Greek Approximations of Square Roots

Transcription

1 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: For the case when the lesser square is 81 (whose root is 9), we have = 63. Then, 9 x 2 = 18 and 63/18 = 3 with a remainder of 9 (whose root is 3). Thus, 144 = = 12. The associated visual representation is: For the case when the lesser square is 64 (whose root is 8), we have = 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 = = 12. The associated visual representation is:

2 The appropriate visual is as follows, where the shaded square represents Theon s fudge at the end. 67 o 4' 55" 67 o 4489 o 268' 3688"40"' 4' 268' 16" 55" 3688"40"' The actual area of the shaded square is 55 x 55 = 3025, while the remaining area in the extraction process was = Thus, Theon s error is = 225 (i.e. 225/ = , which is quite small). He concludes that 4500 is approximately 67 o 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 = 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 into two rectangles, each with area 0.08 x 134 = Unfortunately, the remainder of 0.28 is greater than the desired square with area 0.08 x 0.08 = Thus, inside our original square of area 4500 we have formed visually a square of area and

3 side length plus a new gnomon with remaining area of = Repeat the process by multiplying by 2 to equal because there are two rectangles forming the arms of the gnomon Divide by to get with a remainder of Dividing the area of into two rectangles, each with area x = Theon would then stop the process, noting that the remainder is , which is close to a square with side length Note: Our error is (0.002) 2 = The visual representation would be: And Theon would conclude that 4500 is approximately 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 = m 2 /n 2, or m 2 = 3n 2. This implies that the m 2 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 m 2 = (3k) 2 = 9k 2 = 3n 2, or 3k 2 = n 2. But this implies that n 2 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 m 2 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 = m 2 /n 2, or m 2 = 4n 2. An impasses is reached, as we can continue by assuming that m 2 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 m 2 has 4 as a factor). The proof ends here, but note that it does not prove that 4 is rational.

4 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 ) or on-line (e.g. %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 Investigate Theon s ladder, a simple way to use the pattern: 1 1

5 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 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 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

6 Sesonske, A. and Fleming, N. (eds.) (1965). Plato s Meno: Text and Criticism. Wadswoth Publishing Company.