Exercise 8.6.7

Question

\begin{enumerate}[label=(\alph*)] 
\item Show that $AB$ is a cut and that property (o5) holds.
\item Propose a good candidate for the multiplicative identity (1) on $\mathbf{R}$ and show that this works for all cuts $A \geq O$.
\item Show the distributive property (f5) holds for non-negative cuts.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item For (c1): $-1 \in AB$ so $AB \neq \emptyset$. If $a' \notin A$ and $b' \notin B$, then $a'b' \geq 0$, and for any $a\in A,\ b \in B$, $ab < ab' < a'b' \neq ab$ so $AB \neq \mathbf{Q}$ proving (c1). For (c2): suppose $r \in AB$ and let $q < r$. There are two cases to consider: $r < 0$ and $r = ab$. If $r < 0$, then $q < r < 0$ must be in $AB$. If $r = ab$, let's assume $q \geq 0$ (the case $q < 0$ is trivial), implying $ab > 0$. Then $$q = \left(a \frac{q}{ab}\right)(b) = \left(\frac{q}{b}\right)b$$ where since $\frac{q}{ab} < 1$ and is defined, $\frac{q}{b} < a$ and $\frac{q}{b} \in A$. Hence $q \in AB$, proving property (c2). For (c3): suppose $r \in AB$. If $r < 0$ then $r < r/2 < 0$ is also in $AB$. If $r = 0$ then that implies $0 \in A$ and $0 \in B$, which means we can find $a \in A$ and $b \in B$ where $a,b > 0$ and so $r < ab \in AB$. If $r > 0$ then $r = ab$, and we can find $a < a' \in A$ and $b < b' \in B$ with $ab < a'b' \in AB$, proving (c3) in all cases. For (o5): Note that the definition of $AB$ is of the form $K \cup O$ with $K$ some set, so $O \subseteq AB$ trivially. \item Let the multiplicative identity $I = {p \in \mathbf{Q} : p < 1}$. $1/2 \in I$ and $1/2 \notin O$ so $I \neq 0$. Let $A \geq O$ and let $a \in A$, and let $a < a' \in A$. Then $\frac{a}{a'} < 1$ so $\frac{a}{a'}\in I$, and $a = \left(\frac{a}{a'}\right) a' \in I$ implying $A \subseteq IA$. Now let $r \in IA$. If $r < 0$ then since $A \geq O$, $r \in A$. If $r \geq 0$, then we have $r = ia$ with $i \in I$ and $a \in A$, with $a \geq 0$. Then since $i < 1$, we have $ia < a \in A$ and so $r \in A$. Therefore $A = IA$. \item If $a \in X(Y+Z)$, then either $a < 0$ or $a = x(y+z)$, for some $x \in X$, $y \in Y$, and $z \in Z$. If $a < 0$, since $XY + XZ \geq O$, we have $a \in XY + XZ$. If $a = x(y+z)$, we have $a = xy + xz \in XY + XZ$, so either way $X(Y+Z) \subseteq XY + XZ$. Now suppose $a \in XY + XZ$. Then $a = r_1 + r_2$ where $r_1 \in XY$ and $r_2 \in XZ$. We have three cases to consider: $r_1,r_2 \geq 0$, $r_1,r_2 < 0$, and $r_1 < 0$ while $r_2 \geq 0$. (The case of $r_1 > 0$ and $r_2 < 0$ is equivalent to the last case, with $Y$ and $Z$ flipped, which is an acceptable operation since $Y+Z = Z+Y$.) If $r_1$ and $r_2$ are both negative, then $a < 0$ and $a \in X(Y+Z)$ trivially. If $r_1 \geq 0$ and $r_2 \geq 0$, then we have $a =x_1 y + x_2 z$ where $x_1, x_2 \in X$, $y\in Y$, $z \in Z$, and these four numbers are all non-negative. Let $x = \max\{x_1, x_2\}$, and note that $a \leq xy + xz = x(y+z) \in X(Y+Z)$; therefore $a \in X(Y+Z)$. If $r_1 < 0$ while $r_2 \geq 0$, then $a = r_1 + xz$ where $0 \leq x \in X$ and $0 \leq z \in Z$. If $x = 0$ then $0 > a \in X(Y+Z)$, so let's take $x > 0$. If $a < 0$ then clearly $a \in X(Y+Z)$. Otherwise, define $y = \frac{r_1}{x} < 0 $ and note $y \in Y$. We have $a = xy + xz = x(y+z) \geq 0$, so $y+z \geq 0$ and $a \in X(Y+Z)$. \end{enumerate}

Exercise 8.6.7

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Exercise 8.6.7

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