Greek Approximations of Square Roots
Solution Commentary:
Solution of Main Problem:
1. The visual would be a square with side lengths 10+2, and look like this:
2
10
10
2
100
20
20
4
For the case when the lesser square is 81 (whose root is 9), we have 144 – 81
= 63. Then, 9 x 2 = 18 and 63/18 = 3 with a remainder of 9 (whose root is 3).
Thus, 144 = 9 + 3 = 12. The associated visual representation is:
3
9
9
81
3
27
27
9
For the case when the lesser square is 64 (whose root is 8), we have 144 – 64
= 80. Then, 8 x 2 = 16 and 80/16 = 5 with a remainder of 0 (whose root is 0). A
problem has arisen as the area of 144 has been partitioned into 3 regions, not 4
regions. Thus, we can repeat our division 80/16 = 4 with a remainder of 16
(whose root is 4), and now everything works out nicely. Thus, 144 = 8 + 4 = 12.
The associated visual representation is:
4
8
8
32
64
4
16
32
2. The appropriate visual is as follows, where the shaded square represents
Theon’s “fudge” at the end.
67 o
67 o
4489 o
4'
55"
268'
3688"40"'
4'
55"
16"
268'
3688"40"'
The actual area of the shaded square is 55” x 55” = 3025”’, while the
remaining area in the extraction process was 46”40’ = 2800”’. Thus, Theon’s error
is 3025”’ – 2800”’ = 225”’ (i.e. 225/216000 = 0.00104166…, which is quite small).
He concludes that 4500 is approximately 67o4’55”.
3. In base ten, the steps would be as follows:
• Start with a lesser square, say 4489, whose root is 67
• Take this square with area 4489 and side 67 from the original square of
area 4500, leaving a gnomon (i.e. l-shaped piece) with area 4500-4489
= 11.
• Multiply 67 by 2 to equal 134, because there are two rectangles forming
the arms of the gnomon
• Divide 11 by 134 to get 0.08 with a remainder of 0.28
• Dividing the area of 10.72 into two rectangles, each with area 0.08 x
134 = 5.36.
• Unfortunately, the remainder of 0.28 is greater than the desired square
with area 0.08 x 0.08 = 0.0.0064. Thus, inside our original square of
area 4500 we have formed visually a square of area 4499.7264 and
side length 67.08 plus a new gnomon with remaining area of 0.28 –
0.0064 = 0.2736.
• Repeat the process by multiplying 67.08 by 2 to equal 134.16 because
there are two rectangles forming the arms of the gnomon
• Divide 0.2736 by 134.16 to get 0.002 with a remainder of 0.00528
• Dividing the area of 0.26832 into two rectangles, each with area 0.002 x
67.08 = 0.13416.
• Theon would then stop the process, noting that the remainder is
0.00528, which is close to a square with side length 0.002. Note: Our
error is 0.00528- (0.002)2 = 0.005276.
The visual representation would be:
0.08
67
67
4489
0.002
5.36
0.13416
0.08
0.002
And Theon would conclude that
0.0064
5.36
0.13416
4500 is approximately 67.082.
Extension 1: In base sixty, the first three “fractional” place values are 1/60, 1/3600,
and 1/216000, while in base ten, they are 1/10, 1/100, and 1/1000. Thus, using base
sixty, the iteration reaches a higher level of precision faster than in base ten.
Extension 2: Suppose that 3 was rational, or equal to m/n where m and n are
whole numbers, n not zero, and gcd(m,n)=1. Then 3 = m2/n2, or m2 = 3n2. This
implies that the m2 has 3 as a factor, but 3 a prime number which implies that m itself
has 3 as a factor. Let m = 3k for k a whole number, which implies by substitution that
m2 = (3k)2 = 9k2 = 3n2, or 3k2 = n2. But this implies that n2 has 3 as a factor,
prompting a contradiction of the assumption that gcd(m,n)=1. Thus, 3 is irrational.
Note: Some students may question the validity of the statement “m2 has 3 as a factor,
but 3 a prime number which implies that m itself has 3 as a factor.” For those
students, work out the three cases of m=3k, m=3k+1, and m=3k+2, By squaring the
terms, contradictions for the latter two cases can be reached quickly.
Suppose that 4 was rational, or equal to m/n where m and n are whole numbers, n
not zero, and gcd(m,n)=1. Then 4 = m2/n2, or m2 = 4n2. An impasses is reached, as
we can continue by assuming that m2 contains 4 as a factor, but that does not imply
that m contains 4 as a factor (i.e. m could have 2 as a factor, implying that m2 has 4
as a factor). The proof ends here, but note that it does not prove that 4 is rational.
Extension 3: The result becomes a spiral of right triangles, whose hypotenuses are
the square roots of 2, 3, 4, 5, 6, ….
Some claim that Theodorus stopped at 17 because the next triangle in the spiral
would overlap the first visually. (e.g. Bar-on & Avital, 1984)
Open-Ended Exploration: The intuitive adjustment is to shift from working with areas
to working withy volumes. The mind-shift now involves the idea of a threedimensional gnomon. This approach is explained further by Dauben (2007), though it
is in the context of Chinese root extraction.
Teacher Commentary:
Before exploring Theon’s algorithm, you may want to discuss two stories:
1. Plato’s story of Socrates and the slave boy, as found in Meno. Text versions
are plentiful (e.g. Sesonske & Fleming, 1965, pp. 17-22) or on-line (e.g.
http://www2.una.edu/dburton/Plato%20Meno%20Socrates%20and%20Slave
%20Boy%20Recollection.htm).
2. The “myth-becomes-fact” story of Pythagoras’ disciple Hippasus, who some
claim was the first to discovery that 2 was irrational. The result: Pythagoras
supposedly sentenced Hippasus to death by drowning, in order to preserve his
ideal world of rational numbers.
As a good source of a writing project, students can explore any of the following ideas
relative to Theon’s algorithm for extracting square roots:
• A plethora of proofs exist for proving the irrationality of the 2 , using a wide
range of mathematical techniques. Investigate these proofs and argue which
is (are) the best. Some good sources are Flannery 2006) and the great
resource http://www.cut-the-knot.org/proofs/sq_root.shtml.
• Investigate Theon’s ladder, a simple way to use the pattern:
1
1
•
•
2
3
5
7
12
17
29
41
...
...
to calculate approximations to 2 . It also is extendable to n . Giberson &
Osler (2004) is a great resource as a start.
Morin (2008) claims: “In ancient Greek times, the study of the science of
Quantity was mainly oriented to find the “Natural Order predetermined by the
mind of the world-creating God” (Nichomacus, Ref. IV). So any trace of
‘Natural Order’ they could find in any numerical method was directly related to
the existence of God, whereas all Trial-&-Error methods represented Chaos
and inability to grasp the very essence of God.” Is Theon’s proof a trial-&error technique that leads to chaos and away from God?
In discussing Theon’s root extraction method, Gow (1968) suggests “that his
method was by no means old or familiar, and we must conclude, therefore, in
default of evidence, that the earlier Greeks found square-roots by experiment
only. The process would certainly take a long time, but we have no reason to
suppose that the Greeks were unwilling to spend a long time on a simple
aritgmetical problem.” Does this suggestion make sense, given the existence
of root extraction methods from Babylonia?
Additional References:
Bar-on, E. and Avital, S. (1984). “What is so special about 17 ?” The Journal of
Computers in Mathematics and Science Teaching. Fall, pp. 20-21.
Chabert, J. (ed). (1999). A History of Algorithms: From the Pebble to the Microchip.
Springer.
Dauben, J. (2007). “Chinese mathematics.” In V. Katz (ed.) The Mathematics of
Egypt, Mesopotamia, China, India, and Islam: A Sourcebook. Princeton University
Press.
Flannery, D. (2006). The Square Root of 2: A Dialogue Concerning a Number and a
Sequence. Copernicus Books.
Giberson, S. and Osler, T. (2004). “Extending Theon’s ladder to any square root.”
The College Mathematics Journal, Vol. 35#3, pp. 222-226.
Gow, J. (1968). A Short History of Greek Mathematics. Chelsea Publishing Company
Reprint.
Morín, D. (2008). The Fifth Arithmetical Operation, Arithmonic Mean. Relevant
portion available at http://mipagina.cantv.net/arithmetic/rmdef.htm
Sesonske, A. and Fleming, N. (eds.) (1965). Plato’s Meno: Text and Criticism.
Wadswoth Publishing Company.