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-# COMP182 Homework 2
+# COMP182 Homework 2: Proofs, Sets, and Functions
 
-## Student Responses
+---
 
-<!-- Write your solutions below. Add sections for each problem as needed. -->
-<!-- Use proper mathematical notation where applicable. -->
+## Section 1: Sets
 
-[Your solutions here]
+### Problem 1
 
+**Definition:** The symmetric difference A △ B is the set of elements in A or B but not both:
+A △ B = (A \ B) ∪ (B \ A)
+
+#### Part (a): Prove (A △ B) △ B = A
+
+**Proof** (by element-chasing, using the characterization that x ∈ X △ Y iff x is in exactly one of X, Y):
+
+We consider all four cases based on membership in A and B:
+
+**Case 1: x ∈ A, x ∉ B.**
+- x ∈ A but x ∉ B, so x is in exactly one of {A, B}: thus x ∈ A △ B.
+- x ∈ A △ B but x ∉ B, so x is in exactly one of {A △ B, B}: thus x ∈ (A △ B) △ B. ✓
+
+**Case 2: x ∈ A, x ∈ B.**
+- x ∈ A and x ∈ B, so x is in both: thus x ∉ A △ B.
+- x ∉ A △ B but x ∈ B, so x is in exactly one of {A △ B, B}: thus x ∈ (A △ B) △ B. ✓
+
+**Case 3: x ∉ A, x ∈ B.**
+- x ∉ A but x ∈ B, so x is in exactly one of {A, B}: thus x ∈ A △ B.
+- x ∈ A △ B and x ∈ B, so x is in both: thus x ∉ (A △ B) △ B. ✓
+
+**Case 4: x ∉ A, x ∉ B.**
+- x is in neither A nor B: thus x ∉ A △ B.
+- x ∉ A △ B and x ∉ B, so x is in neither: thus x ∉ (A △ B) △ B. ✓
+
+In Cases 1 and 2, x ∈ A and x ∈ (A △ B) △ B.
+In Cases 3 and 4, x ∉ A and x ∉ (A △ B) △ B.
+
+Therefore, for all x, x ∈ (A △ B) △ B ↔ x ∈ A, which proves **(A △ B) △ B = A**. □
+
+---
+
+#### Part (b): Is A △ (B △ C) = (A △ B) △ C? Prove your answer.
+
+**Yes, symmetric difference is associative.**
+
+**Proof** (by truth table over all 8 membership cases):
+
+For each element x, let a = [x ∈ A], b = [x ∈ B], c = [x ∈ C] ∈ {0, 1}.
+Note that x ∈ X △ Y iff exactly one of x ∈ X, x ∈ Y holds, which corresponds to XOR (addition mod 2).
+
+Thus:
+- x ∈ A △ (B △ C) corresponds to a XOR (b XOR c)
+- x ∈ (A △ B) △ C corresponds to (a XOR b) XOR c
+
+Since XOR is associative (addition mod 2 is associative), we have
+a XOR (b XOR c) = (a XOR b) XOR c
+for all a, b, c ∈ {0,1}.
+
+We can verify all 8 cases:
+
+| a | b | c | B△C | A△(B△C) | A△B | (A△B)△C |
+|---|---|---|-----|---------|-----|---------|
+| 0 | 0 | 0 |  0  |    0    |  0  |    0    |
+| 1 | 0 | 0 |  0  |    1    |  1  |    1    |
+| 0 | 1 | 0 |  1  |    1    |  1  |    1    |
+| 0 | 0 | 1 |  1  |    1    |  0  |    1    |
+| 1 | 1 | 0 |  1  |    0    |  0  |    0    |
+| 1 | 0 | 1 |  1  |    0    |  1  |    0    |
+| 0 | 1 | 1 |  0  |    0    |  1  |    0    |
+| 1 | 1 | 1 |  0  |    1    |  0  |    1    |
+
+In all 8 cases, A △ (B △ C) and (A △ B) △ C agree. 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.**
+
+**Proof:**
+
+Since A ∩ B ⊆ A ∪ B, the inclusion map i: A ∩ B → A ∪ B defined by i(x) = x is injective. For finite sets, an injective map from S to T implies |S| ≤ |T|. Therefore:
+
+|A ∩ B| ≤ |A ∪ B|. □
+
+**When is equality achieved?**
+
+By inclusion-exclusion: |A ∪ B| = |A| + |B| − |A ∩ B|.
+
+Equality |A ∩ B| = |A ∪ B| holds if and only if A ∩ B = A ∪ B.
+
+(Since A ∩ B ⊆ A ∪ B and both sets are finite, |A ∩ B| = |A ∪ B| implies A ∩ B = A ∪ B.)
+
+Now A ∩ B = A ∪ B if and only if A = B:
+- (⇒) If A ∩ B = A ∪ B, then A ⊆ A ∪ B = A ∩ B ⊆ A, so A = A ∩ B, meaning A ⊆ B. By symmetry B ⊆ A, so A = B.
+- (⇐) If A = B, then A ∩ B = A = B = A ∪ B.
+
+**Conclusion:** Equality holds if and only if A = B. □
+
+---
+
+### Problem 3
+
+**Set-based formulation of ordered pairs, and proof.**
+
+**Definition (Kuratowski):** Define the ordered pair (a, b) as the set
+**(a, b) = {{a}, {a, b}}**
+
+We prove: (a, b) = (c, d) if and only if a = c and b = d.
+
+**Proof:**
+
+(⇐) If a = c and b = d, then {a} = {c} and {a, b} = {c, d}, so {{a}, {a, b}} = {{c}, {c, d}}, i.e., (a, b) = (c, d). ✓
+
+(⇒) Suppose (a, b) = (c, d), i.e., {{a}, {a, b}} = {{c}, {c, d}}.
+
+**Case 1: a = b.**
+Then {{a}, {a, b}} = {{a}, {a}} = {{a}}, a set with exactly one element.
+So {{c}, {c, d}} = {{a}} has exactly one element, meaning {c} = {c, d}, so c = d.
+Since {a} is the unique element and {a} = {c}, we get a = c.
+Therefore a = c and b = a = c = d. ✓
+
+**Case 2: a ≠ b.**
+Then {a} ≠ {a, b} (since {a, b} contains b ≠ a), so {{a}, {a, b}} has exactly two distinct elements.
+Thus {{c}, {c, d}} also has exactly two elements, so {c} ≠ {c, d}, which means c ≠ d.
+
+The element {a} (a singleton) must equal either {c} or {c, d} in the outer set.
+- {a} cannot equal {c, d} because {c, d} has two elements (c ≠ d) while {a} has one.
+- Therefore {a} = {c}, giving **a = c**.
+
+Now the other element must match: {a, b} = {c, d} = {a, d}.
+So {a, b} = {a, d}. Since a ≠ b, and a appears in both sets, we need d = b (if d = a, then {c, d} = {a, a} = {a} = {c}, contradicting c ≠ d). Therefore **b = d**. ✓
+
+In both cases, (a, b) = (c, d) implies a = c and b = d. □
+
+---
+
+## Section 2: Functions
+
+### Problem 1
+
+**Compute f ∘ g and g ∘ f where f(x) = x² + 1 and g(x) = x + 2.**
+
+**(f ∘ g)(x) = f(g(x)) = f(x + 2) = (x + 2)² + 1 = x² + 4x + 4 + 1 = x² + 4x + 5**
+
+**(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.**
+
+**Yes, g must be one-to-one.**
+
+**Proof:**
+Suppose g(x₁) = g(x₂) for some x₁, x₂ in the domain of g.
+Then f(g(x₁)) = f(g(x₂)), i.e., (f ∘ g)(x₁) = (f ∘ g)(x₂).
+Since f ∘ g is one-to-one (injective), it follows that x₁ = x₂.
+
+Therefore g is one-to-one. □
+
+*Remark:* Note that we only needed f ∘ g to be injective for this conclusion; the hypothesis that f itself is injective was not required for this direction.
+
+---
+
+### Problem 3
+
+Let f: A → B and S, T ⊆ A. Recall f(S) = {f(x) : x ∈ S}.
+
+#### Part (a): Prove f(S ∪ T) = f(S) ∪ f(T)
+
+**Proof:**
+
+**(⊆)** Let y ∈ f(S ∪ T). Then there exists x ∈ S ∪ T with f(x) = y.
+- If x ∈ S, then y = f(x) ∈ f(S) ⊆ f(S) ∪ f(T).
+- If x ∈ T, then y = f(x) ∈ f(T) ⊆ f(S) ∪ f(T).
+So f(S ∪ T) ⊆ f(S) ∪ f(T).
+
+**(⊇)** Let y ∈ f(S) ∪ f(T).
+- If y ∈ f(S), there exists x ∈ S with f(x) = y. Since S ⊆ S ∪ T, we have x ∈ S ∪ T, so y ∈ f(S ∪ T).
+- If y ∈ f(T), there exists x ∈ T with f(x) = y. Since T ⊆ S ∪ T, we have x ∈ S ∪ T, so y ∈ f(S ∪ T).
+So f(S) ∪ f(T) ⊆ f(S ∪ T).
+
+Therefore **f(S ∪ T) = f(S) ∪ f(T)**. □
+
+---
+
+#### Part (b): Prove f(S ∩ T) ⊆ f(S) ∩ f(T)
+
+**Proof:**
+
+Let y ∈ f(S ∩ T). Then there exists x ∈ S ∩ T with f(x) = y.
+Since x ∈ S ∩ T, we have x ∈ S **and** x ∈ T.
+- Since x ∈ S and f(x) = y: y ∈ f(S).
+- Since x ∈ T and f(x) = y: y ∈ f(T).
+
+Therefore y ∈ f(S) ∩ f(T), which gives **f(S ∩ T) ⊆ f(S) ∩ f(T)**. □
+
+*Note:* The reverse inclusion does not hold in general. For example, let f: {1,2} → {1} with f(1) = f(2) = 1, S = {1}, T = {2}. Then f(S ∩ T) = f(∅) = ∅, but f(S) ∩ f(T) = {1} ∩ {1} = {1} ≠ ∅.
+
+---
+
+### Problem 4
+
+**Given bijections f: A → B and g: A → C, prove there exists a bijection h: B → C.**
+
+**Construction:** Define h: B → C by
+**h = g ∘ f⁻¹**
+
+Since f is a bijection, f⁻¹: B → A is well-defined. Since g: A → C, the composition h = g ∘ f⁻¹: B → C is well-defined.
+
+**Proof that h is a bijection:**
+
+*Injective:* Suppose h(b₁) = h(b₂) for some b₁, b₂ ∈ B.
+Then g(f⁻¹(b₁)) = g(f⁻¹(b₂)).
+Since g is injective: f⁻¹(b₁) = f⁻¹(b₂).
+Since f⁻¹ is injective (it is a bijection): b₁ = b₂. ✓
+
+*Surjective:* Let c ∈ C. Since g is surjective, there exists a ∈ A with g(a) = c.
+Since f is surjective (it is a bijection), there exists b ∈ B with f(b) = a, i.e., f⁻¹(b) = a.
+Then h(b) = g(f⁻¹(b)) = g(a) = c. ✓
+
+Therefore **h = g ∘ f⁻¹ is a bijection from B to C**. □
+
+---
+
+### Problem 5
+
+**If A and B are finite sets of the same size and f: A → B is injective, prove f is surjective.**
+
+**Proof:**
+
+Let |A| = |B| = n.
+
+Since f: A → B is injective, distinct elements of A map to distinct elements of B. Therefore the image f(A) = {f(a) : a ∈ A} has exactly |A| = n distinct elements:
+
+|f(A)| = n.
+
+Since f(A) ⊆ B and |B| = n, we have f(A) ⊆ B with |f(A)| = |B| = n. For finite sets, if S ⊆ T and |S| = |T|, then S = T. Therefore:
+
+f(A) = B.
+
+This means every element of B has a preimage in A, so **f is surjective**. □
+
+---
+
+### Problem 6
+
+**If A and B are infinite sets with a bijection between them, and f: A → B is injective, must f be surjective?**
+
+**No, f need not be surjective.**
+
+**Counterexample:**
+Let A = B = ℤ (the integers). The identity map id: ℤ → ℤ is a bijection, so there exists a bijection between A and B.
+
+Define f: ℤ → ℤ by f(n) = 2n.
+
+- *Injective:* If f(m) = f(n), then 2m = 2n, so m = n. ✓
+- *Not surjective:* The element 1 ∈ ℤ has no preimage: there is no integer n with 2n = 1. ✗
+
+Therefore f is injective but not surjective. □
+
+*Remark:* The key difference from Problem 5 is that infinite sets can have proper subsets of the same cardinality; for example, |2ℤ| = |ℤ| even though 2ℤ ⊊ ℤ. The pigeon-hole principle argument used in the finite case fails for infinite sets.
+
+---
+
+## Section 3: Relations
+
+### Problem 1
+
+**Is R₁ an equivalence relation, where (a,b) R₁ (c,d) iff ad = bc?**
+
+**Yes, R₁ is an equivalence relation.** (Note: this relation on positive integer pairs encodes the equivalence a/b = c/d.)
+
+**Proof:**
+
+*Reflexive:* For any positive integers (a, b), we have ab = ba (commutativity), so (a, b) R₁ (a, b). ✓
+
+*Symmetric:* Suppose (a, b) R₁ (c, d), so ad = bc. Then cb = bc = ad = da, so (c, d) R₁ (a, b). ✓
+
+*Transitive:* Suppose (a, b) R₁ (c, d) and (c, d) R₁ (e, f), so ad = bc and cf = de.
+Multiply the first equation by f: adf = bcf.
+Multiply the second equation by b: bcf = bde.
+So adf = bde. Since d is a positive integer (d > 0), divide both sides by d: af = be.
+Therefore (a, b) R₁ (e, f). ✓
+
+**R₁ is an equivalence relation.** □
+
+---
+
+### Problem 2
+
+**If R₁ and R₂ are equivalence relations on a set A, is R₁ ∩ R₂ an equivalence relation?**
+
+**Yes, R₁ ∩ R₂ is an equivalence relation.**
+
+**Proof:** Let R = R₁ ∩ R₂.
+
+*Reflexive:* Let a ∈ A. Since R₁ is reflexive, (a, a) ∈ R₁. Since R₂ is reflexive, (a, a) ∈ R₂. Therefore (a, a) ∈ R₁ ∩ R₂ = R. ✓
+
+*Symmetric:* Suppose (a, b) ∈ R. 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) ∈ R. ✓
+
+*Transitive:* Suppose (a, b) ∈ R and (b, c) ∈ R. Then (a, b), (b, c) ∈ R₁ and (a, b), (b, c) ∈ R₂.
+Since R₁ is transitive: (a, c) ∈ R₁.
+Since R₂ is transitive: (a, c) ∈ R₂.
+Therefore (a, c) ∈ R. ✓
+
+**R₁ ∩ R₂ is an equivalence relation.** □
+
+---
+
+### Problem 3
+
+**A relation R on A is *circular* if (a,b) ∈ R and (b,c) ∈ R implies (c,a) ∈ R.**
+
+**Claim:** A relation is an equivalence relation if and only if it is reflexive and circular.
+
+**Proof:**
+
+**(⇒) Every equivalence relation is reflexive and circular.**
+
+Let R be an equivalence relation. R is reflexive by definition. For circularity: suppose (a, b) ∈ R and (b, c) ∈ R. By transitivity, (a, c) ∈ R. By symmetry, (c, a) ∈ R. So R is circular. ✓
+
+**(⇐) Every reflexive and circular relation is an equivalence relation.**
+
+Suppose R is reflexive and circular. We must show R is symmetric and transitive.
+
+*Symmetric:* Suppose (a, b) ∈ R. By reflexivity, (b, b) ∈ R. Applying circularity to (a, b) ∈ R and (b, b) ∈ R: (b, a) ∈ R. ✓
+
+*Transitive:* Suppose (a, b) ∈ R and (b, c) ∈ R. By symmetry (just proved), (b, a) ∈ R. Now apply circularity to (b, c) ∈ R and (c, ... )—but instead, apply circularity directly: since (a, b) ∈ R and (b, c) ∈ R, we get (c, a) ∈ R. Then by symmetry (proved above), (a, c) ∈ R. ✓
+
+Since R is reflexive, symmetric, and transitive, **R is an equivalence relation**. □
+
+---
+
+### Problem 4
+
+**Give a poset that has:**
+
+#### Part (a): A minimal element but no maximal element.
+
+**Example:** (ℕ, ≤) where ℕ = {0, 1, 2, 3, ...} with the usual ≤ ordering.
+
+- **Minimal element:** 0, since 0 ≤ n for all n ∈ ℕ (and no element is less than 0).
+- **No maximal element:** For any n ∈ ℕ, n+1 ∈ ℕ and n < n+1, so n is not maximal.
+
+---
+
+#### Part (b): Neither a maximal nor a minimal element.
+
+**Example:** (ℤ, ≤) where ℤ is the set of all integers with the usual ≤ ordering.
+
+- **No minimal element:** For any n ∈ ℤ, n−1 ∈ ℤ and n−1 < n.
+- **No maximal element:** For any n ∈ ℤ, n+1 ∈ ℤ and n+1 > n.
+
+---
+
+### Problem 5
+
+**Can a non-empty poset (A, ≤) with no minimal element be finite?**
+
+**No, such a poset cannot be finite.**
+
+**Proof:**
+
+We prove the contrapositive: every finite, non-empty poset has a minimal element.
+
+Let (A, ≤) be a finite, non-empty partially ordered set. We construct a minimal element as follows:
+
+Pick any element a₁ ∈ A. If a₁ is minimal, we are done. Otherwise, there exists a₂ ∈ A with a₂ < a₁ (strictly). If a₂ is minimal, we are done. Otherwise, there exists a₃ ∈ A with a₃ < a₂. Continuing this process, we obtain a sequence
+
+a₁ > a₂ > a₃ > ···
+
+This is a strictly decreasing sequence of elements. Since < is irreflexive and antisymmetric, no element can repeat in this sequence (if aᵢ = aⱼ with i < j, then aᵢ < aᵢ, which is impossible). Therefore all elements in the sequence are distinct.
+
+Since A is finite, this strictly decreasing sequence cannot continue indefinitely — it must terminate at some element aₖ. By construction, aₖ has no element strictly less than it in the sequence; and since aₖ was chosen because no smaller element had been found, aₖ is minimal in A.
+
+Therefore, **every finite non-empty poset has a minimal element**, and consequently, a non-empty poset with no minimal element must be infinite. □