Traditional Abacus and Bead Arithmetic/Roots/Square root

Theory edit

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


or, in other words, let's define the sequence of integers  


and let's build the sequence   recursively from  


and let   be the integer part of the square root of  


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


For our example we have:

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,   grows as   (two more digits in each step), in fact the sequence  : (0, 400, 456, 456.78, 456.7890, etc.) tends to     or  . By comparison,   , as the integer part of the square root of  , grows only as   (one digit more each step). As   is the largest integer whose square is not greater than   we have   as above but


by definition of  , or


multiplying by  


but as    grows only as  , the second term tends to zero as  .  With this


and   so that we have


For other numbers, the above factors are:   and  , where   is the number of two-digit groups to the left of the decimal point, negative if it is followed by 00 groups (ex.   for  ,   for  , etc.).

This is the basis for traditional manual square root methods.

Procedure edit

We start with  ,  ,  ,  .

First digit edit

Squares table
1 1
2 4
3 9
4 16
5 25
6 36
7 49
8 64
9 81

For   and  , it is trivial to find   such that its square does not exceed   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  .

Rest of digits edit

For  , we have   as defined above and we try to build   in the way:


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




if we write  . Expanding the binomial we have




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   we have 56 on the left and the above expression is


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

Here Knott[1] distinguishes two different approaches:

  • Preparing the divisor
  • Preparing the dividend.

Preparing the divisor edit

This corresponds with the above expression


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


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


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   or   and   from  ; checking that we are not going to negative values and that what remains is less than   (otherwise we would have to revise   up or down). If we do this correctly, what we are left with is the new remainder  . It should be noted that, as we proceed in the calculations (  increasing)   is progressively a smaller and smaller contribution to the divisor  ; 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:  

Napier's bones arranged to help with the third root digit of the example
First 3 digits of   preparing the divisor
Abacus Comment
   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
 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
 423 30990123 309 as remainder
 426 30990123 Double third root digit
 426 30990123 Append next group (90)

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 edit

Starting again with


dividing by 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   we obtain a guess of   by simply dividing the extended half-remainder   by the previous root   (in fact  ).  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   and   from the half remainder, and this makes it convenient to memorize the following table of semi-squares:

Semi-squares table
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 edit

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

Square root of 961 edit

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.
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
 0961 Align the radicand with B
30961 Enter first root digit in A
 -9 Subtract the square of first root digit (9)
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.
0 9 6 1
This method is somewhat faster.
Using division in situ
Abacus Comment
0961 Align the radicand with A
-9 Subtract the square of first root digit (9)
 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:

Abacus Comment
+1 Divide half-remainder B-E by 3. Revising up B
  -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 edit

Abacus Comment
 998001 Enter the radicand
-81 Subtract 9^2=81 from first group
  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
 9987505 C: Rule 8/9>8+8
   -72 Subtract CxB=72 from DE
 998T305 Revise up C
    -405 Subtract 9^2/2=40.5 from F
 9990000 Remainder is 0. Done!
 999 Root is 999

Root of 456.7890123 edit

Our example above...

First 4 digits of  
Abacus Comment
04567890123 Enter x aligning digit pairs from AB, CD, etc.
-4 Subtract 2^2 from first group
  2839450615 Halve remainder and rest of digits pairs
2 2839450615 Enter first root digit in A
+1 Divide BCD by A (revise up B)
  -05 Subtract B^2/2=0.5 from D
21 789450615
 +3 Divide CDEF by AB (revise up C three times)
    -45 Subtract C^2/2=4.5 from F
213554950615 Divide DEFGH by ABC. D: Rule 1/2>5+0
    -5 Subtract DxB=5 from EF
    -15 Subtract DxC=15 from FG
  +2 Revise up D twice
     -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 edit

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 edit

  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.

Further reading edit

External resources edit

  • 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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