 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/2i/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

Starting with

x_{0} = x > 0

a_{n} = ⌊ x_{n} ⌋, x_{n + 1} = \frac{1}{x _{n} - a _{n}}

unless

x_{n}

is already an integer. This gives

x = [a_{0}; a_{1}, a_{2}, \dots] .

Every finite continued fraction is rational, since it is built from integers by finitely many additions and reciprocals. Conversely, if

x = p / q

is rational in lowest terms, then

\frac{p}{q} = a_{0} + \frac{r}{q}, 0 \leq r < q .

When

r \neq = 0

, the next complete quotient is

q / r

. Thus each step is an application of the Euclidean algorithm and replaces the denominator by a smaller nonnegative remainder. The process must terminate. This proves the termination criterion for a simple continued fraction.

For

x = 3

a_{0} = 1, x_{1} = \frac{1}{3 - 1} = \frac{3 + 1}{2}, a_{1} = 1,

and

x_{2} = \frac{1}{x _{1} - 1} = 3 + 1, a_{2} = 2.

Finally,

x_{3} = \frac{1}{x _{2} - 2} = \frac{1}{3 - 1} = x_{1},

3 = [1; \overline{1, 2}] .

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