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

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

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 $863.5$and $864.5$, that is:

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

So that $S$ 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 $746$ the significant figures of our result. So if only three of the six figures in the product $864\cdot 864$  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
$\sqrt{123456789}=11111.11106\cdots$

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: [[../Roots/Square root/|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: [[../Roots/Cube root/|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 $a,b\ll 1$ ${\displaystyle }$

• $(1\pm a)(1\pm b)\approx 1\pm a\pm b$ 
  • 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

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