Traditional Abacus and Bead Arithmetic/Roots/Square root

Theory

Let $x$ 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

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

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

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

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

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

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

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

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

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

For our example we have:

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

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

by definition of $y_{i}$, or

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

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

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

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

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

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

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

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

This is the basis for traditional manual square root methods.

Procedure

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

First digit

Squares table

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

For $i=1$ and $x_{1}=\alpha _{1}$, it is trivial to find $y_{1}$ such that its square does not exceed $x_{1}$ 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 $y_{1}=2$.

Rest of digits

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

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

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

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

or

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

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

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

or

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

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 $i=2$ we have 56 on the left and the above expression is

$$56\geq (40+b)b$$

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

Here Knott distinguishes two different approaches:

• Preparing the divisor
• Preparing the dividend.

Preparing the divisor

This corresponds with the above expression

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

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

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

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

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

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 $(20y_{i-1}+b)b$ or $20y_{i-1}b$ and $b^{2}$ from $100r_{i-1}+\alpha _{i}$; checking that we are not going to negative values and that what remains is less than $2y_{i}+1$ (otherwise we would have to revise $b$ up or down). If we do this correctly, what we are left with is the new remainder $r_{i}$. It should be noted that, as we proceed in the calculations ($i$ increasing) $b$ is progressively a smaller and smaller contribution to the divisor $(20y_{i-1}+b)$; 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, and that you can see described in Square roots as solved by Kojima 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: $x=456.7890123$

Napier's bones arranged to help with the third root digit of the example

First 3 digits of ${\sqrt {456.7890123}}$ 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

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

dividing by 2

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

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 $b$ by simply dividing the extended half-remainder $(100r_{i-1}+\alpha _{i})/2$ by the previous root $y_{i-1}$ (in fact $10y_{i-1}$).  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 $10y_{i-1}b$ 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

$b$	$b^{2}/2$
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).

Using traditional division

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

${\sqrt {998001}}=999$

Abacus	Comment
ABCDEFG
998001	Enter the radicand
-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 ${\sqrt {456.7890123}}=21.3726229\cdots$

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.

Further reading

• 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)

• Treadwell, Steve (2015). "Improvements to the Kato Method for Finding Square Roots" (PDF). 算盤 Abacus: Mystery of the Bead. Archived from the original (PDF) on August 1, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
• Square root (半九九法); the traditional way using traditional division in jccAbacus

External resources

• Murakami's Square root tutor with Kijoho, a JavaScript application that you can run directly in your browser or download to your computer from its GitHub repository. You just have to enter the root in the small input box on the left and repeatedly press the "next" button on the screen to see the development of the process step by step. So you can generate as many examples or exercises as you want.