What is the square root of 230?

Suppose `x > 0` satisfies:

`x = 15+1/(6+1/(15+x))`

Simplifying the right hand side we find:

`x = 15+1/(6+1/(15+x))`

`color(white)(x) = 15+(15+x)/(96+6x)`

`color(white)(x) = (1380+91x)/(91+6x)`

Multiplying both ends by `(91+6x)` this becomes:

`6x^2+91x = 1380+91x`

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

`6x^2 = 1380`

Dividing both sides by `6`, we find:

`x^2 = 230`

So `x = sqrt(230)` and:

`sqrt(230) = 15+1/(6+1/(15+sqrt(230)))`

`color(white)(sqrt(230)) = 15+1/(6+1/(30+1/(6+1/(30+1/(6+1/(30+...))))))`

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) ~~ 15+1/6 = 91/6`

with the property that:

`91^2 = 8281 = 8280+1 = 230 * 6^2 + 1`

For greater accuracy you can terminate the continued fraction later or use this rational approximation `91/6` as the initial approximation in a Babylonian method.