## 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 ${\textstyle l=864\,{\text{mm}}}$ and we want to calculate its area ${\textstyle S}$

${\displaystyle S={l}^{2}=l\cdot l=864\cdot 864=746496\,{\text{mm}}^{2}}$

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 ${\textstyle 863.5}$and ${\textstyle 864.5}$, that is:

${\displaystyle l=864.0\pm 0.5\,{\text{mm}}}$

So that ${\textstyle S}$ will be a value between ${\displaystyle {863.5}^{2}=745632.25}$ and ${\displaystyle {864.5}^{2}=747360.25}$. 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:

${\displaystyle S=746\cdot {10}^{3}\,{\text{mm}}^{2}=7.46\cdot {10}^{5}\,{\text{mm}}^{2}}$

being ${\textstyle 746}$ the significant figures of our result. So if only three of the six figures in the product ${\textstyle 864\cdot 864}$  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[1] hereinafter simply Matemáticas, a small Spanish manual, and see the way these calculations can be done with the abacus.

## Multiplication

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 Normaloperation Abbreviatedoperation

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

6666x1375
Abacus Comment
ABCDEFGHIJKLM
6665     1375 Setup problem
.        . Unit rods
-1 Clear J
6665  6665375
-3 Clear K
6665  8664575
666   8664575 Clear D
-7 Clear L
666   91307 5
66    91307 5 Clear C
-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
-5 Clear J
6665  137 330
-7 Clear I
6665  13 4992
-3 Clear H
6665  1 24987
-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
6665  137330
-7 Clear I
6665  134992
-3 Clear H
6665  124987
-1 Clear G
6665   91637 Result
6665   9164 Result after rounding to 4 figures

## Division

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:

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

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 |  -->   246789|22200                        ------                   24789 11                    2589          _________  ==>   \/123456789 = 11111 Normal operation Abbreviated operation ${\textstyle {\sqrt {123456789}}=11111.11106\cdots }$

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

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_____________ \/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[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 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 ${\displaystyle 3\times 8.7^{2}}$; 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 ${\displaystyle a,b\ll 1}$ ${\displaystyle }$

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

## References

1. 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)
2. Kojima, Takashi (1963), Advanced Abacus: Theory and Practice, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0003-7
3. 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)
4. Cabrera, Jesús (2021). "Newton's method for abacus; square, cubic and fifth roots". jccAbacus. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
5. Knott, Cargill G. (1886), "The Abacus, in its Historic and Scientific Aspects", Transactions of the Asiatic Society of Japan, 14: 18–73

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