Traditional Abacus and Bead Arithmetic, Roots, Square root

Theory

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

x = 456.7890123 = 4 ⋅ 56 ⋅ 78 ⋅ 90 ⋅ 12 ⋅ 30 ⋅ 00 ⋅ 00 ⋯

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

{α}_{1} = 4 , {α}_{2} = 56 , {α}_{3} = 78 , {α}_{4} = 90 , {α}_{5} = 12 , {α}_{6} = 30 , {α}_{7} = 00 , ⋯

and let's build the sequence {x}_{i} recursively from {x}_{0} = 0

{x}_{i} = 100 {x}_{i − 1} + {α}_{i}

and let {y}_{i} be the integer part of the square root of {x}_{i}

{y}_{i} = ⌊ {\sqrt{x}}_{i} ⌋

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} ≥ 0

For our example we have:

Let's see that, by construction, {x}_{i} grows as {}1 {0}^{2 i} (two more digits in each step), in fact the sequence {x}_{i} ⋅ 1 {0}^{4 − 2 i}: (0, 400, 456, 456.78, 456.7890, etc.) tends to x ( {x}_{i} ⋅ 1 {0}^{4 − 2 i} → x ) or x = ( {lim}_{i → ∞} {x}_{i} ⋅ 1 {0}^{4 − 2 i} ). By comparison, {y}_{i} , as the integer part of the square root of {x}_{i}, grows only as {}1 {0}^{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} ≥ 0 as above but

{x}_{i} − ( {y}_{i} + 1 {)}^{2} = {x}_{i} − {y}_{i}^{2} − 2 {y}_{i} − 1 < 0

by definition of {y}_{i}, or

{r}_{i} = {x}_{i} − {y}_{i}^{2} < 2 {y}_{1} + 1

multiplying by {}1 {0}^{4 − 2 i}

{x}_{i} ⋅ 1 {0}^{4 − 2 i} − ( {y}_{i} ⋅ 1 {0}^{2 − i} {)}^{2} < ( 2 {y}_{i} + 1 ) ⋅ 1 {0}^{4 − 2 i}

but as  {y}_{i} grows only as {}1 {0}^{i}, the second term tends to zero as {}1 {0}^{− i}.  With this

{x}_{i} ⋅ 1 {0}^{4 − 2 i} − ( {y}_{i} ⋅ 1 {0}^{2 − i} {)}^{2} → 0

and {x}_{i} ⋅ 1 {0}^{4 − 2 i} → x so that we have

{y}_{i} ⋅ 1 {0}^{2 − i} → y = \sqrt{x}

For other numbers, the above factors are: {}1 {0}^{2 k − 2 i} and {}1 {0}^{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

For i = 1 and {x}_{1} = {α}_{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} = 100 {x}_{i − 1} + {α}_{i} as defined above and we try to build {y}_{i} in the way:

{y}_{i} = 10 {y}_{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} − ( 10 {y}_{i − 1} + b {)}^{2} ≤ 0

or

{x}_{i} − ( a + b {)}^{2} ≤ 0

if we write a = 10 {y}_{i − 1}. Expanding the binomial we have

{x}_{i} − {a}^{2} − 2 a b − {b}^{2} = 100 {x}_{i − 1} + {α}_{i} − ( 10 {y}_{i − 1} {)}^{2} − 2 a b − {b}^{2} ≤ 0

or

100 {r}_{i − 1} + {α}_{i} ≥ 2 a b + {b}^{2} = ( 2 a + b ) b = ( 20 {y}_{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 ≥ ( 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 , ⋯ , 9)?

Here Knott[1] distinguishes two different approaches:

• Preparing the divisor
• Preparing the dividend.

Preparing the divisor

This corresponds with the above expression

100 {r}_{i − 1} + {α}_{i} ≥ ( 20 {y}_{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 = ( 100 {r}_{i − 1} + {α}_{i} ) / ( 20 {y}_{i − 1} + b )

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

b = ( 100 {r}_{i − 1} + {α}_{i} ) / ( 20 {y}_{i − 1} )

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

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

Both steps require subtracting ( 20 {y}_{i − 1} + b ) b or {}20 {y}_{i − 1} b and {b}^{2} from {}100 {r}_{i − 1} + {α}_{i}; checking that we are not going to negative values and that what remains is less than {}2 {y}_{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 ( 20 {y}_{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 Practice2], 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: x = 456.7890123

Napier's bones arranged to help with the third root digit of the example (Image: Jccsvq, CC BY-SA 4.0)

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

100 {r}_{i − 1} + {α}_{i} ≥ ( 20 {y}_{i − 1} + b ) b = 20 {y}_{i − 1} b + {b}^{2}

dividing by 2

( 100 {r}_{i − 1} + {α}_{i} ) / 2 ≥ 10 {y}_{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 {b}^{2} / 2 we obtain a guess of b by simply dividing the extended half-remainder ( 100 {r}_{i − 1} + {α}_{i} ) / 2 by the previous root {y}_{i − 1} (in fact {}10 {y}_{i − 1}).  After that, we need again:

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

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

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.

(Image: Jccsvq, CC BY-SA 4.0) |

(Image: Jccsvq, CC BY-SA 4.0) |

(Image: Jccsvq, CC BY-SA 4.0) |

(Image: Jccsvq, CC BY-SA 4.0) |

(Image: Jccsvq, CC BY-SA 4.0) |

(Image: Jccsvq, CC BY-SA 4.0) | ... |
| | | 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).

• 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.

(Image: Jccsvq, CC BY-SA 4.0) |

(Image: Jccsvq, CC BY-SA 4.0) |

(Image: Jccsvq, CC BY-SA 4.0) |

(Image: Jccsvq, CC BY-SA 4.0) |

(Image: Jccsvq, CC BY-SA 4.0) |

(Image: Jccsvq, CC BY-SA 4.0) | ... |
| | 0 | 9 | 6 | 1 | | |

This method is somewhat faster.

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

Square root of 998001

Root of 456.7890123

Our example above...

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:

• The root, and not its double, replaces the operand (radicand) as in the rest of operations on the abacus.
• (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

• ↑ Knott, Cargill G. (1886), "The Abacus, in its Historic and Scientific Aspects", Transactions of the Asiatic Society of Japan, 14: 18–73
• ↑ Kojima, Takashi (1963), Advanced Abacus: Theory and Practice, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0003-7 {{citation}}: ISBN / Date incompatibility (help)
• ↑ 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.

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. Retrieved August 10, 2021.
• 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. Retrieved August 10, 2021.
• 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.

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