1. **Problem:** Find the power set of $A = \{(-1)^n : n \in \mathbb{N}\}$. The set $A$ consists of values $(-1)^n$ for natural numbers $n$. Since $(-1)^n$ alternates between $-1$ and $1$, $A = \{-1, 1\}$. The power set $P(A)$ is the set of all subsets of $A$. For a set with 2 elements, the power set has $2^2 = 4$ subsets: $$P(A) = \{\emptyset, \{-1\}, \{1\}, \{-1, 1\}\}$$ 2. **Problem:** Symbolic form of "If Yohannes neither studies hard nor attends class regularly then he will fail the test." Given propositions: $p$: Yohannes studies hard $q$: Yohannes passes the test $r$: Yohannes attends class regularly "Neither studies hard nor attends class regularly" means $\neg p \wedge \neg r$. "He will fail the test" means $\neg q$. The conditional statement is: $$ (\neg p \wedge \neg r) \Rightarrow \neg q $$ 3. **Problem:** Truth value of $(\exists x)(\forall y)(xy = x)$ where $U = \mathbb{R}$. We want to check if there exists an $x$ such that for all $y$, $xy = x$. Rewrite: $xy = x \implies x(y - 1) = 0$. For all $y$, this holds only if $x = 0$. Since $x=0$ satisfies this for all $y$, the statement is true. 4. **Problem:** If $(p \lor q) \Rightarrow q$ is false, find truth value of $(p \wedge q) \Leftrightarrow \neg p$. For $(p \lor q) \Rightarrow q$ to be false, antecedent true and consequent false: $p \lor q = T$ and $q = F$. So $q=F$ and $p=T$. Evaluate $(p \wedge q) \Leftrightarrow \neg p$: $p \wedge q = T \wedge F = F$, $\neg p = F$, So $F \Leftrightarrow F$ is true. 5. **Problem:** Truth value of $\neg (\exists x)(p(x) \wedge q(x))$ where $p(x): x \in \mathbb{N} \wedge x \leq 3$, $q(x): x \in \mathbb{N} \wedge x+1$ is odd. Check for $x$ with $1 \leq x \leq 3$: $x+1$ odd means $x$ even. Values: $x=1$: $x+1=2$ even, so $q(1)$ false. $x=2$: $x+1=3$ odd, so $q(2)$ true. $x=3$: $x+1=4$ even, so $q(3)$ false. So $p(2) \wedge q(2)$ is true. Therefore, $(\exists x)(p(x) \wedge q(x))$ is true, so its negation is false. 6. **Problem:** For any two sets $A$ and $B$, if $A \subseteq B$ then find $A \cap B'$. Since $A$ is subset of $B$, all elements of $A$ are in $B$. $B'$ is complement of $B$. So $A \cap B' = \emptyset$. 7. **Problem:** If $z = (-1 - i)^{12}$, find a) $\operatorname{Re}(z)$ b) $\operatorname{Im}(z)$. Express $-1 - i$ in polar form: Magnitude $r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$. Argument $\theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$ (since point is in third quadrant). So $z = r^{12} (\cos 12\theta + i \sin 12\theta) = (\sqrt{2})^{12} (\cos(12 \times \frac{5\pi}{4}) + i \sin(12 \times \frac{5\pi}{4}))$. Simplify: $(\sqrt{2})^{12} = (2^{1/2})^{12} = 2^{6} = 64$. $12 \times \frac{5\pi}{4} = 15\pi$. $\cos 15\pi = \cos(\pi) = -1$, since $15$ is odd. $\sin 15\pi = 0$. Therefore, $$z = 64(-1 + 0i) = -64$$ So a) $\operatorname{Re}(z) = -64$ b) $\operatorname{Im}(z) = 0$ 8. **Problem:** Let $S = \left\{ \frac{(n-4)^2 + 3}{n^2 - 5} : n \in \mathbb{N} \right\}$. Find a) $\operatorname{lub}(S)$ (least upper bound) b) $\operatorname{glb}(S)$ (greatest lower bound). For $n \in \mathbb{N}$, $n \geq 1$. Denominator $n^2 - 5$ is negative for $n=1,2$ and positive for $n \geq 3$. Calculate some values: $n=1$: numerator $(1-4)^2 + 3 = 9 + 3 = 12$, denominator $1 - 5 = -4$, value $= -3$. $n=2$: numerator $(2-4)^2 + 3 = 4 + 3 = 7$, denominator $4 - 5 = -1$, value $= -7$. $n=3$: numerator $(3-4)^2 + 3 = 1 + 3 = 4$, denominator $9 - 5 = 4$, value $= 1$. $n=4$: numerator $0 + 3 = 3$, denominator $16 - 5 = 11$, value $= \frac{3}{11} \approx 0.2727$. $n=5$: numerator $1 + 3 = 4$, denominator $25 - 5 = 20$, value $= 0.2$. As $n$ increases, denominator grows faster, so values approach 0 from above. So $S$ contains negative values (like -7) and positive values approaching 0. Maximum value is at $n=3$ which is 1. Minimum value is $-7$ at $n=2$. Therefore, a) $\operatorname{lub}(S) = 1$ b) $\operatorname{glb}(S) = -7$ 9. **Problem:** Given $\gcd(x, 360) = 30$ and $\operatorname{lcm}(x, 360) = 1800$, find $x$. Use formula: $$x \times 360 = \gcd(x, 360) \times \operatorname{lcm}(x, 360)$$ Substitute values: $$x \times 360 = 30 \times 1800$$ $$x = \frac{30 \times 1800}{360} = \frac{54000}{360} = 150$$ **Final answers:** 1. $P(A) = \{\emptyset, \{-1\}, \{1\}, \{-1,1\}\}$ 2. $(\neg p \wedge \neg r) \Rightarrow \neg q$ 3. True 4. True 5. False 6. $\emptyset$ 7. a) $-64$, b) $0$ 8. a) $1$, b) $-7$ 9. $150$