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## 'Good Approximations' printed from http://nrich.maths.org/

Solve the quadratic equation $x^2 = 7x + 1$. This equation is
equivalent to $x = 7 + \frac{1}{x}$ which has solutions given by
the infinite continued fraction

$$ x = 7 + {1\over\displaystyle 7+ { 1\over \displaystyle 7+ {
1\over \displaystyle 7+ \cdots }}}. $$

This is because, if we think of this last equation as being $x = 7
+ {1\over y}$, then clearly $y = x$. Show that the sequence of
numbers $$7 + {1\over\displaystyle 7}, \quad 7 +
{1\over\displaystyle 7+ { 1\over \displaystyle 7}}, \quad 7 +
{1\over\displaystyle 7+ { 1\over \displaystyle 7+ { 1\over
\displaystyle 7 }}}, \quad \cdots$$ 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
${a\over b}$ , is equal to $\sqrt 53$ correct to six significant
figures.

Now consider $x^2 = 5x + 1$, ...

[See the articles
Continued Fractions I and
Continued Fractions II. ]