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Wednesday, April 14, 2010

Vedic Mathematics Lesson 48: Square Roots 3

In this earlier lesson, we introduced the Vedic Duplex method for finding square roots and solved several problems using the method. We also saw that some problems might create complications for the method. These complications were dealt with in the previous lesson. In this lesson, we will tackle the problem of how to find square roots of numbers that are not perfect squares. Along the way will also tackle the square roots of non-whole 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 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
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 any difference 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
What do these rules mean? They mean that before we start applying the duplex method, we need to rewrite:
  • 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)
Notice that in the previous lessons, we only dealt with whole numbers with no fractional part or zeroes in front of the number, so we were following these rules even though we did not know about them!

Once the number is written according 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 ":")
Following the above rules, we can place the ":" as below in the given numbers:
  • 40 - 40:
  • 240 - 2:40 or 240:
  • 348.4875 - 3:484875, 348:4875, 34848:75, 3484875:
  • 0.10 - 10:
Given these rules, let us now consider the calculation of the square root of 35988001. But instead of putting just 2 digits before the ":", let us take advantage of the rules above, and leave 4 digits after the ":". This gives us the initial figure below:

••|3598: 8 0 0 1
10| :
•G| :
•N| :
-----------------------
••| :
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
118|3481:117
••G| :
••N| :
--------------------------
•••| 59:
This then gives us a gross dividend and net dividend of 1178. The rest of the method 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.

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 course, is the difficulty that comes with division by a larger divisor! And 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:

•••|3598: 8 0 0 1
118|3481:117 116 017 008
••G| :1178116001700081
••N| :1178107900080000
--------------------------
•••| 59: 9 9 0 0
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.

Similarly, consider the calculation of the square root of 413024329. 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 ":".

•|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
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 ":".

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 before 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:

•|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
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 this earlier lesson. But what should we do in the case of numbers with no whole part?

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
Consider the square root of 0.18671041 as we calculated above. This is a purely fractional number, so our square root will have only a fractional part. Moreover, the number of zeroes right after the decimal point in the square is zero (which is an even number). So, we divide that by two and get the number of zeroes right after the decimal point in the square root to be zero also. This then tells us that the square root we are looking for is 0.4321.

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:

•|0009: 6 1
6|0009:0 0
G| :0601
N| :0600
--------------
•| 3: 1 0
We find that our answer line 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.

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 this lesson, we will find the square root to a precision of 3 or 4 digits 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:

•|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
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. When 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.

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 we want to!

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:

••|329: 8 7 0 0 0
36|324:05 22 10 16 12
•G| :058227100160120
•N| :058226088120090
--------------------------
••| 18: 1 6 2 3
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.

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:

•|10: 0 0 0 0 0
6| 9:1 4 3 6 4
G| :1040306040
N| :1039182222
----------------
•| 3: 1 6 1 3
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.

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 result in a larger divisor since this will usually reduce the number of times 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:

••|125: 0 0 0 0 0
22|121:04 18 03 14 10
•G| :040180030140100
•N| :040179014076094
--------------------------
••| 11: 1 8 0 3
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.

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 may 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!

12 comments:

Jeff said...

Thanks for the Vedic math lessons. They are fascinating. In your first lesson, you mentioned you were doing this for your kids. They must be very bright if they are keeping up with you on this. I would be very interested in learning more about their math development -- i.e. what topics you have found the most useful to teach, the hardest to learn, etc., as well as any specific resources you may have used.

Blogannath said...

Thank you for your kind comments, Jeff.

You bring up an interesting point about how it all got started. Yes, my kids are very bright, if I do say so myself, but I don't think a kid needs to be super-bright to follow what is going on in most of these lessons. Because I started writing these lessons at the level at which a middle-schooler could understand (lots of examples, explanations broken down into steps and rules, etc.), and I have kept that standard in mind with all lessons in the series, it would be quite possible for kids the same age as mine to follow along if they wanted to.

But, my kids are not keeping up with me all the way. They have covered most of the arithmetic, but they have not yet gotten into the algebra yet. I decided to stick to a lesson a week so that I would have the motivation to keep learning and writing about this stuff. Eventually, I hope my kids will learn all this and know as much as I do about this subject so that they can teach their kids all this!

As for resources, I have one book on Vedic mathematics by Bharati Krsna Tirthaji. The book covers a lot of topics and is quite comprehensive. But the book is somewhat terse, and I find it not as detailed as my lessons. In fact, the reason I started writing this stuff was so that the contents of this book become more accessible to somebody like my kids. If you are interested, here is the Amazon.com link to this book: http://www.amazon.com/gp/product/8120801636?ie=UTF8&tag=electronicdre-20&linkCode=as2&camp=1789&creative=390957&creativeASIN=8120801636

There are lots of books on the subject that have come out in recent times as the subject has come into vogue, but I have not had a chance to read through or review many of them. There are also several websites that cover parts of this vast subject (mostly the multiplication part since that seems to be the easiest), but I have not come across any websites that are comprehensive or complete. If you have seen any in your searches or surfing on the internet, I would love to know so that I know what I am missing.

Thank you for your interest! Please visit often, and let me know if you have any questions, comments, or concerns, or if there are any topics you would like me to cover or cover in greater detail.

Jeff said...

Thanks,

I'm trying to figure out the best way to supplement my just turned 5 year old's math education. He seems to have some aptitude compared to his peers (i.e. he can add and subtract numbers up to 12, understands place value and the concept of multiplication). So I am really starting from a clean slate and am looking to do things the right way, from the start.

I came across Vedic math when I saw some comments about several mainstream math books being (unattributably)based on Vedic principles. I'm currently reading Secrets of Mental Math by Arthur Benjamin and Michael Shermer, which I got from the library and have a couple of others on their way. I'm getting some good ideas from the book on how to approach teaching my son addition and multiplication.

My earlier question to you about math topics was geared more to your kids overall math education, rather than just Vedic specific.

Blogannath said...

This is just my personal opinion, so take it with a large grain of salt. I think, given that your son is only 5 years old, the more important thing to concentrate on is basic arithmetic like addition and multiplication tables. Learning by rote is often frowned upon these days, but I have never had to stumble over a simple multiplication problem yet because I learned and relearned my tables in my first few years of schooling.

You might also want to add a mechanical enhancement to your son's arithmetic arsenal by trying out abacus classes or something similar (I am sure you saw my comment about the Japanese abacus and my children's training on it in my introduction to Vedic Maths post).

Once your child is doing basic arithmetic without hesitation is when I think Vedic Maths makes sense. In other words, Vedic maths, in my mind, is not a substitute for learning your addition and multiplication tables. I think once a child grasps basic addition and multiplication, they will be ready to try their hand at vedic multiplication, 10's complements, vinculums, maybe even square roots. I think algebra should wait until they are perhaps in 3rd or 4th grade and starting to learn about coordinate graphs, plotting points on a graph and other graphical introductions to the concept of variables and coefficients.

As for my kids, they are well above grade level in maths. They score very high on standardized tests (99th percentile), and can do things their classmates can not, such as solving multiple simultaneous equations, solving quadratic equations, etc. But I think they would not be where they are now if we had not started them out on addition and multiplication tables at home even though schools here do not emphasize that as much as they should.

Again, this is just my uninformed opinion. You should train your son according to what you think he can handle given your knowledge of his strengths and weaknesses. Whatever you decide, I wish you the best of luck! Please don't hesitate to post back if you have other questions or want a sounding board for your opinions.

I would encourage all other readers to participate in this discussion and tell everyone what your opinions and experiences have been. It will help you as well as other readers, so I encourage you to share, and share freely. Thank you.

Anonymous said...

hello... thanks for the great explanations & the great blog too!

I am facing a small issue... finding the square root of 4692 using the method above. The answer up to 5-digits is 68.49817, however I am only able to calculate to 68.48.

Your help wold be appreciated!

Blogannath said...

I am not sure how you got 68.48. Here is the calculation I did. If you have any questions about this, please post back:

••|46: 9 2 0 0 0 0 0
12|36:10 13 20 28 24 28 35
G| :109 132 200 280 240 280 350
N| :109 68 136 120 40 119 86
----------------------------------
••| 6: 8 4 9 8 1 7

As you can see from above, I do get the correct answer to at least 5 digits (sorry about the spacing, I don't see any way to post this comment in a fixed-width font like courier).

Pedro Andrade said...

Hey there :D
First of all, I'd like to congratulate you on your blog! It sure made its point to divulge paced, didactic teaching to all levels of students.

Now, I had an issue while using the vedic algorithm on a non-integer number: 8,25.

If I set 8,25 like "825:", the result I wished to obtain goes smoothly like:

••|825: 0 0
56|784:41 18
•G| :410180
•N| :410131
---------------
••| 28: 7 2

Since sqroot(8,25)= 2,872281323, that result does lead us to a precise result.

However, if I set 8,25 like "8:25", the algorithm appears not to work. It follows:

••|8: 2 5 0
2|4:4 24 146
•G| :422451460
•N| :421641298
-----------------
••|2: 9 9 9

It just keeps on going like "2,999...". Did I operate it correctly?

Thank you for your attention.

Blogannath said...

Hi Pedro, one small error I notice is that the number to the left on the second row of your illustration should be 4 (twice of the first digit you put down on the answer row), not 2. If you try it with 4 there instead of 2, you should get the right answer. I haven't worked it out, but I trust the method! Good luck.

sid sahu said...

You know this is a very good post i hadent thought about this for quite a while and you have like sparked me to look into it further and re educate my self in the subject....thanks,hope to see more of your posts soon
speed math

JATISH KUMAR said...

please tell me what is the square root of -1 and how to find sqrt(negative numbers)

Blogannath said...

Negative numbers don't have real square roots. Their square roots are complex numbers, and the square root of -1 is represented as i. To find the square root of any other negative number, find the square root of the positive number and then add an i to the end of the answer.

Juhi Mehta said...

Hello! Your lessons are very helpful. Could you please teach cube roots too?

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