In the previous lesson, we learned the basics of the Vedic Duplex method for finding the square root of a number. We will expand on the methodology in this lesson and deal with some special cases we might encounter during our application of the Vedic duplex method for finding square roots of numbers.
For a refresher on the Vedic Duplex method, with the square root calculation illustrated for a couple of uncomplicated examples, please read the full lesson here.
The first complication we are likely to encounter when using the duplex method is that sometimes, the quotient might become larger than 9. The solution to this problem is quite simple: limit the quotient to 9 and increase the remainder as necessary to compensate for the reduced quotient.
To illustrate this, consider the calculation of the square root of 36481. Since the number if 5 digits long, we separate out the first digit from the rest of the digits in preparation for the application of the duplex method. We then identify that the first digit, 3, is larger than 1^2 = 1, but smaller than 2^2 = 4. Therefore, the first number on the answer row becomes 1, and our divisor becomes twice of that, which is 2. Our first remainder becomes 3 - 1 = 2, and therefore, our first gross and net dividends become 26. This is illustrated in the figure below:
•|3: 6 4 8 1
We now notice that our divisor, 2, can go into our net dividend, 26, 13 times with no remainder. But in the duplex method, we always restrict our quotients to be single digits. In other words, we add numbers to the answer row one digit at a time. Because of this, we put down 9 on the answer row as the quotient, and put down 8 as our next remainder (remember that 9*2 + 8 = 26). This then gives us a gross dividend of 84, and a net remainder of 84 - the duplex of 9 (which is 81) = 3. This is shown in the figure below:
At this point, our quotient becomes 1 and we get a remainder of 1, making our gross dividend 18 and our net dividend = gross dividend - duplex of 91 (which is 2x9x1 = 18) = 0. This is shown below:
•|3: 6 4 8 1
Our next quotient then becomes zero, and so does our next remainder. We get a gross dividend of 01, and a net dividend of 1 - duplex of 910 (which is equal to 1) = 0. Thus our last quotient again becomes zero, and we are left with a remainder of zero. Since there are no more digits in the square, the procedure is complete. This is shown in the figure below:
•|3: 6 4 8 1
2|1:2 8 1
•|1: 9 1
Our answer line consists of 19100. Since the square has 5 digits before the decimal point, the square root has to contain 3 digits before the decimal point. Thus our square root is 191.00.
•|3: 6 4 8 1
2|1:2 8 1 0
•|1: 9 1 0 0
For more examples on how to deal with limiting the quotient to 9 when using the duplex method, please read the full lesson here.
Thus, our first complication is relatively easy to deal with. Always make sure that the maximum quotient digit is 9. The remainder might become larger than the divisor, but in the duplex method, that is perfectly acceptable.
Now, we move on to the next complication we are likely to encounter when using the duplex method. And that is the problem of our net dividend becoming negative. This can happen when the duplex of the numbers to the right of the colon on the answer line becomes larger than the gross dividend.
Obviously, either the gross dividend has to be increased or the duplex has to be decreased, or both to prevent this from happening. Since the gross dividend is formed from the next digit of the square appended to the remainder from the previous division, one way to increase the gross dividend is to reduce the quotient from the division and increase the remainder. And because of the beauty of the way in which mathematics sometimes works, this also has the effect of reducing the duplex at the same time! Hopefully, this combination will cause the net dividend to turn positive, and allow us to continue with the application of this method to find the square root.
Let us take a few examples to illustrate this. First consider the square root of 20164. We set up the duplex method in the figure below, and go through the first couple of steps. We see that the first net dividend we have is 10, which can be divided by our divisor, 2, 5 times with no remainder. This would then lead to a new gross dividend of 1, and a net dividend of -24 because the duplex of 5 is 25. The figure below shows this (the net dividend is not shown below):
This illustrates clearly the complication we might encounter from time to time in the application of this method and which we must recover from to continue application of the method. Notice that in this respect, the Vedic duplex method is different from some other methods that freely allow the use of vinculums in their solution (such as polynomial division, straight arithmetic division, etc.).
•|2: 0 1 6 4
To get around this problem, we reduce the second quotient to 4 and carry over a remainder of 2 to the next step, which then leads to a gross dividend of 21, and a net dividend of 5 (21 - the duplex of 4, which is 16). This then leads to the full solution as below:
We get an answer line of 14200, which then leads to the final answer of 142.00 since we know that the square root has to contain 3 digits before the decimal point.
•|2: 0 1 6 4
2|1:1 2 1 0
•|1: 4 2 0 0
Next consider the square root of 101761. We have performed the first few steps of the duplex method, and are now faced with a negative net dividend because our gross dividend is 21, and our duplex is 83.
This once again calls for us to reduce the quotient by 1, and increase the remainder to 8, so that our gross dividend becomes 81. We then see that this enables us to solve the problem because the net dividend becomes zero rather than becoming negative. This is illustrated below:
•|10: 1 7 6 1
6| 9:1 5 2 2
•| 3: 1 9 1
We then derive the final answer of 319.00 from the answer line based on the fact that the square root we are looking for contains 3 digits before the decimal point.
•|10: 1 7 6 1
6| 9:1 5 2 8
•| 3: 1 9 0 0
For several more examples of dealing with negative net dividends by adjusting the quotient and remainder, please refer to the full lesson here.
Hopefully, this lesson has given you the tools necessary to handle the computation of most square roots. We still have not dealt with how to find square roots of non-perfect squares. Since this lesson has already become quite long, that will have to wait till the next lesson. In the meantime, I hope you will take the time to practice the duplex method so that you can apply it confidently to any perfect square. Good luck, and happy computing!