Solve the quadratic equation x2 = 7x + 1. This equation is equivalent to
x = 7 +  1

x

which has solutions given by the infinite continued fraction


x = 7 +  1

7+   1

7+   1

7+ …
.

This is because, if we think of this last equation as being x = 7 + 1/y, then clearly y = x. Show that the sequence of numbers
7 +  1

7
,    7 +  1

7+   1

7
,    7 +  1

7+   1

7+   1

7
,     …
gives better and better approximations to one of the solutions of the original quadratic equation. [Note that to find these approximations you can simply repeat the steps: 'take reciprocal, add 7', over and over again].

Find integers a and b, with b less than 400, such that a/b , is equal to √53 correct to six significant figures.

Now consider x2 = 5x + 1, ...

[See the articles on continued fractions first published in May 1999 and June 1999 ]