Complete quotients in the continued-fraction algorithm

Work in the complete ordered field of real numbers (Complete ordered field (least-upper-bound property)). For a real number $\alpha$, set ${\alpha }_0:=\alpha \mathrm{.}$ Given ${\alpha }_n$, the Archimedean property (Every complete ordered field is Archimedean) and the well-ordering principle (The well-ordering principle) produce a unique integer $a_n$ with $a_n\le {\alpha }_n<a_n+1.$

For completeness, here is that construction. Choose positive integers $r,s$ with $-r<{\alpha }_n<s$ by Archimedeanness, and let $T:=\{k\in \mathbb{N}:{\alpha }_n<-r+k\}\mathrm{.}$ The set $T$ is nonempty because $r+s\in T$, so it has a least element $k_0$. Since $-r<{\alpha }_n$, one has $k_0>0$; write $k_0=j+1$. Minimality gives $-r+j\le {\alpha }_n<-r+j+1$, so $a_n:=-r+j$ works. If two integers satisfied the displayed inequalities, discreteness of the integer order would put one at least $1$ above the other and contradict the upper inequality, proving uniqueness. If ${\alpha }_n\mathrm{\ne }{a}_n$, define the next complete quotient ${\alpha }_{n+1}:=\frac{1}{{\alpha }_n-a_n}\mathrm{.}$

Thus the continued-fraction algorithm associates to $\alpha$ its integer parts $a_n$ and its successive complete quotients ${\alpha }_n$. Whenever ${\alpha }_{n+1}$ is defined, one has $0<{\alpha }_n-a_n<1$ and therefore ${\alpha }_{n+1}>1$, so every later digit $a_{n+1},a_{n+2},\dots$ is positive.