1. **Problem:** Given a survey of 25 cars with options air-conditioning (A), radio (R), and power windows (W), find various counts of cars with specific option combinations. **Step 1:** Use the principle of inclusion-exclusion and given data: - $|A|=15$, $|R|=12$, $|W|=11$ - $|A \cap W|=5$, $|A \cap R|=9$, $|R \cap W|=4$ - $|A \cap R \cap W|=3$ **Step 2:** Calculate only power windows: $$|W| - |A \cap W| - |R \cap W| + |A \cap R \cap W| = 11 - 5 - 4 + 3 = 5$$ **Step 3:** Calculate only air-conditioning: $$|A| - |A \cap W| - |A \cap R| + |A \cap R \cap W| = 15 - 5 - 9 + 3 = 4$$ **Step 4:** Calculate only radio: $$|R| - |A \cap R| - |R \cap W| + |A \cap R \cap W| = 12 - 9 - 4 + 3 = 2$$ **Step 5:** Radio and power windows but not air-conditioning: $$|R \cap W| - |A \cap R \cap W| = 4 - 3 = 1$$ **Step 6:** Air-conditioning and radio but not power windows: $$|A \cap R| - |A \cap R \cap W| = 9 - 3 = 6$$ **Step 7:** Only one option: Sum of only A, only R, only W = $4 + 2 + 5 = 11$ **Step 8:** At least one option: Use inclusion-exclusion: $$|A \cup R \cup W| = |A| + |R| + |W| - |A \cap R| - |A \cap W| - |R \cap W| + |A \cap R \cap W|$$ $$= 15 + 12 + 11 - 9 - 5 - 4 + 3 = 23$$ **Step 9:** None of the options: Total cars - at least one option = $25 - 23 = 2$ --- 2. **Problem:** Show that $\sqrt{p}$ is irrational for any prime $p$. **Step 1:** Assume $\sqrt{p}$ is rational, so $\sqrt{p} = \frac{a}{b}$ with integers $a,b$ coprime and $b \neq 0$. **Step 2:** Square both sides: $$p = \frac{a^2}{b^2} \implies a^2 = p b^2$$ **Step 3:** This implies $p | a^2$, so $p | a$ (since $p$ prime). **Step 4:** Let $a = p k$ for some integer $k$. **Step 5:** Substitute back: $$p^2 k^2 = p b^2 \implies p k^2 = b^2$$ **Step 6:** So $p | b^2$ and hence $p | b$. **Step 7:** But $a$ and $b$ share factor $p$, contradicting coprimality. **Step 8:** Therefore, $\sqrt{p}$ is irrational. --- 3. **Problem:** Prove divisibility properties. (a) If $a | b$, then $a | -b$, $-a | b$, and $-a | -b$. **Step 1:** $a | b$ means $b = a k$ for some integer $k$. **Step 2:** Then $-b = a (-k)$, so $a | -b$. **Step 3:** Also, $b = (-a)(-k)$, so $-a | b$. **Step 4:** Similarly, $-b = (-a) k$, so $-a | -b$. (b) If $ac | bc$, then $a | b$. **Step 1:** $ac | bc$ means $bc = ac m$ for some integer $m$. **Step 2:** Simplify: $bc = a c m \implies b = a m$ (assuming $c \neq 0$). **Step 3:** Hence, $a | b$. --- 4. **Problem:** Determine boundedness of sets. (a) $A = \{x : x = \frac{1}{n}, n \in P\}$ where $P$ is primes. - Since primes $n \geq 2$, $x \leq \frac{1}{2}$. - $A$ is bounded above by $\frac{1}{2}$ and bounded below by $0$ (since $1/n > 0$). - So $A$ is bounded, bounded above and below. (b) $B = \{x : x = 3^n, n \in P\}$. - Since primes $n$ grow without bound, $3^n$ grows without bound. - So $B$ is unbounded above. - Also, $3^n > 0$ for all $n$, so bounded below by $0$. - So $B$ is bounded below but unbounded above. (c) $C = \{x : x = \frac{1}{n}, n \in \mathbb{Z}, n \neq 0\}$. - As $n \to \pm \infty$, $x \to 0$. - Values range over positive and negative fractions. - So $C$ is bounded above by $1$ (when $n=1$) and bounded below by $-1$ (when $n=-1$). - Hence, $C$ is bounded, bounded above and below.