You may also like

problem icon

There's a Limit

Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?

problem icon

Not Continued Fractions

Which rational numbers cannot be written in the form x + 1/(y + 1/z) where x, y and z are integers?

problem icon

Comparing Continued Fractions

Which of these continued fractions is bigger and why?

Good Approximations

Stage: 5 Challenge Level: Challenge Level:1

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. ]