3a) Validity check: Let p = "the team is late" Let q = "the team can play the game" Let r = "the referee is here" The given statements can be translated as: 1. If p, then  q (If the team is late, then it cannot play the game) 2. If r, then q (If the referee is here, then the team can play the game) 3. p (The team is late) From statement 1 and 3, we can infer  q (The team cannot play the game). This does not necessarily imply that r (The referee is not here), so the argument is invalid. 3b) Showing equivalence: p ∧ q ⇐⇒ p To show this equivalence, we need to prove two implications: 1. p ∧ q ⇒ p: Assume p and q are true. Then since p is true, p ∧ q is also true. Therefore, p ∧ q implies p. 2. p ⇒ p ∧ q: Assume p is true. Then p ∧ q is true if both p and q are true. Since p is true by assumption, q can be any value. Therefore, p implies p ∧ q. 4) Mathematical induction: Base case: For n = 1, x^1 + y^1 = x + y which is divisible by (x + y). Inductive step: Assume x^n + y^n is divisible by (x + y) for some odd natural number n. Now consider x^(n+2) + y^(n+2): x^(n+2) + y^(n+2) = x^2 * x^n + y^2 * y^n = x^2 * x^n + y^2 * y^n + x^2 * y^n - x^2 * y^n = x^2(x^n + y^n) + y^n(x^2 - y^2) Since x and y are odd numbers, x^2 - y^2 is divisible by (x + y). Therefore, x^(n+2) + y^(n+2) is divisible by (x + y) for all odd natural numbers n ≥ 1. 5) Solving z^6 = 1: z^6 = 1 z = ±1, ±i, ±√3/2 ± i/2 6) Solving for the number of each type of box: Let x be the number of 20 kg boxes and y be the number of 25 kg boxes. We have: x + y = 50 20x + 25y = 1175 Solving these equations simultaneously, we get: x = 25 and y = 25 Therefore, there are 25 boxes of each type.