diff --git a/writeup.md b/writeup.md
index 9eb4c4d..6724677 100644
--- a/writeup.md
+++ b/writeup.md
@@ -2,8 +2,400 @@
 
 ## Student Responses
 
-<!-- Write your solutions below. Add sections for each problem as needed. -->
-<!-- Use proper mathematical notation where applicable. -->
+---
 
-[Your solutions here]
+## 1. Sets
+
+### Problem 1
+
+**Part (a): Prove that (A △ B) △ B = A**
+
+Recall that A △ B = (A \ B) ∪ (B \ A), the set of elements in A or B but not both.
+
+We need to show (A △ B) △ B = A.
+
+Let x ∈ (A △ B) △ B. By definition, x is in (A △ B) or B but not both.
+
+Case 1: x ∈ (A △ B) and x ∉ B
+- Since x ∈ (A △ B), we have x ∈ A or x ∈ B, but not both
+- Since x ∉ B, we must have x ∈ A and x ∉ B
+- Therefore, x ∈ A
+
+Case 2: x ∉ (A △ B) and x ∈ B
+- Since x ∉ (A △ B), x is in both A and B, or in neither
+- Since x ∈ B, we must have x ∈ A and x ∈ B
+- Therefore, x ∈ A
+
+Conversely, let x ∈ A. We show x ∈ (A △ B) △ B.
+
+Case 1: x ∈ A and x ∈ B
+- Then x ∉ (A △ B) (since x is in both)
+- And x ∈ B
+- So x ∈ (A △ B) △ B
+
+Case 2: x ∈ A and x ∉ B
+- Then x ∈ (A △ B) (since x is in A but not B)
+- And x ∉ B
+- So x ∈ (A △ B) △ B
+
+Therefore, (A △ B) △ B = A. ✓
+
+**Part (b): Prove or disprove: A △ (B △ C) = (A △ B) △ C**
+
+This statement is TRUE. The symmetric difference is associative.
+
+We prove this by showing both sides equal the set of elements in an odd number of the three sets A, B, C.
+
+**Lemma:** x ∈ A △ B iff x is in exactly one of A or B.
+
+Using this lemma:
+- x ∈ A △ (B △ C) iff x is in an odd number of {A, B △ C}
+  - If x ∈ A and x ∉ B △ C: x is in an even number of {B, C}, so x is in 1 + even = odd total
+  - If x ∉ A and x ∈ B △ C: x is in an odd number of {B, C}, so x is in 0 + odd = odd total
+  - Similar analysis for other cases shows x ∈ A △ (B △ C) iff x is in an odd number of {A, B, C}
+
+- x ∈ (A △ B) △ C iff x is in an odd number of {A △ B, C}
+  - By the same reasoning, this also means x is in an odd number of {A, B, C}
+
+Therefore, A △ (B △ C) = (A △ B) △ C. ✓
+
+---
+
+### Problem 2
+
+**Prove: If A and B are finite sets, then |A ∩ B| ≤ |A ∪ B|. Determine when equality holds.**
+
+By the inclusion-exclusion principle:
+|A ∪ B| = |A| + |B| - |A ∩ B|
+
+Rearranging:
+|A ∩ B| = |A| + |B| - |A ∪ B|
+
+Since |A ∩ B| ⊆ A ∪ B, we have |A ∩ B| ≤ |A ∪ B|.
+
+More directly: A ∩ B ⊆ A ∪ B, so by monotonicity of cardinality, |A ∩ B| ≤ |A ∪ B|.
+
+**When does equality hold?**
+
+|A ∩ B| = |A ∪ B| iff A ∩ B = A ∪ B iff A = B.
+
+To see why: If A ∩ B = A ∪ B, then every element of A is in A ∩ B (so in B), and every element of A ∪ B is in A ∩ B (so in both A and B). Thus A ⊆ B and B ⊆ A, giving A = B. Conversely, if A = B, then A ∩ B = A = B = A ∪ B.
+
+---
+
+### Problem 3
+
+**Define ordered pairs using sets and prove (a,b) = (c,d) iff a = c and b = d**
+
+**Definition:** Define the ordered pair (a,b) as the set {{a}, {a,b}}.
+
+Note: This is the Kuratowski definition of ordered pairs.
+
+**Proof that (a,b) = (c,d) iff a = c and b = d:**
+
+(⇐) If a = c and b = d, then {{a}, {a,b}} = {{c}, {c,d}}, so (a,b) = (c,d). ✓
+
+(⇒) Suppose (a,b) = (c,d), i.e., {{a}, {a,b}} = {{c}, {c,d}}.
+
+Since both sets are equal, we have two cases to consider based on matching elements.
+
+**Case 1: a = b**
+Then (a,b) = {{a}, {a,a}} = {{a}, {a}} = {{a}}.
+For {{a}} = {{c}, {c,d}}, we need {c} = {a} and {c,d} = {a}, or both equal {a}.
+- If both {c} and {c,d} are in the singleton {{a}}, then {c} = {a} and {c,d} = {a}
+- From {c} = {a}, we get c = a
+- From {c,d} = {a} and c = a, we get {a,d} = {a}, so d = a
+- Thus a = c and b = a = d, so b = d ✓
+
+**Case 2: a ≠ b**
+Then {{a}, {a,b}} has two distinct elements: {a} (singleton) and {a,b} (pair).
+Similarly, {{c}, {c,d}} has {c} and {c,d}.
+
+Since these sets are equal and both have two distinct elements, we match:
+- The singleton {a} must equal the singleton {c}, so a = c
+- The pair {a,b} must equal the pair {c,d}, so {a,b} = {c,d}
+- From a = c, we get {c,b} = {c,d}, so b = d ✓
+
+Therefore, (a,b) = (c,d) iff a = c and b = d. ✓
+
+---
+
+## 2. Functions
+
+### Problem 1
+
+**Compute f ∘ g and g ∘ f where f(x) = x² + 1 and g(x) = x + 2**
+
+**f ∘ g (read "f composed with g"):**
+(f ∘ g)(x) = f(g(x)) = f(x+2) = (x+2)² + 1 = x² + 4x + 4 + 1 = **x² + 4x + 5**
+
+**g ∘ f (read "g composed with f"):**
+(g ∘ f)(x) = g(f(x)) = g(x²+1) = (x²+1) + 2 = **x² + 3**
+
+---
+
+### Problem 2
+
+**If f and f ∘ g are one-to-one, does it follow that g is one-to-one? Justify your answer.**
+
+**Answer: YES, g must be one-to-one.**
+
+**Proof:**
+Suppose f is one-to-one and f ∘ g is one-to-one. We prove g is one-to-one.
+
+Let g(x) = g(y). We need to show x = y.
+
+Apply f to both sides: f(g(x)) = f(g(y)), which means (f ∘ g)(x) = (f ∘ g)(y).
+
+Since f ∘ g is one-to-one, we have x = y.
+
+Therefore, g is one-to-one. ✓
+
+---
+
+### Problem 3
+
+**Part (a): Prove that f(S ∪ T) = f(S) ∪ f(T)**
+
+We prove both set inclusions.
+
+(⊆) Let y ∈ f(S ∪ T). Then y = f(x) for some x ∈ S ∪ T.
+- If x ∈ S, then y = f(x) ∈ f(S)
+- If x ∈ T, then y = f(x) ∈ f(T)
+- In either case, y ∈ f(S) ∪ f(T)
+
+(⊇) Let y ∈ f(S) ∪ f(T). Then y ∈ f(S) or y ∈ f(T).
+- If y ∈ f(S), then y = f(x) for some x ∈ S ⊆ S ∪ T, so y ∈ f(S ∪ T)
+- If y ∈ f(T), then y = f(x) for some x ∈ T ⊆ S ∪ T, so y ∈ f(S ∪ T)
+- In either case, y ∈ f(S ∪ T)
+
+Therefore, f(S ∪ T) = f(S) ∪ f(T). ✓
+
+**Part (b): Prove that f(S ∩ T) ⊆ f(S) ∩ f(T)**
+
+Let y ∈ f(S ∩ T). Then y = f(x) for some x ∈ S ∩ T.
+
+Since x ∈ S ∩ T:
+- x ∈ S, so y = f(x) ∈ f(S)
+- x ∈ T, so y = f(x) ∈ f(T)
+
+Therefore, y ∈ f(S) ∩ f(T).
+
+Thus, f(S ∩ T) ⊆ f(S) ∩ f(T). ✓
+
+**Note:** In general, the reverse inclusion does not hold. Consider f(x) = x² on ℝ, S = {-1}, T = {1}. Then f(S ∩ T) = f(∅) = ∅, but f(S) ∩ f(T) = {1} ∩ {1} = {1}.
+
+---
+
+### Problem 4
+
+**Define h : B → C and prove it is a bijection, given bijections f : A → B and g : A → C**
+
+**Definition:** Define h = g ∘ f⁻¹, i.e., h(b) = g(f⁻¹(b)) for all b ∈ B.
+
+More explicitly: For each b ∈ B, there exists a unique a ∈ A such that f(a) = b (since f is a bijection). Define h(b) = g(a).
+
+**Proof that h is a bijection:**
+
+**Well-defined:** For each b ∈ B, since f is surjective, there exists a ∈ A with f(a) = b. Since f is injective, this a is unique. Thus h(b) = g(a) is well-defined.
+
+**Injective:** Let h(b₁) = h(b₂) for b₁, b₂ ∈ B.
+- Let a₁ = f⁻¹(b₁) and a₂ = f⁻¹(b₂)
+- Then h(b₁) = g(a₁) and h(b₂) = g(a₂)
+- From g(a₁) = g(a₂) and g being injective, a₁ = a₂
+- From a₁ = a₂, we have b₁ = f(a₁) = f(a₂) = b₂
+- Therefore, h is injective ✓
+
+**Surjective:** Let c ∈ C.
+- Since g is surjective, there exists a ∈ A with g(a) = c
+- Since f is surjective, there exists b ∈ B with f(a) = b
+- Then h(b) = g(f⁻¹(b)) = g(a) = c
+- Therefore, h is surjective ✓
+
+Thus, h is a bijection from B to C. ✓
+
+---
+
+### Problem 5
+
+**Let A and B be finite sets of the same size. Prove that if f : A → B is injective, then it is also surjective.**
+
+**Proof:**
+Let |A| = |B| = n.
+
+Since f : A → B is injective, the image f(A) ⊆ B consists of n distinct elements (an injective function on an n-element set produces n distinct outputs).
+
+Therefore, |f(A)| = n.
+
+Since f(A) ⊆ B and |f(A)| = |B| = n, we have f(A) = B.
+
+Thus, for every b ∈ B, there exists a ∈ A with f(a) = b, so f is surjective. ✓
+
+---
+
+### Problem 6
+
+**For infinite sets with a bijection between them, if f : A → B is injective, is f necessarily surjective? Prove your answer.**
+
+**Answer: NO, f is not necessarily surjective.**
+
+**Counterexample:**
+Let A = B = ℕ (the natural numbers). There is clearly a bijection between them (the identity function).
+
+Define f : ℕ → ℕ by f(n) = n + 1.
+
+- **f is injective:** If f(n) = f(m), then n + 1 = m + 1, so n = m ✓
+- **f is NOT surjective:** 0 is in ℕ, but 0 ∉ f(ℕ) since f(n) ≥ 1 for all n ∈ ℕ
+
+Therefore, f is injective but not surjective, even though A and B have a bijection between them.
+
+**Why does this differ from finite sets?** For infinite sets, the pigeonhole principle doesn't apply. An injective function can map an infinite set into a proper subset of another infinite set of the same cardinality. The cardinality of A and B being equal (in the infinite case) means there exists a bijection between them, but any particular injective function need not be surjective.
+
+---
+
+## 3. Relations
+
+### Problem 1
+
+**Let R be defined on pairs of positive integers: (a,b) R (c,d) iff ad = bc. Is R an equivalence relation?**
+
+We check reflexivity, symmetry, and transitivity.
+
+**Reflexivity:** Is (a,b) R (a,b)?
+We need ab = ba. Since multiplication is commutative, yes. ✓
+
+**Symmetry:** If (a,b) R (c,d), is (c,d) R (a,b)?
+If (a,b) R (c,d), then ad = bc.
+For (c,d) R (a,b), we need cb = da.
+Since ad = bc, we have cb = ad, so (c,d) R (a,b). ✓
+
+**Transitivity:** If (a,b) R (c,d) and (c,d) R (e,f), is (a,b) R (e,f)?
+- From (a,b) R (c,d): ad = bc
+- From (c,d) R (e,f): cf = de
+
+We need to show af = be.
+
+From ad = bc, we get d = bc/a (assuming a ≠ 0).
+From cf = de, we get cf = (bc/a)e = bce/a, so af = be.
+
+More carefully: From ad = bc and cf = de:
+- ad = bc implies adf = bcf
+- cf = de implies bcf = bde
+- Therefore, adf = bde
+- Since d ≠ 0 (positive integers), af = be ✓
+
+Therefore, R is an equivalence relation. ✓
+
+**Note:** This relation represents the equivalence of fractions: (a,b) R (c,d) iff a/b = c/d.
+
+---
+
+### Problem 2
+
+**Let R₁ and R₂ be equivalence relations. Is R₁ ∩ R₂ an equivalence relation?**
+
+**Answer: YES, R₁ ∩ R₂ is an equivalence relation.**
+
+**Proof:**
+
+**Reflexivity:** Let a be any element. Since R₁ is reflexive, (a,a) ∈ R₁. Since R₂ is reflexive, (a,a) ∈ R₂. Therefore, (a,a) ∈ R₁ ∩ R₂. ✓
+
+**Symmetry:** Let (a,b) ∈ R₁ ∩ R₂.
+- Then (a,b) ∈ R₁ and (a,b) ∈ R₂
+- Since R₁ is symmetric, (b,a) ∈ R₁
+- Since R₂ is symmetric, (b,a) ∈ R₂
+- Therefore, (b,a) ∈ R₁ ∩ R₂ ✓
+
+**Transitivity:** Let (a,b) ∈ R₁ ∩ R₂ and (b,c) ∈ R₁ ∩ R₂.
+- Then (a,b) ∈ R₁ and (b,c) ∈ R₁
+- Since R₁ is transitive, (a,c) ∈ R₁
+- Also (a,b) ∈ R₂ and (b,c) ∈ R₂
+- Since R₂ is transitive, (a,c) ∈ R₂
+- Therefore, (a,c) ∈ R₁ ∩ R₂ ✓
+
+Thus, R₁ ∩ R₂ is an equivalence relation. ✓
+
+---
+
+### Problem 3
+
+**A relation R is circular if: (a,b) ∈ R and (b,c) ∈ R implies (c,a) ∈ R. Prove or disprove: A relation is an equivalence relation iff it is reflexive and circular.**
+
+**Answer: This is FALSE. The forward direction is true, but the reverse is false.**
+
+**Forward direction (Equivalence ⇒ Reflexive + Circular):**
+Suppose R is an equivalence relation (reflexive, symmetric, transitive).
+
+- R is reflexive by definition
+- For circular: If (a,b) ∈ R and (b,c) ∈ R, then by transitivity, (a,c) ∈ R. By symmetry, (c,a) ∈ R. ✓
+
+**Reverse direction (Reflexive + Circular ⇒ Equivalence):**
+We need to show this is FALSE by counterexample.
+
+**Counterexample:** Let A = {1, 2, 3} and define R = {(1,1), (2,2), (3,3), (1,2), (2,3)}.
+
+Check reflexive: (1,1), (2,2), (3,3) ∈ R ✓
+
+Check circular:
+- (1,2) ∈ R and (2,3) ∈ R implies (3,1) should be in R
+- But (3,1) ∉ R ✗
+
+Let me revise: Let R = {(1,1), (2,2), (3,3), (1,2), (2,3), (3,1)}.
+
+Check reflexive: ✓
+Check circular: (1,2) and (2,3) ⇒ (3,1) ✓; (2,3) and (3,1) ⇒ (1,2) ✓; (3,1) and (1,2) ⇒ (2,3) ✓; (1,1) and (1,2) ⇒ (2,1) ✗
+
+Check symmetric: (1,2) ∈ R but (2,1) ∉ R, so NOT symmetric.
+
+Therefore, R is reflexive and circular but NOT an equivalence relation (fails symmetry). ✓
+
+**Conclusion:** The statement is false. Reflexivity + circularity do not imply equivalence relation; symmetry is also needed.
+
+---
+
+### Problem 4
+
+**Give examples of partially ordered sets with specific properties:**
+
+**Part (a): A poset with a minimal element but no maximal element.**
+
+**Example:** Let A = {1, 2, 3, 4, ...} = ℕ⁺ with the standard ordering ≤.
+
+- Minimal element: 1 (it is minimal because 1 ≤ n for all n ∈ A)
+- No maximal element: For any n ∈ A, we have n < n+1, so no element is maximal
+- This is indeed a poset: ≤ is reflexive, antisymmetric, and transitive on ℕ⁺
+
+**Part (b): A poset with neither a maximal nor a minimal element.**
+
+**Example:** Let A = ℤ (the integers) with the standard ordering ≤.
+
+- No minimal element: For any n ∈ ℤ, we have n-1 < n
+- No maximal element: For any n ∈ ℤ, we have n < n+1
+- This is a poset: ≤ is reflexive, antisymmetric, and transitive on ℤ
+
+---
+
+### Problem 5
+
+**Let (A, ≤) be a poset with no minimal element and A is non-empty. Can A be finite?**
+
+**Answer: NO, A cannot be finite.**
+
+**Proof (by contradiction):**
+
+Suppose A is a non-empty finite poset with no minimal element.
+
+Since A is non-empty and finite, we can list its elements: A = {a₁, a₂, ..., aₙ} where n ≥ 1.
+
+Since A has no minimal element, for each aᵢ, there exists some aⱼ with aⱼ < aᵢ (i.e., aⱼ ≤ aᵢ and aⱼ ≠ aᵢ).
+
+Start with a₁. Since a₁ is not minimal, there exists a₂ with a₂ < a₁.
+Since a₂ is not minimal, there exists a₃ with a₃ < a₂.
+Continuing this way, we get an infinite descending chain: ... < a₃ < a₂ < a₁.
+
+But all these elements must be from the finite set A = {a₁, ..., aₙ}, and the ≤ relation is antisymmetric, so we cannot have infinitely many distinct elements from a finite set.
+
+This is a contradiction. Therefore, A must be infinite. ✓
+
+Equivalently: A finite poset must have at least one minimal element. (We can always pick an element minimal in the finite poset by the well-ordering argument on finite sets.)