Irrational Number is Limit of Unique Simple Infinite Continued Fraction

Theorem
Let $$x$$ be an irrational number.

Then:
 * The simple infinite continued fraction determined from $$x$$ by the Continued Fraction Algorithm converges to $$x$$;
 * No other simple infinite continued fraction converges to $$x$$.

Proof
Let $$x$$ be an irrational number.

The Continued Fraction Algorithm produces a sequence of (non-simple) finite continued fractions as follows:
 * $$x = x_1 = \left[{a_1, x_2}\right] = \left[{a_1, a_2, x_3}\right] = \cdots = \left[{a_1, a_2, \ldots, a_n, x_{n+1}}\right]$$

where:
 * $$\forall n \ge 1: a_n = \left \lfloor {x_n} \right \rfloor$$

and the step from one step to the next is done by:
 * $$x_n = a_n + \frac 1 {x_{n+1}}$$.


 * First we need to show that the ICF built up thus has a limit $$x$$.

To do that we show that the sequence of simple finite continued fractions:
 * $$\left[{a_1}\right], \left[{a_1, a_2}\right], \ldots, \left[{a_1, a_2, \ldots, a_n}\right], \ldots$$

converges to $$x$$.

Using standard notation for numerators and denominators:
 * $$\left[{a_1, a_2, \ldots, a_k}\right] = \frac {p_k} {q_k}$$

we have, for $$n \ge 2$$:
 * $$\left|{x - \frac {p_k} {q_k}}\right| < \frac 1 {q_n q_{n+1}}$$

by Value of Simple Continued Fraction.

We have that the sequence of the denominators $\left \lfloor {q_n} \right \rfloor$ is strictly increasing.

So from Basic Null Sequences and the Squeeze Theorem, $$\frac 1 {q_n q_{n+1}} \to 0$$ as $$n \to \infty$$.

So the SICF determined from $$x$$ by the Continued Fraction Algorithm does converge to $$x$$.


 * Now we need to show it is unique.

This will be achieved by the Second Principle of Mathematical Induction.

Suppose $$\left[{a_1, a_2, a_3, \ldots}\right] = \left[{b_1, b_2, b_3, \ldots}\right]$$ have the same value.

First we note that if $$\left[{a_1, a_2, a_3, \ldots}\right] = \left[{b_1, b_2, b_3, \ldots}\right]$$ then $$a_1 = b_1$$ since both are equal to the integer part of the common value.

This is our basis for the induction.

Now suppose that for some $$k \ge 1$$, we have:
 * $$a_1 = b_1, a_2 = b_2, \ldots, a_k = b_k$$.

Then all need to do is show that $$a_{k+1} = b_{k+1}$$.

Now:
 * $$\left[{a_1, a_2, a_3, \ldots}\right] = \left[{a_1, a_2, \ldots, a_k, \left[{a_{k+1}, a_{k+2}, \ldots,}\right]}\right]$$

and similarly
 * $$\left[{b_1, b_2, b_3, \ldots}\right] = \left[{b_1, b_2, \ldots, b_k, \left[{b_{k+1}, b_{k+2}, \ldots,}\right]}\right]$$.

As these have the same value and have the same first $$k$$ partial quotients, it follows that:
 * $$\left[{a_{k+1}, a_{k+2}, \ldots,}\right] = \left[{b_{k+1}, b_{k+2}, \ldots,}\right]$$.

But now $$a_{k+1} = b_{k+1}$$ as each is equal to the integer part of the value of this SICF.

Hence the result.