Vedic Mathematics Lesson 47: Square Roots 2

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.

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 3Vertically 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 Th! e 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 Equati! ons 4
Cubic Equations
Quartic Equations
Polynomial Division 1
Polynomial Division 2
Polynomial Division 3
Square Roots 1

To refresh our memories, consider the calculation of the square root of 26224641, as shown below. This should reinforce how the gross and net dividends are calculated, how duplexes! are computed, etc.

••|26:  2  2  4  6  4  1
10|25:01 02 01 00 00 00
•G|  :012022014006004001
•N|  :012021010000000000
----------------------------
••| 5:  1  2  1  0  0  0

We see that the answer line contains 5121000. Since the square contains 8 digits before the decimal point, we know that the square root has to contain 4 digits before the decimal point. Therefore, we conclude that our final answer is 5121.000.

Similarly, consider the calculation of the square root of 81378441 as shown below.

••|81:  3  7  8  4  4  1
18|81:00 03 01 00 00 00
•G|  :003037018004004001
•N|  :003037018000000000
--------------------------
••| 9:  0  2  1  0  0  0

Notice how the calculation of the duplex includes the zero right after the colon on the answer line. Normally, we would ignore leading zeroes when ! doing any mathematical operations on a number, but in the dupl! ex metho d, the leading zero after the colon is significant and has to be included in the duplex calculations!

Since the square root needs to have 4 digits before the decimal point because of the length of the square before the decimal point, we get our final answer as 9021.000.

Now, we are ready to deal with possible complications during the use of this method. 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 th! an 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
2|1:2
G| :26
N| :26
-------------
•|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:

•|3: 6 4 8 1
2|1:2 8
G| :2684
N| :2603-------------
•|1: 9

At this point, ! our quot ient 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
2|1:2 8 1
G| :268418
N| :260300
-------------
•|1: 9 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 0
G| :26841801
N| :26030000
-------------
•|1: 9 1 0 0

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 r! oot is 191.00.

Now, consider the calculation of the square root of 358801, as shown below. We see that we have to reduce our quotient to 9 twice during the duplex procedure even though the true quotient could have been higher based purely on the values of the net dividend and the divisor. Each time, we ended up with a remainder that was larger than our divisor, 10.

••|35:  8  8  0  1
10|25:10 18 17 08
•G|  :108188170081
•N|  :108107008000
---------------------
••| 5:  9  9  0  0

At the end of the procedure, we get an answer line that reads 59900. Since the square contains 6 digits before the decimal point, we know that the square root has to contain 3 digits before the decimal point. This then leads us to our correct square root, 599.00.

Thus, our first complication is relatively easy to deal with. Always make sure that the maximum quotient digit is 9. The rem! ainder 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 met! hod 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):

•|2: 0 1 6 4
2|1:1 0
G| :1001
N| :10
---------------
•|1: 5

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 polynomi! al division, straight arithmetic division, etc.).

T! o get ar ound 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:

•|2: 0 1 6 4
2|1:1 2 1 0
G| :10211604
N| :10050000
---------------
•|1: 4 2 0 0

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.

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.

•|10: 1 7 6 1
6| 9:1 5 2 2
G|  :11572621
N|  :115608
---------------
•| 3: 1 9 1

This once again calls for us to reduce the quotient by 1, and i! ncrease 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 8
G|  :11572681
N|  :11560800
---------------
•| 3: 1 9 0 0

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.

Now, consider the calculation of the square root of 73599241, as shown in the figure below. We have stopped the calculation at a point which leads to a negative net dividend:

••|73:  5  9  9  2  4  1
16|64:09 15 06
•G|  :095159069
•N|  :095134
--------------------------
••| 8:  5  8

We see that our duplex is 2x5x8 = 80, while our gross dividend is only ! 69. So, we have to, once again, reduce the quotient by 1 and ! increase our remainder by 16, so that our gross dividend becomes 229, which then allows us to continue the computations without further problems as illustrated below:

••|73:  5  9  9  2  4  1
16|64:09 15 22 15 13 08
•G|  :095159229152134081
•N|  :095134159013008000
--------------------------
••| 8:  5  7  9  0  0  0

This then gives us a square root of 8579.000, based on the fact that our square root needs to have 4 digits before the decimal point.

Now, consider the calculation of the square root of 89401, as shown below. We see that the last quotient shown on the answer line in the figure below could have been 2. However, the duplex of the resulting answer line will be at least 81 (square of the 9 in the middle of the number). So, because we need our gross dividend to be at least 81, we make the quotient 0, and carry over the entire net dividend from this step as a remainder to the ! next step so that we get a gross dividend of 81.

•|8:  9  4  0  1
4|4: 4 13 17 08
G| : 49134170081
N| : 49053008
-------------------
•|2:  9  9  0

We can now complete the figure by calculating the net dividend and the resulting quotient. Being able to estimate the duplex and making adjustments to account for it as we go along is a useful skill to develop as it will prevent us from making frequent corrections to our work based on the net dividend becoming negative.

•|8:  9  4  0  1
4|4: 4 13 17 08
G| : 49134170081
N| : 49053008000
-------------------
•|2:  9  9  0  0

We see that the answer line is now 29900, and based on the fact that we need 3 digits before the decimal point, we conclude that the square root needs to be 299.00.

Notice that in the previous example, we in fact faced both complications during the computations. We had to l! imit our quotients to 9 in the first couple of divisions, and ! then had to reduce the quotient in the division after that to keep the net dividend from becoming negative!

Finally, let us consider the calculation of the square root of 35988001 as shown below.

••|35:  9  8  8  0  0  1
10|25:10 19 27 26
•G|  :109198278260
•N|  :109117116017
--------------------------
••| 5:  9  9  9

We have used our knowledge of how to overcome the first complication by limiting our quotient to 9 in the last 3 divisions and carrying over big remainders because of that. Now, we are left with a net dividend of 17, and a choice of whether we can take advantage of the fact that 10 goes into 17 once with a remainder of 7, or to use a different quotient to make sure no future problems occur in the application of the method.

We notice that the minimum duplex we will get from the answer line once we add the fourth digit to it is 2x9x9 = 162. So, if we make the q! uotient 1 and carry over a remainder of 7, we will end up with a negative net dividend. So, we decide to reduce the quotient by 1, ending up with the figure below:

••|35:  9  8  8  0  0  1
10|25:10 19 27 26 17 08
•G|  :109198278260170081
•N|  :109117116017008000
--------------------------
••| 5:  9  9  9  0  0  0

This then gives us the correct square root of 5999.000!

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!

Finding square roots