Want To Learn Something New

Kids, try asking these questions to your parents;

1) Are they familiar with the word “square root”?

2) Had they ever tried taking the square root of an eight-digit number such as 38,775,529 without the aid of a calculator?

3) Are they familiar with the “long hand division method”?

If they answer “yes” to all my questions, maybe some how, they will agree with me, along with the other parents and teachers that such a kind of mathematics, is a “real burden” for grade school kids like you.

But I devised a much easier method that I called Square Edging (MSM-1 Format), designed specially, for kids like you.

WELCOME TO THE WORLD OF SQUARE EDGING!

Notes from the Author:

For parents and Math teachers who have the ideas about Squaring and Taking the Square Roots of Numbers, start with SE Telegram and if you're interested to know how it works, start reading the topic "New Method of Squaring" to fully understand what "Square-Edging MSM-1 Format" is all about

For Beginners

Kids,do you know that besides adding, subtracting, multiplying and dividing numbers, there are two other very interesting math operations that are also very useful in many activities? If you wish to know the area of a square or the measures of the sides of a right triangle, two special math operations - getting the square values or getting the square roots of numbers, are some of the more advanced math operations that you must learn.

They are advanced because unlike ordinary multiplication and division, their operations (methods of getting the answers), are much, much more tricky and difficult.

SQUARE OF A NUMBER

Getting the ‘square value’ of a number is like doing a special kind of multiplying a number, in which both the multiplicand and the multiplier are equally the same values. Sometimes, it is described as product of a number multiplied by itself.

Examples:

2 x 2 = 4
27 x 27 = 729
146 x 146 = 21,316

The value 4 is sometimes called the ‘square value’ of 2, or simply, “square of 2”. The same way, 729 is the “square of 27” and 21,316, the “square of 146”. Sometimes, instead of writing 2x2, 27x27 or 146x146, “a small number 2 in upper right side” of a given number is used as a symbol, telling you to multiply that number by itself. So, instead of 2x2, we write 2² = 4 and 27x27 as 27² = 729, while 146x146 as 146² = 21,316.

Maybe, you are wondering why it is called ‘square’. Probably, early mathematicians noticed that the measure of the area of a square is always equal to a certain ‘number multiplied by itself’, so they named it, that way.

SQUARE ROOT

On the other hand, getting the square root of a number, need a very different way, of dividing a number. Unlike in ordinary division at which you need to mention the value of the divisor, in getting the square root of a number, both the divisor and the quotient are unknown and the difficult thing is, both divisor and the quotient must be equally in the same values.

Examples: 36 ÷ 3 = 12 36 ÷ 4 = 9 36 ÷ 6 = 6

In the above examples, 36 can be divided by 3 or 4 but the quotient would not be equal or the same with the divisor. Dividing 36 by 6, we can get a quotient equal to 6, which is the same exact value as to the divisor. In this situation, we can say that 6 is a square root of 36.

In doing this special kind of division, the symbol √ is used before a given number (example √144, read as, “the square root of one hundred forty-four’), to tell you to look for a divisor that will give a quotient, equal to that divisor. Dividing 144 by 12, we come up with a quotient equal to 12 (144 ÷ 12 = 12). Showing equal values for both divisor and quotient we can say then, that √144 = 12.

But there are occasions that the given numbers are in large values (example, √139,876). Getting the square root of such large valued numbers requires a very tedious and tricky method called ‘long hand division’. But as a practice, small valued numbers are introduced for grade school children, to make them easier to memorize.

Easy To Memorize “TABLE OF SQUARE ROOTS”

√1 = 1
√4 = 2
√9 = 3
√16 = 4
√25 = 5
√36 = 6
√49 = 7
√81 = 9
√100 = 10

PERFECT SQUARES

Not all numbers from 1 to 100 give a square root in whole exact values. Most of them are in decimal values. Below is a list of examples of numbers, having no whole exact square root values:

√ 2, √3, √10, √99 , √28 , √50

Counting from 1 to 100, there are only ten numbers having square roots in ‘exact whole values’ and they are called perfect squares (or simply call them “PERKS”).

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

(Using a calculator, find the square roots of each numbers from 1 to 100 and write down which numbers have an exact whole numbers)

SET OF “PERFECT SQUARES” = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

The above numbers are called Perks because they are squares of whole numbers.

Easy To Memorize “TABLE OF SQUARES”

1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
10² = 100

Square Roots of Five-Digit/Six-Digit Numbers

Check This Out: A Much Easier Explanation, A Revised Version

 
The ideas behind 3D/4D.SE are very important because they are the ‘basics’ of Square Edging (MSM-1 Format). In dealing with 5D/6D.SE and 7D/8D.SE, the procedures you learned in 3D/4D.SE are always included.
The 5D/6D.SE is the start of the “real square edging”. In this topic, you will soon begin to appreciate the things that you learned from the past (specially, the SSQ Method).

Five-Digit Square Edging
Let us start taking the square root of a five-digit perfect square number.

Given Problem: What is the square root of 73,441?

√73,441 = ?

Using what we learned from 3D/4D.SE, we can proceed as follows;

Step 1

√07’34’41

Step 2 :

√07’34’41

Step 3:

√07’34’41

2 _ 1

_ _ 9

Step 4 :

√07’34’41

2 _ 1

2 _ 9

Now, our problem is how to get the middle digit.

Step 5

Instead of leaving the middle of each possible square roots blanks, simply put a letter “N” between 2 and 1. In the same way, put also an N between 2 and 9.

√07’34’41

2 N 1

2 N 9

Now, to find out the missing digits, apply the SSQ Method

Looking For the Missing Digits

Square Edging is not really the opposite or reverse process of SSQ. In fact, they are the same. The only difference is that in SSQ, what we are looking for is the total sum. While in S.E, the total sum is already given and what we are looking for is the missing digit or digits.

To make our search for the missing digits easy and simple, let us first consider the first incomplete square root.

2N1² = ?

Step 6

Create a PSL

2N1² = 04’nn’01 ← PSL

The capital letter ‘N’, represent the missing digit and the small “nn”, its equivalent index square. We do not really have an idea of what these ‘N’ and ‘nn’ are, but including them in our equations will make things easier for us.

Step 7

Cross multiply N to the digit on its right and multiply again by 2. Just leave it that way. It temporarily represent SP1.

Nx1x2 = ? ← SP1

Total Sum

I mentioned before that the difference of SE from SSQ is that in SE, the total sum is already given. Use the digits in the given number as the total sum. Do not include the square root sign “ √ "

√07’34’41 becomes “ 07’34’41 ”

Step 8

Create a T-Sum

07’34’41 ← TSum

Step 9

Put them all together - the PSL, temporary SP1 and T-Sum. Put a plus sign + to the temporary SP1, to indicate that you wish to add the temporary SP1 to the PSL.

.... 2N1² = 04’nn’01 ← PSL

.+Nx1x2 = ? ← SP1

………… 07’34’41 ← TSum

Link to the Missing Digit

If you look at it closely, the last two-digit of the PSL is 01, while the last two-digit of the T Sum is 41”. This is the link for us to know, the missing middle digit. If the last digit of SP1 must be in the tens decimal place, we should "focus our attention" to the "tens decimal places of the PSL and the T-Sum".

0 ← at tens decimal place of the PSL

4 ← at tens decimal place of the PSL

In SSQ, to get the total sum, we must add the sub-product SP1 to the PSL.

0 + ? = 4

What number we must add to 0 to have a sum equal to 4?

The answer is 4.

So the last digit of SP1 is 4

Nx1x2 = ..4 ← SP1

Use the notation “..” (two dots), to indicate that maybe, there are digits before 4 but we do not know yet, what they are. The important thing is that, we have now an idea of what the last digit of SP1.

Knowing the Middle Digit

In N x 1x 2, we can multiply 1 by 2, but not N. Simply multiply 1 x 2.

N x 1 x 2 = N x 2

N x 2 -= ..4

N represents a single digit number. So what is that single-digit number, when you multiplied by 2, give us a product that ends with 4?

There always two numbers

N = 2, 7

If we replace N by 2, 2x2 = 4

If we replace N by 7, 7x2 = 14

Complementary Multiplicands

Always remember that all sub-products are in even numbers. In multiplication, if the multiplier is an even number, there is always a pair of multiplicands that will give products that ends with the same last digits.

Table of Complementary Multiplicands

(Note: Each pair of digits when multiplied by the same even number give products that ends with the same last digits

0 and 5

1 and 6

2 and 7

3 and 8

4 and 9

Step 9

Instead of N, write down the its equivalent digits

221

272

We have now the first set of possible square roots.

Step 10: Repeat the process for 2N9, start again in step 6.

Check if we have the same data for 2N9

PSL of 2N9 = 04'nn'81

Temporary SP1 = Nx9x2

N x 18 can be written as Nx8

N x 8 = ..6

N = 2,7

Our second set of possible square roots is;

229

279

So now we completed the four possible square roots

221

271

229

279

But only one of them is the true square root of 07’34’41

Which among the four possible square roots is the true square root? We can use the 3D.SSQ, squaring them, one at a time.But it will take a lot of time and effort for you to do that. Don’t worry, there is a better way.

Squares Ending in 25

I think it would be helpful, if I’ll teach you first, a “trick” in squaring two-digit numbers ending in 5 (or numbers in the multiples of 5).

Table of Multiples of Five Squares (M5.Sq)

05² = 0’25

15² = 2’25

25² = 6’25

35² = 12’25

45² = 20’25

55² = 30’25

65² = 42’25

75² = 56’25

85² = 72’25

95² = 90’25

But actually, there is no need to memorize the table above. If you will notice, all squares in the multiples of 5 always end in 25. But check this out;

0 x 1 = 0

05² = 0’25

1 x 2 = 2

15² = 2’25

2 x 3 = 6

25² = 6’25

3 x 4 = 12

35² = 12’25

.

.

.

9 x 10 = 90

95² = 90’25

Easy Trick

Example:

85² = ?

Step 1

Check the first digit of the given number

The first digit in 85² is 8

First digit = 8

Step 2

Multiply that digit to a number next to it, higher by 1

The digit next to 8 is 9

8 x 9 = 72

Step 3

Write down their product. Also write next to it, the digits 25 with the sign ’

72’25

Therefore,

85² = 72’25

That’s how easy to do that trick.

Parameter Check

Prof. Barbosa’s Method (remember that helpful tips in Four-Digit Square Edging?) is only effective (or accurate), in dealing with two possible square roots. But this time, we have four possible square roots.

All the possible square roots (221, 729, 771 and 771) are, in-between 200 and 300. We can use that limitation as our “parameter”, to correctly guess the location of the true square root of 07’34’41. But the limit provided by 700 and 800, is just not enough. We need the middle-half of 700 and 800 to make our search, more precise and accurate.

The ‘actual’ parameter checker, if we consider the possible square roots in their actual values (3-digit numbers) can be represented this way:

Parameter Checker

300² = 09’00’00

250² = 06’25’00

200² = 04’00’00

Looking at the right side values of the parameter checker, the given number, 07’34’41 is less than 09’00’00 but greater than 06’25’00. So, 07’34’41 is in-between 09’00’00 and 06’25’00. It is in the higher limit area (H.L.A.).

Looking at left side of the P-Chk (parameter checker), the ‘true square root’ should also be, in-between 250 and 300 (or in the higher limit area). If we arrange the four possible square roots, from lowest to highest, it will appear this way;

221, 229, 271, 279.

The P. Chk is giving us a clue that the ‘true square root’ is higher than 250 but less than 300. 221 is lower than 250, so we must exclude it. The same way, 229 is also lower than 250. So the only remaining possible square roots are 271 and 279. So, we reduced our possible squares from 4 into only 2 possible square roots.

271

279

The Parameter Checker only gives us an idea, in which area (higher or lower limit area), the true square root might be located. But it is not enough. In H.L.A. (higher limit area), where we assumed the location of the true square root of 7’34’41, two possible square roots still remained (271 and 279). Which among the two is the correct answer? One way to know it is by using the 3D.SSQ. But again, it would be a long process. Don’t worry, there’s another way.

Square Root Locator

We need to further sub-divide the HLA into two parts – upper portion and the lower portion. To do that, we need a middle number between 250 and 300. That middle number will become a boundary that will separate the upper portion from the lower portion of the HLA.

How to know the middle number between 250 and 300? Just follow these instructions:

First: Add 250 and 300

250 + 300 = 550

Second: Divide the sum by two

550 ÷ 2 = 275

The Middle-Half of 250 and 300 = 275

The first group of data for our Square Root Locator will appear as;

300
275

250

Next, square them and write down their equivalent square values;

300² = 9’00’00

275² = ?

250² = 6’25’00

But how we can get the square of 275? It can be done by SSQ but there is an alternate way. Just follow again these similar instructions:

First: Add 9’00’00 and 6’25’00

9’00’00 + 6’25’00 = 15’25’00

Second: Divide the Sum by two

15’25’00 ÷ 2 = 7’62’50

Third: Subtract the quotient by 6’25

7’62’50 – 6’25 = 7’56’25

275² = 7’56’25

Now, we have a complete data for Square Root Locator

300² = 9’00’00

275² = 7’56’25

250² = 6’25’00

Let us sub-divide them into two parts;

(A)

300² = 9’00’00 \ Upper portion, which is between 275 and 300

275² = 7’56’25 /← Middle Boundary that separate the upper and lower portions

250² = 6’25’00

(B)

300² = 9’00’00

275² = 7’56’25 \← Middle Boundary that separate the upper and lower portions

250² = 6’25’00 / Lower portion, which is between 250 and 275

Looking at the right side data of the Sq. Rt. Loc., the given value, 7’34’41 is below 7’56’25 but above 6’25’00, so it belongs to the lower portion of the HLA.

The two possible square roots have also their respective locations.

279 is above 275 but below 300 (ignore their square signs “ ^{2 ”} ) or in the upper portion.

While 271 is between 275 and 250 or in lower portion.

If 7’34’41 is in the lower portion, it must also follow that the corresponding data on its left side might be its equivalent square root. 271 is also in the lower portion.

Therefore, the true square root is 271.

√7’34’41 = 271

SE Telegram

After you become familiar with the basic rules and procedures of Square Edging, you can use the SE Telegram, to ease up the computation. It is a combination of doing some parts mentally and writing some other parts directly without the formal representations, such as the use of square sign, equal sign and omitting some letters and others, not so important.

Example:

√38,775,529

Step 1: Determine first, the two IPS (initial possible square roots). There is no need to write down the middle letters M and N

*At Left Column, 1^st Row

√38,775,529

... 6 ......... 3

... 6 ......... 7

Step 2: Create a modified P CHK.

1) On the first line, write down the notation “ P: ” followed by the first 4 digits of the given problem.

2) On the second line, write down the middle numbers, separated by a colon “ : ” and no square sign. On its right side, draw an arrow, either an up ↑ or down ↓ arrow, based on the following conditions:

Condition 1: If P is less than the middle square value, draw a down ↓ arrow

Condition 2: If P is greater than the middle square value, draw an up ↑ arrow

3)On the third line, write down M, followed by an either a 5↑ or a 4↓, based on the following conditions:

Condition 4: If the arrow on the second line is ↑, write down 5↑, then on its right side, write the notation /_ 750/

Condition 5: If the arrow on the second line is ↓, write down 4↓, then on its right side, write the notation /_ 250/

*At Right Column, 1^st Row

.....P: 38’77

. 65 : 42’25 ↓

....M : 4↓ / 6250/

Step 3: Start computing the SE data.

Let’s begin with 6MN3

1) Looking for the ‘N’ digits

* At Left Column, 2^nd Row

N3 : 09

(Blank)

....... 2 ... ← (take note, the 2 here is the second to the last digit of the given problem)

The letter N is included because we’re looking for that missing digit. Leave the second line blank and ask your self this question:

Q: What number needed to add to 0 to have a sum of 2?

A: 2

Write down on the second line “ Nx6 ” (6 is the double value of 3), followed by a colon :, then the answer “2”. On their right, decide which “pair of multiplicands” is the correct combination:

..... N3 : 09

... Nx6 : 2 . ..... 2, 7

............. 2

2) Looking for the M digits

Since we’re doing some parts mentally, directly write down the two combinations from the digits we gathered:

* At Left Column, 3^rd Row

..... 23 : 4’09

... 2x6 : 1’2 .

............ 5’29

. Mx6 :(Blank)

............ 5 ← (this is the third to the last digit of the given problem)

Leaving some space in the second line blank, give you time to think first of what digit to write, by looking for a number needed to add to come up with the correct sum

..... 23 : 4’09

... 2x6 : 1’2 .

............ 5’29

. Mx6 : 0 . ... 0, 5

............ 5

6023

Underline the 0 since in P Chk, 4↓ indicates that the digits for M is 4 below.

Write down on the sixth line, the first complete digits of the first possible square root

Helpful Tip: Write down first 0, then 23 and then look for the first digit 6, so you would not be confused in writing down 6023

Do the Same procedures for the combination 73

* At Left Column, 4^th Row

..... 73 : 49’09

... 7x6 : 4’2 .

............ 3’29

. Mx6 : 2 . ... 2, 7

............ 5

6273

On the right column, below the P- Chk, do the same procedures starting from Step 3, to complete the data for the second IPS ( 6MN7)

*At Right Column, 2^nd Row

..... N7 : 49

... Nx4 : 8 . ..... 2, 7

............. 2

*At Right Column, 3^rd Row

....... 27 : 4’49

... 2x14 : 2’8 .

.............. 7’29

... Mx4 : 8 . ... 2, 7

.............. 5

6227

*At Right Column, 4^th Row

....... 77 : 49’49

..... 7x14 : 9’8 .

................ 9’29

..... Mx4 : 6 . ... 4, 9

................ 5

6477

Step 4: Arrange the 4 possible square roots from the lowest value to the highest by inserting the letters A, B, C, D on their right sides to avoid writing them again.

A - the lowest possible square root

B - 2^nd to the lowest

C - 2^nd to the highest

D – the highest of all

At Left Column ..... At Right Column

...... 6023 (A) .................... 6227 (B)

...... 6273 (C) ..................... 6477 (D)

Step 5: Use the Square Root locator to determine which of the 4 possible square roots will remain

42’25

(Blank)

36’00 .

78’25 / 2

39’12

Leave the second line (M) blank. Add 42’25 (H) and 36’00 (L).

We come up with a sum of 78’25. Divide by 2

Doing it mentally, always subtract the quotient by 6.

* At Left Column, 5^th Row

42’25

39’06 \ ↓ (Using this middle value as reference, the P: 38’77 is lesser than)

36’00 /

78’25 / 2

39’12

The ↓ arrow indicates that the two higher values (C and D) are eliminated. A and B remained

Step 6: Use the 2^nd Square Root Locator to determine which of the two remaining possible square roots, is the true square root of 38,775,529

39’06

(Blank)

36’00

75’06/2

37’53

Complete the data and determine where P is located

*At Right Column, 5ht Row

39’06\

37’53/ ↑

36’00

75’06/2

37’53

We therefore choose B as the true square root

√38,775,529 = 6,227