Prove that $ \sqrt{5} $ is irrational geometrically Hardy and Wright I am reading Hardy and Wright's Intro to Number Theory chapter about irrational number and there is a section proving that $ \sqrt{5} $ is irrational geometrically, but the proof is confusing me for some parts. The proof, extracted verbatim from the book, is presented below for convenience.
$ \textbf{Proof} $: We argue in term of $ \displaystyle x = \frac{\sqrt{5} - 1}{2} $. Then $ x^2 = 1 - x $. Geometrically, if $ AB = 1, AC = x $, then $ AC^2 = AB.BC $ and $ AB $ is divided in golden section by $ C $. (For the geometric figure, see Fig 4 here.)
If we divide $ 1 $ by $ x $, taking the largest integral quotient, viz. $ 1 $, the remainder is $ 1 - x = x^2 $. If we divide $ x $ by $ x^2 $, the quotient is again $ 1 $ and the remainder is $ x - x^2 = x^3 $. We next divide $ x^2 $ by $ x^3 $, and continue the process indefinitely; $ \textbf{at each stage the ratios of the number divided,} $ 
$ \textbf{the divisor, and the remainder are the same.} $ I am confused about this bold sentence. I work out a few examples and I got:
$ 1 = 1.x + (1 - x) = 1.x + x^2 $
$ x = 1.x^2 + (x - x^2) = 1.x^2 + x^3 $
$ x^2 = 1.x^3 + (x^2 - x^3) = 1.x^3 + x^4 $
and so on, so the ratios of the number divided, the divisor, and the remainder are all $ \displaystyle \frac{1}{x} $, is that what they are implying here?
Continue the proof: 

Geometrically, if we take $ CC_{1} $ equal and opposite to $ BC $, $ AC $ is divided at $ C_{1} $ in the same ratio as $ AB $ at $ C $, i.e. in golden section; if we take $ C_{1}C_{2} $ equal and opposite to $ C_{1}A $, then $ C_{1}C $ is divided in golden section at $ C_{2} $; and so on. Since we are dealing at each stage with a segment divided in the same ratio, the process can never end.

I currently stuck on this paragraph, what do they mean by "$ AC $ is divided at $ C_{1} $ in the same ratio as $ AB $ at $ C $?" At first I thought this means $ \displaystyle \frac{AB}{AC} = \frac{AC}{AC_{1}} $, but this is not true since $ \displaystyle \frac{AB}{AC} = \frac{1}{x} $ while $ \displaystyle \frac{AC}{AC_{1}} = \frac{x}{2x - 1} $.
Continue the proof: 

If $ x $ is rational, then $ AB $ and $ AC $ are integral multiples of the same length $ \delta $, and the same is true for $ C_{1}C, C_{1}C_{2}, \dots $, i.e. all the segments in the figure. Hence, we can construct an infinite sequence of descending integral multiples of $ \delta $, and this is plainly impossible.

If $ x = AC $ is rational, then $ \displaystyle AC = \frac{a}{b} $ and $ AB = 1 $, then how can "$ AB $ and $ AC $ are integral multiples of the same length $ \delta $?"
Any insight is appreciated.
 A: 
$ \textbf{at each stage the ratios of the number divided,} $ 
  $ \textbf{the divisor, and the remainder are the same.} $ I am confused about this bold sentence. 

They mean that
$$\color{red}{\text{red}}:\color{blue}{\text{blue}}:\color{green}{\text{green}}=1:x:x^2$$
always holds in 
$$\color{red}{1}=1\times \color{blue}{x}+\color{green}{x^2}$$
$$\color{red}{x}=1\times \color{blue}{x^2}+\color{green}{x^3}$$
$$\color{red}{x^2}=1\times \color{blue}{x^3}+\color{green}{x^4}$$
$$\vdots$$

I currently stuck on this paragraph, what do they mean by "$ AC $ is divided at $ C_{1} $ in the same ratio as $ AB $ at $ C $?" 

$\qquad\qquad$
In the paragraph, they mean that
$$BC:CA=AC_1:C_1C=CC_2:C_2C_1=C_1C_3:C_3C_2=\cdots$$
where $BC=C_1C=x^2,AC_1=C_2C_1=x^3,CC_2=C_3C_2=x^4,\cdots$.

If $ x = AC $ is rational, then $ \displaystyle AC = \frac{a}{b} $ and $ AB = 1 $, then how can "$ AB $ and $ AC $ are integral multiples of the same length $ \delta $?"

$AB=b\times\frac 1b$ and $AC=a\times\frac 1b$ are integral multiples of $\delta=\frac 1b$.
