even for the above problem. You would never need to add the divisor to the previous remainder to make the product with the multiplier a whole number, so the adjustment is never used, and the answer will be identical to what you would have gotten if you had used the method in the previous lesson. It is probably a good idea to learn the method in this lesson thoroughly in place of the simplified method of the previous lesson since the more general method has a wider applicability than the simplified method.

In any case, the series of remainders then becomes 20, 14, 16, 5, 19, 4, 9, 28, 1, 10, 7, 8, 18, 25, 2. Since the sesanyanka of 31 is 9, we take the last digits after multiplication of these remainders by 9 and get the value of 2/31 as follows: 0.064516129032258 . . ..

Let us solve one more problem to make sure the method is fully understood. We will tackle 6/29 using this method. Our first remainder is 2 (adding a zero after 6 and dividing by 29). Our multiplier is therefore 2/6, which is 1/3. Note that, in this case, not only is our multiplier not a whole number, but the multiplier is less than 1. So, you can see that the multiplier in this method can be any positive number and need not follow any rules such as being a whole number or being greater than 1.

Since the product of 2 and 1/3 is not a whole number, we see that we have to add 2 29's to 2 to get a number whose product with 1/3 is a whole number. That gives us 20 as the next remainder. Once again, we add 2 29's to this remainder so that the product with 1/3 is a whole number. This gives us a remainder of 26. We then continue this process until we get a full series of remainders as follows: 2, 20, 26, 28, 19, 16, 15, 5, 21, 7, 12, 4, 11, 23, 27, 9, 3, 1, 10, 13, 14, 24, 8, 22, 17, 25, 18, 6 (we stop at 6 because that is the numerator of our problem).

Notice that in the derivation of this series, most of the remainders were derived simply by applying the multiplier (1/3) to the previous remainder (after adjustment with the divisor as needed). There was hardly any need to actually cast away 29's since the successive remainders were mostly less than 29 (except in the one step that lead to the derivation of 17 as the remainder). Thus, this method can sometimes be easier than the steps would lead you to believe (which itself is quite easy to begin with!). Since the sesanyanka of 29 is 1, conversion of the series of remainders above to the recurring decimal form is trivially easy, leading to the final result: 6/29 = 0.2068965517241379310344827586 . . ..

I hope this lesson has been interesting for everyone, and useful for those who were struggling with long division to find the answers to division problems that involve large denominators that don't have 2 or 5 as factors. The method in this lesson is a very general method that can be used for solving such problems quite quickly. But, in the next lesson, I will introduce another method that can be used to find the answers to problems like 4/13, 6/29, etc., using just the reciprocals of the denominators (which can be easily derived using the method in either this lesson or the previous lesson). Until then, I hope you will practice the method in this lesson with more examples. Good luck, and happy computing!

## Content From TheFreeDictionary.com

In the News

Article of the Day

This Day in History

Today's Birthday

Quote of the Day

Word of the Day

Match Up
 Match each word in the left column with its synonym on the right. When finished, click Answer to see the results. Good luck!

Hangman

Spelling Bee
difficulty level:
score: -