 |
|
||||
| Home | Blog | My books! | Newsletter | Worksheets | Lessons | Videos | Online math resources | Reviews | Curriculum guide |
|
Tweet |
|
Maria's Math news - a FREE math newsletter for parents and teachers Subscribe below:
Worksheets Various worksheets on TONS of math topics you can generate for free! 
Basic operations worksheet generator is updated! Make worksheets for whole numbers or integers - all four operations, including add & subtract or multiply & divide - horizontal or vertical - add a border and/or extra workspace, use a variable, and more.
Curriculum Advice, reviews, and resources to help you choose a math curriculum! Online math games, tutorials, and activities Games you can play online, interactive tutorials, fun math websites and more. Arranged by topic/level for ease of use. Lessons Learn how to TEACH concepts or about general concerns in math education. Reviews In-depth reviews of math products Math help & tutoring A list of free message boards, math help websites, and online tutoring services. Fun and games I have two games on my site, plus links to many. Game: Choose Math Operation Word guess game (easy hangman) Word guess game (difficult) |
You are here: Home → Articles → Why square root algorithm works
Why the square root algorithm works
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
x + r |
≤ |
√a |
|
(x + r)2 |
≤ |
a |
|
x2 + 2xr + r2 |
≤ | a |
| x2 + (2x + r) r | ≤ | a |
|
(2x + r) r |
≤ | a − x2 |
and the last line corresponds to the step where the user tries different values of r on that empty line so that 2x and something times something is less than the subtraction result.
An example:
Let's say we are trying to find √3150 with the algorithm that resembles long division. At every 'go' of the algorithm we use a pair of digits from the number and will find one digit in the answer. In a sense, at each go we get a better approximation of the actual square root. Since most square roots are irrational numbers, the best one can ever do is just approximate them, usually with a decimal number that has certain number of decimal digits.
From a calculator one finds that to two decimals √3150 = 56.12. The first round of applying the algorithm would produce the digit 5, telling us that √3150 is between 50 and 60. On the second round we'd find the digit 6, or in other words we would know √3150 is between 56 and 57. The third round would give us the decimal digit 1, letting us know √3150 is between 56.1 and 56.2, and so on.
From now on we will concentrate on ONE single round of applying the algorithm. Let's say we've already found the digit 5 for the square root, and are using the algorithm to find the next digit r. Note that r is between 0 and 9 and that the digit 5 means that √3150 is 50 + something. We can say that √3150 is between 50 + r and the next whole number, which is 50 + r + 1. So we know that 50 + r ≤ √3150, and from that one derives that
|
50 + r |
≤ |
√3150 |
|
(50 + r)2 |
≤ |
3150 |
|
502 + 100r + r2 |
≤ | 3150 |
| 2500 + (100 + r) r | ≤ |
3150 |
|
(100 + r) r |
≤ |
3150 − 2500 = 650 |
Now see how the last line (100 + r) r ≤ 650 is used in the algorithm:
First we find a number whose square is less than or equal to the first pair or first number (52 is less than 31).
| 5 | |
| √31 | .50 |
|
|
|
||||||||||||||||||||||||||||||||
| Square the 5, giving 25, write that underneath the 31, and subtract. Bring down the next pair of digits. | Then double the number above the line, 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 hundred-something times something would
be less than or equal to 650. 105 x 5 = 525 106 x 6 = 636 107 x 7 = 749, so 6 works. |
Let's try it once more with the next round. Now we've found √3150 is between 56 and 57, and are looking for the next digit q. Note that q is between 0 and 9 and that √3150 is between 56 + q/10 and 56 + (q+1)/10. So we know that 56 + q/10 ≤ √3150, and from that one derives that
|
56 + q/10 |
≤ |
√3150 |
|
560 + q |
≤ |
10√3150 |
|
(560 + q)2 |
≤ |
100 x 3150 |
|
5602 + 1120q + q2 |
≤ | 315000 |
| 313600 + (1120 + q) q | ≤ |
315000 |
|
(1120 + q) q |
≤ |
315000 − 313600 = 1400 |
You can see how the last line (1120 + q) q ≤
1400
is used in the algorithm.
|
|
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
Calculate 6 x 106, write that below 650, subtract, bring down the next pair of digits (in this case the decimal digits 00). |
Then double the number above the line, and write it down 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 eleven-hundred twenty-something times something would be less than or equal to 1400. 1121 x 1 = 1121 1122 x 2 = 2244, so 1 works. |
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
You don't see the subtraction 315000 − 313600 done as such in the calculations, but instead the subtraction 650 − 636 (corresponding to 65000 − 63600). That is because in the previous 'go' the subtraction 3150 − 2500 was done, corresponding to 315000 − 250000, which has left the difference 65000.
See also:
Should I teach my child to calculate square roots without a calculator?
Another example of using the square root algorithm
A geometric view of the square root algorithm
The fast, FUN, and easy way to learn multiplication. Learn the upper times tales in two sittings using mnemonic stories.
|
Copyright 2003-2013 Maria Miller
About
Privacy policy
Contact
Advertise