# The Root of the Matter: Approximating Roots with the Greeks - Method, Motivation, and Verification

Author(s):
Matthew Haines and Jody Sorensen (Augsburg University)

### Method and Motivation

In his work On Mathematics Useful for the Understanding of Plato [4], Theon proposed a method which can be used to provide ever closer rational approximations of $$\sqrt{2}.$$ This method is sometimes referred to as Theon's ladder. The method starts with $$x_0=1,$$ $$y_0=1.$$ Our estimate of $$\sqrt{2},$$ which will be the ratio $$\displaystyle{\frac {y} {x}},$$ thus starts as 1. The numbers $$x$$ and $$y$$ are sometimes referred to as side and diagonal numbers – more on that to come. [5] To get a better estimate of $$\sqrt{2},$$ we let $$x_1=x_0+y_0$$ and $$y_1 = 2x_0+y_0.$$ Thus we get that $$x_1=2$$ and $$y_1=3,$$ so our new estimate is $$\displaystyle{\frac {3} {2}} = 1.5.$$ The recursive formula is $x_{n+1} = x_n+y_n \ \ \ \ \ \ \ \ y_{n+1} = 2x_n+y_n .$ Continuing in this way for a few more steps gives us the results in Table 1.

$\begin{array}{r|r|r|r} n & x_n & y_n & y_n/x_n \\\hline 0 & 1 & 1 & 1 \\\hline 1 & 2 & 3 & 1.5 \\\hline 2 & 5 & 7 & 1.4 \\\hline 3 & 12 & 17 & 1.4167 \\\hline 4 & 29 & 41 & 1.4138\\\hline 5 & 70 & 99 & 1.4143\\\hline 6 & 169 & 239 & 1.4142\end{array}$

Table 1. Side and diagonal numbers

So it appears that Theon's method approximates $$\sqrt{2} = 1.4142....$$ Note that the denominators (or sides), $$x_n,$$ are the Pell numbers, and their properties are well-studied by number theorists.

We cannot be sure how (or why) Theon came up with this procedure. One possible origin is numerical. [1] If $$\sqrt{2}$$ were rational, then $$\displaystyle\sqrt{2} = {\frac{y}{x}}$$ for some positive integers $$x$$ and $$y.$$ Squaring and simplifying gives us $$2x^2 =y^2.$$ So, does that equation have integer solutions? Let's look at the list of possibilities for $$y^2$$ and $$2x^2$$ in Table 2.

$\begin{array}{r|cccccccc} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline n^2 & 1 & 4 & \fbox{9} & 16 & 25 & 36 & \fbox{49} & 64 \\\hline 2n^2 & 2 & \fbox{8} & 18 & 32 & \fbox{50} & 72 & 98 & 128\end{array}$

Table 2. Values of $$y^2$$ and $$2x^2$$

From this tiny table we don't see any places where a square is equal to two times a square, but we do see some places where they are close. For example if $$x=5$$ and $$y=7,$$ then $$2x^2 = 50 = y^2+1.$$ This indicates that $$\displaystyle{\frac {7}{5}}$$ is a decent approximation of $$\sqrt{2}.$$ If we continue this table and list the possibilities as fractions, we get this list: ${\frac{1}{1}}, \,{\frac{3}{2}},\, {\frac{7}{5}},\, {\frac{17}{12}},\, \dots.$ These are all values of $$x$$ and $$y$$ where $$2x^2=y^2 \pm 1.$$ If you look for a pattern here, you see that the new denominator is the sum of the previous numerator and denominator, or $$x_{n+1} = x_n+y_n.$$ The new numerator is this new denominator plus the old denominator, or $$y_{n+1} = x_{n+1}+x_n = x_n+y_n+x_n = 2x_n+y_n.$$ Testing this out gives the next term as $$\displaystyle{\frac{41}{29}}$$ which satisfies $$2\cdot29^2 = 1682=41^2+1.$$ So Theon could have found this pattern by inspection.

There are also possible geometric motivations for this process, which explain why $$x$$ and $$y$$ are referred to as "side" and "diagonal" numbers. Suppose we start with an isosceles right triangle $$ABC$$ whose leg (side) is of length $$x$$ and hypotenuse (diagonal) of length $$y,$$ as in Figure 1. Note that this figure has the added point $$H$$ to create the square $$ABHC$$ to emphasize the names "side and diagonal numbers." Extend the two sides by $$y$$ to create an isosceles right triangle with sides of length $$x+y.$$

Figure 1. An isosceles right triangle with sides of length $$x+y$$ and diagonal $$\overline{DG}$$ of length $$2x+y$$

We can see that the larger triangle has a hypotenuse of length $$2x+y$$ by noticing that $\triangle BCA \cong \triangle DBE \cong \triangle GCF$ and that $$BCFE$$ is a rectangle. Thus $$DE=FG=x$$ and $$EF=y.$$

This triangle motivates the idea that if the side is extended from $$x$$ to $$x+y,$$ then the diagonal should be extended from $$y$$ to $$2x+y,$$ which fits Theon's method. We should note that no integers $$x$$ and $$y$$ exist that are sides and diagonals of the isosceles right triangle. This approach is simply a motivation and not a proof.

### Verification

Let's prove that Theon's method gives a method for finding better and better rational approximations of $$\sqrt{2}.$$  Suppose we start with $$x$$ and $$y$$ satisfying $$2x^2-y^2 = \pm 1$$ (note that $$x_0=y_0=1$$ satisfies this) and let $$x^*=x+y, y^*=2x+y.$$ Then

\begin{eqnarray*} 2(x^*)^2-(y^*)^2 &=& 2(x+y)^2-(2x+y)^2 \\ &=& -2x^2+y^2 = \mp 1.\end{eqnarray*}

This shows that as we iterate, $$2x^2-y^2$$ remains $$\pm 1.$$ Now if $$2x_n^2-y_n^2 = \pm 1,$$ then  $$\displaystyle 2 = {\frac {y_n^2} {x_n^2}} \pm {\frac {1} {x_n^2}}.$$ If we start with positive values for $$x_0$$ and $$y_0,$$ then as we iterate, $$x_n$$ gets bigger by at least one at each step, so $$x_n$$ goes to $$\infty.$$ Therefore we can say that $$\displaystyle {\frac {1} {x_n^2}}$$ will go to $$0^{+}.$$ This means that $$\displaystyle\left({\frac{y_n} {x_n}} \right)^2$$ will tend to 2. Assuming that $$\displaystyle \lim_{n\rightarrow \infty} {\frac {y_n} {x_n}}$$ exists, then $$\displaystyle {\frac {y_n} {x_n}}$$ goes to $$\sqrt 2,$$ as desired.

So, in about 100 CE, Theon of Smyrna developed (for an unknown reason) an iterative method that can be used to approximate $$\sqrt 2$$ by rational numbers. With our modern mathematical tools, we will look at Theon's method through the lenses of Geometry and Linear Algebra.