## Theory

Let ${\displaystyle x}$  be the number of which we want to obtain the square root ${\textstyle y={\sqrt {x}}}$ ; Let's consider its decimal expansion, for example: ${\textstyle x=456.7890123}$ . Let's separate its digits into groups of two around the decimal point in the following way

${\displaystyle x=456.7890123=4\cdot 56\cdot 78\cdot 90\cdot 12\cdot 30\cdot 00\cdot 00\cdots }$

or, in other words, let's define the sequence of integers ${\displaystyle \alpha _{i}}$

${\displaystyle \alpha _{1}=4,\alpha _{2}=56,\alpha _{3}=78,\alpha _{4}=90,\alpha _{5}=12,\alpha _{6}=30,\alpha _{7}=00,\cdots }$

and let's build the sequence ${\textstyle x_{i}}$  recursively from ${\textstyle x_{0}=0}$

${\displaystyle x_{i}=100x_{i-1}+\alpha _{i}}$

and let ${\displaystyle y_{i}}$  be the integer part of the square root of ${\displaystyle x_{i}}$

${\displaystyle y_{i}=\lfloor {\sqrt {x}}_{i}\rfloor }$

i.e.  ${\displaystyle y_{i}}$  is the largest integer whose square is not greater than ${\displaystyle x_{i}}$ . Finally, let us call remainders to the differences

${\displaystyle r_{i}=x_{i}-y_{i}^{2}\geq 0}$

For our example we have:

${\displaystyle i}$  ${\displaystyle \alpha _{i}}$  ${\displaystyle x_{i}}$  ${\displaystyle y_{i}}$  ${\displaystyle r_{i}}$
0 0 0 0
1 4 4 2 0
2 56 456 21 15
3 78 45678 213 309
4 90 4567890 2137 1121
5 12 456789012 21372 26628

Let's see that, by construction, ${\displaystyle x_{i}}$  grows as ${\displaystyle 10^{2i}}$  (two more digits in each step), in fact the sequence ${\displaystyle x_{i}\cdot 10^{4-2i}}$ : (0, 400, 456, 456.78, 456.7890, etc.) tends to ${\displaystyle x}$  ${\textstyle \left(x_{i}\cdot 10^{4-2i}\rightarrow x\right)}$  or ${\textstyle x=\left(\lim _{i\to \infty }x_{i}\cdot 10^{4-2i}\right)}$ . By comparison, ${\displaystyle y_{i}}$  , as the integer part of the square root of ${\displaystyle x_{i}}$ , grows only as ${\displaystyle 10^{i}}$  (one digit more each step). As ${\displaystyle y_{i}}$  is the largest integer whose square is not greater than ${\displaystyle x_{i}}$  we have ${\displaystyle r_{i}=x_{i}-y_{i}^{2}\geq 0}$  as above but

${\displaystyle x_{i}-(y_{i}+1)^{2}=x_{i}-y_{i}^{2}-2y_{i}-1<0}$

by definition of ${\displaystyle y_{i}}$ , or

${\displaystyle r_{i}=x_{i}-y_{i}^{2}<2y_{1}+1}$

multiplying by ${\displaystyle 10^{4-2i}}$

${\displaystyle x_{i}\cdot 10^{4-2i}-(y_{i}\cdot 10^{2-i})^{2}<(2y_{i}+1)\cdot 10^{4-2i}}$

but as  ${\displaystyle y_{i}}$  grows only as ${\displaystyle 10^{i}}$ , the second term tends to zero as ${\displaystyle 10^{-i}}$ .  With this

${\displaystyle x_{i}\cdot 10^{4-2i}-(y_{i}\cdot 10^{2-i})^{2}\rightarrow 0}$

and ${\displaystyle x_{i}\cdot 10^{4-2i}\rightarrow x}$  so that we have

${\displaystyle y_{i}\cdot 10^{2-i}\rightarrow y={\sqrt {x}}}$

For other numbers, the above factors are: ${\displaystyle 10^{2k-2i}}$  and ${\displaystyle 10^{k-i}}$ , where ${\displaystyle k}$  is the number of two-digit groups to the left of the decimal point, negative if it is followed by 00 groups (ex. ${\displaystyle k=-1}$  for ${\displaystyle x=0.00456}$ , ${\displaystyle k=-2}$  for ${\displaystyle x=0.00000456}$ , etc.).

This is the basis for traditional manual square root methods.

## Procedure

We start with ${\displaystyle i=0}$ , ${\displaystyle x_{0}=0}$ , ${\displaystyle y_{0}=0}$ , ${\displaystyle r_{0}=0}$ .

### First digit

Squares table
${\displaystyle b}$  ${\displaystyle b^{2}}$
1 1
2 4
3 9
4 16
5 25
6 36
7 49
8 64
9 81

For ${\displaystyle i=1}$  and ${\displaystyle x_{1}=\alpha _{1}}$ , it is trivial to find ${\displaystyle y_{1}}$  such that its square does not exceed ${\displaystyle x_{1}}$  through the use of the following table of squares that is easily retained in memory as it is only a subset of the multiplication table. In the case of the example we find ${\displaystyle y_{1}=2}$ .

### Rest of digits

For ${\displaystyle i>1}$ , we have ${\displaystyle x_{i}=100x_{i-1}+\alpha _{i}}$  as defined above and we try to build ${\displaystyle y_{i}}$  in the way:

${\displaystyle y_{i}=10y_{i-1}+b}$

where ${\displaystyle b}$  is a one-digit integer ranging from 0 to 9. To obtain ${\displaystyle b}$  we have to choose the greatest digit from 0 to 9 such that:

${\displaystyle x_{i}-y_{i}^{2}=x_{i}-(10y_{i-1}+b)^{2}\leq 0}$

or

${\displaystyle x_{i}-(a+b)^{2}\leq 0}$

if we write ${\displaystyle a=10y_{i-1}}$ . Expanding the binomial we have

${\displaystyle x_{i}-a^{2}-2ab-b^{2}=100x_{i-1}+\alpha _{i}-(10y_{i-1})^{2}-2ab-b^{2}\leq 0}$

or

${\displaystyle 100r_{i-1}+\alpha _{i}\geq 2ab+b^{2}=(2a+b)b=(20y_{i-1}+b)b}$

The left side of the above expression may be seen as simply the previous remainder with the next two-digits group appended to it, and the parenthesis of the last term as twice the previous root with digit b appended to it. In our example, for ${\displaystyle i=2}$  we have 56 on the left and the above expression is

${\displaystyle 56\geq (40+b)b}$

which holds only for ${\displaystyle b=0}$  or ${\displaystyle b=1}$  so that 1 is our next root digit but, how can we proceed in the general case without having to systematically explore every possibility (${\displaystyle b=0,1,\cdots ,9}$ )?

Here Knott[1] distinguishes two different approaches:

• Preparing the divisor
• Preparing the dividend.

#### Preparing the divisor

This corresponds with the above expression

${\displaystyle 100r_{i-1}+\alpha _{i}\geq (20y_{i-1}+b)b}$

And it is the strategy usually used with paper and pencil and can also be implemented, of course, on the abacus. In the above expression, if we see the left part as the dividend and the parenthesis as the divisor, b is the first digit of the division

${\displaystyle b=(100r_{i-1}+\alpha _{i})/(20y_{i-1}+b)}$

but since we don't know b yet, we approximate it using only the main part of the divisor

${\displaystyle b=(100r_{i-1}+\alpha _{i})/(20y_{i-1})}$

This gives us a guess as to what the value of b might be, but we need:

1. Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
2. Obtain the next remainder to prepare the calculation of the next digit of the root.

Both steps require subtracting ${\displaystyle (20y_{i-1}+b)b}$  or ${\displaystyle 20y_{i-1}b}$  and ${\displaystyle b^{2}}$  from ${\displaystyle 100r_{i-1}+\alpha _{i}}$ ; checking that we are not going to negative values and that what remains is less than ${\displaystyle 2y_{i}+1}$  (otherwise we would have to revise ${\displaystyle b}$  up or down). If we do this correctly, what we are left with is the new remainder ${\displaystyle r_{i}}$ . It should be noted that, as we proceed in the calculations (${\displaystyle i}$  increasing) ${\displaystyle b}$  is progressively a smaller and smaller contribution to the divisor ${\displaystyle (20y_{i-1}+b)}$ ; so the process indicated above will look more and more like a mere division.

This is the method proposed by Takashi Kojima in his second book: Advanced Abacus - Theory and Practice[2], and that you can see described in Square roots as solved by Kojima[3] in Totton heffelfinger’s website, works to which I refer the reader to see explanations and examples. What follows here, for purposes of illustration, is a sketch of how the calculation might be started in our example: ${\displaystyle x=456.7890123}$

First 3 digits of ${\displaystyle {\sqrt {456.7890123}}}$  preparing the divisor
Abacus Comment
ABCDEFGHIJKLM
4567890123 Entering radicand starting in CD (first group)
2 First root digit in B
-4 Subtract square of B from first group
2  567890123 Null remainder
4  567890123 Doubling B. Appending next group (56)to remainder
41 567890123 5/4≈1, try 1 as next root digit
-4 Continue division by 41, subtract 1✕41 from EF
-1
41 157890123 15 as remainder
42 157890123 Double second root digit
42 157890123 Append next group (78)
423157890123 157/42≈3, try 3 as next root digit
-12 Continue division by 423, subtract 3✕423 from E-H
-06
-09
423 30990123 309 as remainder
426 30990123 Double third root digit
426 30990123 Append next group (90)
etc.

As can be seen, twice the root grows to the left of the abacus to the detriment of the radicand / remainder and the groups of two figures still unused. This is contrary to what happens with the rest of the elementary operations on the abacus, where the result sought replaces the operand (or one of them). For this reason, the traditionally preferred method for obtaining square roots seems to have been the following, where we will see the root appear directly on the abacus, not its double .

#### Preparing the dividend

Starting again with

${\displaystyle 100r_{i-1}+\alpha _{i}\geq (20y_{i-1}+b)b=20y_{i-1}b+b^{2}}$

dividing by 2

${\displaystyle (100r_{i-1}+\alpha _{i})/2\geq 10y_{i-1}b+b^{2}/2}$

This modified expression will allow us to directly obtain the square root in the abacus following practically the same procedure as above with only keeping half-remainders on our instrument. Here, with square roots, the change is almost trivial, but it will be more important when dealing with cube roots. As can be seen in the above expression, neglecting the term ${\textstyle b^{2}/2}$  we obtain a guess of ${\displaystyle b}$  by simply dividing the extended half-remainder ${\displaystyle (100r_{i-1}+\alpha _{i})/2}$  by the previous root ${\displaystyle y_{i-1}}$  (in fact ${\displaystyle 10y_{i-1}}$ ).  After that, we need again:

1. Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
2. Obtain the next half-remainder to prepare for the next digit of the root.

This is done by subtracting ${\displaystyle 10y_{i-1}b}$  and ${\textstyle b^{2}/2}$  from the half remainder, and this makes it convenient to memorize the following table of semi-squares:

Semi-squares table
${\displaystyle b}$  ${\displaystyle b^{2}/2}$
1 0.5
2 2
3 4.5
4 8
5 12.5
6 18
7 24.5
8 32
9 40.5

Fortunately, since 2 is a divisor of our base (10), the decimal fractions in the table have a finite expression; which will not happen when we try to extend this procedure to cube roots and we have to deal with thirds of cubes. According to Knott, this makes cube roots a problem that is not well suited to being treated with abacus.

## Examples

Here three examples are presented, for additional examples please see Further reading and specially External resources below.

### Square root of 961

In this example we have two groups of two figures: 09 and 61. The first group informs us that the first digit of the root is 3.

There are two ways to start square roots:

• Aligning the groups to the left of the abacus from column B and using the traditional division to obtain the semi-remainder.
A B C D E ... 0 9 6 1
This is the form used in the old books and also the one used in Murakami's Square root tutor with Kijoho (see below External resources).
Abacus Comment
ABCDE
0961 Align the radicand with B
30961 Enter first root digit in A
-9 Subtract the square of first root digit (9)
30061
30305 Divide the remainder B-E by 2 (帰除法)
• Aligning the groups to the left of the abacus from column A and using in-situ division, as explained in chapter: Division by powers of two, to obtain the semi-remainder.
A B C D E ... 0 9 6 1
This method is somewhat faster.
Using division in situ
Abacus Comment
ABCDE
0961 Align the radicand with A
-9 Subtract the square of first root digit (9)
0061
0305 Divide in situ the remainder by 2
30305 Enter first root digit in A

From here, the state of the abacus coincides and we can continue:

Continuation
Abacus Comment
ABCDE
30305
+1 Divide half-remainder B-E by 3. Revising up B
-3
31005
-05 Subtract b^2/2 =0.5 from D
31000 Half-remainder is 0, Done! Root is 31
31 Root is 31

### Square root of 998001

${\displaystyle {\sqrt {998001}}=999}$
Abacus Comment
ABCDEFG
-81 Subtract 9^2=81 from first group
188001
940005 Halve the remainder in situ
9940005 Enter first root digit into A
9930005 B: Rule 9/9>9+9
-405 Subtract 9^2/2=40.5 from D
9989505
9987505 C: Rule 8/9>8+8
-72 Subtract CxB=72 from DE
998T305 Revise up C
+1
-99
9990405
-405 Subtract 9^2/2=40.5 from F
9990000 Remainder is 0. Done!
999 Root is 999

### Root of 456.7890123

Our example above...

First 4 digits of ${\displaystyle {\sqrt {456.7890123}}=21.3726229\cdots }$
Abacus Comment
ABCDEFGHIJKL
04567890123 Enter x aligning digit pairs from AB, CD, etc.
-4 Subtract 2^2 from first group
567890123
2839450615 Halve remainder and rest of digits pairs
2 2839450615 Enter first root digit in A
+1 Divide BCD by A (revise up B)
-2
-05 Subtract B^2/2=0.5 from D
21 789450615
+3 Divide CDEF by AB (revise up C three times)
-6
-3
-45 Subtract C^2/2=4.5 from F
213154950615
213554950615 Divide DEFGH by ABC. D: Rule 1/2>5+0
-5 Subtract DxB=5 from EF
-15 Subtract DxC=15 from FG
213548450615
+2 Revise up D twice
-426
213705850615
-245 Subtract 7^2/2=24.5 From H
21370560F615 Root so far: 21.37
etc.     etc.

The root 2137… (21.37…) is appearing on the left. See chapter: Abbreviated operations to see how to quickly approximate the next 4 digits.

## Conclusion

The method explained as: Preparing the dividend is known as 半九九法 (Hankukuhou in Japanese, Bàn jiǔjiǔ fǎ in Chinese) that we freely translate here as the Half-remainder method and is by far the most convenient to use, at least for two reasons:

1. The root, and not its double, replaces the operand (radicand) as in the rest of operations on the abacus.
2. (The most important) Dividing by numbers that start with 1 is awkward. Think of the first group of two digits, its value is between 1 and 99 and determines the first figure of the square root. For values ​​of the first pair between 25 and 99 (75% of the cases) the first digit of the root is between 5 and 9 and its double begins with one! Therefore, if we use the method preparing the divisor, we will be dividing by numbers that begin with 1 in 75% of the cases. On the contrary, if we use the method preparing the dividend, only in the case that the first group is 1, 2 or 3 (3% of the cases) will we have to divide by numbers that start with one.

The superiority of the half-remainder or preparing the dividend method is undeniable.

## References

1. Knott, Cargill G. (1886), "The Abacus, in its Historic and Scientific Aspects", Transactions of the Asiatic Society of Japan, 14: 18–73
2. Kojima, Takashi (1963), Advanced Abacus: Theory and Practice, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0003-7
3. Heffelfinger, Totton (2003). "Square Roots as Solved by Kojima". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 1, 2021. Retrieved August 16, 2021.

• Siqueira, Edvaldo; Heffelfinger, Totton. "Kato Fukutaro's Square Roots". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 1, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)