On Lecture 4 - The University of Hong Kong

Transcription

On Lecture 4 - The University of Hong Kong
MATH2001/LE4/MKS/ML/11-12
THE UNIVERSITY OF HONG KONG
DEPARTMENT OF MATHEMATICS
MATH2001: Development of Mathematical Ideas
On Lecture 4 (January 31, 2012)

We began the lecture with a seemingly opposing view from James Joseph Sylvester (1814-1897) on
his experience in studying Euclid’s Elements at school (but also note what he continued to say on the
importance of geometry!)
The French philosopher Bernard Bovier de Fontenelle (1657-1757) went even further to stress the
importance of geometry in other areas of human endeavour.

We then discussed an attempt by the French mathematician Alexis Claude Clairaut (1713-1765) to
reorganize Euclid’s Elements into a more suitable textbook É léments de géométrie (1741; 1753). In
the preface he forsaw the bitter experience most novices may encounter (similar to the experience of
Sylvester). He made use of surveying (land measurement) to motivate various notions and results in
Elements.
To illustrate his point on “searching and discovering” we looked at how he introduced Pythagoras’
Theorem to the readers.

Euclid did not present Pythagoras’ Theorem (Proposition 47 of Book I) this way but deduced it from
the postulates and theorems that come before it. (Question: How does one come up with the
construction line ADG?)
A more “natural” proof (probably how the ancient Greeks arrived at the result) was given later as that
of Proposition 31 of Book VI, from which we may get a hint where the idea of the construction line
ADG comes from.
(Question: Why did Euclid prove the same result twice in Elements? Why did he not prove the result
in Book I as in Book VI? Why not wait till Book VI?)

We continued to discuss different proofs/explanations of Pythagoras’ Theorem so as to exemplify the difference
in mathematical thinking, working and presentation in the Greek (Western) and non-Greek (Eastern) worlds.
 The axiomatic and logical aspect of Euclid’s Elements has long been stressed. We discussed
an alternative feature of the book following the reasoning put forth by S.D. Agashe. Besides
the role played by metric geometry in both the motivation of the design and the presentation,
the algorithmic flavour (so prevalent in the ancient Eastern world as compared to the ancient
Western world) is prominent.
Proposition 14 of Book II says, “To construct a square equal to a given rectilineal figure.” It
seems that the motivating problem of interest is to compare two polygons. The onedimensional analogue of comparing two straight line segments is easy; one simply places one
segment onto the other and checks which segment lies completely inside the other or whether
the two segments are equal to one another. This is in fact what Proposition 3 of Book I tries
to do: “Given two unequal straight lines, to cut off from the greater a straight line equal to the
less.” To justify this result one relies on Postulate 1, Postulate 2 and Postulate 3. The twodimensional case is not as straightforward, except for the special case when both polygons
are squares, in which case one can compare the area through a comparison of the side by
placing the smaller square at the lower left corner of the larger square. Incidentally, one
needs to invoke Postulate 4 in doing that. What Proposition 14 of Book II sets out to do is to
reduce the comparison of two polygons to that of two squares.
Let us look a bit further into the proof of Proposition 14 of Book II. It can be separated into
two steps: (i) construct a rectangle equal (in area) to a given polygon, (ii) construct a square
equal (in area) to a given rectangle. Note that (i) is already explained through Proposition 42,
Proposition 44 and Proposition 45 of Book I, by triangulating the given polygon then
converting each triangle into a rectangle (on a given straight line segment) of equal area, then
finally piling up all the rectangles. Incidentally, one relies on the famous (notorious?)
Postulate 5 on (non)parallelism to prove those results. What about the final kill in (ii)? A
preliminary step is to convert a given rectangle into an L-shaped gnomon of equal area,
which is illustrated in Proposition 5 of Book II that says, “If a straight line be cut into equal
and unequal segments, the rectangle contained by the unequal segment of the whole together
with the square on the straight line between the points of section is equal to the square on the
half.”
Proposition 5 of Book II asserts that a certain rectangle is equal (in area) to a certain gnomon
which is a square ( c 2 ) minus another square ( b 2 ). To accomplish (ii) one wants to construct
a square ( a 2 ) equal to the difference of two squares ( c 2  b2 ), or equivalently, the square ( c 2 )
is a sum of two squares ( a 2  b2 ). This leads naturally to the famous Pythagoras’ Theorem,
which is Proposition 47 of Book I. The Pythagoras’ Theorem epitomizes the relationship
between shape and number, between geometry and algebra.