Geometry for multiplication, division, and roots

Contemporary plane geometry of the sort taught in the standard American high school is most heavily informed by two books and a third mathematician. The first of these is Euclid’s Elements, which is so conceptually tied to planar geometry that it is typically referred to as Euclidean geometry. However, it is only part of the story to suggest that we teach Euclid’s geometry.

I have discussed the al-Jabr of al-Khwarizmi in earlier posts. This was the application of geometry to lay the groundwork of algebra. Rene Descartes took the opposite approach, applying algebra to geometry, in his Geometry. It is largely because of Descartes and the coordinate system that bears his name that geometry has become infused so deeply with algebra.

The third major influence on high school geometry is David Hilbert, who sought to redo Euclid’s axioms more rigorously.

However, it is Descartes that is the interest of this post.

In his first examples, he illustrates the difference between the geometric mind and the algebraic mind. While the algebraic mind tends to view numbers in terms of quantities, the geometric mind views them in terms of lengths. Let us say we wanted to multiply 5 and 3. The standard ways we’re taught in elementary school including making a grid of five rows and three columns (or vice versa), or making five piles of three objects each (or vice versa) and counting the discrete objects.
Here is how the concepts of multiplication and division are seen in geometry, according to Descartes. Let us take the length of AB to be the unit. Measure BD and BC as the numbers to multiplied. Draw AC, then draw DE parallel to AC. Then the length of BE is the product of the lengths of BD and BC.

This diagram and the proof of the relationships between these lengths appears in a standard geometry book today: \[\frac{BD}{BA} = \frac{BE}{BC}\]

If \(BA = 1\), then \(BE = BC \times BD\). But the typical presentation of this is either as a novelty or as a consequence of \(\Delta BAC ~ \Delta BDE\), not in terms of its conceptual significance. For Descartes, this examples both bridges algebra and geometry, and illustrates the difference between their approaches.

Descartes’s second example involves drawing a square root. Let us say we want to find the square root of some length, GH. First, draw a line segment FG which is one unit long. Next, locate K, which is the midpoint of FH. Draw a semicircle with a center of K and FH as a diameter. Finally, draw a line perpendicular to FH through G. The length of IG will be the square root of the length of GH.

This example can be proven algebraically. Let \(GH = x\) and draw KI, which will be a radius of circle K. Then we have these lengths: \[FG = 1 \\ FH = x + 1 \\ KF = KI = \frac{x+1}{2} \\ KG = \frac{x+1}{2} – 1 = \frac{x-1}{2} \]

We now have \(\Delta KGI\), which is a right triangle with legs KG and GI and hypotenuse IK. Using the Pythagorean Theorem, we see that \[IG = \sqrt{\left(\frac{x+1}{2}\right)^2 – \left(\frac{x-1}{2}^2\right)} \\ = \frac{\sqrt{(x^2 + 2x + 1) – (x^2 – 2x + 1)}}{2} = \frac{\sqrt{4x}}{2} = \sqrt{x} \]

This example does not give us a specific, measurable value for square roots: We cannot do so, because with the exception of perfect squares, integers have irrational roots. What it gives us is a visual way to estimate square roots using standard geometric techniques. If we were creating a number line, for instance, with the roots marked off, we could use this technique to do so.

Side note: There’s another technique for locating the square roots. This technique is iterative and hence accumulates error when constructed by hand, but it also reflects another useful concept about square roots:

  1. Locate (1, 0). This is \(\sqrt{1}\), the active square root.
  2. Draw a line perpendicular to the \(x\)-axis through the active square root.
  3. Locate the point one unit up along this line (i.e., (x, 1), where x is the active square root).
  4. Draw an arc starting with the center through the origin and (x, 1) on the circle, then down to the x-axis.
  5. The x-intercept of this arc will be the next integer’s square root (starting with \(\sqrt{2}\)).
  6. Go back to step 2 and repeat.

This works because step 3 creates a right triangle with a leg of 1 and a leg of \(\sqrt{x}\), meaning that the hypotenuse will have a length of \(\sqrt{x+1}\). Note if you do this three times, you’ll get to 2; do it another five times, you’ll get to 3, and so on. This reinforces the crucial relationship between triangles and circles, perhaps even more than Descartes’s technique.


Leave a Comment

Your email address will not be published. Required fields are marked *