Simple Finite Continued Fraction has Rational Value

Theorem

Let $\tuple {a_0, \ldots, a_n}$ be a simple finite continued fraction‎ of length $n$.

Then its value $\sqbrk {a_0, \ldots, a_n}$ is a rational number.

This will be proved by induction on the number of partial denominators.

For all $n \in \N$, let $\map P n$ be the proposition that the continued fraction $\sqbrk {a_0, a_1, \ldots, a_n}$ has a rational value.

$\map P 0$ is true, as $\sqbrk {a_0}$ is an integer and therefore rational.

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

The continued fraction $\sqbrk {a_0, a_1, \ldots, a_k}$ has a rational value.

Then we need to show that the continued fraction $\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} }$ also has a rational value.

Consider the continued fraction $\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} }$.

$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = a_0 + \dfrac 1 {\sqbrk {a_1, a_2, \ldots, a_k, a_{k + 1} } }$

$a_0$ is an integer, as we have seen.

By the induction hypothesis, $\sqbrk {a_1, a_2, \ldots, a_k, a_{k + 1} }$ is a rational number, and so is its reciprocal.

Hence as $a_0 + \dfrac 1 {\sqbrk {a_1, a_2, \ldots, a_k, a_{k + 1} } }$ is the sum of two rational numbers, it is itself rational.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore for all $n \in \N$, the continued fraction $\sqbrk {a_0, a_1, \ldots, a_n}$ has a rational value.

$\blacksquare$