Euclidean geometry, distance and quadrance
While distance is a practically useful linear quantity, the corresponding quadratic quantity is often mathematically more fundamental. Our understanding of distance squared, or quadrance, goes back to ancient Greek mathematics and Euclid’s formulation of Pythagoras’ theorem, and forms a bridge to an exciting development of modern geometry.
In this step, you will

learn how the ancient Greeks viewed Pythagoras’ theorem using quadrance rather than length

see how irrationalities can be avoided with Rational Trigonometry.
Pythagoras’ theorem
A right triangle is a triangle with one right (90 degree) angle. Pythagoras’ theorem states that a square built on the hypotenuse of a right triangle (the triangle’s longest side) has an area which is exactly the sum of the areas of the squares built on the shorter two sides of the triangle.
This is actually how Euclid stated it in Proposition 47 of Book 1 of Euclid’s “The Elements”, presented by David Joyce — in terms of areas of squares, rather than squares of lengths of sides. The reason Euclid preferred this language was that it had the big advantage of avoiding irrationalities.
Avoiding irrationalities
In the triangle \(\overline{\normalsize{A_1A_2A_3}}\) shown, the lines \(\normalsize{A_1A_3}\) and \(\normalsize{A_2A_3}\) are aligned with the grid lines, so we can easily count that the lengths of the segments \(\overline{\normalsize{A_2A_3}}\) and \(\overline{\normalsize{A_1A_3}}\) are \(\normalsize{3}\) and \(\normalsize{5}\) respectively. However the length of the hypotenuse segment \(\overline{\normalsize{A_1A_2}}\) leads to the irrational quantity
\[\Large{\sqrt{3^2+5^2}=\sqrt{34}=5. 830951894 8453004709...}\]which we can never completely determine. It follows that even the “existence” of an “irrational number” such as \(\normalsize{\sqrt{34}}\) is computationally problematic.
Note however that the calculations of the areas of the three squares is an elementary matter – this is why the ancient Greeks preferred to measure segments by constructing squares on them, a process they called quadrature. Following the Greeks, we define the quadrance of a segment, denoted usually by \(\normalsize{Q}\), to be the area of a square built on it.
It is easy to count areas to see that the quadrances \(\normalsize{Q_1}\) and \(\normalsize{Q_2}\) in the diagram above are \(\normalsize{9}\) and \(\normalsize{25}\), respectively. The area \(\normalsize{Q_3}\) of the square on the hypotenuse \(\normalsize{A_1A_2}\) is a bit trickier to determine, but in the first diagram below, you should be able to piece together pairs of triangles into rectangles and count that the total area, or quadrance, is in fact \(\normalsize{34}\). And this is also the sum of \(\normalsize{9+25}\), just as Pythagoras taught us.
Q1 (M): Use the second picture, with \(\normalsize{Q_3}\) inscribed in a larger square with known side lengths, to give an alternate derivation of \(\normalsize{Q_3=34}\).
Rational trigonometry
Quadrance is an important quadratic concept. It scales quadratically, so that if we double a segment, its quadrance multiplies by \(\normalsize{4}\). From a theoretical point of view, the precision of working with quadrance outweighs the disadvantage of dealing with a quadratic (rather than a linear) relation.
There is also an associated rational analog of angle called the spread between lines. This approach to trigonometry was introduced by Norman in his 2005 book Divine Proportions: Rational Trigonometry to Universal Geometry. The first chapter is quite readable for high school students, you can have a look at Divine Proportions book.
Discussion
Pythagoras’ theorem is easily the most important result in mathematics. All students should know at least one proof. How about you? If someone asked you: “Why is Pythagoras’ theorem true?” what would be your answer?
Answers
A1. The big square has area \(\normalsize{(3+5)^2=64}\), and two of the opposite triangles fit together to give a \(\normalsize{3\times5}\) rectangle with area \(\normalsize{15}\). There are two of those, so the remaining area \(\normalsize{Q_3}\) must be \(\normalsize{642\times15=34}\).
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