# janmr blog

## Arithmetic by Geometry April 24, 2010

Today real numbers are most often represented by applying (elementary) functions to (decimal) integers. Throughout history, though, arithmetic and propositions involving (positive) real numbers were often considered from a purely geometrical point of view. Real numbers were identified by the length of some line segment and, e.g., the product of two numbers was identified by the area of a rectangle with side-lengths equal to the two numbers. This made sense from a physical/applied point of view, but it had certain shortcomings.

According to Hawking‘s God Created the Integers, this was changed by René Descartes in the seventeenth century. He treated any (positive) real number as the length of a line segment, also products, square roots, and so on. This article considers some of the constructions he considered in Problems the Construction of Which Requires Only Straight Lines and Circles (included and commented in Hawking's book), namely multiplication, division and root extraction.

### Multiplication

Let A be a point in the Euclidean plane and let two non-parallel lines have one end-point at A and otherwise extend infinitely. Place now the points B and D on one of the lines and C and E on the other, with the restriction that the lines BC and DE should be parallel. See Figure 1.

We see that the triangles ABC and ADE are similar and this implies that

(1)

$\frac{|AD|}{|AB|} = \frac{|AE|}{|AC|}$

If we now place point B such that $|AB|=1$, point C such that $|AC|=x$, point D such that $|AD|=y$, and point E such that BC and DE are parallel, we have

$z = |AE| = \frac{|AC| |AD|}{|AB|} = x y.$

### Division

Division can be performed using the same geometrical construction as for multiplication: Place point B such that $|AB|=1$, point D such that $|AD|=y$, point E such that $|AE|=z$, and point C such that BC and DE are parallel. Then we have

$x = |AC| = \frac{|AE| |AB|}{|AD|} = \frac{z}{y}.$

### Square Root

Let the line segment CD have length $x$ and extend the line from D to a point A such that $|AD|=1$. Draw a semicircle with AC as diameter. Now draw a line through D which is perpendicular to AC and call the intersection between this line and the circle point B. The length of the line segment BD is now equal to the square root of $x$. See Figure 2.

Why is this so? It can be shown by using the Pythagorean theorem and a bit of algebra, but it is most easily seen by considering similar triangles again. First, we argue that the angle at B in the triangle ABC in Figure 3 is a right angle. This is so because the two triangles ABM and BCM are isosceles triangles and the sum of the angles in the triangle ABC is thus $2\alpha+2\beta=\pi$, implying that $\alpha+\beta$ is equal to a right angle. Figure 3. Any triangle inscribed in a semicircle has a right angle.

Consider Figure 4. It is now easy to see that the triangles ADB, BDC, and ABC are all similar. For instance, ABC and BDC both have a right angle and share the angle at C, so the remaining angles must be equal.

We now have

$\frac{|CD|}{|BD|} = \frac{|BD|}{|AD|}$

and if we set $|AD|=1$ and $|CD|=x$ we get

$x = |CD| = |BD|^2 \quad \Rightarrow \quad |BD| = \sqrt{x}.$