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 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 MurGoded Mur, Antonino Matemáticas Spanish Mathematics Zaragoza (Spain) 1945 Compendios CHOP 20-26 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   ————   9163NormaloperationAbbreviatedoperation 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 bookAdvanced Abacus: Theory and Practice Kojima Takashi 1963 Charles E. Tuttle Co., Inc. Tokyo 978-0-8048-0003-7, 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 multiplicationMultifactorial Multiplication 2005 Heffelfinger, Totton Tejón, Fernando 算盤 Abacus: Mystery of the Bead http://totton.idirect.com/soroban/MultiFac/ August 14, 2021 https://web.archive.org/web/20210801083254/http://totton.idirect.com/soroban/MultiFac/ August 1, 2021 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 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    ——                                 8Normal operationAbbreviated 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: + 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 = 11111Normal operationAbbreviated 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 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 2512001Normal operationAbbreviated 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 abacusNewton's method for abacus; square, cubic and fifth roots 2021 Cabrera, Jesús jccAbacus https://sites.google.com/view/jccabacus/abacus-articles/advanced-abacus-algorithms/newtons-method-for-abacus-square-cubic-and-fifth-roots?authuser=0 August 17, 2021 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. KnottKnott, Cargill G. The Abacus, in its Historic and Scientific Aspects Transactions of the Asiatic Society of Japan 14 1886 18-73 https://archive.org/details/in.gov.ignca.26020/page/17/mode/2up (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 ; 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----22707obtain 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   ex: ex: ex: ex: … References Bibliography Roots/Cube root