history of mathematics - National STEM Centre

An Arabic commentary by
al-Naziri on Heron's work
quotes Euclid's Elements,
Book II, propositions 1, 3
and 2 in support of the
results shown as (1), (3) ,„
and (5), respectively. Jjjj
5 Arab mathematics
Given: BL = BP + PL where P is the mid-point of DL.
To show that: BL x DB + (BP) 2 = (DP) 2
Proof: DB x BL = DB x (BP + PL). (1)
Add (BP) 2 to both sides of this equation and substitute DP for PL to get
DP x DB + BP x DB + (BP) = BL x DB + (BP) 2 .
(2)
But BP x DB + (BP) 2 = (DB + BP) x BP = DP x BP
Combining the last two equations gives
DP x BP + DP x DB = BL x DB + (BP) 2
Also DP x BP + DP x DB = DP x (BP + DB) = (DP) 2 .
Combining the last two equations gives the final result
BLxDB + (BP) 2 =(DP) 2
To continue with Abu Kamil's solution, in Figure 5.9a, BL x DB + (BP) 2 = (DP) 2
where (DP) 2 = 5 2 = 25, and BL x DB = area of ABLH = 21.
So(BP) 2 =4orBP = 2.
Therefore DP - BP = DB = x - 5 - 2 = 3 or the area of square ABDG is 9.
To find the other root of x, which is the side of square ABDG in Figure 5.9b, a
similar argument, using another result from Euclid, gives
WBxBD + (BP) 2 =(PD) 2 where (PD) 2 =25, WBxBD = 21 and (BP) 2 =4.
So PD + BP = BD = x = 5 + 2 = 1 or the area of square ABDG is 49.
3 Express Euclid's rule algebraically and explain why it works.
4 Describe the two solutions in your own words.
5 What differences do you notice between the two approaches?
Solution of cubic equations
In his book Algebra, Omar Khayyam explored the possibility of using parts of
intersecting conies to solve cubic equations. Traces of such approaches are found in
the works of earlier writers such as Menaechmus (circa 350 BC) and Archimedes
(287-212 BC), and Omar Khayyam's near contemporary al-Hayatham (circa 965-
1039). They observed, for quantities a, b, c and d, that if — = — = — , then
2 ( c \( d \ c c d a
= \ — \\ — \- — a or c 3 = b 2a . If b = 1 , you y can calculate the cube root of a if c
and d exist so that c
2
= d and d = ac .
Omar Khayyam re-discovered and extended the geometric argument implicit in this
algebra. If you think of c and d as variables and of a as a constant, then the
equations c 2 = d and d 2 = ac are the equations of two parabolas with perpendicular
axes and the same vertex. This is illustrated in Figure 5.10. The two parabolas have
the same vertex B, with axes AB = a and CB = b = 1 , and they intersect at E. In the
rectangle BDEF, BF = DE = c and BD — FE = d . Since AB is a line segment and
the point E lies on the parabola with vertex B and axis AB, the rectangle BDEF has
the property that (FE) = AB x BF or d 2 = ac .
(3)
(4)
(5)
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