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
Mul tiplication 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
D! ivisibility 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
Before we address the issue of finding square roots of numbers that are not perfect squares, we need to deal with another aspect of the Vedic Duplex method that we have not dealt with before. In this earlier lesson, we mentioned the general rule for splitting the square into parts such that the part before the ":" was either 1 digit long or 2 digits long, depending on whether the square had an odd number of digits or even number of digits.
In reality, the duplex method gives one a lot of flexibility in terms of how the given square is split into two parts. There are some rules we have to follow to make sure we get the right square roots though. First right the given square following the rules below:
- If the number is a whole number, then remove the decimal point and any zeroes there may be after it. Also remove any zeroes before the number ! (what would be considered meaningless zeroes that don't make a! ny diffe rence to the value of the number)
- If the number is not a whole number (that is, it has a decimal part), then remove any zeroes before the number (meaningless zeroes that don't make any difference to the value of the number). Add a zero if necessary to end of the number, after the decimal point, so that the number of digits after the decimal point is an even number
- 04857 as 4857 (remove meaningless zeroes before the number)
- 45.2 as 45.20 (add a zero if necessary to make the number of digits after the decimal point even)
- 0.013 as .0130 (combination of both of the above rules)
Once the number is written accordi! ng to the rules above, we need to follow the rules below to split the number into two parts with the ":".
- The number of digits before the ":" has to be at least one. The number before the ":" can not be entirely made of zeroes
- If the ":" is placed in the whole portion of a number with both whole and fractional parts, then the number of digits of the whole part after the ":" has to be even (it can be zero, since zero is a valid even number, so you can replace the decimal point with a ":")
- If the ":" is placed in the fractional portion of a number, then there should be an even number of digits after the ":" (once again, there could be zero digits after the ":")
- 40 - 40:
- 240 - 2:40 or 240:
- 348.4875 - 3:484875, 348:4875, 34848:75, 3484875:
- 0.10 - 10:
This is perfectly legal since there are an even number of digits (4) after the ":". We also know that 60^2 is 3600, so the highest number whose square is less than the part of the square to the left of the ":" (3598) must be 59. So, we put down 59, and it square in the appropriate places in the figure. We also set 2*59 = 118 as our divisor, and 3598 - 3481 = 117 as our remainder as below:
••|3598: 8 0 0 1
10| :
•G| :
•N| :
-----------------------
••| :
This then gives us a gross dividend and net dividend of 1178. The rest of the metho! d proceeds exactly as before. The main difference is that our divisor is much larger, but that may actually be an advantage since we are less likely to encounter the case where the quotient goes over 9. The other important difference is that the number of digits on the answer line to the right of the ":" goes down by at least one, so the duplex calculations are not only likely to be less complicated, but also result in smaller duplexes such that it is unlikely for the net dividend to become negative. And last but not least, because there are fewer digits to the right of the ":" in the square, there are going to be fewer divisions overall (even though each division may be a little more complicated because of the higher divisor.
•••|3598: 8 0 0 1
118|3481:117
••G| :
••N| :
--------------------------
•••| 59:
Thus, a small change in the way we split up the given square into two parts is likely to have very positive impacts on the probability of encountering the complications we spent the previous lesson addressing. The main problem, of cours! e, is the difficulty that comes with division by a larger divi! sor! An d it may also reduce the complexity of the overall calculation by reducing the number of divisions performed.
Proceeding with the method, and completing the figure we started above, we now get the following figure:
We once again get 599900 on the answer line. Note that we did not have to limit the quotient to 9 in any step of the above process or have to reduce the quotient and increase the remainder to prevent the net dividend from becoming negative. Knowing that the square consists of 8 digits before the decimal point, we set aside 4 digits of the answer line before the decimal point, giving us a final answer of 5999.00.
•••|3598: 8 0 0 1
118|3481:117 116 017 008
••G| :1178116001700081
••N| :1178107900080000
--------------------------
•••| 59: 9 9 0 0
Similarly, consider the calculation of the square root of 41302432! 9. Our normal method would have us set aside the given square as 4:13024329, but we could just as easily restructure it as 413:024329 (6 digits after the ":", which is legal since 6 is an even number) or 41302:4329 (4 digits after the ":") or even 4130243:29 or 413024329:. The last 2 are almost pointless since we would spend a lot of time hunting for the highest perfect square below 4130243 or finding the square root of the given square entirely by trial and error.
However, we notice that 20^2 = 400 is a little below 413. Thus the structuring of the given square into 413:024329 is likely to give us some advantages. Below are two figures, with the first one representing our original way of finding the square root, and the second one showing how it is done by having a larger chunk of the square to the left of the ":".
Either way we get an answer line in which the first 5 digits (which is what our square root should contain before the decimal point given that the square contains 9 digits) are 20323, and the rest of the digits are zeroes. Thus our answer is 20323, regardless of whether we choose to leave 8 digits or 6 digits of the given square behind the ":".
•|4: 1 3 0 2 4 3 2 9
4|4:0 1 1 2 1 2 1 0
G| :0113102214231209
N| :0113101302010000
--------------------! ---
•|2: 0 3 2 3 0 0 0 0
••|! 413: 0 2 4 3 2 9
40|400:13 10 13 02 01 00
•G| :130102134023012009
•N| :130093122001000000
--------------------------
••| 20: 3 2 3 0 0 0
Notice that the rules we established at the beginning of this lesson also allow us to find square roots of non-whole numbers (expressed in the form of decimals) without any problem. To illustrate, let us find the square root of 0.18671041. Following the first set of rules, we rewrite the given number as .18671041 after removing the meaningless zero before the decimal point. We are now free to form the figure with 18 be! fore the ":", 1867 before the ":", 186710 before the ":" or the entire number, 18671041, before the ":".
Based on the ease with which we can find square roots that are below the numbers to the left of the ":", we will go with 18 before the ":" since 4^2 = 16 is well-known and easy to calculate mentally. The resulting figure is shown below:
Our answer line now reads 4321000. How do we determine the true square root from this answer line? In the case of numbers that have a whole part in addition to any fractional part (there are valid numbers before the decimal point in the square), we already know that the number of digits of the square root before the decimal point depends on the number of whole digits in the square as explained in thi s earlier lesson. But what should we do in the case of numbers with no whole part?
•|18: 6 7 1 0 4 1
8|16:02 02 02 01 00 00
G| :026027021010004001
N| :026018009000000000
--------------------------
•| 4: 3 2 1 0 0 0
Then we use the following rules to determine where to place the decimal point in the answer line:
- The square root of a purely fractional number can not contain a whole part. The square root is also entirely fractional
- If the number of zeroes right after the decimal point in the square is even, the square root will have half that number of zeroes right after the decimal point. Add zeroes in front of the answer line to get the appropriate number of zeroes if necessary
- If the number of zeroes right after the decimal point in the square is odd, subtract one from it, and then divide by 2 to get the number of zeroes after the decimal point in the square root. Once again, if necessary, add zeroes in front of the answer line to get the appropriate number of zeroes if necessary
Now consider the square root of 0.000961. First we discount the zero before the decimal point as a meaningless zero, and rewrite our number as .000961. We then see that we can not split the number as 00:0961 because the digits before the ":" can not all be zeroes. Thus, we can either split it as 0009:61 or 000961:. We will reject the last one as not very practical since we would then be trying to solve the problem by trial and error. We settle for 0009:61, and we get the figure below:
We find that our ! answer l ine contains 310. Since our square is entirely fractional, our square root will be entirely fractional too. Moreover, since the square started with three zeroes after the decimal point, our square root has to start with one zero after the decimal point (we subtract one from three and divide the result by two). In this case, using the rules above, we arrive at a final answer of 0.031.
•|0009: 6 1
6|0009:0 0
G| :0601
N| :0600
----! ----------
•| 3: 1 0
Now that we have the basics of the method established, let us see how we can apply this to the calculation of some square roots of numbers that are not exact squares. Before we go there though, we need to figure out how the duplex method signifies that the given square is a perfect square. That is, how do we know to stop the duplex method at some point? The simple answer is that we will run out of digits in the number at the same time as our net dividend becomes zero.
The primary indication that a number is not a perfect square comes when we find that the net dividend! does not become zero at the same time as we run out of digits in the given square. When this happens, we have to continue with the procedure by adding 0's to the right of the number on the top line of the figure (the line containing the square). We then proceed with finding the gross dividend and net dividend as before.
Note that getting a remainder of 0 from the last division is not an indication that the algorithm has concluded. The algorithm ends only when a net dividend can be calculated, and it is zero, and there are no more digits in the square. Even if the remainder from a given step is zero, we still have to calculate the gross dividend and net dividend. If the net dividend is not zero at the end of this process, then the square root is not complete yet.
For numbers which are not perfect squares, the procedure will never end, but we can choose to end the process after calculating the square root to the required degree of precision. In th! is lesson, we will find the square root to a precision of 3 or! 4 digit s after the decimal point in most cases. Note that the process of finding duplexes becomes more and more complicated the higher the precision we need from the algorithm since the sequence of digits in the answer line becomes longer and longer.
We will illustrate the procedure with a few examples. First consider the square root of 2. The calculation of this famous irrational number is shown in the figure below:
We have stopped the algorithm with an answer line of 1414213562, and the net dividend has not become zero at any step in this algorithm. Several times during this algorithm, we have limited the quotient so that the net dividend does not become negative. And the duplex has steadily become larger, with the current duplex being 81. Whe! n you stop the algorithm, make sure that the net dividend you leave behind is not negative. That is why we made the last digit of the square root we have found so far, to be 2 rather than 7. A quotient of 7 would have resulted in a negative net dividend in the next step, thus telling us that it is not a valid digit for the square root. We had to reduce the quotient to 2 to get a positive net dividend for the next step, and this tells us that 2 is the right digit for the square root in that position.
•|2: 0 0 0 0 0 0 0 0 0 0
2|1:1 2 2 4 3 4 6 8 10 10
G| :1020204030406080100100
N| :1004120706121822014019
----------------------------
•|1: 4 1 4 2 1 3 5 6 2
Given that our square (2) has one digit before the decimal point, we conclude that the square root of 2 is 1.414213562 to a precision of 9 digits after the decimal point. In general, we will not calculate square roots to that level of precision, but this is an illustration of the fact that the method has no inherent limitation as to the precision with which we can calculate square roots. As long as we are willing to put up with the hassles of calculating duplexes! of longer and longer numbers, we can keep going however far w! e want t o!
By the way, the square of 1.414213562 is 1.999999998944727844, which is close enough to 2 for most practical purposes!
Let us now calculate the square root of 32987 to a precision of 3 digits after the decimal place. We make the decision to split the number up as 329:87 based on our knowledge that 18^2 = 324, which is quite close to 329. This results in the figure below:
We get an answer line of 181623, and since we know that our square root has to have 3 digits before the decimal point, we conclude that the square root we are looking for is 181.623. The actual square root of 32987 is 181.62323639887050531602222963625, and our answer is accurate to the first three decimal places (which is the precision we set out to calculate! the square root to). Also, 181.623^2 is 32986.914129, which is quite close to 32987.
••|329: 8 7 0 0 0
36|324:05 22 10 16 12
•G| :058227100160120
•N| :058226088120090
--------------------------
••| 18: 1 6 2 3
Consider the square root of 0.1 now. To comply with the rules from earlier in the lesson, we rewrite the given square as .10. We then put the ":" at the end of the given number to get the figure below:
We have an answer line of 31613 now. Since our square does not have any digits before the decimal point, and does not have any zeroes immediately after the decimal point, our final answer is 0.31613.
•|10: 0 0 0 0 0
6| 9:1 4 3 6 4
G| :1040306040
N| :1039182222
----------------
•| 3: 1 6 1 3
Finally, consider the number 1.25. We can tackle the task of finding its square root by splitting it up as 1:25 or 125:. We know that 1^2 = 1, and 11^2 = 121, so either way of splitting up the number seems equally convenient. In general, when we have such a choice, it is better to take the choice that will resul! t in a larger divisor since this will usually reduce the numbe! r of tim es we need to make adjustments to the quotient to prevent the net dividend from becoming negative. Therefore, we choose 125:, and the figure below shows how the square root is derived using that split-up of the given square:
We now have 111803 on the answer line. Given that our square has one number in front of the decimal point, we conclude that the square root must be 1.11803.
••|125: 0 0 0 0 0
22|121:04 18 03 14 10
•G| :040180030140100
•N| :040179014076094
--------------------------
••| 11: 1 8 0 3
Hopefully, this and the earlier lessons on square roots have made you confident about the Vedic Duplex method and all the details of how to apply the method. Hopefully the rules in this and the earlier lessons will help you to rewrite squares as appropriate, split them up correctly for the application of the duplex method, and also recover from complications you ma! y face during the application of the method itself. Finally, I hope I have made it clear how to derive the final answer from the digits on the answer line. Good luck, and happy computing!
How to find the square of a number
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