# Traditional Abacus and Bead Arithmetic/Appendix: Abbreviated operations
## Introduction

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 pre-computer era included a chapter on abbreviated operations. The motivation is the following. Suppose we measure the side of a square and we obtain l=864mm{\textstyle l=864\, {\text{mm}}} and we want to calculate its area S{\textstyle S}
$S={l}^{2}=l\cdot l=864\cdot 864=746496\,{\text{mm}}^{2}$
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 863.5{\textstyle 863.5}and 864.5{\textstyle 864.5}, that is:
$l=864.0\pm 0.5\,{\text{mm}}$
So that S{\textstyle S} will be a value between ${863.5}^{2}=745632.25$ and ${864.5}^{2}=747360.25$. 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:
$S=746\cdot {10}^{3}\,{\text{mm}}^{2}=7.46\cdot {10}^{5}\,{\text{mm}}^{2}$
being 746{\textstyle 746} the significant figures of our result. So if only three of the six figures in the product 864⋅864{\textstyle 864\cdot 864} 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 hereinafter simply Matemáticas, a small Spanish manual, and see the way these calculations can be done with the abacus.

## Multiplication

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, where he says:
We can also try multifactorial multiplication or the like; for instance
6666x1375 :
| Abacus | Comment |
| ABCDEFGHIJKLM | |
| 6665 1375 | Setup problem |
| . . | Unit rods |
| -1 | Clear J |
| +6665 | Add 1✕6665 to G-J |
| 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:
6666x1375 Modern multiplication :
| Abacus | Comment |
| ABCDEFGHIJKLM | |
| 6665 1375 | Setup problem |
| +330 | Add 5✕66 to K-M |
| -5 | Clear J |
| 6665 137 330 | |
| +4662 | Add 7✕666=4662 to J-M |
| -7 | Clear I |
| 6665 13 4992 | |
| +19995 | Add 3✕6665=19995 to I-M |
| -3 | Clear H |
| 6665 1 24987 | |
| +6665 | Add 1✕6665=6665 to H-L |
| -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
6666x1375 Traditional multiplication :
| Abacus | Comment |
| ABCDEFGHIJKL | |
| 6665 1375 | Setup problem |
| -5 | Clear J |
| +330 | Add 5✕66 to J-L |
| 6665 137330 | |
| -7 | Clear I |
| +4662 | Add 7✕666=4662 to I-L |
| 6665 134992 | |
| -3 | Clear H |
| +19995 | Add 3✕6665=19995 to H-L |
| 6665 124987 | |
| -1 | Clear G |
| +6665 | Add 1✕6665=6665 to G-K |
| 6665 91637 | Result |
| 6665 9164 | Result after rounding to 4 figures |

## Division

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:
4567.8/95.62 :
| 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

😖 hard to read, right? Also in Spanish …
Example from Matemáticas:
Square root of 123456789 :
| __________    \/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 | --> 246789|22200             ------          24789 11           2589      _________  ==> \/123456789 = 11111 |
| Normal operation | Abbreviated operation |
| √(123456789)=11111.11106⋯{\textstyle {\sqrt {123456789}}=11111.11106\cdots } ||
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 Half-remainder method (半九九法) as explained in chapter: Square root, which requires changing remainder into half-remainder 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.
Square root of 123456789 半九九法 (hankukuhou) :
| 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

Example from Matemáticas:
| 3_____________ \/1234567890123|10727        ------ | 3_____________ \/1234567890123|107    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 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 (see chapter: Cube root).
Obviously, the cube root of 666 is between 8 and 9 because the number is in the range 512-728.
Cube root of 666 :
| Abacus | Comment |
| ABCDEFG | |
| 666 | Enter 666 in BCD |
| + | (Unit rod) |
| -512 | Subtract 83=512 from BCD |
| 154 | |
| 8154 | Enter 8 in A. Divide B-F by 8 (A) |
| 8192500 | Divide B-F by 3 |
| 8641662 | Divide B by 8 (A) |
| 8781662 | Subtract B2=49 from CD |
| 8732662 | Multiply C-F by 3 in C-G |
| 87 9800 | Multiply C-F by 8 (A) in C-G |
| 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 $3\times 8.7^{2}$; We will use Newton's binomial expansion to form the square and we will multiply it by three by adding twice the value obtained.
Caption text :
| 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!) |
| 8733----22707 | 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

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 $a,b\ll 1$ $$
* $(1\pm a)(1\pm b)\approx 1\pm a\pm b$
** ex: $1.005\times 0.996\approx 1.001$
* $(1\pm a)/(1\pm b)\approx 1\pm a\mp b$
* $(1\pm a)^{n}\approx 1+n\cdot a$
** ex: $1.005^{3}\approx 1.015,\,0.995^{3}\approx 0.985$
* ${\sqrt[{n}]{1\pm a}}\approx 1\pm a/n$
** ex: ${\sqrt[{3}]{1.006}}\approx 1.002$
* $1/(1\pm a)\approx 1\mp a$
** ex:$1/1.003\approx 0.997,\,1/0.997\approx 1.003$
* $1/(1\pm a)^{n}\approx 1\mp n\cdot a$

## References

* ↑ Goded Mur, Antonino (1945). Matemáticas (in Spanish). Zaragoza (Spain): Compendios CHOP. pp. 20–26. {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
* ↑ Kojima, Takashi (1963), Advanced Abacus: Theory and Practice, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0003-7
* ↑ 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 parameter |accesdate= ignored (|access-date= suggested) (help)
* ↑ Cabrera, Jesús (2021). "Newton's method for abacus; square, cubic and fifth roots". jccAbacus. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
* ↑ Knott, Cargill G. (1886), "The Abacus, in its Historic and Scientific Aspects", Transactions of the Asiatic Society of Japan, 14: 18–73