Traditional Abacus and Bead Arithmetic/Appendix: Abbreviated operations
Introduction edit
This chapter is special in the sense that its content is not directly related to the traditional method of the abacus, but it is an interesting resource to shorten arithmetic operations, both with the abacus and in written calculation. We include it in this book because we make sporadic use of these abbreviated operations throughout it.
Some arithmetic books of the precomputer era included a chapter on abbreviated operations. The motivation is the following. Suppose we measure the side of a square and we obtain and we want to calculate its area
a result with 6 digits, but if we have measured the side of the square with a measuring tape that only appreciates millimeters, what we can say is that the value of the side is between and , that is:
So that will be a value between and . This means that we only know with certainty the first two digits of the result S (74) and that the third digit is probably a 6; the rest of the digits of the multiplication are meaningless (we say they are not significant) and we should not include them in our result. We should write:
being the significant figures of our result. So if only three of the six figures in the product are significant, why calculate all six? This is what abbreviated operations are for.
In this chapter we will follow the examples that appear in Matemáticas by Antonino Goded Mur^{[1]} hereinafter simply Matemáticas, a small Spanish manual, and see the way these calculations can be done with the abacus.
Multiplication edit
Write the product of the multiplicand by the first figure of the multiplier, write below the product of the multiplicand without its last figure by the second of the multiplier, below the product of the multiplicand without its last two figures by the third of the multiplier and so on.—Translated from Matemáticas
Example from Matemáticas:
6665 x 1375 ——————— 33325 46655 19995 6665 ——————— 9164375 
6665 x 1375 ———— 6665 1999 466 33 ———— 9163 
Normal operation 
Abbreviated operation 
On the abacus this problem can be dealt with in several ways, for example, using Kojima’s Multiplication Beginning with the Highest Digits of the Multiplier and Multiplicand, explained in his second book^{[2]}, where he says:
As the operation starts by multiplying the first digits of the multiplier and multiplicand, it is convenient for approximations.
We can also try multifactorial multiplication^{[3]} or the like; for instance
Abacus  Comment 

ABCDEFGHIJKLM  
6665 1375  Setup problem 
. .  Unit rods 
1  Clear J 
+6665  Add 1✕6665 to GJ 
6665 6665375  
+18  Add 3✕6 to GH 
+18  Add 3✕6 to HI 
+18  Add 3✕6 to IJ 
3  Clear K 
+15  Add 3✕5 to JK 
6665 8664575  
666 8664575  Clear D 
+42  Add 7✕6 to HI 
+42  Add 7✕6 to IJ 
+42  Add 7✕6 to JK 
7  Clear L 
666 91307 5  
66 91307 5  Clear C 
+30  Add 5✕6 to IJ 
+30  Add 5✕6 to JK 
5  Clear M 
66 91637  Result 
. .  Unit rods 
9164  Result after rounding to 4 figures 
But we can also use multiplication methods starting with the last multiplicand digits as modern multiplication:
Abacus  Comment 

ABCDEFGHIJKLM  
6665 1375  Setup problem 
+330  Add 5✕66 to KM 
5  Clear J 
6665 137 330  
+4662  Add 7✕666=4662 to JM 
7  Clear I 
6665 13 4992  
+19995  Add 3✕6665=19995 to IM 
3  Clear H 
6665 1 24987  
+6665  Add 1✕6665=6665 to HL 
1  Clear G 
6665 91637  Result 
6665 9164  Result after rounding to 4 figures 
And even traditional multiplication by clearing first and then adding the partial products shifted one column to the left
Abacus  Comment 

ABCDEFGHIJKL  
6665 1375  Setup problem 
5  Clear J 
+330  Add 5✕66 to JL 
6665 137330  
7  Clear I 
+4662  Add 7✕666=4662 to IL 
6665 134992  
3  Clear H 
+19995  Add 3✕6665=19995 to HL 
6665 124987  
1  Clear G 
+6665  Add 1✕6665=6665 to GK 
6665 91637  Result 
6665 9164  Result after rounding to 4 figures 
Division edit
The first digit of the quotient is found as usual, the remainder is divided by the divisor without its last digit, the new remainder by the divisor without its last two digits and so on.—Translated from Matemáticas
Example from Matemáticas:
4567.8 95.62 743.00 —————— 73.660 47.77 6.7250 .0326 
4567.8 95.62 743.0 —————— 95.6 73.8 4 ————— 95 7.3 7 ——— 9 .1 7 —— 8 
Normal operation  Abbreviated operation 
As can be seen, the potentially infinite sequence of long division steps in which a new quotient figure is obtained is replaced by a finite sequence of divisions by a shrinking divisor in which we obtain only one digit of the quotient. This can be done using our favorite division method; for example, using traditional division and traditional division arrangement:
Abacus  Comment 

ABCDEFGHIJ  
9562 45678  
. .  Unit rods 
4  Rule: 4/9>4+4 
+44  
9562 49678  
20  Subtract 4x5 from GH 
24  Subtract 4x6 from HI 
8  Subtract 4x2 from IJ 
9562 47430  
7  Rule: 7/9>7+7 
+77  
9562 47130  
35  Subtract 7x5 from HI 
42  Subtract 7x6 from IJ 
9562 47738  
7  Rule: 7/9>7+7 
+77  
9562 47708  
35  Subtract 7x5 from IJ 
9562 47773  
7  Rule: 7/9>7+7 
+77  
9562 47770  
+1  Revising up 
7  
9562 47783  
. .  Unit rods 
Square root edit
The current method is followed until half the figures of the root have been exceeded, obtaining the next digits by dividing the remainder followed by the periods not used by the double of the root found, followed by as many zeros as periods have been added.—Translated from Matemáticas
😖 hard to read, right? Also in Spanish …
Example from Matemáticas:
__________ \/123456789 11111  1    023  21x1 21    0245  221x1 221    02467  2221x1 2221    024689 22221x1 22221  02468 
______ \/12345 111  1    023 21x1 21    0245 221x1 221    024  > 24678922200  24789 11 2589 _________ ==> \/123456789 = 11111 
Normal operation  Abbreviated operation 
Withouth going into details, this way of shortening the square root obtention can be justified in several ways, for example using Taylor series development or Newton's method, perhaps not the simplest way but that is interesting to mention especially for what comes below about cubic roots.
In what follows the process will be illustrated using the Halfremainder method (半九九法) as explained in chapter: Square root, which requires changing remainder into halfremainder and double of the root into simply the root in the Matemáticas paragraph above. Note that the second phase, the division, can be done in the form of an abbreviated division since it only makes sense to obtain a limited number of figures from its quotient. As a consequence, obtaining the last figures from the root costs progressively less work and time; so we can call this division the accelerated phase of root extraction.
Abacus  Comment 

ABCDEFGHIJ  
123456789  Setting up the problem as usual 
23456789  Subtracting the square of 1 from first group 
117283945  Halving the remainder 
1 117283945  Entering 1 as first root digit in A 
11 17283945  New root digit 1 in B (revising up) 
1  
5  Subtract half of the square of 1 from D 
11 12283945  
111 2283945  New root digit 1 in C (revising up) 
11  
5  Subtract half of the square of 1 from F 
111 1233945  Now the second or accelerated phase starts 
+1  Divide 111 into 123 
111  
1111 123945  
+1  Divide 11 into 12 
11  
11111 13945  Done, we have now 5 root digits! 
Cube root edit
The current method is followed until half the figures in the root have been exceeded, obtaining the next digits by dividing the remainder followed by the periods not used by the triple of the square of the root followed by as many zeros as the periods have been added.—Translated from Matemáticas
😖😖
Example from Matemáticas:
3_____________ \/123456789012310727  
3_____________ \/1234567890123107 9524  9524890123 3434700  2655490 27 2512001 
Normal operation  Abbreviated operation 
This abbreviation can also be justified in several ways, including Newton's method, by the way and by far the best approximation to obtain cube roots with the abacus^{[4]} even though it is not a traditional technique, it is much more efficient than any traditional method and, if we use it, we can say that, in a certain sense, we are using an abbreviated method from the beginning. Nevertheless, here is an example using a traditional method: the cube root of 666. We will follow here the method explained by Cargill G. Knott^{[5]} (see chapter: Cube root).
Obviously, the cube root of 666 is between 8 and 9 because the number is in the range 512728.
Abacus  Comment 

ABCDEFG  
666  Enter 666 in BCD 
+  (Unit rod) 
512  Subtract 83=512 from BCD 
154  
8154  Enter 8 in A. Divide BF by 8 (A) 
8192500  Divide BF by 3 
8641662  Divide B by 8 (A) 
8781662  Subtract B2=49 from CD 
8732662  Multiply CF by 3 in CG 
87 9800  Multiply CF by 8 (A) in CG 
87 7840  Subtract B3=343 from EFG 
87 7497  Root 8.7, Remainder 7.497 
So we have obtained 8.7 as the root so far, leaving a remainder of 7.497. To apply the shortcut we need to form the divisor ; We will use Newton's binomial expansion to form the square and we will multiply it by three by adding twice the value obtained.
Abacus  Comment 

ABCDEFGHIJKLM  
87 7497  Squaring 8.7 
+49  
112  
+64  
87 7497 7569  Multiplying it by 3 
+14  
+10  
+12  
+18  
87 7497 22707  Proceed to divide 7.497 by 227.07 (can be abbreviated!) 
873322707  obtain only two digits of quotient 
Alternatively, you can also divide twice by 8.7 and then by 3 to get the same result. Compare the result 8.733 to 3666=8,7328917
Other useful abbreviations edit
What follows is a completely different type of abbreviated calculation that may prove useful in practice. They are all a consequence of Taylor's theorem.
For

 ex:

 ex:

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 …
References edit
 ↑ Goded Mur, Antonino (1945). Matemáticas (in Spanish). Zaragoza (Spain): Compendios CHOP. pp. 20–26.
{{cite book}}
: Unknown parametertrans_title=
ignored (transtitle=
suggested) (help)  ↑ Kojima, Takashi (1963), Advanced Abacus: Theory and Practice, Tokyo: Charles E. Tuttle Co., Inc., ISBN 9780804800037
 ↑ Tejón, Fernando; Heffelfinger, Totton (2005). "Multifactorial Multiplication". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 1, 2021.
{{cite web}}
: Unknown parameteraccesdate=
ignored (accessdate=
suggested) (help)  ↑ Cabrera, Jesús (2021). "Newton's method for abacus; square, cubic and fifth roots". jccAbacus.
{{cite web}}
: Unknown parameteraccesdate=
ignored (accessdate=
suggested) (help)  ↑ Knott, Cargill G. (1886), "The Abacus, in its Historic and Scientific Aspects", Transactions of the Asiatic Society of Japan, 14: 18–73