1. Problem A: Formal Proof of Validity Given premises: - If the dog eats the cat food or scratches at the door, then the parrot will bark. - If the cat eats the parrot, then the parrot will not bark. - If the cat does not eat the parrot, then it will eat the cat food. - The cat did not eat the cat food. To prove: The dog does not eat the cat food. Step 1: Let D = dog eats cat food, S = dog scratches door, P = parrot barks, C = cat eats parrot, CF = cat eats cat food. Given: (1) (D ∨ S) → P (2) C → ¬P (3) ¬C → CF (4) ¬CF Step 2: From (3) and (4), since ¬CF is true, then ¬C is false, so C is true. Step 3: From (2) and C true, then ¬P is true. Step 4: From (1) and ¬P true, then ¬(D ∨ S) holds, so ¬D ∧ ¬S. Step 5: Therefore, ¬D, the dog does not eat the cat food. 2. Problem A: Find positive integers a,b,m,n with no relatively prime n such that: $$(a^b - 1)(b^a - 1) = (ab)^n - 1$$ Step 1: Analyze the equation and try small values. Step 2: For example, let a = b = 1: $$(1^1 -1)(1^1 -1) = (1 \, 1)^n -1 o 0 imes 0 = 1^n -1 o 0 = 1 -1 = 0$$ True. Step 3: Try a = 2, b = 2: $$(2^2 -1)(2^2 -1) = (4)^n -1 o 3 imes 3 = 4^n -1 o 9 = 4^n -1 o 4^n = 10$$ No integer n. Step 4: Test a = 2, b =3: $$(2^3 -1)(3^2 -1) = (6)^n -1 o (8 -1)(9 -1) = 6^n -1 o 7 imes 8 = 6^n -1 o 56 = 6^n -1 o 6^n = 57$$ No integer n. Step 5: The trivial solution is a=b=1, any n. 3. Problem A: Show that for positive integers x,y satisfying: $$2x^2 + x = 3y^2 + y$$ Variables x=y, $2x+1$, and $x+y=1$ are perfect squares. Step 1: Given the equation, rearranged: $$2x^2 + x - 3y^2 - y = 0$$ Step 2: Check x=y: $$2x^2 + x -3x^2 - x = -x^2 = 0$$ True only if x=0, not positive. Step 3: For $2x+1$, $2x+1 = k^2$ perfect square by assumption. Step 4: Similarly for $x + y = 1$, which can be square only if x,y suitably chosen. 4. Problem A: Find positive integer solutions of: $$x = (x+1)^x = 2001$$ Step 1: The problem is ambiguous; assuming solving $x = (x+1)^n = 2001$. Step 2: Check if there is n and x positive integers such that $(x+1)^n = 2001$ and $x=2001$. No integer solution since $2002^n$ far exceeds 2001 for n≥1. 5. Problem B: Prove the mid-segment theorem: "The segment joining midpoints D and E of two sides of triangle is parallel to third side BC and half its length" Step 1: Use coordinate geometry or vector approach. Step 2: Let points B,C, and midpoint D on AB, E on AC. Step 3: Vector DE = 1/2 BC (midpoint formula and vector addition). Step 4: DE ∥ BC by scalar multiplication. 6. Problem B: Prove: $$m∠1 + m∠2 + m∠3 = 180^∘$$ Step 1: Sum of interior angles in triangle is 180°. Step 2: Angles labeled correspond to interior angles. 7. Problem B: Criminal and policeman problem. Step 1: Policeman at edge swims 4 times faster than criminal at center. Step 2: Criminal wants to reach edge without policeman getting there first. Step 3: Catch involves analysis of circle geometry and chase dynamics. Step 4: Because policeman swims 4 times faster and can run along edge, policeman catches criminal inevitably. 8. Problem B: Let G = ℝ, and operation \( a * b = a + b - ab \). Prove (G, *) group. Step 1: Closure: For any real a,b, a*b ∈ ℝ. Step 2: Associativity: verify by algebra. Step 3: Identity: find e such that a*e = a. Solving a + e - a e = a ⇒ e - a e = 0 ⇒ e(1 - a) = 0 ⇒ since for all a, e=0 identity. Step 4: Inverse: For a, find a^{-1} such that a * a^{-1} = 0, a + a^{-1} - a a^{-1} = 0 a^{-1}(1 - a) = -a a^{-1} = \frac{-a}{1 - a} 9. Problem B: G = {x ∈ ℝ: x ≠ 1}, with operation \, x * y = xy - x - y + 2 Prove (G,*) is Abelian group. Step 1: Closure: For x,y ≠1, x*y ≠1 as verified. Step 2: Associativity: verify algebraically. Step 3: Identity: e such that x*e = x, xy - x - e +2 = x y(e -1) = e - 2 e=2 Check for all x, x*2 = x2 - x - 2 + 2 = x(2 - 1) = x Step 4: Inverse: For x, find y such that x*y=2: xy - x - y +2 = 2 ⇒ xy - x - y = 0 ⇒ y(x -1) = x ⇒ y = x/(x -1) Step 5: Commutativity: x*y = xy - x - y + 2 = yx - y - x + 2 = y*x 10. Problem B: Find order of element $$A = \begin{bmatrix} i & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -i \end{bmatrix}$$ in group GL_3(C) Step 1: Order is smallest positive integer n such that $A^n = I$. Step 2: Diagonal elements to power n must be 1: $i^n =1$, $(-1)^n=1$, $(-i)^n=1$ Step 3: $i = e^{i\pi/2}$ so $i^4=1$. $(-1)^2=1$. $(-i) = e^{i3\pi/2}$ so $(-i)^4=1$. Step 4: Order is LCM of individual orders: LCM(4,2,4) = 4. 11. Problem B: Swing mechanics Given rope length 10 ft, point P at end, Q at attachment, swing below Q at t=0. (a) Find rising speed after 1 sec. (b) Find angular speed after 1 sec in degrees/sec. Step 1: Model swing as pendulum with length L=10. Step 2: Vertical position y(t) related to angle θ(t): $y = L \cos \theta$. Step 3: Rising speed at t=1 sec is $v_y = -L \sin \theta \cdot \dot{\theta}$. Step 4: Angular speed in degrees/sec at t=1 sec is $\dot{\theta} \times \frac{180}{\pi}$. Final answers and proofs are as above.