Continued fraction form for rational numbers less than $1$ 
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? 
 A: Find the c.f. of the reciprocal, then put that under $1$ in a fraction. For example $3/2=1+1/2$, so
$$\frac{2}{3}=0+\cfrac{1}{1+\cfrac{1}{2}}.$$
A: 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
$$\rm \bigg\langle\frac{3}{8}\bigg\rangle\ \ =\ \ 0\: +\: \cfrac{1}{\bigg\langle\cfrac{8}3\bigg\rangle}\ \ =\ \  0 \:+\: \cfrac{1}{2 + \cfrac{1}{\bigg\langle\cfrac{3}2\bigg\rangle}}\ \ =\ \  0\: +\: \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}2}}  $$
A: 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.
