1. **Problem statement:** We have the set $$S = \{x \in \mathbb{Q} \mid x = \frac{a}{b}, a,b \in \mathbb{Z}, 0 < b \leq 11\} \cap (-1,1)$$. (c) Find the cardinality $$|S|$$ of $$S$$. (d) Show that $$\sup S$$ exists and determine it with proof. (e) Given $$f(x) = |x - \sqrt{2}|$$, find $$x^* \in S$$ that minimizes $$f(x)$$ and prove uniqueness. --- 2. **Step (c): Find cardinality $$|S|$$** - $$S$$ consists of rational numbers $$\frac{a}{b}$$ with integers $$a,b$$, $$0 < b \leq 11$$, and $$x \in (-1,1)$$. - For each denominator $$b$$ from 1 to 11, find all integers $$a$$ such that $$-1 < \frac{a}{b} < 1$$. - This implies $$-b < a < b$$. - Since $$a$$ is integer, $$a \in \{-b+1, -b+2, ..., b-1\}$$. - Number of such $$a$$ for each $$b$$ is $$2b - 1$$. - Total cardinality: $$|S| = \sum_{b=1}^{11} (2b - 1) = 2 \sum_{b=1}^{11} b - \sum_{b=1}^{11} 1 = 2 \cdot \frac{11 \cdot 12}{2} - 11 = 11 \cdot 12 - 11 = 132 - 11 = 121$$. 3. **Step (d): Show $$\sup S$$ exists and find it** - $$S \subset (-1,1)$$, so $$S$$ is bounded above by 1. - Since $$S$$ contains rationals arbitrarily close to 1 but less than 1 (e.g., $$\frac{b-1}{b}$$ for $$b \leq 11$$), the supremum is 1. - However, 1 is not in $$S$$ because $$x < 1$$. - Therefore, $$\sup S = 1$$. - Proof: For any $$\epsilon > 0$$, choose $$b$$ such that $$\frac{b-1}{b} > 1 - \epsilon$$ and $$b \leq 11$$, so $$\frac{b-1}{b} \in S$$ and $$\frac{b-1}{b} > 1 - \epsilon$$. - Hence, 1 is the least upper bound. 4. **Step (e): Find $$x^* \in S$$ minimizing $$f(x) = |x - \sqrt{2}|$$ and prove uniqueness** - $$\sqrt{2} \approx 1.4142$$, but $$S \subset (-1,1)$$, so all $$x \in S$$ satisfy $$x < 1$$. - The closest $$x \in S$$ to $$\sqrt{2}$$ is the maximum $$x$$ in $$S$$, which is $$\sup S = 1$$. - Since $$1 \in S$$ (because $$1 = \frac{1}{1}$$ and $$1 < 1$$ is false, so actually 1 is not in $$S$$ because $$x < 1$$), so 1 is not in $$S$$. - The maximum element in $$S$$ is the largest rational less than 1 with denominator $$\leq 11$$, which is $$\frac{10}{11} \approx 0.9091$$. - So $$x^* = \frac{10}{11}$$ minimizes $$f(x)$$. - Proof of minimality: For any $$x \in S$$, $$x \leq \frac{10}{11}$$, so $$|x - \sqrt{2}| \geq |\frac{10}{11} - \sqrt{2}|$$. - Uniqueness: Suppose another $$x' \in S$$ also minimizes $$f$$, then $$|x' - \sqrt{2}| = |x^* - \sqrt{2}|$$. - Since $$\sqrt{2} > 1$$ and $$x^*$$ is the closest rational less than 1, any other $$x' < x^*$$ will have larger distance. - Hence, $$x^* = \frac{10}{11}$$ is unique minimizer. --- **Final answers:** - (c) $$|S| = 121$$ - (d) $$\sup S = 1$$ - (e) $$x^* = \frac{10}{11}$$ uniquely minimizes $$f(x)$$.