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 ## Student Responses
 
-<!-- Write your solutions below. Add sections for each problem as needed. -->
-<!-- Use proper mathematical notation where applicable. -->
+---
 
-[Your solutions here]
+## 1. Algorithmic Thinking [20 pts]
+
+Wayne Zachary's 1977 paper "An information flow model for conflict and fission in small groups" is a seminal work on predicting social network structure changes. Here is the analysis through the five-step algorithmic thinking process:
+
+**Understanding the Problem:**
+Zachary observed a karate club at a university that experienced a real conflict and eventually split into two groups. His question was: can we predict which members would join which group after the split based on the social network structure before the split? The motivation was to understand how information flow and social ties influence group dynamics and fission. He collected data on the social relationships (friendships) among the 34 club members and recorded which group each member joined after the split. He thought of the problem in terms of network structure—members with stronger ties to influential people should follow those people to their respective groups.
+
+**Formulating the Problem:**
+Zachary formulated this as a graph partitioning problem. The members are vertices in a graph, the friendships are weighted edges (with weights based on interaction strength), and the goal is to partition the vertices into two groups such that the partition matches the actual split observed. This is equivalent to the community detection problem: find the optimal division of a network into clusters that minimizes the edges between groups and maximizes edges within groups.
+
+**Algorithm Design:**
+Zachary used a spectral clustering technique based on network modularity. He computed structural properties of the network and used information flow models to determine which members should be in which group. The algorithmic technique employed was spectral graph partitioning—analyzing the eigenvalues and eigenvectors of the graph Laplacian to identify natural divisions in the network structure.
+
+**Algorithm Implementation:**
+The algorithm was implemented by calculating information flow patterns and network structure metrics for each member. Zachary used graph-based measures to score the likelihood of each member belonging to each group, then assigned members to maximize agreement with the observed split.
+
+**Solving the Original Problem:**
+Zachary's findings showed that his network-based algorithm could predict the club split with very high accuracy (over 93% of members were correctly predicted). This validated his hypothesis that social network structure and information flow patterns strongly determine how groups fragment. The paper demonstrated that algorithmic approaches could effectively model real social phenomena, making it foundational for network analysis and community detection algorithms.
+
+---
+
+## 2. Propositional Logic [30 pts]
+
+### Problem 2.1: Converting Sentences to Propositional Logic
+
+**(a) "Neither the wind nor the rain affected the house."**
+
+Propositions:
+- p = wind affected the house
+- q = rain affected the house
+
+Compound proposition: **¬p ∧ ¬q** (equivalently: ¬(p ∨ q))
+
+Explanation: "Neither...nor" means both conditions are false.
+
+**(b) "If the pandemic is not controlled, more people will die."**
+
+Propositions:
+- p = pandemic is controlled
+- q = more people will die
+
+Compound proposition: **¬p → q**
+
+Explanation: This is a direct conditional statement where the antecedent is the negation of pandemic control.
+
+**(c) "Pedestrians should neither jaywalk nor cross a red light, or they will get a ticket."**
+
+Propositions:
+- j = pedestrians jaywalk
+- c = pedestrians cross a red light
+- t = pedestrians get a ticket
+
+Compound proposition: **(j ∨ c) → t**
+
+Explanation: "Should neither...nor, or they will get a ticket" means if they do either prohibited action, they will face consequences.
+
+**(d) "A plant is not poisonous unless it is tropical."**
+
+Propositions:
+- p = plant is poisonous
+- t = plant is tropical
+
+Compound proposition: **p → t** (equivalently: ¬t → ¬p)
+
+Explanation: "Unless" indicates a conditional—a plant is only poisonous if it is tropical.
+
+**(e) "Growing in the tropic is necessary but not sufficient for a plant to be poisonous."**
+
+Propositions:
+- t = plant grows in the tropic
+- p = plant is poisonous
+
+Compound proposition: **p → t** (with the understanding that t ↛ p, meaning tropical growth does not guarantee poisoning)
+
+Explanation: Necessary means if poisonous then tropical (p → t). Not sufficient means tropical does not guarantee poisonous (¬(t → p)).
+
+---
+
+### Problem 2.2: Satisfiability and Tautology of XOR Propositions
+
+**(a) p XOR p**
+
+**Answer: Unsatisfiable**
+
+**Proof:**
+XOR is true when exactly one operand is true. For p XOR p:
+- When p = T: p XOR p = T XOR T = F (both true, not exactly one)
+- When p = F: p XOR p = F XOR F = F (both false, not exactly one)
+
+Since p XOR p is false under all interpretations, it is unsatisfiable.
+
+**(b) p XOR (NOT p)**
+
+**Answer: Satisfiable and Tautology**
+
+**Proof:**
+- When p = T: p XOR ¬p = T XOR F = T (exactly one is true)
+- When p = F: p XOR ¬p = F XOR T = T (exactly one is true)
+
+Since p XOR ¬p is true under all interpretations, it is a tautology (and therefore satisfiable).
+
+**(c) (p XOR q) OR (p XOR (NOT q))**
+
+**Answer: Satisfiable and Tautology**
+
+**Proof:**
+For any values of p and q:
+- When p = T, q = T: (T XOR T) OR (T XOR F) = F OR T = T
+- When p = T, q = F: (T XOR F) OR (T XOR T) = T OR F = T
+- When p = F, q = T: (F XOR T) OR (F XOR F) = T OR F = T
+- When p = F, q = F: (F XOR F) OR (F XOR T) = F OR T = T
+
+Since the expression is true under all interpretations, it is a tautology.
+
+---
+
+### Problem 2.3: Logical Equivalence of Compound Propositions
+
+**(a) (p AND q) → r and (p → r) AND (q → r)**
+
+**Answer: Not Logically Equivalent**
+
+**Proof by counterexample:**
+Let p = T, q = F, r = F:
+- (p ∧ q) → r = (T ∧ F) → F = F → F = **T**
+- (p → r) ∧ (q → r) = (T → F) ∧ (F → F) = F ∧ T = **F**
+
+Since the two expressions have different truth values under this interpretation, they are not logically equivalent.
+
+**(b) (p OR q) → r and (p → r) OR (q → r)**
+
+**Answer: Not Logically Equivalent**
+
+**Proof by counterexample:**
+Let p = T, q = F, r = F:
+- (p ∨ q) → r = (T ∨ F) → F = T → F = **F**
+- (p → r) ∨ (q → r) = (T → F) ∨ (F → F) = F ∨ T = **T**
+
+Since the two expressions have different truth values under this interpretation, they are not logically equivalent.
+
+**(c) p → (q ↔ r) and (p → q) ↔ (p → r)**
+
+**Answer: Logically Equivalent**
+
+**Proof by truth table:**
+
+| p | q | r | p→(q↔r) | (p→q)↔(p→r) |
+|---|---|---|---------|------------|
+| T | T | T | T→T = T | T↔T = T |
+| T | T | F | T→F = F | T↔F = F |
+| T | F | T | T→F = F | F↔T = F |
+| T | F | F | T→T = T | F↔F = T |
+| F | T | T | F→T = T | T↔T = T |
+| F | T | F | F→F = T | T↔F = F |
+| F | F | T | F→F = T | F↔T = F |
+| F | F | F | F→T = T | F↔F = T |
+
+Wait, checking row 5: (p → q) ↔ (p → r) = (F → T) ↔ (F → T) = T ↔ T = T ✓
+Row 6: (F → T) ↔ (F → F) = T ↔ T = T, but p→(q↔r) = F→(T↔F) = F→F = T ✓
+Row 7: (F → F) ↔ (F → T) = T ↔ T = T, but p→(q↔r) = F→(F↔T) = F→F = T ✓
+
+Checking all rows: both columns are identical, so the propositions are **logically equivalent**.
+
+---
+
+## 3. Predicate Logic [30 pts]
+
+### Problem 3.1: Truth Values of Predicate Statements
+
+**(a) There exists x such that x³ = -1**
+
+**Truth Value: TRUE**
+
+**Explanation:** In the real numbers, x = -1 satisfies the equation: (-1)³ = -1. Therefore, the statement is true.
+
+**(b) For all x, (-x)² = x²**
+
+**Truth Value: TRUE**
+
+**Explanation:** For any real number x, we have (-x)² = (-1·x)² = (-1)²·x² = 1·x² = x². This is true for all real x.
+
+**(c) For all x, 2x > x**
+
+**Truth Value: FALSE**
+
+**Explanation:** This is false because when x = 0, we have 2(0) = 0, which is not greater than 0. Also, for negative x, 2x < x. So the statement fails for multiple values.
+
+**(d) For all x, there exists y such that x·y = y·x**
+
+**Truth Value: TRUE**
+
+**Explanation:** For any real number x, we can choose y = 1 (or any real number, by commutativity). Then x·y = x·1 = x and y·x = 1·x = x, so x·y = y·x. Multiplication of real numbers is commutative.
+
+**(e) For all x, for all y, there exists z such that z = (x + y)/2**
+
+**Truth Value: TRUE**
+
+**Explanation:** For any two real numbers x and y, the midpoint z = (x + y)/2 always exists and is a real number. This is the arithmetic mean of x and y.
+
+**(f) There exists x such that for all y, x ≤ y²**
+
+**Truth Value: TRUE**
+
+**Explanation:** Choose x = 0. For all real numbers y, we have y² ≥ 0, so 0 ≤ y². The statement is true.
+
+**(g) There exists x such that for all y, x ≤ y**
+
+**Truth Value: FALSE**
+
+**Explanation:** This would require x to be less than or equal to every real number. No such x exists. For any candidate x, we can find y = x - 1, which gives x > y, violating the requirement.
+
+---
+
+### Problem 3.2: Logical Equivalence of Quantified Expressions
+
+**(a) ∀x(P(x) ∨ Q(x)) and (∀xP(x)) ∨ (∀xQ(x))**
+
+**Answer: Not Logically Equivalent**
+
+**Proof by counterexample:**
+Let the domain be the set of integers. Define:
+- P(x): "x is even"
+- Q(x): "x is odd"
+
+For the left side: ∀x(P(x) ∨ Q(x)) means every integer is either even or odd, which is **TRUE**.
+
+For the right side: (∀xP(x)) ∨ (∀xQ(x)) means either all integers are even or all integers are odd, which is **FALSE**.
+
+Since the two expressions have different truth values under this interpretation, they are not logically equivalent.
+
+**(b) ∀x(P(x) ∧ Q(x)) and (∀xP(x)) ∧ (∀xQ(x))**
+
+**Answer: Logically Equivalent**
+
+**Proof:**
+- Left side: ∀x(P(x) ∧ Q(x)) is true iff for every element in the domain, both P(x) and Q(x) are true.
+- Right side: (∀xP(x)) ∧ (∀xQ(x)) is true iff P(x) is true for all x AND Q(x) is true for all x.
+
+These two statements express the same condition—that both predicates hold for every element. Therefore, they are logically equivalent. Both sides have the same truth value under any interpretation.
+
+**(c) (∀xP(x)) ∨ (∀xQ(x)) and ∀x∀y(P(x) ∨ Q(y))**
+
+**Answer: Not Logically Equivalent**
+
+**Proof by counterexample:**
+Let the domain be {0, 1}. Define:
+- P(x): "x = 0"
+- Q(x): "x = 1"
+
+For the left side: (∀xP(x)) ∨ (∀xQ(x)) means either all elements equal 0 or all elements equal 1, which is **FALSE** (we have both 0 and 1).
+
+For the right side: ∀x∀y(P(x) ∨ Q(y)) means for every pair of elements x and y, either x = 0 or y = 1. Let me check all pairs:
+- x=0, y=0: P(0) ∨ Q(0) = T ∨ F = T
+- x=0, y=1: P(0) ∨ Q(1) = T ∨ T = T
+- x=1, y=0: P(1) ∨ Q(0) = F ∨ F = **F**
+
+Since we have a pair (x=1, y=0) where P(x) ∨ Q(y) is false, the right side is **FALSE**.
+
+Both sides are false in this case, so let me try another example:
+
+Let domain be {0, 1} with:
+- P(x): "x = 0"
+- Q(x): "x ≥ 0"
+
+Left side: (∀x(x=0)) ∨ (∀x(x≥0)) = F ∨ T = **T**
+
+Right side: ∀x∀y(P(x) ∨ Q(y)) = ∀x∀y(x=0 ∨ y≥0)
+- All y satisfy y ≥ 0 in our domain, so Q(y) is always true
+- Thus P(x) ∨ Q(y) is always true
+- Right side is **T**
+
+Let me try: P(x) = "x = 0", Q(x) = "x = 2"
+
+Left: (∀x(x=0)) ∨ (∀x(x=2)) = F ∨ F = **F**
+
+Right: ∀x∀y((x=0) ∨ (y=2))
+- When x=1, y=0: (1=0) ∨ (0=2) = F ∨ F = **F**
+
+So both are false. Try: P(x) = "x is even", Q(x) = "x > 0" on domain {0,1}
+
+Left: (∀x(x even)) ∨ (∀x(x>0)) = (T ∧ F) ∨ (F ∧ T) = F ∨ F = **F**
+
+Right: ∀x∀y(P(x) ∨ Q(y)) where P(x) = "x is even", Q(y) = "y > 0"
+- x=0: even (T), x=1: odd (F)
+- y=0: not >0 (F), y=1: >0 (T)
+- Check: x=1,y=0: (F∨F) = **F**
+
+So both false again. Actually, right side is weaker (easier to satisfy). Let me think more carefully:
+
+Left: (∀xP(x)) ∨ (∀xQ(x)) requires one predicate to hold everywhere
+Right: ∀x∀y(P(x) ∨ Q(y)) requires that for any x and y, at least one is true
+
+Right is clearly weaker. When is right true but left false?
+
+If right is true: ∀x∀y(P(x) ∨ Q(y)), meaning for any x, if P(x) is false, then Q(y) is true for all y.
+If left is false: ¬(∀xP(x)) ∧ ¬(∀xQ(x)), meaning some x has ¬P(x) and some y has ¬Q(y).
+
+These can both be true. Example with domain {0,1}: P(x) = "x=0", Q(x) = "x=1"
+
+Right: ∀x∀y((x=0) ∨ (y=1))
+- x=0: P(0)=T, so T∨Q(y)=T for all y ✓
+- x=1: P(1)=F, so need Q(y)=T for all y. But Q(0)=F ✗
+
+So right side is FALSE.
+
+Let me try a working counterexample: domain {a,b}, P(x)="x=a", Q(x)="x=b"
+
+Right: ∀x∀y((x=a) ∨ (y=b))
+- x=a: True for all y
+- x=b, y=a: (b=a) ∨ (a=b) = F ∨ F = F ✗
+
+Still doesn't work. The right side requires that if something is not P, then everything must be Q.
+
+Actually, I realize: Right side is ∀x∀y(P(x) ∨ Q(y)) is true iff: for each x where P(x) is false, there must be a y where Q(y) is true. But it's for ALL y, so we need Q(y) true for all y when any P(x) is false.
+
+This means: if any P(x) is false, then Q must be true everywhere. So right is true iff [P is true everywhere] OR [Q is true everywhere], which is the left side!
+
+Wait, that can't be right. Let me reconsider:
+
+∀x∀y(P(x) ∨ Q(y)) is true iff there do not exist x, y such that P(x) is false and Q(y) is false.
+
+∀x: ¬(∃y: ¬Q(y)) OR ∀y: ¬(∃x: ¬P(x))
+Which means: (∃y: ¬Q(y)) implies ∀x: P(x)
+
+Actually, simpler: ∀x∀y(P(x) ∨ Q(y)) ≡ (∀xP(x)) ∨ (∀yQ(y))
+
+Let me verify:
+- Suppose ∀xP(x) is true. Then P(x) ∨ Q(y) is true for all y.
+- Suppose ∀yQ(y) is true. Then P(x) ∨ Q(y) is true for all x.
+- Conversely, suppose ∀x∀y(P(x) ∨ Q(y)). If ∀xP(x) is false, then ∃x: ¬P(x). For this x, P(x) ∨ Q(y) must hold for all y, so Q(y) must hold for all y. Thus ∀yQ(y).
+
+So they are logically equivalent! But the problem asks if they are equivalent. Let me re-read: "(∀xP(x)) ∨ (∀xQ(x))" vs "∀x∀y(P(x) ∨ Q(y))"
+
+The right side has different variable names: x in first part, y in second. But they're both universally quantified over the same domain. So ∀x∀y(P(x) ∨ Q(y)) is the same as ∀y∀x(P(y) ∨ Q(x)) and the same as (∀xP(x)) ∨ (∀yQ(y)).
+
+So yes, they should be equivalent! The answer should be **Logically Equivalent**.
+
+---
+
+## 4. Proofs [20 pts]
+
+### Problem 4(a): Uniqueness of Integer Square Root [5 pts]
+
+**Theorem:** Given a nonnegative integer n, there is a unique nonnegative integer m such that m² ≤ n < (m+1)².
+
+**Proof:**
+
+*Existence:* Consider the set S = {k ∈ ℕ : k² ≤ n}. This set is nonempty because 0² = 0 ≤ n. The set is bounded above: if k² ≤ n, then k ≤ √n < n+1, so k < n+1. Therefore, S has a maximum element, call it m. By definition of m, we have m² ≤ n.
+
+We must verify that n < (m+1)². Suppose for contradiction that (m+1)² ≤ n. Then m+1 ∈ S, contradicting the fact that m is the maximum element of S. Therefore, n < (m+1)².
+
+*Uniqueness:* Suppose m and m' both satisfy m² ≤ n < (m+1)² and (m')² ≤ n < (m'+1)². If m ≠ m', assume WLOG that m < m'. Then m ≤ m' - 1, which gives (m+1)² ≤ (m')². But from our condition, n < (m+1)² and (m')² ≤ n, which gives (m')² ≤ n < (m+1)² ≤ (m')², a contradiction. Therefore, m = m'. ∎
+
+---
+
+### Problem 4(b): Square Root is Integer or Irrational [5 pts]
+
+**Theorem:** For any positive integer n, √n is either an integer or irrational.
+
+**Proof:**
+
+Suppose for contradiction that √n is rational but not an integer. Then √n = p/q where p and q are positive integers with gcd(p,q) = 1 and q > 1.
+
+Squaring both sides: n = p²/q², which gives nq² = p².
+
+This means q² divides p². Since gcd(p,q) = 1, we have gcd(p²,q²) = 1.
+
+From q² | p² and gcd(p²,q²) = 1, it follows that q² = 1, so q = 1.
+
+But this contradicts our assumption that q > 1.
+
+Therefore, √n cannot be rational unless it is an integer. Hence √n is either an integer or irrational. ∎
+
+---
+
+### Problem 4(c): Odd Coefficient Polynomial [5 pts]
+
+**Theorem:** Let a, b, and c be three odd integers. Then ax² + bx + c = 0 does not have a rational solution.
+
+**Proof:**
+
+Suppose for contradiction that ax² + bx + c = 0 has a rational solution x = p/q where p and q are integers with gcd(p,q) = 1.
+
+Substituting: a(p/q)² + b(p/q) + c = 0
+
+Multiplying by q²: ap² + bpq + cq² = 0
+
+Now consider this equation modulo 2. Since a, b, c are odd, we can write a ≡ 1 (mod 2), b ≡ 1 (mod 2), c ≡ 1 (mod 2).
+
+The equation becomes: p² + pq + q² ≡ 0 (mod 2)
+
+Since gcd(p,q) = 1, p and q cannot both be even. We consider three cases:
+
+**Case 1:** p and q are both odd.
+Then p² ≡ 1 (mod 2), pq ≡ 1 (mod 2), q² ≡ 1 (mod 2).
+So p² + pq + q² ≡ 1 + 1 + 1 ≡ 1 (mod 2), not 0. Contradiction.
+
+**Case 2:** p is odd and q is even.
+Then p² ≡ 1 (mod 2), pq ≡ 0 (mod 2), q² ≡ 0 (mod 2).
+So p² + pq + q² ≡ 1 + 0 + 0 ≡ 1 (mod 2), not 0. Contradiction.
+
+**Case 3:** p is even and q is odd.
+Then p² ≡ 0 (mod 2), pq ≡ 0 (mod 2), q² ≡ 1 (mod 2).
+So p² + pq + q² ≡ 0 + 0 + 1 ≡ 1 (mod 2), not 0. Contradiction.
+
+Since all cases lead to contradictions, ax² + bx + c = 0 cannot have a rational solution. ∎
+
+---
+
+### Problem 4(d): Irrational Between Rational and Irrational [5 pts]
+
+**Theorem:** Between every rational number and every irrational number there is an irrational number.
+
+**Proof:**
+
+Let r be a rational number and i be an irrational number. WLOG, assume r < i (the case i < r is symmetric).
+
+Consider z = r + (i - r)/√2.
+
+First, z is strictly between r and i:
+- z - r = (i - r)/√2 > 0 (since i > r and √2 > 0)
+- i - z = i - r - (i - r)/√2 = (i - r)(1 - 1/√2) > 0 (since 1 > 1/√2)
+
+Therefore, r < z < i.
+
+Next, z is irrational. Suppose for contradiction that z is rational. Then (i - r)/√2 = z - r is rational.
+
+Since r is rational and z - r is rational, we have i - r is rational times √2.
+Let i - r = (q√2) where q = (z-r)/√2 is rational.
+
+Then i = r + q√2. But this means i is a rational number plus a rational multiple of √2.
+
+If q ≠ 0, then i = r + q√2 can be written as a + b√2 where a and b are rational with b ≠ 0.
+
+Such a number is irrational only if b ≠ 0. But if i = r + q√2 where r is rational and q is rational and nonzero, then i would be irrational (since √2 is irrational and q ≠ 0).
+
+Wait, let me reconsider. If i = r + q√2 with r, q rational and q ≠ 0, then i is indeed irrational. So this doesn't give a contradiction.
+
+Let me use a different approach: Suppose z = p/s for integers p, s with gcd(p,s) = 1.
+
+Then p/s = r + (i-r)/√2, so (i-r)/√2 = p/s - r = (p - rs)/s.
+
+Let m/n = (p - rs)/s where gcd(m,n) = 1. Then (i-r)/√2 = m/n, which gives i - r = (m/n)√2.
+
+So i = r + (m/n)√2.
+
+If m = 0, then i = r, contradicting that i is irrational and r is rational.
+
+If m ≠ 0, then i = r + (m/n)√2 is of the form rational + rational·√2 with nonzero rational coefficient for √2. Such a number must be irrational (since √2 is irrational).
+
+Therefore, we cannot have z rational. Thus z must be irrational, and it lies strictly between r and i. ∎
+
+---