A note from Beer: This algorithm, and it's related HTML & CSS were borrowed from HomeSchoolMath.net. The original page, in its original form can be found by clicking here.
How to calculate a square root without a calculator
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| 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).
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 is good exercise in basic operations for middle school students, and studying the logic behind it can be a good thinking exercise for high school students.
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 |
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| 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. |
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| 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. |
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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. |
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Thus to one decimal place, √645 = 25.4
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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:
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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 |
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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 |