Hidden Harmonies

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Hidden Harmonies Page 3

by Ellen Kaplan


  One of Neugebauer’s followers14 suggested that the obscure fourth column recorded squared cosines or tangents of the angle opposite the short side, and that the rows decrease roughly one degree at a time. This would be remarkable, since the Old Babylonians had little concept of angles other than right angles (even these showed a fair amount of wobble), and none at all of measuring them, much less calculating their cosines or tangents.15

  Which brings us back to The Irascible Scholar—but in a roundabout way. In 1980 an urbane mathematician named R. Creighton Buck wrote a brilliant paper, “Sherlock Holmes in Babylon”,16 in which he looked again at Plimpton 322 and at Bruins’ work (perhaps some of the bile had dissipated by then). He cites the proposals of a D. L. Voils, which (you find in his references) were soon to appear in Historia Mathematica. In the nearly three decades since then, no paper by Voils has surfaced, there or elsewhere. He is unknown to Wikipedia. The Web produced a phone number, which we called. Whoever answered told us: “There is no Mr. Voils.” Thus had we heard, and did in part believe. He is The Invisible Man.

  Invisible Men write Disappearing Documents. The ghost of D. L. Voils haunted our days, until at last diligence tracked him down in Florida. He had written the paper, he said, which was for some reason rejected, and he chose not to rewrite it. When he moved from Wisconsin he left it behind with other papers, which his son subsequently sold. Since matter is neither created nor destroyed, only changed in form, the atoms of his argument wait somewhere to be reassembled.

  What Buck says that Voils suggests, based on Bruins, is that Plimpton 322 was a teacher’s trot for making up igi-igi-bi problems!at These were, as you saw, a stock-in-trade of the schools, since they taught the solution of what we call quadratics by manipulating the Babylonian Box. Some were of the form , a constant, some (like the one you saw above),. While tables of reciprocals were a commonplace, what a teacher needed to know was which c yielded nice solutions—and here was the answer.

  Between the three of them (we can say ‘between’, because Voils is the invisible middle term), this is their explanation. Column II involves the differences the teacher would like to set his problem equal to (like the 7 in the illustration we gave), column III bears on the sums, and IV serves to check an intermediate step. But where did they get their values from?

  Take, our trio says, any numbers you like, a and b (perhaps listed in columns on the missing third of the tablet), and set , the igi, so that will be its reciprocal, igi-bi (let’s write it xR, with R for ‘reciprocal’, so that we won’t be wedded to some particular power of 60 in the numerator). Then

  and that difference in the numerator, a2 − b2, is what is listed in column II.

  To see how it works, take the entries in row 1: 119 is in the difference column, II, and if we play around a bit, we’ll find that a could be 12 and b, 5: for then a2 = 144 and b2 = 25, so a2 − b2 is indeed 119. The teacher can now set up his problem: find x and xR if x − xR= 7 (since this is 12 − 5). He knows the answers will be and —or, since he doesn’t deal in such fractions, he expects his students to proceed in the same way we showed above with, and come up with 12 and 5 as the solution—for in fact, this is the very same problem! He could also have made up the problem x + x R= 7, since 122 + 52 = 169, as he finds in column III. Notice that the a and b and the sums and differences of their squares are his secret knowledge: the student knows only the cleaned-up statement of the problem, and the instructions for using the diagram:

  The igi-bi over the igi is 7 beyond

  The igi and the igi-bi are what?

  You:

  The 7 that the igi-bi over the igi is beyond, to two you break, then 3½

  3½ with 3½ let them eat each other, then 12½

  To 12½ that came up for you, 60, the field,

  add then 60 and 12¼

  The equalside of 60 12¼ is what? 8½

  8½ and 8½ its equal lay down

  Then 3½ the holder

  From one tear out

  To one add

  One is 12 [8½+ 3½]

  The second 5 [8½− 3½]

  12 is the igi-bi, 5 is the igi.17

  Be glad you went to school three millennia later.

  If you wonder at the stubborn butting of heads against calculations by hand, in cuneiform, with a base 60 system, encumbered by awkward notation, of such quantities—so do we. Conceivably Plimpton 322, with its large values (although 12,709 in row 4 of column II, for example, only needs a to be 125 and b, 54), marks a transition from a geometric to a more formulaic style. But for all that squares and their sums and differences are in play, of Pythagorean triples we haven’t a shadow cast by the slimmest of gnomons—and that, Watson, is what we wanted to know.

  Since Bruins is undeniably behind this explanation, it’s as if someone with disgusting habits were punching you in the stomach while screaming the truth in your ear. You’d hate to admit he was right. But then, we do have him at two removes, and Buck’s manner is charming. Eleanor Robson, his severest critic, takes him to task (in “Neither Sherlock Holmes nor Babylon”)18 for being a gentleman-amateur and therefore trying to show up the plodding professionals; for speaking of Babylon at large, when the tablet really came from the city of Larsa; for being (somehow) part of the colonization, appropriation, and domestication of the pre-Islamic Middle East by Western society; and for misinterpreting Voils’ theory—which she knows of only through him. She concludes by agreeing with his interpretation. The criticism of her paper by a fellow scholar you have already seen.

  Neugebauer’s unicorn has turned out in the end to be indeed a fabulous beast. A slab delivered to earth by superior beings might have had such triples on it, but not a bit of clay worked by Babylonians. There is, however, one more thread to follow that might yet lead to the Pythagorean Insight in ancient Mesopotamia. Rather than a slab, this is a tablet the size of your palm, numbered YBC 7289, which sits snugly in Yale’s Babylonian Collection—and on it is incised a square scored with two diagonals into four congruent isosceles right triangles. Across one of these diagonals is written, in cuneiform, “1 24 51 10”—i.e., : which translated into our base 10 is 1.414212962—an approximation correct to five decimal places for . Here is that tablet, full size:

  Quite apart from the question of how they managed to calculate so precisely, a right triangle with sides 1, 1, is certainly not gnomonic. Mustn’t they have had the Pythagorean Insight after all in order to know that the diagonal had length ?

  The answer lies twelve hundred miles west and thirteen hundred years in the future from the Old Babylonians. This diagram, differently scaled, plays a key role in the dialogue Meno, which Plato wrote in fifth-century B.C. Athens. In it, Socrates is shown eliciting from a slave-boy what the mithartum (as those Babylonians would have said)—what the side-length—must be of a square whose area is 8. Implicit in the text is this figure:

  Plato chose to infer the soul’s immortality from having a mortal slave recollect an immortal truth; we infer that to this end he himself recalled what he knew from Babylonian lore. Socrates draws a square of side 2 and has the boy deduce that its area is 4. When he asks him for the side of a square with area 8, the boy proposes that the side should be double that of the first square (making a square of four 2 × 2 squares). Disabused of this by Socrates, he is then led to see that the square on the diagonals of these four squares has half their area, .19

  Notice that Plato chooses his question so that the boy’s erroneous middle step—constructing a square of area 16—will lead to the insight, even though the answer won’t be the more fundamental , but 2.

  The Old Babylonian tablet also seems to set a leading but much less interesting question, for the inscribed “1 24 51 10” is in a small, neat hand, as is a “30” along one side, meaning , or ½. In a sprawling, student hand, beneath the constant for , is written “42 25 35”—i.e., : exactly half of the constant which the teacher will have found in one of those pedagogical helps they had even then—a ‘table of coefficients’. Can
’t we hear the teacher asking: if this is the diagonal when the side is 1, what will it be when the side isn’t 1 but ½? A relatively easy question, but in a setting which might have been meant to prepare for deeper issues.au

  What matters for us is that this diagram—scaled up by Plato, down by whoever the teacher scribe may have beenav—shows by its simple symmetries that the area of a square built in this way on what Socrates calls “the line from corner to corner” must be 2: half of the enclosing square’s 4. In neither case are triangles, or the Pythagorean Theorem, invoked (although Plato would have known the Theorem very well).

  Our reconstruction, we admit, is by the enjoyably fallacious post hoc ergo propter hoc, with antecedents taken as causes (since Plato did this, a Babylonian must have done the same, before him). We warned you that our telling would be postmodern. But is the argument as farfetched logically as it is in time? If we have similar footprints, eons apart though they are, can we not plausibly argue that similar creatures made them?

  Let’s probe how the Old Babylonians came up with that astonishingly accurate approximation for , to see what this might tell us about the antiquity of the Pythagorean Insight.

  When a courtier leaned over to Bach while Frederick the Great was conducting his orchestra and whispered, “What rhythm!”, Bach whispered back: “You mean, what rhythms!” These inveterately calculating Babylonians will disappoint us, as you’ll soon see, by having devised many ways of evaluating square roots, and producing widely varying answers—that surprisingly accurate one among them. We know that accuracy and application weren’t paramount concerns of theirs,20 but did a variety of answers to a single question mean that for them reality (at least as far as measurement went) had no definite rhythm? Or are we seeing a jumble of different approaches from different times, places, and schools? We’ll look at all of these in the context of a right triangle with short side a, long side b, and diagonal c—whether or not the Babylonians failed to separate such a figure from a completing rectangle; and we’ll test each against that value on YBC 7289 for .

  A fascinating late Old Babylonian fragment (recognized in the 1980s to be made up of two yet smaller pieces in different museums)21 contains two waysaw of trying to find each of a, b, and c, given the other two, with a always equal (in our number system) to 10, b to 40, and c taken as 41¼. Perhaps it tells us more than it meant to.

  The first method takes . This isn’t a particularly good approximation. For a = 10 and b = 40 it does indeed give us c = 41.25, which isn’t far from the correct 41.231 . . . , and for our right triangle with a = 1 and b = 1, comes out as 1.5. But this is the least of the problems with this method. It is meant to yield a and b as well. Given b and c it does indeed produce a = 10. But for b there is a real difficulty: as the problem on the tablet reads (translating to our numbers):

  The breadth is 10, the diagonal is 41¼. What is the height?

  You: No.

  This machine can’t be run backward—at least by them: they would have to have applied the famous quadratic formula (discovered at least two, indeed more like three, millennia later) to get

  This lack of reversibility—and hence of generality—carries a message for us, which we’ll soon read.

  The second method on this tablet is c = b + 2a2b: a very poor approximation indeed, and one with which, again, they don’t even try solving for b—although we know that . For their c = 41.25, this method produces the absurd .2052 . . . , and in a right triangle with legs 1, the hypotenuse would be 3 rather than . Was this formula a scribal error—but if so, for what? Robson generously points out that “its accuracy is dependent on the size of the triangle, as well as its shape . . . the error is less than 10% for triangles with hypotenuse between 0.53 and 0.72.”22 This message goes into the same envelope with the last.

  Since a subtractive companion for the first method above, shows up long after, in Alexandria, some scholars would conjure it up in Babylon.23 It gives the useless estimate ½ for . All three of these methods, however, are one-shot affairs, so were not only makeshifts in the absence of insight, but couldn’t possibly approach what we know are irrational square roots, like . For this, iteration from a first good guess is needed. The Babylonians must have had this approach to get the value for on the Yale tablet (just when you’re ready to give up on them, they startle you with their ingenuity). We find three fascinating candidates if we read back, once again, from what was for them the distant future.

  The most famous of these ways of closing in on any square root is Heron’s Approximation. Make a rough guess, x1, for the square root you seek, and then find your second approximation, x2, by calculating

  How close, and how quickly, will this get us to = 1.414212962, if we start with the reasonable guess x1 = 1, and in general take

  Although for the Old Babylonians this would have involved division into the “irregular” number 17 (a number whose divisors aren’t 2, 3, or 5), they could have managed it, and after five tedious iterations would have gotten 1.414213562: agreement to the first five places. Each further iteration deals with progressively larger fractions, but brings the overestimate steadily down. You wonder, though, just how much in their style is Heron’s Approximation. Why took ye from us our loved Old Babylonian Box?

  A second adventure into post hocery brings us to Theon of Smyrna, who lived a generation after Heron, from about A.D. 70 to 136. You’ll see his clever idea most clearly if you begin with our isosceles right triangle with both legs a1 = 1 (so we’ll call both a1), and suppose for starters that its hypotenuse c1 is also 1.

  Now extend each leg by c1 = 1, giving us legs a2 = 2, and c2 = 2a1 + c1 = 3.

  It may not surprise you that successive approximations take on the same values as Heron’s, but half as fast, so that Theon’s estimates stroll in a more leisurely way toward the Yale tablet’s value for . Can we picture the Babylonian scribes having had the time for these peregrinations? Yes: their society was bureaucratic, but lacked efficiency experts. Will they have had the patience, though, to carry out ever more tedious computations? Surely for each, weariness will at some point have made them hear their mothers calling—without wondering whether or when (this year, next year, sometime, never) the process would conclude. They were very unlikely to have thought of reality itself as shaky: that would have to wait thirty-seven hundred years, for enough angst to kick in. But “approximation” may not have meant to them what it does to us. They may have thought of values differently arrived at the way we think of variant spellings: there is no ideal that ‘color’ or ‘colour’ better realizes. In any event, unlike the problem with triangles whose sides are, say, 10 and 40, the square built on 2 gave them the advantage of always being able to test their latest value simply by squaring it. Once again, however, Theon’s cleverness doesn’t seem theirs (the diagram of it we gave was produced by a nineteenth-century German mathematician).24

  We’ve saved for last the magnificent idea of Ptolemy (who lived about half a generation after Theon, and not all that far away, in Roman Egypt)—not only because it gets the exact value on YBC 7289 in three iterations, but because its method comfortably derives from the Babylonian Box!25

  Say you wanted to find . Make a square box with that side—

  whose area is therefore 4500—and fit into its upper left-hand corner the largest square with an integer side that you can. Since the largest integer with a square less than 4500 is 67, that will be the side of our square, and its area will be 672 = 4489; we draw the square, and see that the area of the remaining gnomon is 4500 − 4489 = 11.

  Now fit into this gnomon a smaller gnomon, hugging the square we just drew in, whose width (let’s call it ) will have x be the largest possible integer that will still keep this new gnomon inside the big, initial square. We obviously haven’t drawn this to scale: we’ve made the 4489 square much too small, in order to leave room for the gnomons coming up.

  Digression: are you bothered that the width is , not just x? Where did that denominator of 60 come from? We�
��re getting successively better approximations to 4500 , having found the first approximation ~ 67. We would now want to find the next decimal place ~67.x , where the integer ‘x’ lies for us in the tenths place, hence stands for the numerator in . But since the Babylonians used 60 instead of 10 as the basis for their place-value system, their successive denominators will be powers of 60.ax

  We repeat this process to find y for the next term: (see, if you like, the upcoming figure). This will be conceptually easy (being the same process as before) but calculationally hard: just what the Old Babylonians relished.

  For what’s left of our original square is a slender gnomon whose area is 4500 − (4489 + the area of the ‘x’ gnomon). With x = 4, its area (including

  Stepping back, you see not only how brilliant but how very Old Babylonian Ptolemy’s method is (like Heron and Theon, he was reputed to have been in touch with old knowledge). Only one last snag has to be removed if we are to apply it—as someone, so long ago, must have—to come up with that little clay tablet’s approximation to .

  For what if, at the very outset, the first square you fitted into the box whose side you wished to calculate was so small as to leave a gnomon with an area greater than or equal to its own? You would be stymied. This is the problem that confronts us in approximating , since the largest square with an integer side that will fit into a square with area 2 has area 1, leaving a gnomon of area 1:

  Babylonian cleverness to the rescue again: instead of dealing in sixtieths, multiply your whole calculation by 60, thus avoiding the problem with too big an initial box. As long as you remember to divide your answer by 60, you’ll be safe.

 

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