In this example we see continued fractions used to give rational approximations to irrational numbers.

The following solution was done by Ling Xiang Ning, Raffles Institution, Singapore.

Using the quadratic formula to solve the equation

$x^2 = 7x + 1$

$x^2 - 7x - 1 = 0$

I find that $x$ is $(7 \pm \sqrt{53} )/2$. The positive solution is approximately $7.140054945$

This equation is equivalent to $x = 7 + 1/x$ and hence to the sequence of continued fractions mentioned in the problem. These continued fractions give better and better approximations to the positive root of the quadratic equation and I shall do them one by one.

$7+\frac{1}{7} = \frac{50}{7} = 7.142857142$

$7+\frac{1}{7+\frac{1}{7}} = \frac{357}{50} = 7.14$

$7+\frac{1}{7+\frac{1}{7+\frac{1}{7}}} = \frac{2549}{357} = 7.140056022$

$7+\frac{1}{7+\frac{1}{7+\frac{1}{7 + \frac{1}{7}}}} = \frac{18200}{2549} = 7.140054923$

To find a rational approximation to $\sqrt{53}$ we take, as above, $\frac{7 + \sqrt{53}}{2} \approx \frac{2549}{357}$

which gives $\sqrt{53} \approx 2 (\frac{2549}{357}) - 7 \approx \frac{2599}{357}.$

$5+\frac{1}{5} = \frac{26}{5} = 5.2$

$5+\frac{1}{5+\frac{1}{5}} = \frac{135}{26} = 5.192307692$

$5+\frac{1}{5+\frac{1}{5+\frac{1}{5}}} = \frac{701}{135} = 5.192592592$

$5+\frac{1}{5+\frac{1}{5+\frac{1}{5+\frac{1}{5}}}} = \frac{3640}{701} = 5.192582025$

Similarly, using the equation, $x^2 = 5x + 1$, which has solutions $\frac{5\pm \sqrt{29} }{2}$ , we can find a rational approximation to $\sqrt{29}$. The positive root is approximately $5.192582404$. The sequence of continued fractions is:

As you can see, the sequence of continued fractions gives better and better approximations to the positive root of the quadratic equation. \par Using $\frac{5\pm \sqrt{29}}{2} \approx \frac{3640}{701}$ gives $\frac{3775}{701}$ as a rational approximation to $\sqrt{29}$.