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Historia Mathematica 41 (2014) 103–106
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Cube roots of integers. A conjecture about
Heron’s method in Metrika III. 20
Christian Marinus Taisbak
Copenhagen University, Denmark
Available online 17 December 2013
Abstract
Did Heron (or his teachers) use sequences of differences to find an approximate value of the cube root of an integer? I venture a
conjecture of his heuristics and a couple of possible mathematical proofs of his method.
© 2013 Elsevier Inc. All rights reserved.
MSC: 01A20; 01A03; 11B25; 40A03
Keywords: Heron; Sequences; Differences; Cube root
In Metrika III. 20 Heron describes how to find an approximation to the cube root of 100. In A History of
Greek Mathematics, vol. II, p. 341, Thomas Heath translated:
Take the nearest cube numbers to 100 both above and below; these are 125 and 64.
Then 125 − 100 = 25, and 100 − 64 = 36.
Multiply 5 into 36; this gives 180. {Multiply 4 into 25; this gives 100}.1
Add 100, making 280. {Divide 180 by 280}2; this gives 9/14. Add this to the side of the smaller cube; this
gives 4 914 . This is as nearly as possible the cube root (“cubic side”) of 100 units.
There cannot be much doubt that the method described can be presented by the formula, first (?) suggested
by Wertheim in 1899 (according to Heath, A History of Mathematics, vol. II): Generally, if a3 < A <
(a +1)3, suppose that A−a3 = d1, and that (a +1)3 −A = d2. Then Heron’s formula for the approximated
cube root was
a + [(a + 1)d1
]
/
[
(a + 1)d1 + ad2
]
E-mail address: marinus@mail.dk.
1 My addition.
2 Heath’s addition.
0315-0860/$ – see front matter © 2013 Elsevier Inc. All rights reserved.
http://dx.doi.org/10.1016/j.hm.2013.10.003
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mailto:marinus@mail.dk
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104 C.M. Taisbak / Historia Mathematica 41 (2014) 103–106
1. Heuristics
So far I follow Thomas Heath (and Wertheim). But how did Heron arrive at this expression? I venture
a conjecture, based on sequences of differences. In the table below, N increases by the constant differ-
ence �x = 1, while N3 increases by increasing first and second differences, �y and �y2. Only the third
difference sequence, �y3, is constant.
�x 1 1 1 1 1 1 1 1
x N 1 2 3 4 5 6 7 8 9
y N3 1 8 27 64 125 216 343 512 729
�y 7 19 37 61 91 127 169 217
�y2 (6) 12 18 24 30 36 42 48
�y3 6 6 6 6 6 6 6
Let us suppose that Heron (or another Greek mathematician), whether concerned about the sequences �y2
and �y3 or not, investigated the sequence �y, looking for ratios. (I believe that Greek mathematicians
were always and every day on the look-out for ratios.) A trained mind might see the following ratios of
successive first differences (the symbol >≈ denotes “is slightly greater than”):
7 : 19 >≈ 1 : 3
19 : 37 >≈ 1 : 2(= 2 : 4)
37 : 61 >≈ 3 : 5
61 : 91 >≈ 2 : 3(= 4 : 6)
91 : 127 >≈ 5 : 7
127 : 169 >≈ 3 : 4(= 6 : 8)
169 : 217 >≈ 7 : 9
If instead of the least couples (e.g. 2 : 3, cf. Euclid’s Elements VII.20) Heron decided for the least but one
(e.g. 4 : 6, in parentheses), he could not help observe that the members of the ratio are the nearest x’es to
the left and right of the couple �y’s considered, one x being skipped. What does that mean? Could it tell
something about cube roots between integers? I think it could.
An example: Between 27 (33) and 125 (53) there is one cube of an integer, 64 (43). The total �y-value
is 37 + 61 = 98, but the two summands must be weighted differently, since each of them produces (or is
produced by) an increase of 1 in the �x sequence. The 37 (let us call them sinistra) weigh as much as
the 61 (dextra) – and the list of ratios above tells us how much: 3 sinistra count as much as 5 dextra. To
compare those values we may introduce a common weight – say: a pondus, and define one sinistrum =
five pondera, and one dextrum = three pondera. If we use the pondera to measure �y, we find that the
increase from 33 to 53 is 185 + 183 pondera, so as to create equal increases in �x. (We must accept a
small deficiency of 2 in the lesser interval – caused by the approximations in the table of ratios, but that
deficiency will have decreasing weight as the numbers grow unlimited.) Thus, the interval from 33 = 27 to
53 = 125 is 368 pondera, and the interval from 27 to 64, measured in pondera, is about half of that. We
conclude that the cube root of 64 is also about half-way between 3 and 5, thus “approximately” 4.
Now it’s almost obvious what Heron is doing when he finds the cube root of 100, that is: He looks for
a non-integral root between 4 and 5. The lower �y-value (100 − 64) is 36, and the upper (125 − 100) is
25. To calculate the local pondus he takes each of the 36 to be worth 5 pondera (from the upper x-value),
C.M. Taisbak / Historia Mathematica 41 (2014) 103–106 105
and each of the 25 to be worth 4 pondera (from the lower x-value), by analogy with the example analyzed
above. That is:
The pondera in the interval (100 − 64) are 5 times 36 = 180, and in the interval (125 − 100) are 4 times
25 = 100, in total 280. Therefore the value of the cube root of 100 is 4 + 180/280 = 4 + 9/14 (only 1.27
per thousand too great).
Another example (they are legion) will corroborate the method: Calculate the cube root of 250. Nearest
cubes are 216 (63) and 343 (73). The lower �y is 250 − 216 = 34, the upper �y is 343 − 250 = 93. The
local pondera are 7 times 34 = 238 and 6 times 93 = 558, in total 796. Therefore the value of the cube
root of 250 is 6 + 238/796 = 6 + 119/398 >≈ 6 + 17/57. Extra: By dividing by 5 you get the cube root of
2 <≈ 1 + 1/4 + 1/100, an interesting result for cube-doublers.
2. Proofs
With all deference to G. Eneström and his proof of Wertheim’s formula (by “quite elementary consid-
erations”, Heath, A History of Mathematics, vol. II), it is safe to say that the “elementary considerations”
probably were out of reach for Heron, in so far as they seem to depend on algebraic symbols and methods
that were outside the scope of Greek mathematics at the time. Moreover, Heron did not need any other
corroboration than the fact that the method works, and that the separate results are easily confirmed by
multiplication. I suggest proofs (hardly necessary) inspired by doings with ratios in Euclid’s style, even if
presented in algebraic terms. I am sure that Diophantos (for one) could put it into Greek prose. (Again,
please read the symbol >≈ “is slightly greater than”.)
First part: Let a − 1, a, a + 1 be three successive positive integers. To find the ratio between the differ-
ences of their cubes.
Assertion: If S = a3 − (a − 1)3 and D = (a + 1)3 − a3 then S : D >≈ a − 1 : a + 1.
Proof:
a3 − (a − 1)3 = a3 − (a3 − 3a2 + 3a − 1) = 3a2 − 3a + 1 = 3a(a − 1) + 1 = S.
(a + 1)3 − a3 = a3 + 3a2 + 3a + 1 − a3 = 3a2 + 3a + 1 = 3a(a + 1) + 1 = D.
If we ignore the +1, which becomes even more insignificant for increasing a, and cancel 3a, then (since D
is the greater) we conclude that S : D >≈ a − 1 : a + 1.
Second part: To determine an approximate cube root of a non-cube integer A. Suppose it to be a, whose
integer neighbors are a − f and a + g (where f and g are numbers < 1).
Assertion: If P = a3 − (a − f )3 and Q = (a + g)3 − a3 then P : Q >≈ f (a − f ) : g(a + g).
Proof:
a3 − (a − f )3 = a3 − (a3 − 3a2f + 3af 2 − f 3) = 3a2f − 3af 2 + f 3 = 3af (a − f ) + f 3 = P.
(a + g)3 − a3 = (a3 + 3a2g + 3ag2 + g3) − a3 = 3a2g + 3ag2 + g3 = 3ag(a + g) + g3 = Q.
If we ignore the +f 3 and +g3, which are insignificant cubes of numbers < 1, we have
P : Q >≈ f (a − f ) : g(a + g).
Using the concept of “compound ratio” [equivalent to multiplying by (a + g) : (a − f )] we get
(a + g)P : (a − f )Q >≈ f : g.
106 C.M. Taisbak / Historia Mathematica 41 (2014) 103–106
Since all numbers on the left side are given (Heron: a + g = 5, a − f = 4, P = 36, Q = 25), the ratio
f : g is given (Heron, 180 : 100), and therefore the ratiof : (f + g) (Heron, 180 : 280). Since f + g = 1,
f is given as a fraction < 1 (Heron, 180/280 = 9/14), and so a is given.
3. Envoi
Is this conjecture historically acceptable? Did the Ancients know and use sequences of differences? At
least I might claim it to be a possible derivation, because the “so-called Pythagoreans” (to quote Aris-
totle) took an interest in gnomons, particularly square gnomons, i.e. differences between squares. It was
well known that the increments constitute an arithmetical progression with constant difference 2, in fact
successive odd numbers (Heath, A History of Mathematics, I.77).
Further, Hypsicles (150–120 B.C.), in his Anaphorikos assumes that rising times of stars form a linear
sequence, increasing and decreasing with constant difference (O. Neugebauer, HAMA 715). In fact, it is
tempting to ascribe Heron’s procedure to Hypsicles, who was a competent mathematician and particularly
known for his treatise on regular solids (book xiv of the Elements).
	Cube roots of integers. A conjecture about Heron's method in Metrika III. 20
	1 Heuristics
	2 Proofs
	3 Envoi