Rational Numbers and Simple Finite Continued Fractions are Equivalent

Theorem
Every simple finite continued fraction‎ has a rational value.

Conversely, every rational number can be expressed as a simple finite continued fraction‎.

Proof that SFCF has a Rational Value
First we prove that any simple finite continued fraction‎ has a rational value.

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 expansion $\left[{a_1, a_2, a_3, \ldots, a_n}\right]$ has a rational value.

Basis for the Induction

 * $P(1)$ is true, as $\left[{a_1}\right]$ 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 $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

The continued fraction expansion $\left[{a_1, a_2, a_3, \ldots, a_k}\right]$ has a rational value.

Then we need to show that the continued fraction expansion $\left[{a_1, a_2, a_3, \ldots, a_k, a_{k+1}}\right]$ also has a rational value.

Induction Step
Consider the continued fraction expansion $\left[{a_1, a_2, a_3, \ldots, a_k, a_{k+1}}\right]$.

By the first continued fraction identity:
 * $\left[{a_1, a_2, a_3, \ldots, a_k, a_{k+1}}\right] = a_1 + \dfrac 1 {\left[{a_2, a_3, \ldots, a_k, a_{k+1}}\right]}$

$a_1$ is an integer, as we have seen.

By the induction hypothesis, so is $\left[{a_2, a_3, \ldots, a_k, a_{k+1}}\right]$, and so is its reciprocal.

Hence as $a_1 + \dfrac 1 {\left[{a_2, a_3, \ldots, a_k, a_{k+1}}\right]}$ 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 expansion $\left[{a_1, a_2, a_3, \ldots, a_n}\right]$ has a rational value.

Proof that Every Rational Number can be expressed as a SFCF
Let $\dfrac a b$ be a rational number expressed in canonical form.

That is $b > 0$ and $a \perp b = 1$.

By the Euclidean Algorithm, we have:

Thus from the system of eqns at the right hand side, we get:

This shows that $\dfrac a b$ has the SFCF $\left[{q_1, q_2, q_3, \ldots, q_n}\right]$.

Note
It can be seen from this proof that there is a close connection between continued fractions and the Euclidean Algorithm.