## Mar 4, 2009

### Welcome!!! This is my first post: Calculate square root without a calculator

I hope to share my joys in Mathematics.

Calculate square root without a calculator

http://www.homeschoolmath.net/teaching/square-root-algorithm.php

# How to calculate a square root without a calculatorand should your child learn how to do it

Many school books seem to think that since calculators can find square roots, that kids don't need to learn how to find square roots using any pencil-and-paper method. But learning at least the "guess and check" method for finding the square root will actually help the student UNDERSTAND and remember the square root concept itself!

So even though your math book may totally dismiss the topic of finding square roots without a calculator, you can consider to let them practice at least the first method presented here. This method, "guess and check", actually works around what the square root is all about, so I would consider exercises with it as essential to help children understand the concept of square root.

Depending on the child, it might be good to concentrate on teaching the concept of square root without taking the time for paper-pencil calculations. In this case, you can study the guess and check method with the help of a simple calculator that doesn't calculate square roots but can quickly do the multiplications.

## Finding square roots by guess & check method

One simple way to find a decimal approximation to, say √2 is to make an initial guess, square the guess, and depending how close you got, improve your guess.  Since this method involves squaring the guess (multiplying the number times itself), it actually uses the definition of square root, and so can be very helpful in teaching the concept of square root.

### Example: what is √20 ?

Children first learn to find the easy square roots that are whole numbers, but quickly the question arises as to what are the square roots of all these other numbers.  You can start out by noting that (dealing here only with the positive roots) since √16 = 4 and √25 = 5, then √20 should be between 4 and 5 somewhere.

Then is the time to make a guess, for example 4.5. Square that, and see if the result is over or under 20, and improve your guess based on that.  Repeat the process until you have the desired accuracy (amount of decimals).  It's that simple and can be a nice experiment for children.

### Example: Find √6 to 4 decimal places

Since 22 = 4 and 32 = 9, we know that √6 is between 2 and 3. Let's just make a guess of it being 2.5. Squaring that we get 2.52 = 6.25. That's too high, so make the guess a little less. Let's try 2.4 next. To find approximation to four decimal places we need to do this till we have five decimal places, and then round the result.

 Guess Square of guess High/low 2.4 5.76 Too low 2.45 6.0025 Too high but real close 2.449 5.997601 Too low 2.4495 6.00005025 Too high, so between 2.449 and 2.4495 2.4493 5.99907049 Too low 2.4494 5.99956036 Too low, so between 2.4494 and 2.4495 2.44945 5.9998053025 Too low, so between 2.44945 and 2.4495.

This is enough since we now know it would be rounded to 2.4495 (and not to 2.4494).

## Finding square roots using an algorithm

There is also an algorithm that resembles the long division algorithm, and was taught in schools in days before calculators.  See the example below to learn it. While learning this algorithm may not be necessary in today's world with calculators, working out some examples can be used as an exercise in basic operations for middle school students, and studying the logic behind it can be a good thinking exercise for high school students.

### Example: Find √645 to one decimal place.

First group the numbers under the root in pairs from right to left, leaving either one or two digits on the left (6 in this case).  For each pair of numbers you will get one digit in the square root.
To start, find a number whose square is less than or equal to the first pair or first number, and write it above the square root line (2).

 2 √6 .45
 2 √6 .45 - 4 2 45
 2 √6 .45 - 4 (4 _) 2 45
 2 √6 .45 - 4 (45) 2 45
Square the 2, giving 4, write that underneath the 6, and subtract.  Bring down the next pair of digits. Then double the number above the square root symbol line (highlighted), and write it down in parenthesis with an empty line next to it as shown. Next think what single digit number something could go on the empty line so that forty-something times something would be less than or equal to 245.
45 x 5 = 225
46 x 6 = 276, so 5 works.
 2 5 √6 .45 .00 - 4 (45) 2 45 - 2 25 20 00
 2 5 √6 .45 .00 - 4 (45) 2 45 - 2 25 (50_) 20 00
 2 5 . 3 √6 .45 .00 - 4 (45) 2 45 - 2 25 (503) 20 00
Write 5 on top of line.
Calculate 5 x 45, write that
below 245, subtract,  bring down the next pair of digits (in this case the decimal digits 00).
Then double the number
above the line (25), and write the doubled number (50) in parenthesis with an empty line next to it as indicated:
Think what single digit number  something could go  on the empty line so that five hundred-something
times something would be  less than or equal to 2000.
503 x 3 = 1509
504 x 4 = 2016, so 3 works.
 2 5 . 3 √6 .45 .00 .00 - 4 (45) 2 45 - 2 25 (503) 20 00 - 15 09 4 91 00
 2 5 . 3 √6 .45 .00 .00 - 4 (45) 2 45 - 2 25 (503) 20 00 - 15 09 (506_) 4 91 00
 2 5 . 3 9 √6 .45 .00 .00 - 4 (45) 2 45 - 2 25 (503) 20 00 - 15 09 (506_) 4 91 00
Calculate 3 x 503, write that
below 2000, subtract,  bring down the next digits.
Then double the 'number' 253 which is above the line (ignoring the decimal point), and write the doubled number 506 in parenthesis with an empty line next to it as indicated: 5068 x 8 = 40544
5069 x 9 = 45621, which is less
than 49100, so 9 works.

Thus to one decimal place, 645 = 25.4

 I read your suggestion for calculating square root without a calculator. I teach Math for Elementary Teachers and developmental math courses (algebra) to adults. I feel that the focus should be on understanding the number rather than an exercise in following a memorized algorithm. I suggest you have the student determine the pair of perfect squares the number falls between. For example, if finding the sqrt of 645, it falls between the sqrt of 625 which equals 25 and the sqrt of 576 which equals 26. So the sqrt of 645 has to be between 25 and 26. Where does it fall between? There are 50 numbers between 576 and 525. 645 is 20 numbers beyond 625, so 20/50 = 0.4 So the sqrt of 645 is very close to 25.4 This method provides the student with a process that improves their understanding of numbers without expecting them to memorize an algorithm, and it provides an answer to the nearest tenth. Andrea S. Levy, Ed.D.

 Im currently a student at MCC I'm taking a course that is for Elementary Math Teachers. We are supposed to do a lesson plan so that we can teach elementary children how to use the Pythagorean theorem. I need to learn how to break down Pythagorean theorm for an elementary child. I got stuck at the square rooting part. Read my answer to this question.

The method you show in the article is archaic. There is a MUCH more efficient algorithm. (This is the algorithm actually used behind the scenes inside a calculator when you hit the square root button.)

1. Estimate the square root to at least 1 digit.
2. Divide this estimate into the number whose square root you want to find.
3. Find the average of the quotient and the divisor. The result becomes the new estimate.

The beauty of this method is that the accuracy of the estimate grows extremely rapidly. Each cycle will essentially double the number of correct digits. From a 1-digit starting point you can get a 4-digit result in two cycles. If you know a square root already to a few digits, such as sqrt(2)=1.414, a single cycle of divide and average will give you double the digits (eight, in this case).

In addition to giving a way to find square roots by hand, this method can be used if all you have is a cheap 4-function calculator. If students can get square roots manually, they will not find square roots to be so mysterious. Also, this method is a good first example of an itterative solution of a problem.

David Chandler

This other way is called Babylonian method of guess and divide, and it truly is faster. It is also the same as you would get applying Newton's method. See for example finding the square root of 20 using 10 as the initial guess:

 Guess Divide Find average 10 20/10 = 2 average 10 and 2 to give new guess of 6 6 20/6 = 3.333 average 3.333 and 6 gives 4.6666 4.666 20/4.666= 4.1414 average 4.666,4.1414= 4.4048 4.4048 20/4.4048=4.5454 average = 4.4700 4.4700 20/4.4700=4.4742 average = 4.4721 4.4721 20/4.4721=4.47217 average = 4.47214 This is already to 4 decimal places 4.47214 20/4.47214=4.472132 average =4.472135 4.472135 20/4.472135=4.472137 average = 4.42136

 The poster asserts that the article's method is "archaic" and that the "Babylonian Method" is more efficient. At first glance, this would appear to be so, because the poster's example finds the square root of the two digit whole number 20 instead of the article's example of 645. However, I actually worked out the article's example (square root of 645) using both methods and found that the Babylonian Method required 9 "cycles of divide and average" to arrive at the answer. Also, the Babylonian Method requires the student to perform 5 digit long division - no small feat for an elementary or middle school student. The article's method, on the other hand, only requires the student to perform one 4 step, long division problem by working out at the most a half a dozen or so 4 digit x 1 digit multiplication problems. It is therefore reasonable to conclude that the Babylonian Method is more suitable to solve by calculator or solve by computer, while the article's method is more suitable to solve by pencil-and-paper. Since the subject of the article was how to teach an elementary or middle school student to easily find square roots with a paper-and-pencil method, the article's "archaic" method seems to be the most fitting. Alex

 In response to Alex's post, How did it take you 9 cycles to produce 25.4 using the Babylonian Method on 645? It takes 1.5 steps if you use your guess as 25 1)645/25=25.8 (25+25.8)/2=25.4 2)645/25.4=~25.39 The Babylonian method is very effective if one already knows many perfect squares to approximate the original value. I find that students cannot follow the reasons behind the algorithm in this post, while the divide and average method seems to be more intuitive if they have worked with averages before. Daniel

 I am doubtful about teaching the long division method for extracting square roots. The Babylonian method is easier to remember and understand, and it affords just as much practice in basic arithmetic. More importantly, it has clear connections to topics such as Newton's method and recursive sequences that will be encountered in calculus and beyond. The long division method is somewhat faster for manual calculation, but it leads to no other important topics -- it is a dead end. David

 I was trained on old computer circuitry and binary hardware algorithms. The method used to calculate the root of 645 is the method used in high performance binary calculations since it only requires shift, subtract, and compare which are all single cycle/stage instructions or are diverted to a co-processor. Convert a number to binary, split it into 2 bit groups, and use the above routine. Multiply and divide require 10's to hundreds of cycles/stages and kill preformance and pipelines. It is faster to perform a square root than a divide since divide works through 1 bit per cycle/stage and square root steps through 2 bits per cycle. Brad

 Can we find the nth root by division method. if yes then please tell me ? or tell how we can 3rd, 4th, root by division method. Amar Deep Yes, we can. It looks quite tedious to do by hand, but the algorithm exists for any root and is similar to the square root one. See these links: an example of using division method for finding cube root, and info about the nth root algorithm (or paper-pencil method).

what is the square root of -1?

Tamara Yardley

-1 cannot have a square root (at least, not a real one) because any two numbers with the same "sign" (+/- positive or negative), when multiplied, will equal a positive number. Try it: +2 × +2 = 4    and     -2 × -2 = 4.

Since a square root of a number must equal that number when multiplied by itself. When you multiply this number by itself, and set it up as a full equation ( n * n = x ), the two factors (n and n) are either both positive or both negative since they are the same number. Therefore, their product will be positive. No real number multiplied by itself will equal a negative number, so -1 cannot have a real square root.

Blake

Square root of -1 is not a real number. It is denoted by i and called the imaginary unit. From i and its multiples we get pure imaginary numbers, such as 2i, 5.6i, -12i an so on. It leads to a whole new number system of complex numbers where numbers have a real part and an imaginary part (for example 5 + 3i or -20 - 40i). And there is a lot of fascinating mathematics done with this number system!