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
"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."

Example from Matemáticas:

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:

"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 or the like; for instance

But we can also use multiplication methods starting with the last multiplicand digits as modern multiplication:

And even traditional multiplication by clearing first and then adding the partial products shifted one column to the left

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."

Example from Matemáticas: 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:

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."

😖 hard to read, right? Also in Spanish …

Example from Matemáticas: 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.

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."

😖😖

Example from Matemáticas:

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.

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.

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$$