1. **Problem Statement:** Justify the properties in $\mathbb{R}$ using ordered field axioms: (i) Show that $x(-y) = -xy$. (ii) Show that if $0 < a < b$, then $0 < \frac{1}{a+b}$. (iii) Show that for $a \in \mathbb{R}$, $|x| \leq a$ if and only if $-a \leq x \leq a$. 2. **Property (i): $x(-y) = -xy$** - Recall the distributive law: $x(y + (-y)) = xy + x(-y)$. - Since $y + (-y) = 0$, we have $x \cdot 0 = xy + x(-y)$. - But $x \cdot 0 = 0$, so $0 = xy + x(-y)$. - Adding $-xy$ to both sides gives $-xy = x(-y)$. 3. **Property (ii): If $0 < a < b$, then $0 < \frac{1}{a+b}$** - Since $a > 0$ and $b > a > 0$, their sum $a + b > 0$. - The reciprocal of a positive number is positive, so $\frac{1}{a+b} > 0$. 4. **Property (iii): $|x| \leq a$ iff $-a \leq x \leq a$** - By definition, $|x| = x$ if $x \geq 0$, and $|x| = -x$ if $x < 0$. - If $|x| \leq a$, then: - If $x \geq 0$, $x \leq a$. - If $x < 0$, $-x \leq a$ implies $x \geq -a$. - Combining, $-a \leq x \leq a$. - Conversely, if $-a \leq x \leq a$, then $|x| \leq a$ by definition. 5. **Problem (b): Let $A$ and $B$ be bounded subsets of $\mathbb{R}$. Prove that if every $b \in B$ is an upper bound of $A$, then $\sup A \leq \inf B$.** - Since every $b \in B$ is an upper bound of $A$, $a \leq b$ for all $a \in A$ and $b \in B$. - By definition, $\sup A$ is the least upper bound of $A$, so $\sup A \leq b$ for all $b \in B$. - $\inf B$ is the greatest lower bound of $B$, so $\inf B \leq b$ for all $b \in B$. - Combining, $\sup A \leq b$ for all $b \in B$ and $\inf B \leq b$ for all $b \in B$. - Hence, $\sup A \leq \inf B$. **Final answers:** (i) $x(-y) = -xy$ by distributive law and additive inverses. (ii) If $0 < a < b$, then $0 < \frac{1}{a+b}$ since $a+b > 0$. (iii) $|x| \leq a$ if and only if $-a \leq x \leq a$ by definition of absolute value. (b) If every $b \in B$ is an upper bound of $A$, then $\sup A \leq \inf B$ by properties of sup and inf.