Continued fraction form for rational numbers less than $1$

Question

How could we convert a rational number (less than $1$) to the continued fraction form?

This is probably an extension of this question. After reading Bill Dubuque's answer here and here, I got curious about continued fractions and tried myself to convert rationals into continued fractions. I noticed that for rational numbers greater than $1$ we could use extended Euclid theorem for gcd computation (manually), but I am not sure how to do the same for rationals less than $1$. Any ideas?

Answer 1

The recursion for computing the continued fraction expansion $\rm\:\langle x \rangle\:$ of a real $\rm\:x > 0\ $ is as follows

$$\rm \langle x\rangle\ =\ \lfloor x\rfloor\ + \dfrac{1}{\langle (x-\lfloor x\rfloor)^{-1}\rangle}\:,\qquad where\ \ \ \lfloor x\rfloor\ :=\ floor(x)\ $$

For example

$$\rm \bigg\langle\frac{3}{8}\bigg\rangle\ \ =\ \ 0\: +\: \cfrac{1}{\bigg\langle\cfrac{8}3\bigg\rangle}\ \ =\ \ 0 \:+\: \cfrac{1}{2 + \cfrac{1}{\bigg\langle\cfrac{3}2\bigg\rangle}}\ \ =\ \ 0\: +\: \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}2}} $$

Answer 2

Find the c.f. of the reciprocal, then put that under $1$ in a fraction. For example $3/2=1+1/2$, so

$$\frac{2}{3}=0+\cfrac{1}{1+\cfrac{1}{2}}.$$

Answer 3

Using the Euclidean algorithm for $\frac{3}{8}$ we have, $$ \begin{align*} 3&=8(0)+3\\ 8&=3(2)+2\\ 3&=2(1)+1\\ 2&=1(2)+0 \end{align*} $$ and we take the quotients as our convergents $q_1=0$, $q_2=2$, $q_3=1$, and $q_4=2$. Sometimes it's useful to look at only the first few convergents. For hand computation I've been using the following (albit odd) method.

$ \begin{array}{cc|ccc} \,&\,&0&2&1&2\\ \hline 0&1&0&1&1&3\\ 1&0&1&2&3&8 \end{array} $

The top row lists the quotients. The 0 in row two is computed by $0(0)+1=0$. Then moving right, $2(0)+1=1$. Then, $1(1)+0=1$. Then, $2(1)+1=3$. The third row always starts with 1. Here, $0(0)+1=1$. Then, $2(1)+0=2$ and $1(2)+1=3$, finally $2(3)+2=8$.

In general, taking the top and bottom number under one of the quotients (or convergents) gives the fraction up to that point. That is, $\frac{1}{3}=0+\cfrac{1}{2+\cfrac{1}{1}}$ and using them all gives the orignal fraction $\frac{3}{8}$.

Finally, the last chunk of 4 numbers, "take the diagonal" note $1(8)-3(3)=-1$ and $3(3)-1(8)=1$. In general for a fraction one of those diagonals is the (positive) GCD.