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
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:
or
if we write . Expanding the binomial we have
or
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:
- 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 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:
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 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:
- 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 and from the half remainder, and this makes it convenient to memorize the following table of semi-squares:
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.
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:
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 edit
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 edit
Our example above...
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 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:
- 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 edit
- ↑ 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
- ↑ 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
- Siqueira, Edvaldo; Heffelfinger, Totton. "Kato Fukutaro's Square Roots". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 1, 2021.
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- 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.
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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.