Sample Solutions to Homework 1 2. Answer the questions in the following examples regarding logical fallacies. (a) During their argument, Anthony tells Marie that he hates her and wished that she would get hit by a car. Later that evening, Anthony receives a call from a friend who tells him that Marie is in the hospital because she was struck by a car. Anthony immediately blames himself and reasons that if he hadn’t made that comment during their fight, Marie would not have been hit. What logical fallacy has Anthony committed? Faulty cause/effect (questionable cause) (b) Louise is running for class president. In her campaign speech she says, “My opponent does not deserve to win. She is a smoker and she cheated on her boyfriend last year.” What fallacy has Louise committed? appeal to motive, or ad hominem. (Fallacies of relevance) (c) Bill: “I think capital punishment is wrong.” Adam: “No it isn’t. If it was wrong, it wouldn’t be legal.” Of which fallacy is this an example? Begging the question (Improper premise) (d) Ricky is watching television when he sees a commercial for foot cream. The commercial announcer says, “This is the best foot cream on the market because no other foot cream makers have been able to prove otherwise!” What fallacy has the announcer committed? Appeal to ignorance (e) Maria has been working at her current job for more than 30 years at the same wage. She desperately wants a raise so she approaches her boss to ask for one. She says, “You are one of the kindest people I know. You are smart and good-looking and I really love your shoes.” What type of fallacy is this? Appeal to emotion. 3. Let A = {0, 1}. P(A) = ? P(P(A)) = ? 푃(퐴) = {휙, {0}, {1}, {0, 1}} 푃(푃(퐴)) = {휙, {휙}, {{0}}, {{1}}, {{0, 1}}, {휙, {0}}, {{1}, 휙}, {{1}, {0}}, {{1}, {0}, 휙} , {{0, 1}, 휙} , {{0, 1}, {0}}, {{0, 1}, {0}, 휙}, {{0, 1}, {1}}, {{0,1}, {1}, 휙 }, {{0,1}, {1}, {0}}, {{0, 1}, {1}, {0}, 휙}} Solution: Let 푆(푥, 푦) denotes the relation that “푥 shaves 푦,” and 푏 be the only barber, 퐴 = {푥 |푠(푏, 푥)} be the people shaved by b, 퐵 = {푥|푠(푥, 푥) = 푓푎푙푠푒} be the set of the people who don’t shave themselves. From assumption, A=B: assume 푏 ∈ 퐵 ⇒ 푠(푏, 푏) = 푓푎푙푠푒 and 퐴 = 퐵 ⇒ 푏 ∈ 퐴 ⇒ 푠(푏, 푏) = 푡푟푢푒 thus reaches a contradiction. Assume 푏 ∉ 퐵 ⇒ 푠(푏, 푏) = 푡푟푢푒 푎푛푑 퐴 = 퐵 ⇒ 푏 ∉ 퐴 ⇒ 푠(푏, 푏) = 푓푎푙푠푒 thus reaches a contradiction. Proof: The set in the question is a product of two countable sets. Since the product of two countable sets is countable, this result implies the set in the question is countable. If we cannot use the result, a direct proof goes as follows: To find a function 푓: 푁 × {푎, 푏, 푐} ↦ 푁 that is injective, then 푁 × {푎, 푏, 푐} is countable. 0 푥 = 푎 Let 푔: 푁 × {0, 1, 2} ↦ 푁, and ℎ: {푎, 푏, 푐} ↦ {0, 1, 2}, ℎ(푥) = {1 푥 = 푏 thus 푓(푛, 푙) = 2 푥 = 푐 푔(푛, ℎ(푙)), thus if it is shown a function 푔(푛, ℎ) is injective, and ℎ is apparently a injective function, then 푓 is injective. Let 푔(푛, ℎ) = (푛 − 1) ∗ 3 + ℎ it would satisfy the injective condition: 1. 푖푛푗푒푐푡푖푣푒: ∀푥1∀푥2 (푥1 ≠ 푥2 → 푓(푥1) ≠ 푓(푥2)) ∀푥1∀푥2 (푥1 ≠ 푥2 → 푓(푥1) ≠ 푓(푥2)) ⟷ ∀푥1∀푥2 (푓(푥1) = 푓(푥2) → 푥1 = 푥2) Let 푔(푛1, ℎ1) = 푔(푛2, ℎ2) ⇒ (푛1 − 1) ∗ 3 + ℎ1 = (푛2 − 1) ∗ 3 + ℎ2 ⇒ 3푛1 + ℎ1 = 3푛2 + ℎ2 ⇒ 3(푛1 − 푛2) = ℎ2 − ℎ1 ⇒ −2 ≤ 3(푛1 − 푛2) ≤ 2 2 2 ⇒ − ≤ (푛 − 푛 ) ≤ since 푛 , 푛 푎푟푒 푛푎푡푢푟푎푙 푛푢푚푏푒푟푠, thus 푛 − 푛 is an integer. 3 1 2 3 1 2 1 2 ⇒ 푛1 − 푛2 = 0 ⇒ 푛1 = 푛2 ⇒ ℎ1 = ℎ2, i.e. ∀푥1∀푥2 (푓(푥1) = 푓(푥2) → 푥1 = 푥2) is true, i.e. 푔(푛, ℎ) is injective. The function 푓(푛, ℎ(푙)) is injective, where 푛 ∈ 푁, 푙 ∈ {푎, 푏, 푐}, and set 푁 × {푎, 푏, 푐} is countable. Proof: ∀푘 ∈ ℕ(ℕ푘 푖푠 푐표푢푛푡푎푏푙푒). Prove with induction: When 푘 = 1: 푁푘 = 푁 , and 푁 is countable, so is for 푁푘. Assume 푁푘 is countable where 푘 > 1, to show 푁푘+1 = 푁푘 × 푁 is countable, since 푁푘 and 푁 both are countable, thus 푁푘+1 = 푁푘 × 푁 is countable. Proof: Solution 1: Let want to show is countable: 푎0 푥 = 0 st : define as 푓(푥) = { , then express as (푎0, 푎1) where the 1 element in tuple is 푎1 푥 = 1 nd 푓(0), and 2 element is 푓(1), thus F = {(푎0, 푎1)| 푎0 ∈ ℕ, 푎1 ∈ ℕ} = ℕ × ℕ thus is countable. Solution 2: Let want to show is countable: 푎 푥 = 0 : define as 푓(푥) = { 0 , then express as where the 1st element in 푎1 푥 = 1 tuple is input, and 2nd element in tuple is the output, thus thus is countable. Proof: if A and B are countable, then there exists 푓퐴 ∶ 퐴 ↦ ℕ , 푓퐵 ∶ 퐵 ↦ ℕ , and 푓퐴, 푓퐵 are injective define 푓: 퐴 ∪ 퐵 ↦ ℕ as 2푓퐴(푥) 푤ℎ푒푟푒 푥 ∈ 퐴 푓(푥) = { 2푓퐵(푥) + 1 푤ℎ푒푟푒 푥 ∈ 퐵\퐴 Want to show 푓(푥) is injective: ∀푥1∀푥2 (푓(푥1) = 푓(푥2) → 푥1 = 푥2) If 푓(푥1) = 푓(푥2) 푖푠 푎푛 푒푣푒푛 푛푢푚푏푒푟 ⇒ 2푓퐴(푥1) = 2푓퐴(푥2) ⇒ 푓퐴(푥1) = 푓퐴(푥2) ⇒ 푥1 = 푥2 If 푓(푥1) = 푓(푥2) 푖푠 푎푛 표푑푑 푛푢푚푏푒푟 ⇒ 2푓퐵(푥1) + 1 = 2푓퐵(푥2) + 1 ⇒ 푓퐵(푥1) = 푓퐵(푥2) ⇒ 푥1 = 푥2 For both cases, we have shown ∀푥1∀푥2 (푓(푥1) = 푓(푥2) → 푥1 = 푥2) is true, i.e. 푓(푥) 푖푠 푖푛푗푒푐푡푖푣푒. 1 푥 = 푎 Proof: let ℎ: {푎, 푏, 푐} ↦ {1, 2, 3} such that ℎ(푥) = {2 푥 = 푏, then define 푓: {푎, 푏, 푐}∗ ↦ ℕ such 3 푥 = 푐 0 푦 = 휀 that 푓(푦) = { 푘 푖 where 푦 = 푥1 ∘ 푥2 … ∘ 푥푘, this function is ∑푖=1 ℎ(푥푖) ∗ 10 표푡ℎ푒푟푤푖푠푒 ∗ ∗ injective from {푎, 푏, 푐} ↦ ℕ , since ∀푦1 ∀푦2 (푓(푦1) = 푓(푦2) → 푦1 = 푦2) . thus {푎, 푏, 푐} is a countable set. .