Simple Finite Continued Fraction has Rational Value

Theorem
Let $n \geq 0$ be a natural number.

Let $(a_0, \ldots, a_n)$ be a simple finite continued fraction‎ of length $n$.

Then its value $[a_0, \ldots, a_n]$ is a rational number.

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

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition that the continued fraction $\left[{a_0, a_1, \ldots, a_n}\right]$ has a rational value.

Basis for the Induction

 * $P(0)$ is true, as $\left[{a_0}\right]$ is an integer and therefore rational.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

The continued fraction $\left[{a_0, a_1, \ldots, a_k}\right]$ has a rational value.

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

Induction Step
Consider the continued fraction $\left[{a_0, a_1, \ldots, a_k, a_{k+1}}\right]$.

By definition of value of a finite continued fraction:
 * $\left[{a_0, a_1, \ldots, a_k, a_{k+1}}\right] = a_0 + \dfrac 1 {\left[{a_1, a_2, \ldots, a_k, a_{k+1}}\right]}$

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

By the induction hypothesis, so is $\left[{a_1, a_2, \ldots, a_k, a_{k+1}}\right]$, and so is its reciprocal.

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

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

Therefore for all $n \in \N^*$, the continued fraction $\left[{a_0, a_1, \ldots, a_n}\right]$ has a rational value.

Also see

 * Correspondence between Rational Numbers and Simple Finite Continued Fractions