# 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 $$(1.8)$$.

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

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 $$R=\{x\in\mathbb {Q} \mid x and then compare the cuts $$\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. 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 $$\rho'\geq \alpha$$.