About rational numbers as Dedekind cuts. I am reading "An Introduction to Calculus" by Kunihiko Kodaira.
There is Theorem 1.3 in this book and I am very confused.
We identify a rational number $r$ with a Dedekind cut $(R, R')$ where $R = \{q \in \mathbb{Q} | q < r\}, R' = \{q \in \mathbb{Q} | r \leq q\}$.  
In the following proof of Theorem 1.3, $\alpha = (A, A')$ is a Dedekind cut and the elements of $A$ and $A'$ are also Dedekind cuts.  
Is it ok?  
Are the elements of $R$ and $R'$ Dedekind cuts or original rational numbers?
Assume that $s \in R$ and $s' \in R'$ are Dedekind cuts.
Then, are the elements of $S$ and $S'$ Dedekind cuts or original rational numbers, where $s = (S, S')$?  

Theorem 1.3
  For any real number $\alpha = (A, A')$,
$$A = \{r \in \mathbb{Q} | r < \alpha \}, A' = \{r \in \mathbb{Q} | r \geq \alpha \}. (1.8)$$ 

Proof:
To compare a rational number $r$ with a real number $\alpha$, we consider $r$ as a real number and represent it as a Dedekind cut:
$$r = (R, R'), R = \{s \in \mathbb{Q} | s < r\}, R' = \{s \in \mathbb{Q} | s \geq r \}.$$
Since $A \cup A' = \mathbb{Q}, A \cap A' = \emptyset$, it is sufficient to prove that if $r \in A$, then $r < \alpha$ and if $r \in A'$, then $r \geq \alpha$ to prove $(1.8)$.  


*

*Suppose that $r \in A$. If $s < r, s \in \mathbb{Q}$, then $s \in A$. So $R \subset A$. Because $r \notin R$, so $R \neq A$. $\therefore r < \alpha$.  

*Suppose that $r \in A'$. If $s \geq r, s \in \mathbb{Q}$, then $s \in A'$. So $R' \subset A'$. So $A \subset R$. $\therefore r \geq \alpha$.  

 A: The theorem mentioned in your post is almost trivial. It says that if $\alpha=(A, A') $ is a Dedekind cut then $A$ contains all rationals smaller than $\alpha$ and the rest of rationals (those that are greater than or equal to $\alpha$) lie in $A'$.
The technical issue here is that to compare a rational number with a Dedekind cut requires you to replace the rational number $r$ by the corresponding Dedekind cut $\rho=(R, R') $ where $$R=\{x\in\mathbb {Q} \mid x<r\}, R'=\{x\in\mathbb {Q}\mid x\geq r\} =\mathbb{Q} - R$$ and then compare the cuts $\alpha$ and $\rho$.
By definition of a Dedekind if $\alpha=(A, A') $ and $r\in A$ then there is another rational $s\in A$ with $r<s$. And because of this the cut $\rho=(R, R') $ corresponding to $r$ described in last paragraph will satisfy $\rho<\alpha$. Similarly if $r'\in A'$ and $\rho'$ is the cut corresponding to $r'$ then $\rho'\geq \alpha$.
The theorem is thus an easy/trivial consequence of the definition of Dedekind cut. 
A: A Dedekind cut is always a set of "original" rational numbers, never of other Dedekind cuts. However, in order to consider the rational numbers as a subset of the real numbers, we have to characterise the Dedekind cuts which correspond to rational numbers. That is what they are doing here.
