A) INTRODUCTION

In grade schools, children are taught to learn the four fundamental math operations - Addition, Subtraction, Multiplication and Division. But as to my personal opinion (although I'm not 100% sure), knowing how to get the squares and square roots of numbers (particularly, large numbers), are seldom 'tackled' except on special cases that a child must be introduced that the square root of 2 (or written as √2), is equivalent to 1.4142, without explaining how such value was taken (or how to get such value). Even me, personally, has having hard time explaining to a child how to get the square root of a number using the old classical "long hand division method". 


LONG HAND DIVISION

Okay, let us start in explaining how the square root of 2 come into such value as 1.4142  (the process behind it), showing the step by step way of doing it.

Question:  What is square root of  2 ?

......
√2  = ?



STAGE 1

Step 1: Put two zeroes after 2 (Don't forget the decimal point)
 ..........
√2.00

Step 2 : Find a number that when "being multiplied by itself" will bring out a product  that is near but lesser to that given problem number.

..1.......
√2.00       

 we choose 1 because 1 x 1 = 1 ( or 1² = 1)

Step 3: Subtract the square value of our first chosen number (Then bring down the two zeroes)

 .1........
√2.00 
 -1......
 .1 00   (Remainder)

 Now, this is the "very difficult part" to explain....

Step 4: Find a number that will give a value that is less than to the 'difference' that we got (1.00, or 1 00, the result of subtracting 1 from 2). This 'value' must be:

Condition 1: The product of the first number that we already know (that is,1) multiplied by 20
Condition 2: Then to be multiplied again by an 'unknown' number we're trying to guess...
Condition 3: Then must be added to the square value of that 'unknown' number (that unknown number multiplied by itself), to come up to such value that either be equal or less than 1 00

1 x 20 x ? + (? x ?) = equal or less than 1 00


TRIAL AND ERROR METHOD

This stage, more than the difficulty of explaining 'why the need to multiply things by 20' and then to add the square value of such so number, is also the stage of trying to 'experiment' a number (although limited from numbers 1 to 9 only) and do it by trial and error. The truth is that, we have nine numbers we need to 'try' to sample and see to it if the resulting value will be correctly enough to be accepted as the rightful value. It is a very tedious and slow process.

SAMPLE 1:  Let's try the highest number... 9

1 x 20 x 9 + (9 x 9) =?
20 x 9 = 180..... 9 x 9 = 81
180 + 81 = 261    

261 is over 1 00 and it's too far. If a child is good enough in 'estimating values' (although in reality and sad to say, not all of them), he might try another number which is much lower, at least half that number (half of 9 could either be 5 or 4).

SAMPLE 2:  Let's try  5...
 
1 x 20 x 5 + (5 x 5) =?
20 x 5 = 100..... 5 x 5 =25
100 + 25 = 125     

Take note, 125 is 'over' 1 00. So, 5 is not the required number. 

SAMPLE 3:  Let's try  4..

1 x 20 x 4 + (4 x 4) =?
20 x 4 = 80..... 4 x 4 = 16
80 + 16 = 96      

In this case, we accept 4 as the next number to include in our answer     
  
 .1. 4.......
√2.00'00
 -1......
 .1 00
- .. 96..     (Subtract 96 from 100)
.......4 00  (Remainder)

ALTERNATIVE APPROACH

There is this another more practical way of lessening the burden of guessing which among the numbers from 1 to 9 is the possible candidate as our next number for our answer...

1 x 20 x ? + (? x ?) = equal or less than 1 00     

1) Consider the first number (1) and then, multiply by 20

1 x 20 = 20

2) Divide 100 by 20  (As much as possible, keep the values rounded off )

100 ÷ 20 = 5

3) Try 5 or numbers below it  (4, 3... etc)



STAGE 2    

The process is simply the repetition of step 4 of stage 1, except there are some modifications  

Condition 1: The product of the first numbers that we already know (THIS TIME, it will be 14) multiplied by 20
Condition 2: Then to be multiplied again by an 'unknown' number we're trying to guess...
Condition 3: Then must be added to the square value of that 'unknown' number (that unknown number multiplied by itself), to come up to such value that either be equal or less than 4 00

14 x 20 x ? + (? x ?) = equal or less than 4 00    

Using the ALTERNATIVE APPROACH

14 x 20 = 280

Let's round off 280 into 200 (rounding off 280 to 300 seems not divisible)

400 ÷ 200 = 2 

SAMPLE 1:  Let's try  2...

14 x 20 x 2 + (2 x 2) =?
 280 x 2 = 560 ... 2x2 = 4
560 + 4 = 564

Take note, 564 is 'over' 4 00. So, 2 is not the required number. 

SAMPLE 2:   Let's try  1...
 
14 x 20 x 1 + (1 x 1) = ?
280 x 1 = 280..... 1 x 1 = 1
280 + 1 = 281

 In this case, we accept 1 as the next number to include in our answer         

 .1.  4   1..
√2.00'00'00
 -1......
 .1 00
- .. 96.. 
.......4 00   
- .....2 81  
.......1 19'00  (Remainder)



STAGE 3   

Repeating Step 4...

141 x 20 x ? + (? x ?) = equal or less than 4 00 

141 x 20  = 2820   

Let's round off both 11900 and 2820 in a more comfortable values that can easily be divided

12000 for 11900
3000 for 2820

12000 ÷ 3000 = 4

SAMPLE 1:  Let's try 4...

141 x 20 x 4 + (4x 4) = equal or less than 1 19'00
 2820 x 4 = 11280 .... 4 x 4 = 16
11280 + 16 = 11296  

Take note that 11296 is less than 11900...

But ARE WE SURE of it ? What if  5 is really the required number instead of 4?

SAMPLE 2:  Okay, let's try 5...

141 x 20 x 5 + (5x 5) = equal or less than 1 19'00
2820 x 5 = 14100 ....  5 x 5 = 25 

Checking the the result of multiplying 2820 by 5 resulting into 14100, you might notice that...

14100 is over 11900

 There's no need to continue computing, using 5 as our sample. This simply prove that the next candidate is really the number 4.


.1.  4   1   4    .
√2.00'00'00'00
 -1......
 .1 00
- .. 96.. 
.......4 00   
- .....2 81  
.......1 19'00
-......1 12'96   
............6'04'00  (Remainder)


STAGE 4

Repeating Step 4...

1414 x 20 x ? + (? x ?) = equal or less than 6'04'00  

1414 x 20 = 28280

Again, let's round off 60400 and 28280 in a more comfortable values that can easily be divided

60000 for 60400
30000 for 28280

60000 ÷ 30000 = 2

SAMPLE 1:   Let's try 2...

1414 x 20 x 2 + (2x 2) = equal or less than 6'04'00
28280 x 2 = 56560 ..... 2 x 2 = 4
56560 + 4 = 56564

Take note that 56564 is less than 60400...

In that case, 2 could be the possible next number to be included in our answer.

But again, we need to also check  3 as our 'sample' to be sure of it...

SAMPLE 2:    Let's try 3...

1414 x 20 x 3 + (3x 3) = equal or less than 6'04'00
28280 x 3 = 84840 ..... 3 x 3 = 9

Inspecting 84840, there's no need to continue computing  since...

84840 is over 60400

So, it only prove that 3 is not really the next candidate.


.1.  4   1   4   2   .
√2.00'00'00'00
 -1......
 .1 00
- .. 96.. 
.......4 00   
- .....2 81  
.......1 19'00
-......1 12'96      .
............6'04'00   
............5'65'64 .
...............38'36  (Remainder)

Using the long hand division method, you now have an idea  why  √2  = 1.4142

But my real question is this - "Do you think that kids in grade school levels can perform or do such kind of computation process?"