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 Spec! ial 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
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
Qua! dratic E quations 3
Quadratic Equations 4
Cubic Equations
Quartic Equations
Polynomial Division 1
Polynomial Division 2
Polynomial Division 3
Before we delve into the arithmetic of calculating square roots, we make the following observations regarding square roots:
- If a number consists of n digits before the decimal point, where n is odd, then its square root will contain (n+1)/2 digits before the decimal point. Thus, if the number has 3 digits before the decimal point, its square root will contain 2 digits before the decimal point
- If a whole number is not an exact square of another whole number, its square root is always irrational (that is, the decimal expression of the square root of the number will contain an infinite number of digits after the decimal point, and there will be no repeating pattern in the digits after the decimal point)
- If a number with n digits after the decimal point is squared, the resulting answer will contain 2n digits ! after the decimal point. Thus, if a number contains 2n digits! after t he decimal point, and is the result of squaring a rational number, its square root will contain n digits after the decimal point
- If a number contains n digits after the decimal point, where n is odd, then it can not be the result of squaring a rational number. Therefore, the square root of such a number will always be an irrational number
- An exact square never ends in 2, 3, 7 or 8
- If a number ends in 2, 3, 7 or 8, its square root will always be an irrational number
- If an exact square ends in 1, its square root ends in 1 or 9
- If an exact square ends in 4, its square root ends in 2 or 8
- If an exact square ends in 5, its square root ends in 5
- If an exact square ends in 6, its square root ends in 4 or 6
- If an exact square ends in 9, its square root ends in 3 or 7
- Notice in the above observations, how the ending digits of the square root are 10's complements of each other (thus 1 and 9 are 10's complements o! f each other, 2 and 8 are 10's complements of each other, and so on)
- If a perfect square is an odd number, the square root is also an odd number
- If a perfect square is an even number, the square root is also an even number
- A whole number that ends with an odd number of 0's is never the square of a whole number
- An exact square never ends in a 6 if the penultimate digit is even (thus, exact squares can not end in 26, 46, 86, etc.)
- An exact square never has an odd penultimate digit unless the final digit is a 6 (thus, exact squares can not end in 39, 71, etc.)
- An exact square never ends with an even number when the last two digits taken together are not divisible by 4 (thus, no exact square can end in 22, 34, and other non-multiples of 4 if the last digit is even)
That is quite a lot of observations there!
There are lots of different ways to find the square root of a given number. None of them are easy ! or trivial. In most cases, they require at least some messy c! alculati ons. The vedic method of finding a square root is called the Vedic Duplex method. As the name implies, the method involves a concept called the duplex of a number. So, we will start with explaining what the duplex is first.
The duplex of a number is also called the Dvandva Yoga of a number. The duplex of a number is calculated as below:
- For a single digit number, the duplex is simply the square of the number. Thus the duplex of 2 is 4, the duplex of 6 is 36 and so on
- For a 2-digit number, the duplex is simply twice the product of the 2 digits of the number. Thus, the duplex of 16 is 2x1x6 = 12, the duplex of 90 is 2x9x0 = 0, the duplex of 43 is 2x4x3 = 24, and so on
- For n-digit numbers, the duplex is calculated as the sum of several individual duplexes. Pair up the first digit with the nth digit of the number and find the duplex of the resulting 2-digit number. Similarly pair up the second digit with the (n-1)th digit and! find the duplex of the resulting 2-digit number. Continue this process until no more 2-digit pairs can be formed. If a middle digit (that could not be paired with anything else exists at the end of the process, find its duplex also individually. Then add up all the duplexes found. The resulting sum of the duplex of the n-digit number
Some duplexes are shown below for illustration and to make sure the calculation of the duplex is fully understood.
2: 2^2 = 4
5: 5^2 = 25
48: 2x4x8 = 64
91: 2x9x1 = 18
314: 2x3x4 + 1^2 = 25
725: 2x7x5 + 2^2 = 74
350: 2x3x0 + 5^2 = 25
4466: 2x4x6 + 2x4x6 = 96
1398: 2x1x8 + 2x3x9 = 70
9357: 2x9x7 + 2x3x5 = 156
35832: 2x3x2 + 2x5x3 + 8^2 = 106
62787: 2x6x7 + 2x2x8 + 7^2 = 165
8947548937: 2x8x7 + 2x9x3 + 2x4x9 + 2x7x8 + 2x5x4 = 112 + 54 + 72 + 112 + 40 = 390
84738932833: 2x8x3 + 2x4x3 + 2x7x8 + 2x3x2 + 2x8x3 + 9^2 = 48 + 24 + 112 + 12 + 48 + 81 = 325
Notice the following facts about duplexes:
!
- The duplex of a number is the same when the number is reversed from front to back
- The duplex of an n-digit number is always even when n is even
- The duplex of an n-digit number is odd when n is odd, and the middle digit of the number is odd
Now that we know how to calculate the duplex of a number, how do we use it in calculating the square root of a number? We will explain using a simple example. Let us take the number 144, for instance.
Now, almost everyone knows, without any need for calculations, that the square root of 144 is 12. But, let us try to compute that using the vedic duplex method. The method we use can be extended to numbers much larger than 144.
The first step to doing this is to count the number of digits in the given number. If the number of digits is odd, separate the first digit from the rest of the digits. If the number of digits is even, separate the first two digits from the rest of the digits. ! In the case of 144, the number contains 3 digits, which is an odd number of digits. Therefore, we separate the first digit, 1, from the rest of the digits. We then create an initial figure that looks like the below:
•|1: 4 4
•|______
•|_:____
We have separated the 1 from the rest of the digits with a colon. We have also provided spaces in front of each of the remaining digits of the number just as if we want to perform straight division.
Now, identify the single-digit number whose square is less than or equal to the digit(s) before the colon on the first line of the figure we have above. Because of the procedure we have used, the number of digits before the ":" is at most 2, and the highest number whose square is less than or equal to a 2-digit number is always going to be a single-digit number.
In this problem, the digit before the ":" is 1, so the highest single-digit number whose squa! re is less than or equal to this is 1. We put the 1 before th! e ":" on the answer line of the figure above. We also calculate twice of this number (in this case, twice of 1 is 2), and put that number to the left of the "|" on the second line of the figure above. Now, calculate the square of the first digit of our answer, and subtract it from the number before the ":" on the first line. Write the difference on the second line, just before the next digit of the square. The resulting figure is shown below:
•|1: 4 4
2|1:0
---------
•|1:
The 2 to the left of our figure is called the divisor in this procedure. In each subsequent step of the procedure, we will get a gross dividend and a net dividend. What are these quantities?
Our gross dividend is the number we have written after the colon on the second line appended in front of the next digit of the square. Thus, our gross dividend is 04. The net dividend is equal to the gross dividend at this point since there are no di! gits to the right of the ":" on the answer line. If there are any digits on the answer line to the right of the ":", then the net dividend is equal to the gross dividend - the duplex of the digits to the right of the ":" on the answer line. Note that we calculate duplexes only for digits to the right of the ":" on the answer line. We never include the first digit of the answer line in the duplex calculations.
The figure below has been expanded with lines that denote the gross dividend (in the line labeled G), and the net dividend (in the line labeled N), but these are not necessary when doing the procedure mentally (one can calculate them mentally and proceed with the calculation without writing them down as shown).
•|1: 4 4
2|1:0
G| :04
N| :04
---------
•|1:
At this point, we divide the net dividend by our divisor and write the quotient in the answer line. Write the remainder of such a div! ision in front of the next digit of the square on the second l! ine of t he figure. In our case, the net dividend is 4, and dividing that by our divisor, 2, gives us a quotient of 2 and a remainder of 0. The resulting figure is shown below:
•|1: 4 4
2|1:0 0
G| :04
N| :04
---------
•|1: 2
Our next gross dividend is 04 (the remainder from the division we just performed appended in front of the next digit of the square). The next net dividend is 4 - the duplex of 2 (the number to the right of the ":" on the answer line). Since the duplex of 2 is 4, the next dividend becomes 4 - 4 = 0. This is shown in the figure below:
•|1: 4 4
2|1:0 0
G| : 04
N| : 00
---------
•|1: 2
Dividing the net dividend by our divisor, 2, gives us a quotient of 0, and a remainder of 0. Coincidentally, we have also run out of digits in the square. This signifies that we are done with finding the square root (if the remainder had n! ot been zero, we would have added zeroes to the square line and continued the process until we get a remainder of zero or we get tired of continuing the procedure. We will deal with this situation when we try to find the square root of a number that is not an exact square). The final figure we have is below:
•|1: 4 4
2|1:0 0 0
G| :0404
N| :0400
---------
•|1: 2 0
Now, take the number on the answer line. It is 120. Based on our observations from earlier, since the square is 3 digits long before the decimal point, we know that the square root has to be 2 digits long. Thus we take the number in our answer line and put a decimal point after 2 digits, giving us the final square root of 12.0. Obviously, this is correct, and therefore we get some confidence that the method is correct too.
Now, let us tackle a more challenging problem. Let us try to find the square root of 1521. Since the numbe! r of digits in the square is 4, which is even, we separate out! the fir st 2 digits of the square for our figure below:
•|15: 2 1
•|
G|
N|
-----------
•| :
Now, we find that the largest number whose square is less than 15 is 3 since 3^2 = 9 (less than 15), and 4^2 = 16 (more than 15). So we put 3 down on the answer line, and twice of 3, which is 6, as our divisor. The difference between 15 and the square of 3 is 15 - 9 = 6. This is put down on the second line of our figure, giving us our gross dividend of 62 and net dividend of 62 (since we don't have any digits on the answer line to the right of the ":", there is no duplex to calculate and deduct from the gross dividend). This is shown in the figure below:
•|15: 2 1
6|09:6
G| :62
N| :62
-----------
•| 3:
When we divide the net dividend, 62, by our divisor, 6, we get a quotient of 10 and a remainder of 2. In this procedure, we find the square ro! ot one digit at a time, so instead of using 10 as the quotient (which would add 2 digits to the answer line), we use 9 as the quotient, and 8 as the remainder (since 9*6 + 8 = 62). So, the first digit after the ":" on the answer line would be 9. We now get a gross dividend of 81 in our next step. We calculate the duplex of the digit to the right of the ":" on the answer line (9^2 = 81), and subtract it from the gross dividend to get our net dividend. Dividing the net dividend by the divisor gives us a quotient of 0 and remainder of 0. We also run out of digits in the square, signifying that our procedure is complete. We get the final figure below:
•|15: 2 1
6|09:6 8 0
G| :6281
N| :6200
-----------
•| 3: 9 0
We take the number in the answer line, 390, and put a decimal point after the second digit since we know that the square root of a 4-digit number has to have 2 digits before the decimal point. ! Thus, we get a final answer of 39.0. One can verify that thi! s is cor rect by checking to make sure that 39x39 = 1521.
Now, let us try to find the square root of 179776. Since the number contains an even number of digits, we separate out the first two digits. We also find that the first digit of the answer is 4 since 4^2 is 16 (less than 17), and 5^2 is 25 (more than 17). Thus, our initial figure looks like this:
•|17: 9 7 7 6
8|16:1
G| :19
N| :19
------------
•| 4:
The next digit on the answer line becomes 2 since 19 divided by 8 gives us a quotient of 2 and a remainder of 3. Our next gross dividend is therefore 37, and our next net divisor is 37 - the duplex of 2 = 37 - 4 = 33. This is illustrated below:
•|17: 9 7 7 6
8|16:1 3
G| :1937
N| :1933
------------
•| 4: 2
Dividing 33 by 8 gives us a quotient of 4 and a remainder of 1. This leads to the next digit on the answer line being 4.! The next gross dividend becomes 16, and the next net dividend is 16 - the duplex of 24, which is 16 - 16 = 0. Dividing 0 by 8 gives us a quotient of 0 and a remainder of 0. At the same time, we have exhausted all the digits of the square, so we are done with the procedure. We get a final figure as below:
•|17: 9 7 7 6
8|16:1 3 1
G| :193716
N| :193300
------------
•| 4: 2 4 0
We place a decimal point after third digit of the answer line to get our final answer of 424.0. We can verify that 424^2 is 179776, so we have performed the calculations correctly.
We will end this lesson with one more example. Let us try to find the square root of 326041 using our method. I will write down how the final figure looks without the intermediate steps since those steps, by now, would be familiar to all.
••|32: 6 0 4 1
10|25:07 06 01 00
•G| :076060014001!
•N| :076011000000
------------------
&#! 8226; 226;| 5: 7 1 0 0
The final answer is 571.00 (since we need to place a decimal point after the 3rd digit of the answer row. We can verify that 571^2 = 326041.
In this lesson, we have not dealt with any special cases that we are likely to encounter when we start dealing with larger numbers. We also found square roots only of perfect squares.
In the next lesson, we will expand on the basics we have covered in this lesson to deal with complications that are likely to result when using this method. We will also apply this method to some numbers that are not perfect squares and see how the method can be used to derive the square root to arbitrary precision. Until then, good luck, and happy computing!
How to find the square of a number
No comments:
Post a Comment