The Teaching of Geometry Part 25

Besides Proclus, Plutarch testifies to the fact that Pythagoras was the discoverer, saying that "Pythagoras sacrificed an ox on the strength of his proposition as Apollodotus says," but saying that there were two possible propositions to which this refers. This Apollodotus was probably Apollodorus, surnamed Logisticus (the Calculator), whose date is quite uncertain, and who speaks in some verses of a "famous proposition" discovered by Pythagoras, and all tradition makes this the one. Cicero, who comments upon these verses, does not question the discovery, but doubts the story of the sacrifice of the ox. Of other early writers, Diogenes Laertius, whose date is entirely uncertain (perhaps the second century A.D.), and Athenaeus (third century A.D.) may be mentioned as attributing the theorem to Pythagoras, while Heron (first century A.D.) says that he gave a rule for forming right triangles with rational integers for the sides, like 3, 4, 5, where 3^2 + 4^2 = 5^2. It should be said, however, that the Pythagorean origin has been doubted, notably in an article by H. Vogt, published in the _Bibliotheca Mathematica_ in 1908 (Vol. IX (3), p. 15), ent.i.tled "Die Geometrie des Pythagoras," and by G. Junge, in his work ent.i.tled "Wann haben die Griechen das Irrationale entdeckt?" (Halle, 1907). These writers claim that all the authorities attributing the proposition to Pythagoras are centuries later than his time, and are open to grave suspicion. Nevertheless it is hardly possible that such a general tradition, and one so universally accepted, should have arisen without good foundation. The evidence has been carefully studied by Heath in his "Euclid," who concludes with these words: "On the whole, therefore, I see no sufficient reason to question the tradition that, so far as Greek geometry is concerned ..., Pythagoras was the first to introduce the theorem ... and to give a general proof of it." That the fact was known earlier, probably without the general proof, is recognized by all modern writers.

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Pythagoras had studied in Egypt and possibly in the East before he established his school at Crotona, in southern Italy. In Egypt, at any rate, he could easily have found that a triangle with the sides 3, 4, 5, is a right triangle, and Vitruvius (first century B.C.) tells us that he taught this fact. The Egyptian _harpedonaptae_ (rope stretchers) stretched ropes about pegs so as to make such a triangle for the purpose of laying out a right angle in their surveying, just as our surveyors do to-day. The great pyramids have an angle of slope such as is given by this triangle. Indeed, a papyrus of the twelfth dynasty, lately discovered at Kahun, in Egypt, refers to four of these triangles, such as 1^2 + (3/4)^2 = (1 1/4)^2. This property seems to have been a matter of common knowledge long before Pythagoras, even as far east as China.

He was, therefore, naturally led to attempt to prove the general property which had already been recognized for special cases, and in particular for the isosceles right triangle.

How Pythagoras proved the proposition is not known. It has been thought that he used a proof by proportion, because Proclus says that Euclid gave a new style of proof, and Euclid does not use proportion for this purpose, while the subject, in incomplete form, was highly esteemed by the Pythagoreans. Heath suggests that this is among the possibilities:

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[triangles]_ABC_ and _APC_ are similar.

[therefore] _AB_ _AP_ = (_AC_)^2.

Similarly, _AB_ _PB_ = (_BC_)^2.

[therefore] _AB_(_AP_ + _PB_) = (_AC_)^2 + (_BC_)^2,

or (_AB_)^2 = (_AC_)^2 + (_BC_)^2.

Others have thought that Pythagoras derived his proof from dissecting a square and showing that the square on the hypotenuse must equal the sum of the squares on the other two sides, in some such manner as this:

[Ill.u.s.tration: FIG. 1]

[Ill.u.s.tration: FIG. 2]

Here Fig. 1 is evidently _h_^2 + 4 [triangles].

Fig. 2 is evidently _a_^2 + _b_^2 + 4 [triangles].

[therefore] _h_^2 + 4 [triangles] = _a_^2 + _b_^2 + 4 [triangles], the [triangles] all being congruent.

[therefore] _h_^2 = _a_^2 + _b_^2.

The great Hindu mathematician, Bhaskara (born 1114 A.D.), proceeds in a somewhat similar manner. He draws this figure, but gives no proof. It is evident that he had in mind this relation:

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_h_^2 = 4 _ab_/2 + (_b_ - _a_)^2 = _a_^2 + _b_^2.

A somewhat similar proof can be based upon the following figure:

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If the four triangles, 1 + 2 + 3 + 4, are taken away, there remains the square on the hypotenuse. But if we take away the two shaded rectangles, which equal the four triangles, there remain the squares on the two sides. Therefore the square on the hypotenuse must equal the sum of these two squares.

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It has long been thought that the truth of the proposition was first observed by seeing the tiles on the floors of ancient temples. If they were arranged as here shown, the proposition would be evident for the special case of an isosceles right triangle.

The Hindus knew the proposition long before Bhaskara, however, and possibly before Pythagoras. It is referred to in the old religious poems of the Brahmans, the "Sulvasutras," but the date of these poems is so uncertain that it is impossible to state that they preceded the sixth century B.C.,[79] in which Pythagoras lived. The "Sulvasutra" of Apastamba has a collection of rules, without proofs, for constructing various figures. Among these is one for constructing right angles by stretching cords of the following lengths: 3, 4, 5; 12, 16, 20; 15, 20, 25 (the two latter being multiples of the first); 5, 12, 13; 15, 36, 39; 8, 15, 17; 12, 35, 37. Whatever the date of these "Sulvasutras," there is no evidence that the Indians had a definite proof of the theorem, even though they, like the early Egyptians, recognized the general fact.

It is always interesting to a cla.s.s to see more than one proof of a famous theorem, and many teachers find it profitable to ask their pupils to work out proofs that are (to them) original, often suggesting the figure. Two of the best known historic proofs are here given.

The first makes the Pythagorean Theorem a special case of a proposition due to Pappus (fourth century A.D.), relating to any kind of a triangle.

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Somewhat simplified, this proposition a.s.serts that if _ABC_ is _any_ kind of triangle, and _MC_, _NC_ are parallelograms on _AC_, _BC_, the opposite sides being produced to meet at _P_; and if _PC_ is produced making _QR_ = _PC_; and if the parallelogram _AT_ is constructed, then _AT_ = _MC_ + _NC_.

For _MC_ = _AP_ = _AR_, having equal bases and equal alt.i.tudes.

Similarly, _NC_ = _QT_.

Adding, _MC_ + _NC_ = _AT_.

If, now, _ABC_ is a right triangle, and if _MC_ and _NC_ are squares, it is easy to show that _AT_ is a square, and the proposition reduces to the Pythagorean Theorem.

The Arab writer, Al-Nair[=i]z[=i] (died about 922 A.D.), attributes to Th[=a]bit ben Qurra (826-901 A.D.) a proof substantially as follows:

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The four triangles _T_ can be proved congruent. Then if we take from the whole figure _T_ and _T'_, we have left the squares on the two sides of the right angle. If we take away the other two triangles instead, we have left the square on the hypotenuse.

Therefore the former is equivalent to the latter.

A proof attributed to the great artist, Leonardo da Vinci (1452-1519), is as follows:

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The construction of the following figure is evident. It is easily shown that the four quadrilaterals _ABMX_, _XNCA_, _SBCP_, and _SRQP_ are congruent.

[therefore] _ABMXNCA_ equals _SBCPQRS_ but is not congruent to it, the congruent quadrilaterals being differently arranged.

Subtract the congruent triangles _MXN_, _ABC_, _RAQ_, and the proposition is proved.[80]

The following is an interesting proof of the proposition:

Let _ABC_ be the original triangle, with _AB_ < _bc_.="" turn="" the="" triangle="" about="" _b_,="" through="" 90,="" until="" it="" comes="" into="" the="" position="" _a'bc'_.="" then="" because="" it="" has="" been="" turned="" through="" 90,="" _c'a'p_="" will="" be="" perpendicular="" to="" _ac_.="" then="">

1/2(_AB_)^2 = [triangle]_ABA'_,

and 1/2(_BC'_)^2 = [triangle]_BC'C_,

because _BC_ = _BC'_.

[therefore] 1/2((_AB_)^2 + (_BC_)^2) = [triangle]_ABA'_ + [triangle]_BC'C_.

[therefore] 1/2((_AB_)^2 + (_BC_)^2) = [triangle]_AC'A'_ + [triangle]_A'C'C_