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Sunday, June 28, 2009

Vedic Mathematics Lesson 1: A Spectacular Illustration

You can read about how I got interested in Vedic Mathematics in my previous post. In this post, I will try to justify the praise I had for the mental shortcuts Vedic Mathematics has taught me by using a spectacular illustration. In this lesson, we will learn to find the reciprocals of two-digit numbers ending in 9. Thus, we are interested in finding the decimal equivalents of 1/19, 1/29, etc. I will show you 2 methods to work these problems out, and both these methods produce the exact same solution.

Now, note that all these reciprocals are long series of repeating digits in decimal form. There is no closed-form decimal form to any of these reciprocals. The series of repeating digits can be quite long in many cases (18 digits in the case of 1/19, 28 digits in the case of 1/29 and 42 digits in the case of 1/49, for instance). As such, finding the full set of digits that repeat is almost impossible in most calculators that are limited to 8 to 12 digits. Doing the set of long divisions that will result in finding all the repeating digits of the decimal representation involves 18 steps for 1/19, 28 steps for 1/29, 42 steps for 1/49, etc. It is a long and laborious process that consumes a lot of time and paper, and is prone to errors just because of the amount of time and drudgery involved. The only calculator I have found that can show all the repeating digits of these answers is the Microsoft Power Toy Calculator, which is part of the Microsoft Power Toys For Windows XP, which I mentioned in my post on free software suggestions.

With the methods illustrated here, you will be able to find the entire set of repeating digits in one long string without having to do any long division. You can choose to either find the digits from right to left or left to right depending on whether you are more comfortable with simple single-digit multiplications or single-digit divisions. I will illustrate and explain both methods and you can choose whichever method you are more comfortable with when the need arises.

The relevant vedic sutra that addresses this problem simply reads: Ekadhikena Purvena. Literally translated, it says, by one more than the previous one. This is typical of most vedic sutras which are cryptic, and rely on a guru to expound on the full meaning and applications of it to his students (sishyas). The tradition is then carried forward over subsequent generations in what is referred to as the Guru-Sishya parampara. The meaning of the sutra will become apparent as we work through some examples.

Let us take 1/19 first. As mentioned earlier, there are two methods of working out the answer. One of them involves basic multiplication and is worked out from right to left. The other method involves basic division and is worked out from left to right.

Method 1:

We first apply the literal meaning of the sutra and take one more than the number previous to the right-most digit of the denominator. In this case, we get 1 more than 1, which is 2. Call this number our Multiplier.

  1. The method starts out by putting a 1 as the right most digit of the answer. So, our answer so far is 1.
  2. Now take the last number we wrote and multiply it by our Multiplier. We get 1x2 = 2. Write this digit to the left of 1 as part of the answer. Our answer so far is therefore 21.
  3. Now take the last number we wrote down and multiply it by our Multiplier. We get 2x2 = 4. Write this digit to the left of the answer to get 421 as the answer so far.
  4. After we repeat the above process once more, we get 8421 as the answer.
  5. Now, when we repeat the process above, we get 8x2 = 16 which is a two-digit number. When the result is a two-digit number, write down the last digit of the 2-digit number and remember the other digit as a carry-over. Thus, our answer becomes 68421, and we have a carryover of 1.
  6. Repeat the process again. We get 6x2 = 12. Since we have a carryover number, add that to the answer to get 12+1 = 13. Since this is a two-digit number, repeat the process outlined above: write down 3 as part of the answer and keep 1 as the carryover number. Our answer becomes 368421 so far.
  7. Repeating the process again, we get 3x2 = 6, add the carryover number to it to get 6+1 = 7. Write that to the left of the answer to get 7368421.
  8. In the next step, the answer becomes 47368421 with a carryover number of 1.
  9. In the next step, the answer becomes 947368421 (multiply 4 by 2 and add the carryover number).
  10. Continue this process until you start getting repeating digits (starting with 1 which is the last digit of the answer). Then you know that you have found the entire series of digits comprising the answer. Put a decimal point in front of the series of digits and denote that the entire series of digits repeats indefinitely. That is your final answer. In the case of 1/19, the answer is .052631578947368421, repeated indefinitely.

Method 2:

Once again, we first apply the literal meaning of the sutra and take one more than the number previous to the right-most digit of our denominator. In this case, we get 1 more than 1, which is 2. Call this number our Divider.

  1. First put down a decimal point. Then take the numerator and divide it by our Divider. We get 1 divided by 2, which has a quotient of 0 and a remainder of 1. Write the quotient to the right of the decimal point, and keep the 1 as a carryover number. Our answer so far is .0
  2. Now, prepend the carryover number of 1 to the last quotient, 0, to get 10. Divide this number by our Divider, 2, to get 5. Write this to the right of our answer so far as the next digit of the answer. We now have .05
  3. Divide this last quotient by our Divider to get a quotient of 2 and a remainder of 1. Write the quotient to the right of the answer so far. Keep the remainder as a carryover number. Our answer so far becomes .052
  4. As before, prepend the carryover number (1) to the quotient (2) to get 12. Divide this number by our Divider, 2, to get a quotient of 6 and a remainder of 0. 6 becomes the next digit of our answer. Our answer so far becomes .0526
  5. Continue this process (always write the quotient as part of the answer and prepend the remainder to the quotient to get the next number to divide by the Divider) until we start getting repeating digits (0526, etc.)
  6. Remove the repeating digits, denote the remaining digits as repeating indefinitely, and you have your final answer. As in method 1, the final answer from method 2 is also .052631578947368421 repeated indefinitely.

Now, let us see how to apply this method to something like 1/9, for instance. Note that 1/9 can be written as 1/09 to make the denominator a 2-digit number ending in 9. Our Multiplier or Divider becomes 1 more than the first digit of the denominator, 0, which gives us 1. Using method 1, we put down 1 as the first digit of the answer, then multiply it by our multiplier which gives us 1 again. Since the digit has started repeating, we know that our answer is .1 repeated indefinitely.

Similarly, with method 2, we divide our numerator, 1, by our Divider, 1. The quotient is 1. So, we get the first part of our solution as .1. Dividing the quotient by 1 again leads to 1 which is a repeating digit. So, our final answer is .1 repeated indefinitely. I use this simple problem so that you can actually punch the numbers into a basic calculator and verify that the answer is indeed what we found out just now, even though verifying the answer that way for 1/19 or 1/49 is close to impossible. Those reciprocals may require long division by hand to verify the results we get using the vedic methods.

For a more challenging problem, let us now apply this method to find the decimal form of 1/69. Our Multiplier or Divider is 1 more than the first digit of the denominator, 6, so we get 7 as the Multiplier or Divider.

After step 1 of Method 1, we have an answer of 1.
After step 2, we have an answer of 71.
After step 3, we have an answer of 971, with a carryover number of 4.
After step 4, we have an answer of 7971 with a carryover number of 6.
Following the remaining steps of Method 1, we get subsequent answers of 57971 with a carryover of 5, 057971 with a carryover of 4, 4057971 with no carryover, 84057971 with a carryover of 2, and so on.
Carrying on until we get repeating digits, we get the final answer as .0144927536231884057971 repeated indefinitely.

Using method 2, after step 1, we have an answer of .0 with a carryover of 1.
After step 2, we have an answer of .01 with a carryover of 3 (remember that our Divider is 7 and we have to prepend the carryover number to the previous quotient).
After step 3, we get .014 with a carryover of 3.
After step 4, we get .0144 with a carryover of 6.
Subsequent steps lead to .01449 with a carryover of 1, .014492 with a carryover of 5, .0144927 with a carryover of 3, and so on.
After repeating the steps as long as is required, we get the answer as .0144927536231884057971 repeated indefinitely.

When we apply the methods above to 1/99, we get a multiplier or divider of 10 and the answer turns out to be .01 repeated indefinitely. 1/09 and 1/99 are special cases with short repeating sequences because the multipliers and dividers in these special cases are powers of 10 (remember, 1 is 10 raised to the power 0).

The procedure can be used to find the reciprocal of any number ending with 9, not just 2-digit numbers. But the multipliers and dividers get too large to manipulate comfortably when the denominator becomes too large. There are extensions and corollaries to this method to allow one to handle such reciprocals easily. There are also extensions and corollaries to this method to allow division of any number by any number ending in 9 rather just finding the reciprocals of numbers ending in 9. There are further extensions that result in a general rule that will allow one to divide any number by any number without having to do long division at all.

I will post about these extensions and corollaries as I continue learning the inner workings of Vedic Mathematics. In the meantime, hopefully this lesson whets your appetite for the intricacies of arithmetic explored by these methods. Remember, practice makes perfect, so I will leave it as an exercise to the reader to find the reciprocals of 29, 39, 49, 59, 79 and 89 using both method 1 and method 2 discussed in this lesson. The basic idea is that with enough practice, you should be able to do the required calculations in your head and reel off the answers with no hesitation when confronted with a given problem. Good luck and happy calculating!

4 comments:

Anonymous said...

i have a question about Method 1: It says "We first apply the literal meaning of the sutra and take one more than the right-most digit of our denominator. In this case, we get 1 more than 1, which is 2. Call this number our Multiplier"

The right most digit of the denominator (19) is 9 and not 1. Is this correct? please clarify.

Blogannath said...

You are absolutely right. I got the wording wrong. What I meant to say was: We first apply the literal meaning of the sutra and take one more than the number previous to the right-most digit of the denominator. In this case ...

I will make the correction in the article so that future visitors are not confused. Thank you very much for pointing it out to me. Hope I have not embarassed myself by sprinkling such errors too liberally throughout these posts. Thanks once again.

Anonymous said...

FYI wolframalpha dot com can be used to display repeating decimals with several hundreds of decimal places of accuracy!

Blogannath said...

Great information, thank you. I will check out wolframalpha and its capabilities.

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