1. ## Fractions

Continued fractions:
X= a+b y=

2. a) $x-a=a+\frac{1}{1+\frac{1}{b_1+\frac{1}{b_2+\dots}}}-a=\frac{1}{1+\frac{1}{[b_1,b_2,\dots]}}=\frac{1}{1+\frac{1}{\beta}}=\frac{\beta}{\beta+ 1}$ $\Rightarrow(\beta+1)(x-a)=\beta$ (1)
$y-a=a+\frac{1}{1+b_1+\frac{1}{b_2+\frac{1}{b_3+\dots }}}-a=\frac{1}{1+\beta}$ $\Rightarrow(\beta+1)(y-a)=1$ (2)
(1) $+$ (2) $\Rightarrow x+y=2a+1\Rightarrow -x=y-(2a+1)$ $\stackrel{2a+1\in\mathbb{Z}}{=}$ $[a-(2a+1),1+b_1,b_2,b_3,\dots]=[-a-1,1+b_1,b_2,b_3,\dots]$ and $-y=x-(2a+1)=\dots$
b) Apply a):
$x=[1,1,2,2,3,3,\dots],y_0:=[1,1+2,2,3,3,\dots]$ $\stackrel{\text{a)}}{\Longrightarrow}-x=y_0-(2\times1+1)=y_0-3=\dots$.
$y=[1,2,3,4,\dots],x_0:=[1,1,2-1,3,4,\dots]\Rightarrow-y=x_0-(2\times1+1)=\dots$
c) .. I think I don't understand the notation you are using..

3. ## continued fraction

the continued fraction of x, y], where x and y are positive integers.

4. Hmm.. I still don't understand. You mean the problem asks to put $-[a,b,a,b,a,b,a,\dots]$ in the form of a continued fraction?!

5. ## yes

where a and b are positive integers, for example - [1, 2, 1,2,1,2...]

6. ## error

sorry the example was supposed to be -[1,2, 1, 2, 1, 2 ....]

7. Well, in this case, just apply again a):
$y:=[a,b,a,b,\dots]=[a,1+(b-1),a,b,\dots]$
$x:=[a,1,b-1,a,b,\dots]$
$\Rightarrow-y=x-(2a+1)=[-a-1,1,b-1,a,b,\dots]$