# What is the square root of 230?

Jun 27, 2018

$\sqrt{230} = 15 + \frac{1}{6 + \frac{1}{30 + \frac{1}{6 + \frac{1}{30 + \frac{1}{6 + \frac{1}{30 + \ldots}}}}}}$

#### Explanation:

Suppose $x > 0$ satisfies:

$x = 15 + \frac{1}{6 + \frac{1}{15 + x}}$

Simplifying the right hand side we find:

$x = 15 + \frac{1}{6 + \frac{1}{15 + x}}$

$\textcolor{w h i t e}{x} = 15 + \frac{15 + x}{96 + 6 x}$

$\textcolor{w h i t e}{x} = \frac{1380 + 91 x}{91 + 6 x}$

Multiplying both ends by $\left(91 + 6 x\right)$ this becomes:

$6 {x}^{2} + 91 x = 1380 + 91 x$

Subtracting $91 x$ from both sides, this becomes:

$6 {x}^{2} = 1380$

Dividing both sides by $6$, we find:

${x}^{2} = 230$

So $x = \sqrt{230}$ and:

$\sqrt{230} = 15 + \frac{1}{6 + \frac{1}{15 + \sqrt{230}}}$

$\textcolor{w h i t e}{\sqrt{230}} = 15 + \frac{1}{6 + \frac{1}{30 + \frac{1}{6 + \frac{1}{30 + \frac{1}{6 + \frac{1}{30 + \ldots}}}}}}$

Since this continued fraction does not terminate, we can conclude that $\sqrt{230}$ is not expressible as a terminating fraction. In other words, it is irrational.

Terminating before the first occurrence of $30$ we get an efficient first rational approximation:

$\sqrt{230} \approx 15 + \frac{1}{6} = \frac{91}{6}$

with the property that:

${91}^{2} = 8281 = 8280 + 1 = 230 \cdot {6}^{2} + 1$

For greater accuracy you can terminate the continued fraction later or use this rational approximation $\frac{91}{6}$ as the initial approximation in a Babylonian method.