<s>[INST] Find the simple continued fractions for both $\pm \frac{39}{25}$... Find the simple continued fractions for both $\pm \frac{39}{25}$?


*

*So far for $\frac{39}{25}$ I have:


$39 = 1 \times 25 + 14  $
$ 25 = 1\times 14 + 11 $
$14  = 1 \times 11 + 3$
$11 = 3 \times 3 + 2$
$3  = 1 \times 2 + 1$
$2 = 2 \times 1 + 0$
Giving us [1 ; 1, 1, 3, 1,  2 ]


*

*Then for $ - \frac{39}{25}$ I have:


$ -39 = 1 \times -25 - 14   $
$ - 25 = 1 \times -14 -11 $
$ -14 = 1 \times -11 -3 $
$ -11 = 3 \times -3 -2$
$-3 = 1 \times - 2 - 1 $
$-2 =  2 \times -1 + 0 $
Giving us [1; 1,1,3,1,2] 
-I think I have this problem figured out but could someone please confirm or suggest changes. Help is very much appreciated!
 [/INST] I believe that for $-\frac{39}{25}$ you should first rewrite it as $-2+\frac{11}{25}$. Then
$11=0\times25+11$
$25=2\times11+3$
$11=3\times3+2$
$3=1\times2+1$
$2=2\times1+0$
This gives $-\frac{39}{25}=[-2+0;2,3,1,2]=[-2;2,3,1,2]$
Note that recently there has developed a way of representing negative continued fractions in the form
\begin{equation}
x=a_0-\frac{1}{a_1-\frac{1}{a_2-\frac{1}{\ddots}}}
\end{equation}
In this form one takes the ceiling of each improper fraction.
See the thesis of Alex Eustis.
This approach to negative continued fractions is new to me but if I understand it correctly, then
\begin{equation}
-\frac{39}{25}=-1-\frac{1}{2-\frac{1}{5-\frac{1}{3}}}=[1;2,5,3]_-
\end{equation}
</s>