Exercise 8.6.9

Question

Consider the collection of so-called ``rational'' cuts of the form
$$C_r = \{t \in Q: t < r\}$$
where $r \in \mathbf{Q}$. (See Exercise 8.6.1.)
\begin{enumerate}[label=(\alph*)] 
\item Show that $C_r + C_s = C_{r+s}$ for all $r, s \in \mathbf{Q}$. Verify $C_rC_s = C_{rs}$ for the case when $r, s \geq 0$.
\item Show that $C_r \leq C_s$ if and only if $r \leq s$ in $\mathbf{Q}$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item Let $a + b \in C_r + C_s$ satisfying $a \in C_r$ and $b \in C_s$. Then $a + b < r + s$ so $a + b \in C_{r+s}$. Now let $c \in C_{r+s}$. Let $\delta = r+s - c > 0$. Then $d = r - \delta/2 \in C_r$, $e = s - \delta / 2 \in C_s$, and $c = d + e \in C_r + C_s$. Hence $C_r + C_s = C_{r+s}$. Let $x \in C_r C_s $. Since $rs \geq 0$, if $x < 0$ then $x \in C_{rs}$. If $x \geq 0$, $x = ab$ where $a \in C_r$ and $b \in C_s$. Then $a < r$ and $b < s$, implying $ab < rs$ and $x \in C_{rs}$. Now let $c \in C_{rs}$. If $c < 0$ then $c \in C_r C_s$ trivially; otherwise note $rs > 0$ and define $q = \frac{c+rs}{2}$, so $0 \leq c < q < rs$. Define $0 \leq \delta_1 = \frac{c}{q} < 1$ and $\delta_2 = \frac{q}{rs} < 1$. Then $\delta_1r \in C_r$, $\delta_2s \in C_s$, and $c = \delta_1 r \delta_2 s \in C_{rs}$. Therefore $C_rC_s = C_{rs}$. \item Suppose $s < r$. Define $a = \frac{s+r}{2} < r \in C_r$ but $a > s$ so $a \notin C_s$ and $C_r \nleq C_s$. Equivalently, $C_r \leq C_s$ implies $r \leq s$. Now suppose $r \leq s$. Then $a \in C_r$ implies $a < r \leq s$ so $a < s$, $a \in C_r$, and $C_r \subseteq C_s$. \end{enumerate}

Exercise 8.6.9

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Exercise 8.6.9

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