Universal Square Edging (MSM-3 Format)

At the beginning, I thought only of creating or designing a method that would help grade school children, in their studies involving square roots of large numbers. The fact that I was been challenged by these following statements:

“For some reason, extracting the square roots of large numbers was more difficult than expected. For forty minutes or so, I labored to explain and demonstrate the proper method in square root extraction. The process can be very tedious. Attention to detail is necessary. Patience is definitely helpful. Yesterday was undoubtedly one of the harder days for this class”… http://www.mygba.org/extracting-square-roots/

“…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.”

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

Somehow, I agree with them. Why do you think a child should take (or extract), the square root of a large number, in millions, only to find out that there is “no end” in their attempt to find out the square root of such number?

Square Edging, in a sense, is more ‘logical’ and ‘practical’. Children are more interested in knowing “results” which are more reasonable than ‘trivial’ answers (such as, knowing the square root of 6,4967,028). So, MSM-1 and MSM-2 are practically designed, obviously, for “perfect square” numbers in large values only, or else, use a calculator.

MSM-3, has its own “special” place.

Lets take this example:

Given:

√20

Using the traditional method, we come up with the following procedures;

( http://www.itl.nist.gov/div897/sqg/dads/HTML/squareRoot.html )

....4.

√20.00

- 16

.... 4 00

(4 x 20 x ?) + ? ² ≤ 400

Let ? = 5

(4x20x5) + 5² ≤ 400

400 + 25 ≤ 400

425 (over) ≤ 400

Let ? = 4

(4x20x4) + 4² ≤ 400

320 + 16 ≤ 400

336 ≤ 400 Accepted

....4. 4

√20.00

- 16

.... 4 00

.... 3 36 = (4x20x4) + 4²

....... 64 00

This time,

(44x20x ?) + ?² ≤ 64 00

Let us use “7”

(44x20x7) + 49 ≤ 64 00

6160 + 49 ≤ 64 00

6209 ≤ 64 00 Accepted

To make sure, let’ use 8

(44x20x8) + 64 ≤ 64 00

7040 + 64 ≤ 64 00

7104 (over) ≤ 64 00

....4. 4 .7

√20.00 00 00

- 16

.... 4 00

.... 3 36 = (4x20x4) + 4²

....... 64 00

....... 62 09 = (44x20x7) + 49

…......1 91 00

Repeating the same process over and over again makes the number of digits involved to further increase (this time, from two digits, 44, into three, 447), while guessing the next possible digit seems to become very difficult to attain

. (447x20x ?) + ?² ≤ 19100

This is the “main problem” of the traditional method of extracting the square roots of numbers ;

1) The repeated “trial and error” method to come up with the right digit.

2) As you try to add another digit, the longer the multiplication and subtraction of numbers involved

APPLYING THE UNIVERSAL SQUARE EDGING METHOD

Given:

√20

Step 1:

Like in any rule, look for a square, equal or nearest the given number. Write down its square root value.

....4.

√20.00

Step 2:

Create a “P-Chk”

P: 20

4.5 : 20.25 ↓

4↓

The down arrow, indicates that the next digit is below 5 (or 4↓)

Step 3:

Create A Digit Locator. (Note: it is similar to the Square Root Locator, except this time we’re simply looking for the next digit after 4) Please review :

http://easysqrtsforkids.blogspot.com/2010/08/five-digitsix-digit-square-edging.html

4, 3 / 20.25 \

...... \ 18.06 / 20

......... 16 ....

......... 36.25 / 2

......... 18.12

The given number 20 is in-between 20.25 and 18.06. On the left side of the equation are the digits 3 and 4

Step 4:

Do the Digit Sampling

Let start with “4”

4.4² = 16.16

4x8 = ...3 2 .

.......... 19.36 ≤ 20

4.4² = 19.36 Accepted

Step 5:

Write down the 4 as our second digit

....4. 4

√20.00

L.A.L. Multiplication Technique

Do you still remember the R.A.R. multiplication pattern? (See: http://easysqrtsforkids.blogspot.com/2010/08/new-method-of-squaring-numbers.html )

R.A.R. stands for, Reference and All the digits to the Right but this time, you will learn another technique:

L.A.L. stands for Last digit and All the digits to the Left. I will not explain here how I got it (except if requested). It is still based on the same equation (A+B)² = A² + 2AB + B².

We know that 4.4² = 19.36

If we put a digit next to 4.4, example, a digit “2”

4.42² = ?

Instruction 1:

Write down the square of the original value (4.4²) = 19.36 then write next to it the square of 2

4.42² = 19.36’04

Instruction 2:

Following the LAL multiplication technique, multiply the last digit (2) to all to the left digits ( that is, 44, ignore the decimal point).

44 x 2 = 88

Instruction 3:

Don’t forget the DTP, "double the product”

88 x 2 = 176

As a practice, always double the value of the last digit before multiplying to all digits to the left.

44x4 = 176

Instruction 4:

To make it clear, the square of 44.2 must end in 4. So, the last digit of 176 must be aligned to the next digit to the left of the last digit of 19.36’04

.... 4.42² = 19.36’04

(+) 44x4 = .... 17’6 .

.... 4.42² = 19.53’64 (which is the true square root of 4.42, check in your calculator)

Now, let’s go back to our main discussion.

We already know that the second digit of our square root is 4, so our “partial answer” will be equal to “4.4”. The next question is…what will be the next digit?

4.4 ?

Step 6:

Create a 2^nd P-Chk for the next digit after 4.4

In data above, the square value for 4.4 is

4.4² = 19.36

2^nd P-Chk

P: 20.00 00

4.45² = 19.36’25

‘’x10 = .... 44 .... (Ignore the decimal point)

............ 19.80’25 ↑ (the up arrow shows that P: 20 is in upper area)

5↑

The 5↑ indicates that the next digit is above 5 (There are four possible digits, 6, 7, 8 or 9 and only one is acceptable).

Step 7:

To easily determine which group the next digit belongs (9, 8 group or 7, 6 group), use a 2^nd Digit Locator

2^nd Digit Locator

(4.50²) = 20.25’00 (already given, it is the same as to 4.5² = 20.25)

(M.V.) = 20.02’62

(4.45²) = 19.80’25

.............. 40.05’25 / 2

.............. 20.02’62

The M.V. (Middle Value) shows that it is greater than 20.

Therefore, the possible digit belongs to “7, 6”group.

Let us simplify the equation of the 2^nd Locator

2^nd Digit Locator

9, 8 / 20.25

.......\ 20.02 \ ↓

........ 19.80 / 7, 6 (20 is between 20.2 and 19.80)

........ 40.05’/ 2

........ 20.02

The 2^nd digit locator gives us a clue that the next digit could either be 7 or 6

Step 8:

Again, use 7 or 6 as sample, (do the Digit Sampling)

.... 4.47² = 19.36’49

(+)’’x14 = .... 61’6 .

.... 4.47² = 19.98’09

“There is no need to solve for 6, because its answer would be much lesser than 4.47”

Making Sure of Your Choice of Digit:

To avoid the possibility that the 7 is not yet the correct digit, choose 8 as a ‘sample’

.... 4.48² = 19.36’64

(+)’’x16 = .... 70’4 .

.... 4.48² = ... w/out completing the addition, you will notice that if we add 3 to 7, the carry 1 makes the 19 to become 20, giving us a clue that the 4.48² > 20.

This result give as an assurance that 7 is the correct, next digit after 4.4

Therefore, we accept this equation;

4.47² = 19.98’09

Step 9:

For the next digits, simply repeat these three basic rules of square edging

1) Create a parameter checker (P-Chk)

2) Determine the correct digit by using the Digit Locator

3) To make sure which digit is the appropriate digit, do the Digit Sampling

3^rd P Chk

P: 20.00 00

4.475² = 19.98’09’25

‘’x10 = ........ 4’47 .... (Ignore the decimal point)

.............. 20.02’56’25 ↓ (the down arrow shows that P: 20 is in lower area)

4↓

3^rd D Loc

20.02

20.>. \ ↓ (The symbol “20.>” means, 20 point “something”)

19.98 / 2, 1 (the digit locator is giving us a hint that the next digit could either be 2 or 1)

40.00 / 2

20.00

Digit Sampling

4.472² = 19.98’09’04

....’’x4 = ...... 1’78’8 .

4.472² = 19.99’87’84 (acceptable)

To make sure, check digit “3”

4.473² = 19.98’09’09

....’’x6 = ...... 2’68’2 .

4.473² = 20...0.......... > 20 (over)

We are now sure that the next digit is 2, therefore;

4.472² = 19.99’87’84

P-Chk

4.4725² = 19.99’87’84’25

.......................... 44’72 ...

.................20.00’32’56’25 ↓

4↓

D-Loc

20.00’33 (round-off value)

20.00’10 \ ↓

19.99’88 / 2, 1

40.00’21 / 2

20.00’10

D. Sampling

4.4722² = 19.99’87’84’04

‘’x 4..... = ..........17’88’8 .

................ 20. ... 04’ ...... > 20 (over)

4.4721² = 19.99’87’84’01

‘’x 2..... = ........... 8’94’4 .

................ 19.99’96’78’41 (acceptable)

Therefore,

√20 = 4.4721 (approximately)

If you want, you can continue the process by repeatedly doing the parameter checking, digit locating and digit sampling.