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