Understanding Logical Implications and Inductive Proofs in Math
School
Florida State University**We aren't endorsed by this school
Course
MAD 2104
Subject
Mathematics
Date
Dec 11, 2024
Pages
4
Uploaded by LieutenantNightingaleMaster785
Quiz 3AnswersDiscrete Mathematics I1.Find all the truth values for p, q, and r that make the LHS of this logical implicationtrue:[(p ⊕ q) ⊕ r] ∧ (p → q) =⇒ p ∨ q.Is this a valid implication? Why or why not? 2.Prove by a verbal argument (no truth table) thatp ∧ (q ⊕ r) =⇒ (p ∧ q) ∨ (p ∧ r).Can this implication be reversed to conclude that these are logically equivalent? If so give a proof, if not explain why not. 3.Answer the following.a) Are these two sets equal or not equal: ∅, {∅}?Explain your answer.b) Find the elements ofsets A and B when all of the following are true:A –B = { 1, 2, 7, 8 }B –A = { 5, 10 } A ⋂B = { 0, 3, 6, 9 } Specify each set by listing elements. 4. This problem focuses on the structure of an inductive proof.Consider the task of proving the following statement by mathematical induction:For any positive integer n,∑(2𝑖 − 1) = 𝑛2𝑛𝑖=1Note that the summation notation ∑(2𝑖 − 1)𝑛𝑖=1represents the sum, 1+3+5+…+(2𝑛 − 1)Answer the following questions about the proof of this statement by mathematical induction. a.What value of nshould be used in the basis step of this proof?b.What statement must be proven in the basis step?
c.Which of the following is the correct statement to prove in the inductive stepfor some arbitrary integer k ≥ 1? ∑(2𝑘 − 1) = 𝑘2→ ∑(2𝑘 − 3) = (𝑘 + 1)2𝑘+1𝑘=1𝑘𝑘=1∑(2𝑖 − 1) = 𝑛2→ ∑(2𝑖 − 1) = 𝑘2𝑘𝑖=1𝑛𝑖=1∑(2𝑖 − 1) = 𝑘2→ ∑(2𝑖 − 1) = (𝑘 + 1)2𝑘+1𝑖=1𝑘𝑖=1d.You are not completing this proof; however, assuming both the basis step and the inductive step proof can be completed, state the final conclusionof this inductive proof. Answers.1.The student may make out a truth table and read off the truth assignments that make the LHS T. Here I give a logical analysis. Analyzing 1. When the LHS is T both (p⊕q)⊕r and p → q are T. Let’s examine in turn what this implies about the truth values of p and q when (1) r is T and when (2) r is F. (1)rT: In this case, since (p⊕q)⊕r is T, we conclude that p⊕q is F, which means either pTq T or pFq F. Both these truth assignments for p and q make p → q T, so we find two assignments that make the LHS T, namely pTq TrT and pFq FrT.Notice that the second of these, pFq FrT makes the RHS: p ∨ q F so this is not a valid implication. (2)rF: In this case p ⊕ q is T which occurs for the two assignments pTq F and pFq T. However, since p → q is T, the first of these is ruled out, so there is only one assignment in this case that makes the LHS T, namely pFq TrF.In summary, we have three assignments of truth values that make the LHS T: pTq TrT, pFq FrT, pFq TrF,and the middle one falsifies the implication. 2.Proof. Assume the LHS is T. Then p is T and q ⊕ r is T. This means that we have either (1) q TrF or (2) q FrT. In case (1), both p and q are T implying that p ∧ q is T,
which makes the RHS T. In case (2), both p and r are T implying that p∧r is T, which makes the RHS T. Hence the two assignments that make the LHS T also make the RHS T, and this proves the validity of the implication. The implication cannot be reversed since the assignment pTq TrT makes the RHS T and the LHS F. 3a. Are these 2 sets equal or not equal: ∅, {∅}? Explain your answer. No, ∅is the empty set; it contains no elements. {∅}is a set containing 1 element which is the empty set. b.Indicate what sets A and B will contain when all of the following are true:A –B = {1, 2, 7, 8} B –A = {5, 10} A ⋂B = {0, 3, 6, 9} Specify each set by listing elements. Explain how your sets satisfy each criteria listed. A = { 0, 1, 2, 3, 6, 7, 8, 9 } B = { 0, 3, 5, 6, 9, 10 } Element in A but not in B are 1,2,7,8. (A-B) Elements in B but not in A are 5,10. (B-A) Elements in both A and B are 0,3,6,9. (A⋂B)4. Answer the following questions about the proof of this statement by mathematicalinduction.a.What value of nshould be used in the basis step of this proof? n = 1b.What statement must be proven in the basis step?For n = 1,∑(?𝒊 − ?) = ???𝒊=?c.Which of the following is the correct statement to prove in the inductive stepfor some arbitrary integer k ≥1?∑(2𝑖 − 1) = 𝑛2→ ∑(2𝑖 − 1) = 𝑘2𝑘𝑖=1𝑛𝑖=1Expressions on left and right of →are equivalent ∑(2𝑖 − 1) = 𝑘2→ ∑(2𝑖 − 1) = (𝑘 + 1)2𝑘+1𝑖=1𝑘𝑖=1Correct P(k) →P(k+1) ∑(2𝑘 − 1) = 𝑘2→ ∑(2𝑘 − 3) = (𝑘 + 1)2𝑘+1𝑘=1𝑘𝑘=1Wrong expression in summation on right of →d.You are not completing this proof; however, assuming both the basis stepand the inductive step proof can be completed, state the final conclusionofthis inductive proof.
𝒏𝒊=?By PMI, for any positive integer n,∑(?𝒊−?)=𝒏?