1. **Problem:** Determine which statement about rational and irrational numbers is true. 2. **Step 1:** Recall definitions: - Rational numbers can be expressed as a fraction of two integers. - Irrational numbers cannot be expressed as a simple fraction. 3. **Step 2:** Evaluate each option: - A: The product of two irrational numbers is not always irrational (e.g., $\sqrt{2} \times \sqrt{2} = 2$, which is rational). - B: The sum of any two rational numbers is always rational (sum of fractions is a fraction). - C: The product of two irrational numbers is not always rational (e.g., $\sqrt{2} \times \sqrt{3} = \sqrt{6}$, irrational). - D: The sum of two irrational numbers is not always irrational (e.g., $\sqrt{2} + (-\sqrt{2}) = 0$, rational). 4. **Answer:** B is true. 1. **Problem:** Identify the true assertion applying mathematical induction. 2. **Step 1:** Check each formula: - A: $1 + 2 + 2^2 + ... + 2^n = 2^{n+1} - 1$ is a known geometric series sum. - B: $3 + 5 + 7 + ... + (2n + 1) = n^2$ is false; the sum of odd numbers starting at 1 equals $n^2$, but here it starts at 3. - C: $2 + 3 + 4 + ... + n = \frac{n(n+1)}{2} + 1$ is false; sum from 1 to n is $\frac{n(n+1)}{2}$. - D: $n! \geq 2^n$ for $n \geq 0$ is false for small n. 3. **Answer:** A is true. 1. **Problem:** Find the number of subsets of set $A = \{ x | x \text{ is an even number between 17 and 37} \}$. 2. **Step 1:** List even numbers between 17 and 37: $18, 20, 22, 24, 26, 28, 30, 32, 34, 36$ (10 elements). 3. **Step 2:** Number of subsets of a set with $n$ elements is $2^n$. 4. **Answer:** $2^{10} = 1024$ subsets, so option D. 1. **Problem:** Given $n(A \cup B) = 50$, $n(A \cap B) = 15$, and $n(A) = 10$, find $n(B \backslash A)$. 2. **Step 1:** Use formula: $$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$ 3. **Step 2:** Substitute known values: $$50 = 10 + n(B) - 15 \Rightarrow n(B) = 55$$ 4. **Step 3:** Find $n(B \backslash A) = n(B) - n(A \cap B) = 55 - 15 = 40$. 1. **Problem:** Find the truth value of $\neg ( \neg p \Rightarrow q) \lor ( \neg q \land p)$ given $q$ is true and $p$ is any. 2. **Step 1:** Since $q$ is true, $\neg q$ is false. 3. **Step 2:** $\neg p \Rightarrow q$ is always true because $q$ is true. 4. **Step 3:** So $\neg (\neg p \Rightarrow q)$ is false. 5. **Step 4:** $\neg q \land p$ is false (since $\neg q$ is false). 6. **Step 5:** The whole expression is $false \lor false = false$. 1. **Problem:** Find the number of eggs between 50 and 60 that leave remainder 2 when divided by 3 and remainder 4 when divided by 5. 2. **Step 1:** List numbers between 50 and 60. 3. **Step 2:** Check each for $x \equiv 2 \pmod{3}$ and $x \equiv 4 \pmod{5}$. 4. **Step 3:** 59 satisfies both conditions. 5. **Answer:** 59 eggs. 1. **Problem:** If $x$ is odd and $LCM(x, 80) = 30800$, find $x$. 2. **Step 1:** Factor 80: $80 = 2^4 \times 5$. 3. **Step 2:** Factor 30800: $30800 = 2^4 \times 5^2 \times 7 \times 11$. 4. **Step 3:** Since $x$ is odd, it has no factor 2. 5. **Step 4:** $x$ must include factors $5^2 \times 7 \times 11 = 25 \times 7 \times 11 = 1925$. 6. **Answer:** $x = 1925$. 1. **Problem:** Find $\sup(A)$ and $\inf(A)$ for $A = \{ 2 + \frac{(-1)^n}{3n} \}$. 2. **Step 1:** When $n$ is even, $(-1)^n = 1$, so terms are $2 + \frac{1}{3n}$ decreasing to 2 from above. 3. **Step 2:** When $n$ is odd, $(-1)^n = -1$, so terms are $2 - \frac{1}{3n}$ increasing to 2 from below. 4. **Step 3:** So $\sup(A) = 2 + \frac{1}{3 \times 2} = 2 + \frac{1}{6} = \frac{13}{6}$ (largest term at $n=2$). 5. **Step 4:** $\inf(A) = 2 - \frac{1}{3 \times 1} = 2 - \frac{1}{3} = \frac{5}{3}$ (smallest term at $n=1$). **Final answers:** 8. B 9. A 10. D Part II: 1. 40 2. false 3. 59 4. 1925 5. $\sup(A) = \frac{13}{6}$, $\inf(A) = \frac{5}{3}$