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Pagina 1
Vedi nel PDF(si apre in una nuova finestra)A Neolithic Oral Tradition for the
van der Waerden/Seidenberg Origin of Mathematics
JEROLD MATHEWS
Communicated by A. SEIDENBERG
Intreduction
In his recent book [19] VAN DER WAERDEN “ventured a tenative reconstruction
of a mathematical science which must have existed in the Neolithic Age, say
between 3000 and 2500 B.C., and spread from Central Europe to Great Britain,
to the Near East, to India, and to China.” He argues that a common origin* is
highly probable. These views are consistent with those of SEIDENBERG [12], who
says, “What was this older, common [to Pythagorean and Old-Babylonian mathematics] source like? I think its mathematics was very much like what we see in
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V(>a; MATHEW \ABS
the Sulvasutras.’” SEIDENBERG does not mean the common source was necessarily
the Sulvasutras, rather that it “is to be sought either in the Vedic mathematics
or in an older mathematics very much like it.” VAN DER WAERDEN bases much of
his reconstruction on the Chiu Chang Suan Shu (Nine Chapters on the Mathematical Art), a Chinese collection of mathematical problems, written during the
Han-period (200 B.C. to 220 A.D.). The Chiu Chang is, according to VAN DER
WAERDEN, a more systematic and richer source than the Babylonian texts. SEIDEN-
BERG’S primary source is the Sulvasutras, ancient Indian sacred works on altar
constructions, in addition to EucLID's Elements and the Babylonian mathematical
tablets, Both VAN DER WAERDEN and SEIDENBERG discuss the important and difficult questions relating to the dating of their primary source materials and the
likely dependencies among them. Basing themselves on their answers to these questions and their analysis of much of the oldest known mathematical thought,
both men argue for a common source for the Indian, Babylonian, Greek, and
Chinese mathematics.
Here I shall give a small, coherent, and basic core of geometry concerning
* On page 10 of [19] the discussion of a common origin occurs in the context of
Pythagorean triples and the Theorem of PYTHAGORASs. Later, on page 33, VAN DER WAERDEN speaks of “a common mathematical doctrine from which these ideas were derived.”
The ideas in question are “‘the mathematical and religious ideas current in England in the
Neolithic Age, in Greece, in India, and in China ...”
Pagina 2
Vedi nel PDF(si apre in una nuova finestra)rectangles and their parts, including the right triangles on their diagonals and the
gnomons in their corners, which may serve as what VAN DER WAERDEN has
called an ‘‘oral tradition current in the Neolithic age.” I hope to give this hypothesized ancient core some credence through its relation to the Chiu Chang and
its explanatory power.
My work owes much to SEIDENBERG’s papers [9-13], VAN DER WAERDEN’S
earlier book Science Awakening [17], and NEUGEBAUER & Sachs’ work [8] on
Babylonian mathematics. Its present form arose out of a study of VOGEL’s translation [16] of the Chiu Chang.
I have tried to avoid both the use of modern algebraic symbolism, which I
feel is misleading, and any suggestion that, for example, those ancients who spoke
of the difference of two geometric squares may be credited to some extent with
knowledge of the identity
a? — b? = (a — b)(a + b).
I see no necessity for
taking any such position. If, however, by algebra one means certain recurrent
patterns of thought or, even, motor activity (I am here thinking of patterns intrinsic
to mental or written arithmetic calculations or kinesthetically experienced patterns
in the moving of stones, beads, rods, or dust in a computing device), then one may
credit the Indians, Babylonians, Greeks, and Chinese with some knowledge of
algebra. One may conjecture that an algebraic tradition, including viewpoint,
technique, and notation, arose out of attempts to codify the active parts of those
numerical algorithms which began as geomtric relations.
I. Gnomons
The idea of a gnomon or gnomon-construction may have arisen from dualisms
or equivalences occurring among the rituals of the Vedic religion. In [13] SEIDENBERG notes that the circle and square were regarded as dual figures, and further,
circle : square = Gärhapatya : Ahavaniya (circular fire altar: square fire altar) =
earth: sky = human:
divine. The principal expression of the equivalence of two
geometric figures or two altars is the equality of their areas. In addition to the
circle/square equivalence, the pair oblong/square was also considered. SEIDENBERG
finds in the needs of ritual a motivation for squaring the oblong. He notes that in
the Sulvasutras are found the following directions:
If you wish to turn an oblong [ABCD —see Figure 1] into a square take the
tiryanmäni, i.e., the shorter side of the oblong, for the side of a sqaure [AEFD],
divide the remainder [EBCF] into two parts and inverting join these two parts
to the sides of the square. Fill the empty space by adding a small piece. It
has been taught how to deduct it.
The last sentence refers to use of the Theorem of PYTHAGORAS, given elsewhere
in the Sulvasutras, for finding a square equal to the difference of two squares.
It is clear from the Figs. 1(2) or 1(3) that the Vedic literature includes clear
references to the gnomon figure. * SEIDENBERG observes that the desire to square
* The carpenter’s square along the right-side and bottom of Fig. 1(3).
Pagina 3
Vedi nel PDF(si apre in una nuova finestra)CD
F
(1)
195
D
(2)
(3)
Fig. 1
an oblong provides a context in which the idea of a gnomon may have originated
as well as the need to subtract one square from another. I shall return to this last
very interesting point in Section III on the Theorem of PYTHAGORAS.
I shall
propose
Proposition I,
Book II,
of Euctip’s Elements*
(EI,
hereafter) as the first result in the neolithic core.
I.1 In Figure 2(1), if the side BC of the rectangle BCHG is cut into any number
of segments, the rectangle with sides BG and BC is equal (in area) to the
rectangles with sides BG and each of the segments.
In Figure 2(2) is given a rectangle ABCD with diagonal AC. If any point I
on AC is chosen, two rectangles** AI andIC are determined. Each of these
rectangles determines a gnomon: AI is the gnomon-corner and EF and HG are
the gnomon-arms of gnomon ABFIGDA, while IC is the gnomon-corner and
(again) EF and HG are the gnomon-arms of gnomon CDHIEBC. The rectangles
EF and HG are often called gnomon-complements or complements (of one
another). The rectangles AF and AG (HC and EC) are called the gnomonbases of gnomon ABFIGDA (CDHIEBC).
B
DEC
A
E
B
H
Fig. 2
1.2 The gnomon-complements are equal (in area).
This result may be interpreted as a result on similar triangles (AAEI —AICG
and so AE: EI = GC: GT). I shall avoid this interpretation on the grounds that
the equality of the complements is a less sophisticated result. A simple argument
* All references to EucLID’s Elements are to HEATH’s translation [2].
** T shall refer to rectangles, such as EBFI, by their diagonals, here EF.
Pagina 4
Vedi nel PDF(si apre in una nuova finestra)for I.2 may be based on the assumption that the diagonal of a rectangle divides
it into two equal pieces.* Since one-half of rectangle AI and rectangle HG and
one-half of rectangle IC is equal to one-half of rectangle AI and rectangle EF and
one-half of rectangle IC, we see that
rectangle HG is equal to rectangle EF.
From the equality of the complements it follows that the gnomon-bases AF
and AG are equal. I
list this as
1.3 The gnomon-bases are equal (in area).
U. Equivalents to Oblongs and Sums and Differences
of Squares Using Gnomons
Closely connected to the discovery and subsequent use of the gnomon is the
problem of forming the difference of two squares. There are at least two ways in
which the difference of squares may be considered. One way has already been
mentioned, to find (the side of) a square equal to the difference. A second is to
arrange the two squares so that their difference may be seen as the area inside the
larger and outside the smaller. I shall discuss the latter first because it is simpler
and leads to a solution of the former, that is, to the Theorem of PYTHAGORAS.
In Figure 3 are given several configurations showing the difference in area of
two squares.
**
(1)
(2)
(3)
Fig. 3
A figure closely related to that in Figure 3(2) is found in an ancient Chinese
work, the Chou Pei, and the gnomon in Figure 3(3) is found in both China and
India (for China, see Lam & SHEN [5]; for India the idea of a gnomon is clearly
present in the Sulvasutras, as noted on page 194, above). I start with the gnomonfigure.
In Figure 4(1) there is a square ABCD and a gnomon-construction. I shall
tefer to each of the gnomon-corner squares AI and IC as the complement of
the other.
The first result is evident.
II.1 In Figure 4(1), the difference between (the areas of) the full square (AC)
and either of the gnomon-corners is equal to each of
* This is one of
anumber of more elementary results which must have been part of
the ancient core.
** The kind of configuration shown in Figure 3(2) limits the size of the inner square.
Pagina 5
Vedi nel PDF(si apre in una nuova finestra)A (Sum Form): the sum of the (areas of) complementary gnomon-corner
and the gnomon-complements;
B (Product Form): the (area of) rectangle formed by aligning the complementary gnomon-corner and the gnomon-complements.
B
ELA
A
tlh
€
G
C
L
D
B__D
VI,
E
(1)
F
(2)
Fig. 4
The Product Form occurs in EUCLID II.6. EUCLID’s figure is shown in Figure 4(2).
In Figure 4(2), in which BC is equal to CA, EII.6 asserts that rectangle AM together with square HE is equal to square DE. Rectangle AM will be called the
aligned gnomon.
A second configuration of a square within a square is given in Figure 3(1),
and repeated in Figure 5(1) and 5(2) with two possible dissections. In Figure 5(3)
there is a rearrangement of Figure 5(2).
It will be convenient to use the following conventions. In Figure 6, if AB and
CD are given lengths, there are several elementary, associated lengths: the sum
EH of AB and CD, their difference GF, their arithmetic mean EM, and FM,
which is half their difference. I shall denote the last two of these by HS(AB, CD),
and HD(AB, CD). It is evident that the longer of AB and CD is the sum of HS
and HD while the shorter is the difference.
11.2 In Figure 5(1), the difference of the squares with sides AB and CD is
four times the rectangle with sides HS(AB, CD) and HD(AB, CD).
A
B
€
D
(1)
(2)
(3)
Fig.
5
A
—iB
u
ea:
Pagina 6
Vedi nel PDF(si apre in una nuova finestra)This result suggests that HS and HD are important in the ancient tradition.
This will become ever clearer in the remainder of this paper.
Although there is implicit in Figure 5(2) yet another way of expressing the
difference of two squares, I have given this figure for the purpose of showing
that a further dissection of Figure 5(1) followed by a rearrangement gives Figure 5(3).
This will be useful later.
In Figure 7 are given two arrangements for adding two squares. Although
other arrangements are possible—one surely would think of a figure like that
in Figure 3(1)—a
search has not yielded any other arrangements that are attested
in any ancient tradition (to my knowledge), are closely related to the ancient
tradition proposed here, or appear to lead to useful equivalences. Nonetheless, and
because it appears to me that the sum of squares lies a bit deeper than the difference of squares, I may have overlooked other possibilities.
(1)
(2)
Fig. 7
My discussion of the sum of two squares using the arrangement in Figure 7(2)
is motivated by Problem 11 of Chapter IX of the Chiu Chang Suan Shu, which I
will discuss later. An analysis of this problem suggests that in adding the squares
with sides AB and AC, one should consider the square with side HS(AB, AC).
This leads to the dissection in Figure 8, wherein
AD = HS(AB, AC).
:
Since adding the squares on AB and AC together counts all rectangles of the
dissection in Figure 8 once excepting AG, which is counted twice, while doubling
the square on AD may be tallied by the check marks shown —note that instead
of checking the rectangle beneath BD twice, a check was put into its equal, just
adjacent, and likewise elsewhere in the dissection
— the following result is evident:
11.3 In Figure 8, the sum of the squares on AB and AC is the sum of twice
the square on HS(AB, AC) and twice the square on HD(AB, AC).
A
B
vw
D
Cc
“lv
6
Y
Y
Fig. 8
Pagina 7
Vedi nel PDF(si apre in una nuova finestra)The counting argument of 11.3 may be used in the case where D does not
divide BC equally. It follows, then, that
square on AB and square on AC = square on (AB and BD joined) and
square on (AB and DC joined) and
twice the rectangle EF.
This result appears to explain Problem 12 in Chapter IX of the Chiu Chang. We
shall come back to this later.
There is one result remaining to be discussed in this section, the relation
between oblongs and differences of squares. Although I have put it last, this result
is part of a procedure given at the beginning of my discussion of the ancient tradition, in and just prior to Figure 1, namely, how to square an oblong. One first
transforms the oblong into a gnomon, recognizes the gnomon as the difference of
two squares, and then uses the Theorem of PYTHAGORAS to get the side of the desired
square. The result at hand comes from this procedure by omitting the last step.
The procedure which transforms an oblong into a gnomon-construction contains
within its dynamics, structures, and results a significant fraction of the entire ancient tradition and, in embryo, the greater part of this tradition.
11.4 The area of the rectangle with sides AD and DC (in Figure 1) is the difference of the squares with sides HS(DC, AD) and HD(DC, AD).
This seminal result is intimately related to II.1A and 11.1B. First, 11.4 as stated
is equivalent to II.1B. Secondly, from noting that the rectangle with sides AD and
DC is equal to the square on AD and the rectangle with sides AD and the difference between sides DC and AD, II.4 may be seen as equivalent to IL.IA.
The proposed ancient core will be completed in Section III, wherein the Theorem of PYTHAGORAS will be added to the eight results given thus far. Since the
discussion of the Theorem of PYTHAGORAS will lead us away from these eight
results it is appropriate to comment now upon some connections between them and
EucLIb’s Elements. The eight results 1.1-11.4 arose from my attempts to understand
the received arithmetic in Chapter IX of the Chiu Chang. Given this, it is of interest to note that the proposed ancient core is very nearly all of EucLip’s Book II
together with one or two propositions from Book I.
The correspondence between the proposed ancient core and EucLID’s Elements
is shown below.
Ancient Core
EUCLID
i)
Il
EII.1
11)
1.2
El.43
iii)
1.3
EI.43
iv)
TIA
EII.4, EII.7
v)
11.1B
EII.6
vi)
11.2
EII.8
vii)
11.3
EII.9, EII.10
vili)
11.4
EIL.6, EIL.14
Pagina 8
Vedi nel PDF(si apre in una nuova finestra)The ancient core, with the Theorem of PYTHAGORAS (III.1 here; EI.47 in
EUCLID), comprises EI.43 and EI.47 of Book I together with all but three (Propositions 11-13) of the fourteen propositions in Book II.
The propositions EII.2 and EII.3 are special cases of EII.1, which corresponds
exactly to 1.1. The correspondence ii) is clear—I note only that EUCLID”s proposition is stated for parallelograms instead of rectangles. As to iti), 1.3 follows easily
from 1.2.
The results EII.4 and EII.7 are slightly different ways of relating the pieces of
the dissection of a square, shown in Figure 4(1). EIT.4 states that the full square
AC is equal to the two gnomon-corners and the gnomon-complements, while
EII.7 states that the full square AC and either gnomon-corner is equal to the other
gnomon-corner and twice the gnomon-base. The correspondence iv) notes yet
another view of the result underlying all of EII.4, EII.7, and IL.1A. It is reasonable
to expect both redundancy and the separate statement of special cases in the development of mathematics. It is common knowledge that EUCLID’s Elements has
instances of redundancy as well as separately stated special cases. Accordingly,
I have not tried to refine the ancient core proposed here, but have left it as it
emerged from the Chiu Chang.
I noted correspondence v) just after II.1B, calling attention also to EUCLID’S
figure (see Figure 4(2)), which explicitly displays what I have called the aligned
gnomon.
I have included 11.2 among the results in the ancient core even though it does
not appear to be explicitly related to the problems of Chapter IX of the Chiu
Chang. It was included because of its intimate relationships to II.3 and to the
Theorem of PYTHAGORAs. The result II.2 and the corresponding EII.8 given in
EUCLID are the same. The figure used here (Figure 5(1)) is not in the Elements and
allows a somewhat shorter proof. HEATH [2, page 391, vol. 1] also gives a proof
using Figure 5(1). EucLp’s figure for EIT.8 is that shown in Figure 8, which is
used here in 11.3. I have noted earlier that each of Figures 8 and 5(2)— the latter is
very nearly the same as Figure 5(1)—is a rearrangement of the other.
In his commentary on EII.9 and EII.10, HEATH [2, pp. 394-398, vol. 1] notes
that in the proof of EII.9 EucLip has used the Theorem of PYTHAGORAS for
the first time in Book II, that his proof and accompanying diagram are not in
the style of the first eight propositions in this book, and that each of EII.9 and EII.10
may be proved independently of the Theorem of PYTHAGORAS, entirely within the
methods and style of EII.1-8. The counting argument for II.3 sketched here is
well within the spirit of Book II; indeed the diagram used here for the proof
occurs in EII.8.
The corresponding viii) is clear. The two figures — EUCLID’S (given in Figure 4(2))
and Figure 1—are strongly related. The two constructions are the same, step by
step. As to EII.14, which is ““to construct a square equal to a given rectilineal figure,”
EUCLID proves it using EII.5 and EI.47 (Theorem of PYTHAGORAS). The purpose
and the construction procedure given in the Sulvasutras for squaring an oblong,
given here on page 194, are the same as in the Elements.
Finally, 111.1 and El.47 are the same result. I will give a detailed dicussion
IIT.1 in the next section.
Pagina 9
Vedi nel PDF(si apre in una nuova finestra)IH. The Diagonal Square
I do not suppose the term “diagonal square” which I shall use in this section
to denote the square whose side is the diagonal of a given rectangle or the hypotenuse of a right triangle will have wide appeal. Nonetheless, I shall use it for brevity
and, most of all, to avoid using the inappropriate but universally used “Theorem
of PYTHAGORAS” or ‘Pythagorean Theorem.” I shall talk about the diagonal square
and the Diagonal Square Theorem.
If one writes out a conjectured ancient tradition there is in the lineal nature
of writing a first result, a second, and so on. The very procedure one uses to prepare a piece of writing has within it a powerful impetus which orders the output.
Those who write pieces on the history of mathematics are subject to a second
ordering impetus, that which arises out of our view of proper mathematics, wherein
one deduces one result from another so that there always is an ordering of results
leading from axioms to the result at hand.
These remarks form a background to this third section, which is third for
reasons I believe have some validity. Nonetheless, insights and new results which
ultimately form a mature field of mathematics are not necessarily discovered in
an order which turns out at maturity to be acceptable.
The religion-based motivation for squaring an oblong leads to the problem
of finding a square equal (in area) to the difference of two squares, that is, to the
Diagonal Square Theorem. SEIDENBERG [13] gives the diagram shown in Figure 9(1)
and makes the comment, “In trying to subtract a square from a square, one would
place the smaller square into the larger and look at the difference. This could well
lead one to the contemplation of something like [Fig. 9(1)].”
(1)
(2)
Fig. 9
SEIDENBERG goes on to say, “This figure unquestionably was contemplated in
ancient times. The Chou Pei, an ancient Chinese work, has ... [Fig. 9(2)] ...”
SEIDENBERG notes that a proof of the Diagonal Square Theorem (DST)
follows easily from Figure 9(1). (He sketches a proof; see also the argument
below.) Thus he can argue that ‘‘the geometry of the Sulvasutras stems from the
philosophy of equivalence through area.”
There are many paths to the Chou Pei figure. One may consider adding or
subtracting squares using one or the other of Figures 3(1), 7(1), or 7(2). In each of
these arrangements the sides of the two squares are the given quantities. Through
dissection, each of Figures 3(1) and 7(2) can be made to have the form shown in
Figures 5(1) or 5(2). (It is useful to bear in mind that Figure 5(3) is a rearrangement
Pagina 10
Vedi nel PDF(si apre in una nuova finestra)of Figure 5(2).) Now, in Figure 10 I have added the four diagonals of the rectangles
on the outside of Figure 5(1) and included the dotted line segments which take
Figure 5(1) to 5(2).
F
C
Fig. 10
Independently of how Figure 10 came about, it is clear that the large square
(AC) can be dissected in two ways: either
(1) square on DG plus square on ED plus twice rectangle EG
or
(2) square on EG plus twice rectangle EG.
The DST follows immediately, that is,
IIT.1 In Figure 10, the square on DG plus the square on ED is equal to the
diagonal square EG.
It is easy to check that the “given quantities’
— call them XY and WZ in
Figures 3(1) and 7(2)—do not appear in the DST arising from Figure 10, which
came from a “natural” dissection or rearrangement of these figures; rather it is
the quantities HD(XY, WZ) and HS(XY, WZ) which appear.
The part of Chou Pei in which the figure like that in Figures 9(2) or 11(2) appear
has a passage relating to the figure. I quote it in full, from NEEDHAM [7].
Thus, let us cut as rectangle, and make the width 3 wide, and the length 4 long.
The diagonal between the corners will then be 5 long. [Rectangle G*E* in
Figure 11(2) may be taken as the rectangle in question, though the relative
sizes are not the same as in the Chou Pei. Next follows a procedure for completing the figure, from rectangle G*E*.] Now after drawing a square on this diagonal [square E*F*], circumscribe it by half-rectangles like that which has been
left outside, so as to form a plate [the square on side A*B*]. Thus the outer
half-rectangles of width 3, length 4, and diagonal 5, together make two rectangles; then the remainder is of area 25. [Since A*B* would have length 4
plus 3, the square on A*B* would have area 49; this square minus two rectangles (which lie outside of square E*F*, in the form of 4 triangles) gives the
area of E*F* as 49 minus 24, or 25.]
The last square-bracket insert is just (2) of the proof of the DST, appearing just
before IIE.1.
Pagina 11
Vedi nel PDF(si apre in una nuova finestra)This discussion has left out Figure 7(1), which perhaps is the most natural
arrangement for adding two squares. The segments AB and BC are the given
quantities. If we complete the figure by adding lines as shown in Figure 11(1),
where the square DB is equal to square BE, then a rearrangement (suggested by
the small numbers) leads to Figure 11(2), which is labeled as in Figure 10. It is
clear that E*D* is BC and D*G* is AB. The given quantities are, in this case,
part of the DST.
A
5
iO
2
7
13188!
tt.
c
AS
A
6
Di
1
ge
213
IG
p*
als!
1
,
5
8
3
7 e
3
(1)
(2)
(3)
Fig. 11
On the basis of the natural arrangement of two squares in Figure 7(1) and the
direct occurrence of the given quantities in the DST, one might conjecture that
the DST was found through adding squares using 7(1), a rearrangement into the
Chou Pei figure, and an argument similar to that given just prior to III.1. SEIDENBERG has pointed out (private communication), however, that this conjecture has
several flaws. There is in the Vedic rituals a motive for seeking a means of constructing the side of a square whose area is the difference of two squares, and there is
a plausible argument for how one might then come to Figure 9(1) and thence to
the Chou Pei figure in Figure 9(2). Neither SEIDENBERG nor I can give a similar
motive for adding two squares. Even if there were such a motive, there appears
to be no textually-based reason for adding the diagonals in the constructions
leadings to Figure 10, other than, of course, that such leads to the Chou Pei
figure.
Among the Vedic rules for altar construction is “The cord which is stretched
across a square produces an area of double the size.” (See THIBAUT [15, page 233].)
No proof is given in the Sulvasutra. An immediately suggested figure, see Figure
11(3), provides a transparent and compelling argument for this rule. The diagram
in Figure 11(3) is a special case of the Chou Pei diagram in Figure 9(2).
IV. The Problems of the Ninth Chapter of the Chiu Chang Suan Shu
I have now described the hypothesized ‘‘small, coherent, and ancient geometric
tradition.” I used the ninth chapter of the Chiu Chang as a source of ideas for the
nine components of this ancient tradition. In the balance of this paper J shall
discuss the problems of Chapter IX in terms of these nine components. Because to
Pagina 12
Vedi nel PDF(si apre in una nuova finestra)a considerable extent I constructed/conjectured an ancient tradition using the Chiu
Chang, one would naturally expect explanations of the Chiu Chang based on such
a hypothesized tradition to be reasonably coherent and well fitted. Until I can
thoroughly test this conjecture on, say, the Babylonian corpus, I can argue for
the merits of my conjecture only on such grounds as the simplicity of explanation
it allows, or its congruence with received results or figures. I shall use my translation into English of VoGEL’s [16] translation. I shall compare my reconstruction
with those of VOGEL, VAN DER WAERDEN [19], WANG & NEEDHAM [20], and SwETZ
& Kao [14]. VOGEL notes that there are no diagrams in Chapter IX of the Chiu
Chang.
Problems 1, 2, and 3. These three problems deal with the (3, 4, 5) right triangle.
Problem 1 will serve to illustrate all three. ‘Now we have a horizontal side
of 3 feet and a vertical side of 4 feet. Question: How long is the hypotenuse ? Answer:
5 feet. Rule: Multiply each of the horizontal and vertical sides by itself, add these
products, and take the square root of the sum.”
Problems 4 and 5. These two problems are mathematically the same as the
first three but are two steps harder. First, the problems are stated in non-mathematical terms, so that the reader is required to understand that the problem can
be formulated as one concerning a right triangle. Secondly, the given numerical
quantities are no longer whole numbers; however, the quantities are always
chosen (here, as well as elsewhere in Chapter IX) so that all necessary square
roots are rational. In addition to III.1, the solution of Problem 4 uses the result
that an angle inscribed in a semicircle is a right angle and that of Problem 5 uses
the result that a cord wrapped around a vertical cylinder and rising uniformly may
by viewed as the hypotenuse of a right triangle wrapped around the cylinder,
with one side of the triangle remaining horizontal and equal in length to the circumference of the cylinder.
Problems 6-10. The received solutions of this group of problems depend upon
the result 11.1, which gives alternative expressions to the difference of two squares.
I shall give several of these problems in detail.
Problem 6. “We now have a square reservoir, with side 1 chang [1 chang =
10 ch’ih; these are length units]. A reed is growing in the center and its top is
1 ch’ih out of the water. If the reed is inclined towards the shore, the top just
reaches the water of the shore. Question: What are the depth of the water and
the length of the reed? Answer: The water is 1 chang, 2 ch’ih and the reed is 1
chang, 3 ch’ih.
Rule: Multiply half of the side of the reservoir by itself; decrease this by the
product of the length of the reed above the water with itself; divide the
difference by twice the length of the reed above the water. This gives the depth
of the water. Add to this the length of the reed above the water. This gives the
length of the reed.”
Pagina 13
Vedi nel PDF(si apre in una nuova finestra)205
ey
Fig. 12
Neither the hypotenuse nor the vertical side of the right triangle A’B’C’ in
Figure 12(1) is known; their difference D’C’ and the horizontal side of the triangle
are known. Now referring to Figures 4(1) and 12(1), identifying AD with A’B’
(or its equal B’D’) and IG with C’B’, and using III.1 and IL.1A gives directly that
square on A’C’ = square on A’B’ minus square on B’C’
= square on D’C’ and twice the rectangle
with sides D’C’ and C’B’.
From this it is clear that the unknown side C’B’ is given by
(b:b-a-a+Q-:a=6°5-1:'1+2°-)1)
= 12 ch’ih
I
1 chang, 2 ch’ih.
VOGEL’s comment on this problem is that the initial relation is
x? + b?— where x = B’C’—from which x = (6? — a?)/(2a),
(x + a)? =
as in the received
arithmetic. This leaves the transition from the initial relation to the recipe unexplained.
Problem 7. ‘We now have an upright pole and a cord attached to the top of
the pole. If the rope hangs naturally, 3 ch’ih of its length lies upon the ground.
If the end of the cord is placed 8 ch’ih from the pole, the [taut] rope is in a
straight line from the top. Question: What is the length of the rope? Answer:
1 chang, 23; ch’ih.*
Rule: Multiply the distance from the foot of the pole to the end of the rope by
itself and divide the result by the amount of rope lying on the ground. Add
to this result the amount of rope lying on the ground and halve this result.
This is the length of the rope.””
* It should be 24 ch’ih.
Pagina 14
Vedi nel PDF(si apre in una nuova finestra)In Figure 12(2) is given a diagram for Problem 7. It is obvious from Figures 12(1)
and 12(2) that Problems 6 and 7 are similar. If we apply IIL.1 and 11.1B, and
identify A’B’ with AD and A’C’ with IG, it follows directly that
square on B’C’ = the rectangle with sides D’B’ and twice
A’B’ subsequently decreased by D’B’.
Thus A’B’ is
(Gb)
+ a) + a) + 2 = (8-8)
+ 3) + 3) +2
= 214+ 3)+2
= 241 - 2
= 124 = 1 chang, 24 ch’ih.
I shall omit any discussion of Problems 8-10; although their settings vary,
the mathematical model and solution techniques are the same as for Problem 7.
Problems 11 and 12. These problems appear to be slightly more advanced
than the first ten problems and their solutions require a result using the sum of
squares rather than the difference.
Problem 11. “We now have a door whose height is 6 ch’ih, 8 ts’un [10 ts’un =
1 ch’ih] more than its width. The distance between the [diagonal] corners is
1 chang. Question: What are the height and width of the door? Answer:
The width is 2 ch’ih, 8 ts’un and the height is 9 ch’ih, 6 ts’un.
Rule: Multiply 1 chang by itself; this is the beginning amount. Multiply half
of the difference by itself, double the result and subtract this from the beginning amount. Halve the remainder and take the square root. Decrease this result
by half of the difference—
this is the width of the door. Increase the result by
half the difference—this is the height of the door.”
Pagina 15
Vedi nel PDF(si apre in una nuova finestra)From Figure 13 it is seen that the new feature here is that the diagonal is given,
not a side as in Problems 6-10, When we identify B’C’ of Fig. 13 with AB of Fig. 8
and A’C’ with AC and use 11.3 and HI.1, it follows directly that
square on B’A’ = the sum of twice the square on
HS(B’C’, A’C’) and twice the square
on HD(B’C’, A’C)).
The side HD(B’C’, A’C’) is just b + 2. The side HS(B’C’, A’C’) corresponds
to AD in Figure 8; AD is a kind of ““mean-square” between the squares on AB
and AC of Figure 8. If the latter are added it is natural to suppose the sum is
related to twice the mean-square. Perhaps this is one way of viewing II.3. In any
case the above result allows the calculation of the mean-square:
2-square on
AD = a-a — 2((b + 2): (6+ 2)),
and now “halve the remainder [difference] and take the square root.” The side
AD is now known:
AD = Y(10 : 10 — 2(3.4 + 3.4)) + 2 = 6 ch’ih, 2 ts’un.
From Figure 8 it is clear that
width = AD — BD = 6.2 — 3.4 = 2 ch’ih, 8 ts’un,
height = AD + BD = 6.2 + 3.4 = 9 ch’ih, 6 ts’un.
These calculations fit the Rule very well indeed.
Now turning to Problem 12, we have some indication that it was given as a
generalization of Problem 11. For one thing, it is the immediately following
problem; also, if 11.4 was in fact used in Problem 11, a straight-forward adaptation of this result fits the received arithmetic of Problem 12.
Problem 12. “We now have a door whose height and width are not known;
both are shorter than a bamboo pole whose length is unknown. If the bamboo
is held horizontally, it comes out 4 ch’ih longer than the width; if held vertically
it comes out 2 ch’ih longer; if laid in the diagonal it comes out even. Question:
What are the height, the width, and the diagonal of the door? Answer: The
width is 6 ch’ih, the height is 8 ch’ih, and the diagonal is 1 chang.
Rule: Multiply the deficiencies in the vertical and horizontal together, double
the product and take the square root of the result. Add to this the deficiency
in the vertical to give the width of the door; add to it the deficiency in the
horizontal to give the height of the door; add to it both deficiences to give the
diagonal of the door.”
In Figure 14(1) is given a sketch of the context of the problem. Each of A’B’,
BC”, and C’A’ is unknown. In Figure 14(2) is given a figure similar to that of
Pagina 16
Vedi nel PDF(si apre in una nuova finestra)c'
a
A
A ==
ni
A
B
DC
N
b
x
a-x
N
E
\
VA
/
\
E
d
/
F
\
B
:
Y
1
i
|
x
ñ
C'
|
I
y
A
(1)
(2)
(3)
Fig. 14
Figure 8 (which accompanies the result 11.3 used in Problem 11). Make the identifications: A’B’ is AC, BD is deficiency b, and DC is deficiency a. The argument
used to justify II.3 and the analogy between Figures 8 and 14(2) suggest that the
side AB, which is neither the width, height, nor diagonal of the door, will play a
role in the calculations. From Figure 14(2) it follows, using the argument for 11.3
as a suide, that
square on AB and square on AC = square on (AB and BD joined) and
square on (AB and DC joined) and
twice the rectangle EF.
Since (AB and BD joined) is the height of the door, (AB and DC joined) its width,
and AC its diagonal, it follows, using III.1, that
square on AB = twice rectangle EF.
Now, using the given numerical data, we have
square on AB = 2-4-2 = 16,
AB = 4,
AB and DC joined = 4 + 2 = 6 ch’ih = width,
AB and BD joined = 4 + 4 = 8 ch’ih = height,
AB and BC joined = 4 + 2 + 4) = 10 ch’ih = 1 chang = diagonal.
VOGEL’s comments on this problem [16, pp. 96, 121] are as follows (my translation): From the right triangle [see Figure 14(1) and denote A’B’ by z] we have
2? = 222 — 2:(a+ 5)-z + a? + b?.
Adding 2ab to both sides gives
2ab =
z-(a+b)® or z— (a+b)
= V2ab. Hence z — a = b + W2ab and z —
b=a+ V2ab. VOGEL notes that the intermediate calculations are not given by
the Chiu Chang and suggests that the first equation was completed through the
formula a? + 2ab + b? = (a + by?, which VOGEL views as geometrically based.
Pagina 17
Vedi nel PDF(si apre in una nuova finestra)Problem 13. “Now we have a bamboo pole with a height of 1 chang. The end
was broken and touched the earth 3 ch’ih from the root. Question: How high
is the break? Answer: 4% ch’ih.
Rule: Square the distance from the root; the result should then be divided
by the height. From the height of the pole subtract the result and take half of
the remainder. This is the height of the break.”
In Figure 14(3) is given a figure for this problem. We use 11.4 and 111.1. Through
considering the first and third diagrams in Figure 1 and noting from Figure 14(3)
that it is a that is given, we let AD be A’B’ decreased by B’C’ and DC be A’C’.
From 11.4 and 111.1 we have
square on C’D’ = the diffference of the squares on B’D’ and B’C’
Il
the rectangle with sides DC and AD.
This gives the arithmetic
(3-3)
— 10= 10
— 2-B'C'
or
B'C’ = 45 ch’ih.
Problem 14. In this problem (and in Problem 21) a notion of speed is introduced.
In this context ‘‘Pythagorean triples” are calculated, perhaps in a manner similar
to that proposed by VAN DER WAERDEN [17] for the triples in Plimpton 322. I shall
propose a way of calculating these triples which is based on the ancient core, fits
the received arithmetic, and agrees closely with VAN DER WAERDEN’S original
idea.
Problem 14. “Now we have two people starting from the same place. The
speed of A is 7 and the speed of B is 3. B goes east and A first goes 10 pu south
and then towards the north-east until he meets B. Question: How far did each
of A and B travel? Answer: B went 104 pu east. A went 144 pu in the diagonal
direction until he met B.
Rule: Multiply each of 3 and 7 by itself, add the products and halve the sum.
Take this result as the coefficient of the diagonal way of A. Subtract the coefficient of the diagonal way from the product of 7 with itself. The difference is
the coefficient of the southern way. Multiply 7 by 3; the product is the coefficient of the eastern way of B. Lay out the 10 pu towards the south and multiply
it by the coefficient of the diagonal path of A. Again lay out the 10 pu and
multiply it by the coefficient of the eastern way of B. Each product is a dividend. Divide the dividends by the coefficient of the southern way. Each time
we get the amount of the way.”
So as to not interrupt the discussion of the rule itself I shall first discuss the
use of III.1 and 11.4. If we refer to Figure 15(1) and use III.t, the product
Pagina 18
Vedi nel PDF(si apre in una nuova finestra)(1)
(2)
Fig. 15
of bwith itself is the difference of the products of cand
a with themselves. From II. 4,
which uses Figure 1, this difference of squares may be thought of as a product
of sides AD and DC. Moreover, c and a are seen to be HS(AD, DC) and HD(AD,
DC). It is also clear from 11.4 and Figure 1 that € + a = DC and c — a = AD.
Let DC and AD be measured by numbers p and g, respectively; these results may
be summarized in
b:b=c'c—a:a=p'q,
c=(ptq)~2,
a=(p=4q)=2,
c+a=p,
c=a=q.
Problem 14 is that of two persons A and B moving at different rates during
the same time period, that is, by reference to Figure 15(2), the distances c + a
and b are completed in equal times. Hence (c + a) — 7 =b-=3. From b:b=
(c — a)
(c+ a) and (c+ a) +7 = b +3 it follows that (c + a) = (c — a) =
7:7-3-:3.* In a triangle similar to the desired triangle one will have
c+4=7'7=p and c-a=3:3=q. It then follows that—here begins
the received arithmetic
—
ec=(7:74+3:3-2=2, a=p-=c=7:7-29=20, and b=7:3=21,
Since these numbers are proportional to the lengths of the sides of the triangle, it
remains to use the given information that the southern way has length 10 pu.
Thus, multiply all sides by 10 and divide by a. This gives the diagonal and eastern
ways, (10-29) — 20 = 144 pu and (10-21) + 20 = 104 pu.
Problem 15. This problem is the first of two problems on inscribed figures in
right triangles.
* Once the geometry of the problem is translated into arithmetic equalities such as
these, one may infer the statement (*) as follows: 1) From 6-5 = (ce — a)-(c +a)
it follows that ((c — a) + b)- ((e + a) + b) = 1; hence the two factors are reciprocals.
2) From (c + a) + 7 = b = 3 it follows that (c + a) + b =7 + 3. Putting 1) and 2)
together shows that (c
— a) — b—3 +7. Hence (ec +a +(e —-a=7-°7+3-3.
This argument is due in part to SEIDENBERG (private communication).
Pagina 19
Vedi nel PDF(si apre in una nuova finestra)Problem 15. ‘Now we have a horizontal side of 5 pu and a vertical side of 12 pu.
Question: How large is the square inscribed in the right triangle? Answer: The
side of the square is 3% pu.
Rule: Add the horizontal and vertical sides; this is the divisor. Multiply the
horizontal and vertical sides; this is the dividend. Divide the dividend by the
divisor. The result is the side of the square.”
A
A
A
b
D
Ca
L en
BA
D
5
E
Ce
BC
B
e
8
C
(4)
(3)
(2)
(1)
17
15
D
Fig.
16
A figure is given in Figure 16(1). The side of the square CD is to be found. If
the figure is cut along the dotted line from D to C, leaving an uncut hinge at C
the figure may be opened as in Figure 16(2). The dotted lines make four rectangles,
each of which is bisected by a diagonal. It is clear from I.1 that the area of AE is
the sum of the areas of the two rectangles IC and GB. Arithmetically,
(6 + a) «side of square = b-a.
This is the received arithmetic.
The idea of the dissection along CD is due to VAN DER WAERDEN [19, p. 54].
VOGEL [16, p. 97] states that the formula
b - a/(b + a)
follows from similarity
or the equality of the gnomon complements (1.2). The gnomon VOGEL has in mind
must be that in Figure 16(3), obtained from Figure 16(1) by drawing through A
and B lines parallel to CB and CA, respectively, and extending the sides of the
inscribed square. Since
ba = sum of the areas of rectangles AD, DE, DB, and
CD, and the areas of DE and CD are equal, it follows that
6-a = areas of
rectangles AF and CG, that is,
side of square.
b-a = b-side of square + a
Each of these explanations of Problem 15 lies within the proposed ancient
core. I have come to prefer VAN DER WAERDEN’S idea because it is applicable to
the next problem, while it appears that VOGEL’s approach is not.
Problem 16. This problem asks for the diameter of the circle inscribed in an
(8, 15, 17) right triangle. See Figure 16(4). VOGEL [16] suggests that the given rule
may be based on a dissection of the figure. Although VOGEL gives few details it
would appear that he had in mind something quite similar to VAN DER WAERDEN’S solution for Problem 15. If in Figure 16(4) cuts are made from the center
of the inscribed circle to the three vertices, leaving uncut hinges at B and C, the
figure may opened out on either side of BC, so that A, B, and C lie in a straight
line. From this figure, which is analogous to Figure 16(2), the received arithmetic
((15 + 8 + 7) : (diameter — 2) = 15-8)
follows directly.
Pagina 20
Vedi nel PDF(si apre in una nuova finestra)Problems 17-21. Each of these problems may be called a “city problem”.
The first three may be seen as variations on 1.2, the equality of gnomon-complements. The fourth is more difficult, appearing to require two results from the
ancient core, 1.3 and 11.4. Problem 21 is yet another variation, combining a “city
problem” with a “speed problem.” This is consistent with the trend of the problems,
from simple to complex.
Problem 18. “Now we have a city with a rectangular boundary. From east
to west it measures 7 li and from north to south 9 li. In the middle of each
side is an open door. If we go out the east door 15 li we come to a tree. Question: How many steps out the south door must we go to see the tree? Answer: 315 pu.
Rule: Multiply the number of steps from the east door towards the south,
just up to the corner, by the number of steps from the south door east, just up
to the corner. The product is the dividend. Take the number of steps the tree
is from the door as divisor. Divide the dividend by the divisor.”
1
0
C
ì
2
a
~
8
D
b
o
15
A
B
|
(1)
(2)
met 4
(3)
Fig. 17
Using I.2 and Figure 17(1)—and converting li to pu (1 li = 300 pu = 300
“‘steps’’)— gives immediately that
AB «4500 = 1050 - 1350 and so AB = 315 pu.
The solution to Problem 19—see Figure 17(2)—is a variation on that of
Problem 18 in that one of the gnomon-complements is a square with unknown
side AB and the other has known sides.
Problem 20. ““Now we have a city with a square boundary. We do not know
the length of the side. In the middle of each side is an open door. A person
goes out of the north door 20 steps and finds a tree. If a person goes out the
south door 14 steps and then turns towards the west 1775 steps, then he can see
the tree. Question: What is the length of the side of the city? Answer: 250 pu.
Rule: Multiply the number of steps taken from the north door by the number
of steps towards the west. Double this and let it be the shih. Add the number
Pagina 21
Vedi nel PDF(si apre in una nuova finestra)of steps taken from the south and north doors and let it be the tsung-fa. Extract
the square root to give the desired side.” *
Referring to Figure 17(3) and using I.3 we see clearly that
20-1775 = (half of unknown side) - (unknown side and 20 and 14)
or
2-20
1775 = (unknown side) - (unknown side and 20 and 14).
The Rule contains instructions on computing
2-20-1775
and
20 + 14
and
then either invokes a procedure related to square roots or, if one assumes the Rule
means one is to compute VQ * 20 - 1775)/34, is completely wrong. The answer
of 250 pu is correct. I shall give an explanation for a solution within the ancient
tradition.
By use of II.4 the last equation may be written as
2-20-1775 = difference of the squares with sides HS (unknown side and 20
and 14, unknown side) and HD (unknown side and 20 and 14, unknown side).
This leads to the arithmetic (note
17 = (20 + 14) + 2)
2-20-1775 + 17:17
= HS-HS
or
V2 -20-1775 + 17-17
— 17 = 250 pu.
This solution is not described in the Rule. Certainly one may observe in both
places the partial results 2-20-1775 and 20 + 14 and note that both refer to a
square root operation. I note that this problem is the only one in Chapter IX for
which the arithmetic recipe is not elementary and complete within the problem.
I will comment further upon this problem in Section V.
Problem 21. This is another “city problem,” but one having features of two
earlier problems. One part of the problem involves velocities; the handling of
this in the solution is the same as in Problem 14. The geometry is similar to that
in Problem 18; 1.3 is used instead of I.2.
Problems 22-24, Each of these problems describes a situation involving a remote object. In the first problem the distance to a tree is wanted; given are data
relating the tree to a known, nearby figure. In Problem 23 the height of a mountain is wanted; given are data on its relationship to a nearby, vertical pole. One
problem is based on I.2, the other 1.3. Problem 24 involves a well of unknown
depth; given are data obtained at ground level. Each of these three problems uses
either 1.2 or 1.3 in a straightforward manner.
* Shih and tsung-fa are technical terms occurring in the square root algorithm. See
Section V.
Pagina 22
Vedi nel PDF(si apre in una nuova finestra)I now give a table summarizing the usage of the proposed ancient core in the
24 problems of Chapter IX of the Chiu Chang. I note that the problems are
grouped and, within a group, tend to become more complex, sometimes combining
in one problem features of several problems. In Chapter IX as a whole the problems tend from very simple applications of the DST to quite difficult, multistage problems. The last three problems are relatively stmple compared to Problem 21.
VogeL [16, p. 120] argues that the Chiu Chang is not a textbook, rather a
collection of problems, since the methods of calculation are only explained in
detail in a few problems. WANG & NEEDHAM [20, p. 351] note—though on a
different Chapter of the Chiu Chang—that ‘The extreme brevity of the text is
presumably due to the custom of oral teaching by qualified mathematicians, and
to the literary desire of avoiding undue repetition.”
It seems to me that Chapter IX represents a deliberate and fully considered
attempt to order a set of problems, with features of repetition, generalization, and
the blending of problem contexts and solution techniques from several problems
into one. This more nearly characterizes a textbook than a problem collection.
Problem
Ancient Core
1
11.1 (DST)
2
111.1
3
111.1
Category
Purely geometric/arithmetic
4
Il!
Sawing plank from a log
5
11.1
Vine-helix problem
6
111.1 & ILIA
Reed in reservoir
7
T1.1 & IL.1B
Cord on pole
8
IILi & 11.1B
Beam leaning on wall
9
II.1 & IT.1B
Cutting a log
10
11.1 & H.1B
Door problem
1]
111.1 & I1.3
Door problem
12
111.1 & 11.3 generalized
Door problem
13
11.1 £ 11,4
Broken bamboo
14
11.1 & 11.4
Velocities
15
(IT.)* & Li
Inscription problem
16
(IT.1)* & LI
Inscription problem
17
1.2
City problem
18
1.2
City problem
19
1.2
City problem
20
13 & 11.4
City problem
21
I11&13& 114
City problem & velocities
22
1.2
Remote object problem
23
1.3
Remote object problem
24
Remote object problem
* DST not explicit.
Pagina 23
Vedi nel PDF(si apre in una nuova finestra)V. Square Roots, Problem IX.20, and Horner’s Method
I particularly wish to make an argument supporting my reconstruction of
Problem 20 since my use of II.4 is not consistent with the views of WANG & NEEDHAM [20] and D. B. WAGNER (private communication). These distinguished scholars
have an extraordinarily impressive knowledge of both the ancient Chinese language and Chinese mathematics. Given this,
I am very hesitant in advancing explanations at variance with their opinions.
WANG & NEEDHAM’S paper is “‘Horner’s Method in Chinese Mathematics.”
Their main **... new contribution ... is the attempt to show that the mathematicians
of the Ist century B.C. in China understood the essentials [of HoRNER’s method].”
They also say, “It is now generally accepted by historians of mathematics that the
method of Horner (1819) for solving higher numerical equations appeared in a
substantially equivalent form in the works of the Sung algebraists such as Ch’in
Chiu-shao (1247). This method took its origin much earlier in the procedures for
root extraction of the Han dynasty. These are to be found in the ‘Nine Chapters
of Mathematics’ (Chiu Chang Suan Shu) but the text has long been very obscure,
and it is the purpose of the present paper to attempt an explanation of its meaning.”’
Their reconstruction/exposition of the Chiu Chang method of extracting the square
root of number is meticulously done and very clear. Some fifteen years later LAM
Lay YONG [3] wrote a paper, based on a work by YANG Hui in 1261, supporting
WANG & NEEDHAM and also presenting an argument for a geometric basis for the
arithmetic root extraction techniques given in the Chiu Chang. My own explanation, arrived at independently of LAM Lay YoNG but influenced by certain Babylonian procedures and a somewhat obscure and, in my opinion, questionable paper
by GANGULI [1] on Vedic mathematics, parallels that of Lam Lay YONG.
The surviving Chinese procedures on square root extraction are described in
terms of operations on a counting board or calculating device. Although it happens to be a base 10 device, the underlying explanation does not depend upon the
choice of base.* In 1968 VOGEL stated his opinion that the Chiu Chang method
was based upon an algebraic identity. I believe, however, with SEIDENBERG and
VAN DER WAERDEN, that the basis of much of Babylonian, Chinese, and Vedic
mathematics is at bottom geometric. The computational algorithms resting
on the underlying geometry are, however, by and large what have survived and
are the raw material with which we work.
The result II.1B is sufficient to explain the received arithmetic, allowing for
subsequent adaptation to a computing device and an inevitable codification of the
geometric procedure into an algorithm, complete with names for the various
numbers appearing in the computation.
T assume that a numerical measure N of the side of a square with given area
was sought and that the metrological system was sufficiently developed that N
was thought of as, for example, A degrees, B minutes, and C seconds (Babylonian)
or a chang, b ch’ih, and c ts’un. For simplicity I will continue as if there were
* The algorithm is, in fact, the same as the well known “‘divide and average” algorithm, which is the same as ‘NEWTON’S method”.
Pagina 24
Vedi nel PDF(si apre in una nuova finestra)only three digits in N. The method of the algorithm is not restricted to a fixed
number of places.
Using WANG & NEEDHAM’S example, we seek the side N of a square with given
area, say 55,225. We therefore seek digits a, b, and € for which (a - 100 + 6-10
+ c):(a°100 + 5-10 + c) = 55,225.
55,225;
By trial—(3 - 100) - (3 - 100) = 90,000
>
(2-100) : (2 : 100) = 40,000 < 55,225)—we find
a = 2.
From ILIB,
the difference of the squares on TZ (55,225) and TU (40,000) in Figure 18 is the
aligned gnomon. If we neglect the very thin outer gnomon and the small square
with diagonal UV, we have
55,225 — 40,000 > 2 : (100 - a) : (10 - 5)
or
15,225 > 4,000 : db.
100a
10b
ra
Fig. 18
This gives a good estimate of b; in this case b is less than 4. We may check if
b = 3 is correct (is
(2 + 100 + 3 - 10) -(2-100 + 3-10) < 55,2252). The algorithm now repeats the same reasoning for the remaining digits:
55,225 — 40,000 — 30 : (2 : (200) + 30) > 2 -230-c,
2,325 > 460 -c.
This shows that
c< 6.
It follows, as before, that
c = 5
is correct. Indeed,
235 :235 = 55,225,
The arithmetic suggested by II.1B agrees in detail and order with that of the
Chiu Chang, as clarified by WANG & NEEDHAM. LAM LAY YONG'S arguments are
the same and are based directly upon YANG Hur's commentary of 1261 on the
Chiu Chang. YANG Hur's work contains a diagram identical to that in Figure 18—
recall that the Chiu Chang as received has no diagrams.
If HorNER’s method for polynomials is applied to finding a square root (so
that the polynomial is taken to be x? — n, where n is the measure of the square
whose side is wanted), it is easy to check (WANG & NEEDHAM give detailed examples) that the arithmetical work agrees with the Chiu Chang method. Recalling
that HORNER’s method for polynomials is attested in China in 1247, we naturally
suppose that the thirteenth century Chinese based their method on the earlier
Chiu Chang
square root algorithm. Nobody disagrees with this. WANG & NEEDHAM
Pagina 25
Vedi nel PDF(si apre in una nuova finestra)go further, however, and argue that in Problem 20 of the Chiu Chang we have
“distinct evidence that at least towards the end of the pre-Christian era the Chinese
applied their method [for calculating the square roots of numbers] to solve a quadratic equation of the type x? + ax = 5b...” The principal evidence is (1) the
presence in Problem 20 of two technical words which appear in the Chiu Chang
square root procedure and (2) the absence of a detailed recipe which is elementary
and complete within the problem.
I can not account for either of these features of Problem 20 and am not qualified to discuss the linguistic evidence in the manuscript. I wish to observe, however, that Problem 20 is unique in that its Rule does not give a detailed, utterly
explicit recipe for manipulating the given numbers to obtain the answer. Moreover,
this problem is one of a cluster of several similar problems, the “city problems,”
all others of which have explicit Rules, which, in my view, are based on the equality
of the gnomon-complements. Finally, I have shown that II.4—which was also used
in Problem 14— provides a straightforward solution to the problem.
VI. Summary
VAN DER WAERDEN's and SEIDENBERG’S theories on the origin of mathematics
continue work which at least for pre-Greek mathematics, began with NEUGEBAUER
& SACHS [8], who hypothesized that the contents of the ‘‘geometrical algebra”
in Book II of Euctip’s Elements utilized results known in Mesopotamia. VAN DER
WAERDEN and SEIDENBERG recently have hypothesized a development beginning
well before 2000 B.C., initially centered upon an oral tradition of geometrical
constructions and including the “‘Theorem of PYTHAGORAS”; this tradition subsequently divided into two great traditions -one geometric or constructive, the
other algebraic or computational
— which were transmitted separately to the Indians, Babylonians, Chinese, and Greeks. The first and last are thought to have
received the geometrical tradition while the second and third received the algebraic
tradition.
My own work was centered upon Chapter IX of the Chiu Chang, which I have
used as a guide in my attempt to recover the oral tradition. I was also influenced
by the work of NEUGEBAUER, SACHS, and VAN DER WAERDEN on Babylonian
mathematics and by SEIDENBERG’S recent studies of Indian religious works on
altar constructions.
The oral tradition I have proposed fits the Chiu Chang well in the sense that
the core of nine geometric results leads one to the received arithmetic in the 24 problems. I have noted that this agreement is in part assured since the problems were
used in choosing the results. Although I have not yet completed a systematic
examination of the Babylonian materials, the oral tradition proposed here provides
some explanation of them, including the generation of Pythagorean triples and the
so-called normal problems.* In his recent book [19] VAN DER WAERDEN stresses
* NEUGEBAUER has referred to problems in which two numbers are sought and one is
given their product and either their sum or difference as quadratic problems in normal
form.
Pagina 26
Vedi nel PDF(si apre in una nuova finestra)the many similarities between Babylonian and Chinese algebra, noting in both
kinds of algebra the solution of quadratic equations; that sets of linear equations
were solved by eliminating one unknown after the other; and the presence in both
countries of numerical methods for calculating square and cube roots. I believe
that the congruences with Books I and IT of EucLID, the Chou Pei diagram, and
the results given in
the Sulvasutras provide further evidence that the proposed Neolithic oral tradition has some explanatory power beyond the Chiu
Chang.
If one were to use, say, 11.4 repeatedly on problems including numerical
measures of sides, the pattern imposed on the arithmetical reckoning by 11.4
would be sufficiently strong to induce in the mind of the user an algorithm on
the numbers themselves, an algorithm which would come to stand on its own,
with the geometric result gradually assuming less importance. This is my view
of the emergence of algorithmic/algebraic processes from the geometric. Without
at all wishing to enter the geometric algebra controversy, I have tried in this
paper to avoid using modern notation for algebraic identities implicit in geometric
results. It seems to me that such use introduces an unnecessary and misleading
bias into the discussion.
I have discussed some of VOGEL’S comments on Chapter IX of the Chiu
Chang, particularly when I felt his explanations of the problems were misleading
or incomplete. In Problem 12, for example, VoGEL’s explanation appears to me
to be too algebraic for attribution to the autbors or students of the Chiu Chang.
If VoGEL intends to suggest through the algebra corresponding geometric results,
then the suggested approach appears overly complex. For the most part our
explanations are consistent. VOGEL often uses modern algebraic notation in his
explanations and suggests the presence of such algebraic phenomena as factoring
or completing the square. I find this contrary to my own reading. At the same time
VOGEL also uses geometric figures and mentions in several places a geometric
origin for the solution techniques.
My explanations are more complete (at the cost of being more speculative)
than VoGEL’s and in most problems exactly match the arithmetic given in the
Chiu Chang. I have disagreed with VOGEL, WAGNER, and WANG & NEEDHAM
on Problem 20; my discussion is given in Section V.
I believe my remark that the figure suggested by the Vedic result on doubling
a square is a special case of the Chou Pei diagram—
the figure also is given in
PLATO’s Meno, in the dialogue between SOCRATES and a boy — together with the
interplay between Figures 7(1), 8,
11(1), 11(2), 11(3), and 9(2) discussed in Section III
provide a useful summary of some bits and pieces related to the DST.
VAN DER WAERDEN (private communication) has raised the point that denoting
line segments by two letters is not appropriate in describing an oral tradition.
In an oral tradition one would use words such as hypotenuse, shorter leg, side, ...,
or simply point to the line in question in a diagram or model. This had not occurred to me although I did work at the somewhat analogous matter of avoiding
algebraic notation. I agree with VAN DER WAERDEN’S view but have decided not
to change the present paper.
To write down a proposed neolitic oral tradition (or ancient core) is, of course,
speculative. Virtually by definition, written records were not produced in neolithic
Pagina 27
Vedi nel PDF(si apre in una nuova finestra)times. What I have tried to do is to infer what was likely to have been the common
oral heritage of the historical Chinese, Babylonian, Indian, and Greek mathematics. I have assumed that SEIDENBERG’S and VAN DER WAERDEN’S views on the origin of mathematics are more nearly correct than those who believe mathematics
was independently discovered by several peoples. I have not tried to give a detailed
discussion of their views here; I have tried to support their views by drawing
together a reasonably complete oral tradition. I have argued that it explains the
Chiu Chang well, matches Book II of EucLID’s Elements closely, and, leaning
principally on VAN DER WAERDEN’S papers and books, said that it appears to
explain some of the Babylonian corpus. I am working on a paper which will
attempt to go much further with the Babylonian materials.
Note added in proof. 1 have received a letter from Professor VoLKov (Department
of the History of Mathematics, Moscow) in which he discusses the meanings of the
terms shih and tsung-fa associated with Problem 20 (see pp. 212-213). He argues that
the received Rule for Problem 20 is not incomplete; rather, the two technical terms
refer to a well known procedure for solving a quadratic equation numerically. I have
stated that I am not competent in the Chinese language and therefore can not form
an opinion on the merits of Professor VoLkov's (and others) arguments. Certainly they
must be given some weight since they rest directly upon the received text. My own
views are given in Section V. They rest upon the singularity of the Rule for Problem 20
and the observation that II.4 (which was used in Problem 14) provides a simple solution, one consistent with the remainder of the problems.
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Department of Mathematics
Towa State University
Ames JA 50011
(Received March 20, 1985)