In Section 5.1.2: Calculating Square Roots we studied one method for calculating square roots. In this question we’ll discover another, which is actually an example of the Newton-Raphson iterative method, and is equivalent to a method used by the ancient Babylonians to find square roots.

Suppose we want to find the square root of ten. Perhaps we want to know what side length a square field should have if we want its area to be ten square kilometres. Since $3^2 = 9$, and $4^2 = 16$, we know that the side length will be three and a bit, so we might start by imagining a square of side length $3$ inside our square of area $10$. The area not covered by the smaller square is $10 - 3^2 = 1$. What we need to find out is the extra length, marked with ‘$a$’ in the diagram.

The region with area $1$ can be considered as two rectangles and a little square.

Since the little square is much smaller than the rectangles, we might decide to ignore it, and get an approximation for $a$ by just considering the combined area of the two rectangles to be $1$. What would $a$ be in this case? Show hint

The area of each rectangle is $3a$.Each rectangle has one side with length $3$, and another with side $a$, so they each have an area of $3a$. When combined, that makes $6a$ which has to equal $1$, so $a=1/6$.

$3+a$ is now a decent approximation for $\sqrt{10}$. Is $(3+a)^2$ bigger or smaller than $10$ and by how much? Show answer

There are a couple ways we can work this out. We could reason that when we tried to make the area $10$, we didn’t include the little square, which has an area of $\frac{1}{6}\times\frac{1}{6} = \frac{1}{36}$, so $(3+a)^2$ is bigger than $10$ by $\frac{1}{36}$.

Or we can work out what $3+a$ is and square it: \begin{align*} 3+a &= 3 + \frac{1}{6}\\ &= \frac{19}{6}\\ (3+a)^2 &= \frac{361}{36}\\ &= 10 + \frac{1}{36} \end{align*} so again we find that $(3+a)^2$ is bigger than $10$ by $\frac{1}{36}$.

To get an even better approximation, we can repeat this process. This time we need to reduce the side length by a small amount, which I’ll call $b$. Find $b$ so that $3+a-b\approx \sqrt{10}$ by considering the area of the little square, $b^2$, to be negligible. Show hint

Your answer to the last question is the area of the two rectangles and the little square, which is approximately $2(3+a)b$.We want the area of the rectangles and little square to be approximately $1/36$, as this will make $(3+a-b)^2$ closer to $10$. This time we don’t know the side length of the rectangles, but we know the length of a rectangle plus the little square: it’s $3+a = \frac{19}{6}$. Hence the area of each rectangle is $\frac{19}{6}b - b^2$, so the total area of the rectangles and the little square is really $2\times\frac{19}{6}b - b^2$, but since $b^2$ is very small, we can ignore it, and approximate the area to $2\times\frac{19}{6}b$, which should be $1/36$. Solving for $b$, we get $1/228$.

We now have an excellent approximation for $\sqrt{10}$: \[3 + \frac{1}{6} - \frac{1}{228} = \frac{721}{228}.\] To six decimal places, this is $3.162281$, while the true value of $\sqrt{10}$ is $3.162278$ to six decimal places.