Continued Fraction Algorithm/Proof 1

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The Continued Fraction Algorithm is a method for finding the continued fraction expansion for any irrational number to as many partial quotients as desired.

Let $x_0$ be the irrational number in question.

The steps are:

\(\text {(1)}: \quad\) \(\ds k\) \(=\) \(\ds 0\) initialise
\(\text {(2)}: \quad\) \(\ds a_k\) \(=\) \(\ds \floor {x_k}\) the $k$th partial quotient (that is, $a_k$) is the integer part of $x_k$
\(\text {(3)}: \quad\) \(\ds x_{k + 1}\) \(=\) \(\ds \frac 1 {x_k - a_k}\) the subsequent term is the reciprocal of the fractional part of $x_k$
\(\text {(4)}: \quad\) \(\ds k\) \(=\) \(\ds k + 1\) increase $k$ by $1$ then go to step $(2)$

Then $x_0 = \sqbrk {a_0, a_1, a_2, \ldots}$ is the required continued fraction expansion.


Let $x_0$ be an irrational number.

We seek $a_0, a_1, \ldots \in \Z$ such that $x_0 = \left[{a_0, a_1, \ldots}\right]$.

From Bound for Difference of Limit of Simple Infinite Continued Fraction with Convergent:

$x_1$ lies strictly between any successive pair of its convergents.

So, for a start, it has to lie between $C_0 = a_0$ and $C_1 = a_0 + \dfrac 1 {a_1}$.

In particular, as $a_1 \ge 1$, we know that $a_0 < x_0 < a_0 + 1$.

So $a_0 = \left \lfloor {x_0} \right \rfloor$ where $\left \lfloor {x_0} \right \rfloor$ is the floor function of $x_0$.

We note, in particular, that $a_0$ is therefore determined by $x_0$ alone.

Now we write:

$x_0 = \left \lfloor {x_0} \right \rfloor + \left\{{x_0}\right\}$

where $\left\{{x_0}\right\}$ is the fractional part of $x_0$.


$x_0 = a_0 + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \cfrac 1 {\ddots}}} = \left \lfloor {x_0} \right \rfloor + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \cfrac 1 {\ddots}}} = \left \lfloor {x_0} \right \rfloor + \dfrac 1 {\left[{a_1, a_2, a_3, \ldots}\right]}$

From Real Number minus Floor:

$0 \le \left\{{x_0}\right\} < 1$

But because $x_0$ is irrational, $\left\{{x_0}\right\} \ne 0$.

So $0 < \left\{{x_0}\right\} < 1$.

So: $\left[{a_1, a_2, a_3, \ldots}\right] = \dfrac 1 {x_0 - \left \lfloor {x_0} \right \rfloor} = \dfrac 1 {\left\{{x_0}\right\}}$.

Now we write $x_1 = \dfrac 1 {\left\{{x_0}\right\}}$.

Then $x_1 = \left[{a_1, a_2, a_3, \ldots}\right]$.

As $\left\{{x_0}\right\} < 1$ we have that $x_1 = \dfrac 1 {\left\{{x_0}\right\}} > 1$.

So $x_1$ is an irrational number greater than $a_1$ which is positive.

Repeating the argument leads to $a_1 = \left \lfloor {x_1}\right \rfloor$ and so $a_1$ is determined uniquely from $x_1$ and hence from $x_0$.

In the same way, $x_2 = \dfrac 1 {\left\{{x_1}\right\}}$ and so $x_2 = \left[{a_2, a_3, a_4, \ldots}\right]$.

And so on.