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:
 * $$\left[{a_1, a_2, a_3, \ldots, a_k, a_{k+1}}\right] = a_1 + \frac 1 {\left[{a_2, a_3, \ldots, a_k, a_{k+1}}\right]}$$.

$$a_1$$ 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 $$a_1 + \frac 1 {\left[{a_2, a_3, \ldots, a_k, a_{k+1}}\right]}$$ is the sum of two rational numbers, it is itself rational.

So $$P \left({k}\right) \Longrightarrow P \left({k+1}\right)$$ 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 $$\frac a b$$ 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 equations at the right hand side, we get:

$$ $$ $$ $$ $$

This shows that $$\frac a b$$ 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.