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Traditional Abacus and Bead Arithmetic

An open textbook from Wikibooks, a chapter a page, in reading order.

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1번째 페이지CC BY-SA 4.0
Traditional Abacus and Bead Arithmetic, Introduction
3번째 페이지CC BY-SA 4.0
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6번째 페이지CC BY-SA 4.0
These rules: - From an operational point of view, these rules should be read or interpreted slightly differently depending on whether we use the traditional (**TDA**) or the modern (**MDA**) division arrangement. when using **MDA**, the rule **a/b>q+r** must be read: “write **q** as interim quotient
7번째 페이지CC BY-SA 4.0
Even so, **TDA** has been used for centuries in conjunction with the traditional method of division while **MDA** seems to have been deprecated until modern times and the adoption of the modern abacus, even though **MDA** is the first idea that would occur to us if we tried to adapt the old division
9번째 페이지CC BY-SA 4.0
| Abacus | Comment | | --- | --- | | ABCDEFGHIJKLM | | | 36525 1000000 | Use rule: 1/36525>2+26950 on column G | | 36525 2000000 | change 1 in G into 2 | | +26950 | add 26950 to H-L | | 36525 2269500 | Use rule: 2/36525>5+17375 on column H | | 36525 2569500 | change 2 in H into 5 | | +17375 | add 17
10번째 페이지CC BY-SA 4.0
| | --+--- | unit rod | We only need to have the corresponding fractions tabulated or memorized, as in the table below. ## Powers of two In the past, both in China and in Japan, monetary and measurement units were used that were related by a factor of 16[1][2][3], a factor that begins with one which
11번째 페이지CC BY-SA 4.0
But these few are still a lot ... especially if we think that we can also choose whether or not to enter multiplicand and multiplier in the abacus and where to start adding the partial products with respect to said operands. In what follows we will focus on this last aspect. ## Inverse operations Ad
12번째 페이지CC BY-SA 4.0
By comparison, ${y}_{i}$ , as the integer part of the square root of ${x}_{i}$, grows only as ${}1 {0}^{i}$ (one digit more each step). As ${y}_{i}$ is the largest integer whose square is not greater than ${x}_{i}$ we have ${r}_{i} = {x}_{i} − {y}_{i}^{2} ≥ 0$ as above but $$ {x}_{i} − ( {y}_{i} + 1
13번째 페이지CC BY-SA 4.0
0, 400, 456, 456.789, 456.789012, etc. tends to x (${x}_{i} 1 {0}^{3 − 3 i} → x$). By comparison, ${y}_{i}$ , as the integer part of the cube root of ${x}_{i}$, grows only as ${}1 {0}^{i}$ (one digit more each step). As ${y}_{i}$ is the largest integer whose square is not greater than ${x}_{i}$ we h
14번째 페이지CC BY-SA 4.0
Suppose we measure the side of a square and we obtain $l = 864 \ \text{mm}$ and we want to calculate its area S $$ S = {l}^{2} = l ⋅ l = 864 ⋅ 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 millimet
15번째 페이지CC BY-SA 4.0
Traditional Abacus and Bead Arithmetic, Resources for contributors