Uniqueness of 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:

$x_n = a_n + \dfrac 1 {x_{n+1}}$.

• 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:

$\left[{a_1, a_2, \ldots, a_k}\right] = \dfrac {p_k} {q_k}$

we have, for $n \ge 2$:

$\left|{x - \dfrac {p_k} {q_k}}\right| < \dfrac 1 {q_n q_{n+1}}$

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, $\dfrac 1 {q_n q_{n+1}} \to 0$ 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.

$\blacksquare$