For me, it has always been satisfying to be able to demonstrate, to myself and my students, a geometrical foundation beneath many algebraic ideas. So I was pleased to notice a single diagram which tended to illustrate several introductory algebraic ideas. I begin with some history notes since it was in a history book that I found the image.
It seems pretty well known and agreed that, although the well known theorem takes its name from the ancient Greek mathematician Pythagoras (569 B.C.?-500 B.C.?) born on the island of Samos in Greece, it was known well before Pythagoras walked the earth in perhaps four cultures.
There is an Egyptian papyrus called Berlin 6619, written between 2000 and 1786 BC, which includes a problem whose solution is a Pythagorean triple.
There are also several Babylonian tablets in museums and universities around the world which have some connection with Pythagoras's theorem. Although there are still a few math historians who disagrees, most maintain that the Babylonians were familiar with Pythagoras's theorem as early as 1900 BCE, over 1000 years before his birth.
A translation of a Babylonian tablet which is preserved in the British museum goes as follows:-
4 is the length and 5 the diagonal. What is the breadth ?
Its size is not known.
4 times 4 is 16.
5 times 5 is 25.
You take 16 from 25 and there remains 9.
What times what shall I take in order to get 9 ?
3 times 3 is 9.
3 is the breadth.
By about 800 years before the common era, there were Sulba Sutras which mention methods of using the "Pythagorean" Theorem for constructing alters for religious practice, with names for diagonals to construct the doubler (dvi-karani), and tripler of a given square. I will return to one of these constructins in a moment since it prompted this whole post. The most important of these documents are the Baudhayana Sulbasutra written about 800 BC and the Apastamba Sulbasutra written about 600 BC, again, all before the presumed dates of Pythagoras.
The Zhou Bi Suan Jing, or Chou Pei Suan Ching, is one of the oldest and most famous Chinese mathematical texts. The title literally means The Arithmetical Classic of the Gnomon and the Circular Paths of Heaven. This book dates from the period of the Zhou Dynasty (1046 BCE—256 BCE), yet its compilation and addition of materials continued into the Han Dynasty (202 BCE – 220 CE). One of the elements in the book, which is generally believed to predate Pythagoras by at least 400 years, is itself a geometric proof of the theorem.
But on to the point of interest. While recently reading Kim Plofker's excellent Mathematics in India, I came across the following which is included, if I read his notes correctly, in the Baudhayana Subasutra, about 250 years before Pythagoras and another 250 before Euclid.
The texts contain a method of finding the square whose area is equal to the sum of two other squares, and also the difference of two squares. There were no diagrams in the sutras, just verse, but fortunately Professor Plofker shows both constructions in a single diagram. (although my interest is primarily in the second).
Here is my recreation of the Image from his book (figure 2.3, page 22) with a few additional lines:
The figure shows that BE is the hypotenuse of a triangle with sides BC ,a side of square B in Green, and CE, a side of square A in Yellow. So the square constructed on BE has an area that is the sum of the areas of squares A and B by the Pythagorean Theorem.
But if we construct an arc from point B that is equal to a side of square A, then the length at BN is the same length. And looking again at BC, we see that the leg CN completes the right triangle, thus \(BN^2 = BC^2 + CN^2 \) or expressed as the difference of two squares, we have that \( BN^2 - BC^2 = CN^2 \) Thus the difference of the area of squares A and B is given by the square on CN.
With a brief expansion of the sketch above to complete the semi-circle of radius BH, we are ready to add a foundation to a common algebra relationship, \(x^2 - y^2 = (x+y)(x-y) \)
Euclid's Book II proposition XIV To construct a square equal to a given rectilinear figure is very brief because it leans on two previous proofs, Book I, Prop 47 (The Pythagorean Thm.) and Book II, Prop 5, Constructing a square equal to the difference of two squares.
If the diagram is extended to complete the semicircle of radius equal to a side of square A, we can add the illustration of all three ideas. An all important right triangle inscribed in the semi-circle is also added.
Because the radius of circle is equal to the side of square A, the radii BJ and BM are both equal to a (the length of a side of Square A) . So the segment JC is just JB + BJ (or a + b). The segment CM, which is BM - BC just becomes (a-b) .
At this point we simply return to the Pythagorean Theorem , and the three similar triangles that make up right triangle JNM and we can prove that the altitude to the right angle of a right triangle has a length that when squared, will equal the product of the two sections it divides the hypotenuse into....(wow, no wonder we like math notation, all that to say \( CN^2 = CM*CJ \) ) and indeed, \(ax^2 - b^2 = (a+b)(a-b) \) .