diff --git a/writeup.md b/writeup.md
index 22c0a80..975ba3a 100644
--- a/writeup.md
+++ b/writeup.md
@@ -2,8 +2,401 @@
 
 ## Student Responses
 
-<!-- Write your solutions below. Add sections for each problem as needed. -->
-<!-- Use proper mathematical notation where applicable. -->
+---
 
-[Your solutions here]
+## Problem 1: Algorithmic Thinking [20 pts]
 
+Wayne Zachary's 1977 paper "An information flow model for conflict and fission in small groups" studies how and why small social groups fracture. Below are the five steps of Algorithmic Thinking as applied to his work.
+
+### Understanding the Problem
+
+**Question:** Zachary wanted to answer: can the ultimate loyalties of members of a social group—specifically, which sub-group each member will join after a split—be predicted *before* the split occurs, using only observable interaction data?
+
+**Motivation:** Social scientists had long debated whether group fission was driven by external forces or by internal social structure. Zachary believed the internal communication/information-flow network contained enough signal to make precise predictions.
+
+**Data collected:** Zachary observed a university karate club of 34 members over roughly two years (1970–72). He recorded 8 types of social interactions between all pairs of members (e.g., attending the same classes, socializing outside the club, etc.), producing a weighted graph of pairwise interaction strengths.
+
+**Initial approach:** He hypothesized that information flows through the social network much like flow through a physical network, and that the "weakest links" in the network would reveal the natural fault lines along which the group would divide.
+
+### Formulating the Problem
+
+Zachary modeled the social group as a *weighted undirected graph* where:
+- Nodes represent the 34 club members.
+- Edge weights represent the total strength of interaction between each pair of members (aggregated across 8 interaction contexts).
+
+The conflict was between two key nodes: the club president (John A.) and the instructor (Mr. Hi). The mathematical problem became: **find the minimum-capacity cut** in this network that separates the two leaders, thereby partitioning all members into two groups. The min-cut corresponds to the "weakest" set of social ties whose removal divides the network between the two authority figures.
+
+### Algorithm Design
+
+Zachary employed **network flow** (specifically the max-flow/min-cut theorem). By computing the maximum flow from one leader's node to the other's, the min-cut is obtained as a byproduct; this cut identifies which members fall on which side of the fundamental social division.
+
+### Algorithm Implementation
+
+Zachary implemented the Ford-Fulkerson max-flow algorithm on the 34-node weighted graph, computing the minimum cut that separates the two leader nodes. The algorithm was run on the interaction-weight matrix he had assembled from his field observations.
+
+### Solving the Original Problem
+
+Zachary's findings were striking: the minimum cut of the network correctly predicted the faction alignment of **33 out of 34 members** after the club did in fact split (the one mis-predicted member was near the boundary and could have gone either way). This validated the hypothesis that the internal information-flow structure of a social network encodes enough information to forecast group fission, demonstrating that network-theoretic tools can model and predict real social phenomena.
+
+---
+
+## Problem 2: Propositional Logic [30 pts]
+
+### Problem 2.1
+
+**(a) "Neither the wind nor the rain affected the house."**
+
+Let:
+- p = "The wind affected the house"
+- q = "The rain affected the house"
+
+Compound proposition: **¬p ∧ ¬q** (equivalently, ¬(p ∨ q))
+
+---
+
+**(b) "If the pandemic is not controlled, more people will die."**
+
+Let:
+- p = "The pandemic is controlled"
+- q = "More people will die"
+
+Compound proposition: **¬p → q**
+
+---
+
+**(c) "Pedestrians should neither jaywalk nor cross a red light, or they will get a ticket."**
+
+Let:
+- p = "Pedestrians jaywalk"
+- q = "Pedestrians cross a red light"
+- r = "Pedestrians get a ticket"
+
+The sentence means: either pedestrians refrain from both actions, or they get a ticket.
+
+Compound proposition: **(¬p ∧ ¬q) ∨ r**
+
+(Equivalently: (p ∨ q) → r, since (¬p ∧ ¬q) ∨ r ≡ ¬(p ∨ q) ∨ r ≡ (p ∨ q) → r.)
+
+---
+
+**(d) "A plant is not poisonous unless it is tropical."**
+
+Let:
+- p = "A plant is poisonous"
+- q = "A plant is tropical"
+
+"A is not B unless C" means: if not C, then not A, i.e., ¬q → ¬p.
+
+Compound proposition: **¬q → ¬p** (equivalently, p → q)
+
+---
+
+**(e) "Growing in the tropics is necessary but not sufficient for a plant to be poisonous."**
+
+Let:
+- p = "A plant grows in the tropics"
+- q = "A plant is poisonous"
+
+- *Necessary*: p is necessary for q means q → p (you cannot be poisonous without growing in the tropics).
+- *Not sufficient*: p alone does not guarantee q, i.e., ¬(p → q).
+
+Compound proposition: **(q → p) ∧ ¬(p → q)**
+
+---
+
+### Problem 2.2
+
+**(a) p XOR p**
+
+Truth table:
+
+| p | p XOR p |
+|---|---------|
+| T | T XOR T = **F** |
+| F | F XOR F = **F** |
+
+The proposition is always **false** — it is a **contradiction** and therefore **not satisfiable**.
+
+---
+
+**(b) p XOR (¬p)**
+
+Truth table:
+
+| p | ¬p | p XOR ¬p |
+|---|-----|---------|
+| T | F   | T XOR F = **T** |
+| F | T   | F XOR T = **T** |
+
+The proposition is always **true** — it is a **tautology** and therefore **satisfiable**. ✓
+
+---
+
+**(c) (p XOR q) OR (p XOR (¬q))**
+
+We evaluate all four cases:
+
+| p | q | p XOR q | p XOR ¬q | (p XOR q) OR (p XOR ¬q) |
+|---|---|---------|----------|--------------------------|
+| T | T | F | T | **T** |
+| T | F | T | F | **T** |
+| F | T | T | F | **T** |
+| F | F | F | T | **T** |
+
+The proposition is always **true** — it is a **tautology** and therefore **satisfiable**. ✓
+
+**Algebraic proof:**
+
+- p XOR q ≡ (p ∧ ¬q) ∨ (¬p ∧ q)
+- p XOR ¬q ≡ (p ∧ q) ∨ (¬p ∧ ¬q)
+
+Their disjunction:
+
+(p ∧ ¬q) ∨ (¬p ∧ q) ∨ (p ∧ q) ∨ (¬p ∧ ¬q)
+= p ∧ (¬q ∨ q) ∨ ¬p ∧ (q ∨ ¬q)
+= p ∧ T ∨ ¬p ∧ T
+= p ∨ ¬p
+= **T** ✓
+
+---
+
+### Problem 2.3
+
+**(a) (p ∧ q) → r  vs.  (p → r) ∧ (q → r)**
+
+**Not logically equivalent.**
+
+**Counterexample:** Let p = T, q = F, r = F.
+- Left: (T ∧ F) → F = F → F = **T**
+- Right: (T → F) ∧ (F → F) = F ∧ T = **F**
+
+Since the two sides differ, they are not logically equivalent.
+
+**Intuition:** The left side requires r only when *both* p and q hold simultaneously. The right side requires r whenever *either* p or q holds alone—a strictly stronger condition.
+
+---
+
+**(b) (p ∨ q) → r  vs.  (p → r) ∨ (q → r)**
+
+**Not logically equivalent.**
+
+**Counterexample:** Let p = T, q = F, r = F.
+- Left: (T ∨ F) → F = T → F = **F**
+- Right: (T → F) ∨ (F → F) = F ∨ T = **T**
+
+Since the two sides differ, they are not logically equivalent.
+
+**Intuition:** The left says r must hold whenever at least one of p, q holds—this fails if p=T and r=F even when q=F. The right says at least one of the conditionals holds; when q=F, (q → r) is vacuously true, so the whole disjunction is true regardless of r.
+
+---
+
+**(c) p → (q ↔ r)  vs.  (p → q) ↔ (p → r)**
+
+**Logically equivalent.**
+
+**Proof:**
+
+We consider two cases based on the truth value of p.
+
+*Case 1: p = F.*
+- Left: F → (q ↔ r) = **T** (vacuously true)
+- Right: (F → q) ↔ (F → r) = T ↔ T = **T**
+Both sides are true.
+
+*Case 2: p = T.*
+- Left: T → (q ↔ r) = **q ↔ r**
+- Right: (T → q) ↔ (T → r) = q ↔ r
+
+Both sides reduce to **q ↔ r**.
+
+In both cases the two expressions have identical truth values, so they are logically equivalent. ✓
+
+---
+
+## Problem 3: Predicate Logic [30 pts]
+
+### Problem 3.1
+
+Domain: **ℝ** (real numbers)
+
+**(a) ∃x: x³ = −1**
+
+**TRUE.** Take x = −1. Then (−1)³ = −1. ✓
+
+---
+
+**(b) ∀x: (−x)² = x²**
+
+**TRUE.** For any real x: (−x)² = (−1)² · x² = 1 · x² = x². ✓
+
+---
+
+**(c) ∀x: 2x > x**
+
+**FALSE.** Counterexample: x = 0. Then 2(0) = 0 = 0, so 2x > x is 0 > 0, which is false. (Also x = −1 gives 2(−1) = −2 < −1.) ✗
+
+---
+
+**(d) ∀x: ∃y: x·y = y·x**
+
+**TRUE.** Multiplication of real numbers is commutative; for any x, choose y = 0 (or any real number), and x·y = y·x holds trivially. ✓
+
+---
+
+**(e) ∀x: ∀y: ∃z: z = (x + y)/2**
+
+**TRUE.** For any real numbers x and y, the expression (x + y)/2 is a well-defined real number (ℝ is closed under addition and division by nonzero constants). Setting z = (x + y)/2 satisfies the condition. ✓
+
+---
+
+**(f) ∃x: ∀y: x ≤ y²**
+
+**TRUE.** Take x = 0. For any real y, y² ≥ 0 = x. ✓
+
+---
+
+**(g) ∃x: ∀y: x ≤ y**
+
+**FALSE.** For any candidate x, the real number y = x − 1 satisfies y < x, so x ≤ y is false. The real numbers have no minimum element. ✗
+
+---
+
+### Problem 3.2
+
+Assume all quantifiers range over the same nonempty domain D.
+
+**(a) ∀x(P(x) ∨ Q(x))  vs.  (∀x P(x)) ∨ (∀x Q(x))**
+
+**Not logically equivalent.**
+
+**Counterexample:** Let D = {1, 2}, P(1) = T, P(2) = F, Q(1) = F, Q(2) = T.
+
+- Left: P(1) ∨ Q(1) = T ∨ F = T, and P(2) ∨ Q(2) = F ∨ T = T. So ∀x(P(x) ∨ Q(x)) = **T**.
+- Right: ∀x P(x) = F (since P(2) = F); ∀x Q(x) = F (since Q(1) = F). So F ∨ F = **F**.
+
+The right-to-left direction holds: if every element satisfies P or every element satisfies Q, then certainly every element satisfies P ∨ Q. But the left-to-right fails; elements can "alternate" which disjunct they satisfy. ✗
+
+---
+
+**(b) ∀x(P(x) ∧ Q(x))  vs.  (∀x P(x)) ∧ (∀x Q(x))**
+
+**Logically equivalent.**
+
+**Forward (⇒):** Assume ∀x(P(x) ∧ Q(x)). For any x in D, P(x) ∧ Q(x) holds, so P(x) holds and Q(x) holds. Since this is true for every x, we get ∀x P(x) and ∀x Q(x), hence (∀x P(x)) ∧ (∀x Q(x)). ✓
+
+**Backward (⇐):** Assume (∀x P(x)) ∧ (∀x Q(x)). For any x in D, P(x) holds (from ∀x P(x)) and Q(x) holds (from ∀x Q(x)), so P(x) ∧ Q(x) holds. Hence ∀x(P(x) ∧ Q(x)). ✓
+
+---
+
+**(c) (∀x P(x)) ∨ (∀x Q(x))  vs.  ∀x ∀y(P(x) ∨ Q(y))**
+
+**Logically equivalent.**
+
+**Forward (⇒):** Assume (∀x P(x)) ∨ (∀x Q(x)).
+- *Case 1:* ∀x P(x). For any x, y in D, P(x) is true, so P(x) ∨ Q(y) = T.
+- *Case 2:* ∀x Q(x). For any x, y in D, Q(y) is true (y is also in D), so P(x) ∨ Q(y) = T.
+
+In both cases, ∀x ∀y(P(x) ∨ Q(y)) holds. ✓
+
+**Backward (⇐):** Assume ∀x ∀y(P(x) ∨ Q(y)), but for contradiction suppose ¬(∀x P(x)) ∧ ¬(∀x Q(x)).
+
+- From ¬(∀x P(x)): ∃a ∈ D with ¬P(a).
+- From ¬(∀x Q(x)): ∃b ∈ D with ¬Q(b).
+
+By hypothesis (with x = a, y = b): P(a) ∨ Q(b) must be true. But P(a) = F and Q(b) = F, so P(a) ∨ Q(b) = F. Contradiction. Therefore (∀x P(x)) ∨ (∀x Q(x)) holds. ✓
+
+---
+
+## Problem 4: Proofs [20 pts]
+
+### (a) Uniqueness and existence of the integer square root [5 pts]
+
+**Claim:** For every nonnegative integer n, there exists a unique nonnegative integer m such that m² ≤ n < (m+1)².
+
+**Proof:**
+
+*Existence:* Let S = {k ∈ ℤ≥0 : k² ≤ n}. The set S is nonempty since 0 ∈ S (0² = 0 ≤ n). S is bounded above: if k ≥ n + 1, then k² ≥ (n+1)² = n² + 2n + 1 > n, so k ∉ S. Hence S is a finite nonempty set of nonnegative integers and has a maximum; let m = max(S).
+
+By definition, m ∈ S so m² ≤ n. We claim n < (m+1)²: if (m+1)² ≤ n, then m+1 ∈ S, contradicting m = max(S). Therefore m² ≤ n < (m+1)². ✓
+
+*Uniqueness:* Suppose m₁, m₂ are nonnegative integers satisfying the condition, and assume without loss of generality m₁ < m₂. Then m₁ + 1 ≤ m₂ (since they are integers), so
+
+(m₁ + 1)² ≤ m₂² ≤ n.
+
+But this contradicts n < (m₁ + 1)². Hence m₁ = m₂, establishing uniqueness. ✓
+
+---
+
+### (b) √n is an integer or irrational [5 pts]
+
+**Claim:** For any positive integer n, √n is either an integer or irrational.
+
+**Proof by contradiction:** Suppose √n is rational but not an integer. Then we can write √n = p/q where p, q are positive integers with gcd(p, q) = 1 and q > 1 (the denominator is greater than 1 because √n is not an integer).
+
+Squaring both sides: n = p²/q², so nq² = p².
+
+Since q > 1, there exists a prime r that divides q. Then r | q², so r | p² (from nq² = p²). Since r is prime, r | p. So r divides both p and q, contradicting gcd(p, q) = 1.
+
+Therefore no such rational non-integer expression exists, and √n must be either an integer or irrational. ✓
+
+---
+
+### (c) ax² + bx + c = 0 has no rational solution when a, b, c are odd [5 pts]
+
+**Claim:** Let a, b, c be odd integers. Then ax² + bx + c = 0 has no rational solution.
+
+**Proof by contradiction:** Suppose x = p/q is a rational solution, where p, q are integers with gcd(p, q) = 1 and q > 0. Since gcd(p, q) = 1, p and q cannot both be even.
+
+Substituting: a(p/q)² + b(p/q) + c = 0. Multiplying through by q²:
+
+ap² + bpq + cq² = 0.
+
+We consider all cases modulo 2, using the fact that a ≡ b ≡ c ≡ 1 (mod 2):
+
+**Case 1: p odd, q odd.**
+
+ap² ≡ 1·1 = 1, bpq ≡ 1·1·1 = 1, cq² ≡ 1·1 = 1 (mod 2).
+
+Sum: 1 + 1 + 1 = 3 ≡ 1 (mod 2) ≠ 0. Contradiction.
+
+**Case 2: p even, q odd.**
+
+ap² ≡ 1·0 = 0, bpq ≡ 1·0·1 = 0, cq² ≡ 1·1 = 1 (mod 2).
+
+Sum: 0 + 0 + 1 = 1 ≡ 1 (mod 2) ≠ 0. Contradiction.
+
+**Case 3: p odd, q even.**
+
+ap² ≡ 1·1 = 1, bpq ≡ 1·1·0 = 0, cq² ≡ 1·0 = 0 (mod 2).
+
+Sum: 1 + 0 + 0 = 1 ≡ 1 (mod 2) ≠ 0. Contradiction.
+
+(The case p even, q even is impossible since gcd(p, q) = 1.)
+
+In every admissible case we reach a contradiction. Therefore ax² + bx + c = 0 has no rational solution. ✓
+
+---
+
+### (d) Between every rational and every irrational there is an irrational [5 pts]
+
+**Claim:** Let r be rational and t be irrational with r ≠ t. Then there exists an irrational number s strictly between r and t.
+
+**Proof:** Without loss of generality, assume r < t (if t < r, the argument is symmetric with the roles swapped).
+
+Define:
+
+s = (r + t) / 2.
+
+*s lies strictly between r and t:* Since t > r, we have t − r > 0, so (t − r)/2 > 0. Then:
+
+s = r + (t − r)/2 > r     and     s = t − (t − r)/2 < t.
+
+So r < s < t.
+
+*s is irrational:* Suppose for contradiction that s is rational. Then 2s is rational (product of rationals), and r is rational, so
+
+t = 2s − r
+
+would be a difference of rationals, hence rational. This contradicts the assumption that t is irrational.
+
+Therefore s = (r + t)/2 is an irrational number satisfying r < s < t. ✓