COS1501 Assignment 3 (DETAILED ANSWERS) 2024 - DISTINCTION GUARANTEED

COS1501 Assignment 3 (DETAILED ANSWERS) 2024 - DISTINCTION GUARANTEED - DISTINCTION GUARANTEED - DISTINCTION GUARANTEED Answers, guidelines, workings and references .... Question 1 Complete Marked out of 2.00 Question 2 Complete Marked out of 2.00 QUIZ Suppose U = {1, 2, 3, 4, 5, a, b, c} is a universal set with the subset A = {a, b, c, 1, 2, 3, 4}. Which one of the following relations on A is NOT functional? a. {(1, 3), (b, 3), (1, 4), (b, 2), (c, 2)} b. {(a, c), (b, c), (c, b), (1, 3), (2, 3), (3, a)} c. {(a, a), (c, c), (2, 2), (3, 3), (4, 4)} d. {(a, c), (b, c), (1, 3), (3, 3)} Suppose U = {1, 2, 3, 4, 5, a, b, c} is a universal set with the subset A = {a, b, c, 1, 2, 3, 4}. Which one of the following alternatives represents a surjective function from U to A? a. {(1, 4), (2, b), (3, 3), (4, 3), (5, a), (a, c), (b, 1), (c, b)} b. {(a, 1), (b, 2), (c, a), (1, 4), (2, b), (3, 3), (4, c)} c. {(1, a), (2, c), (3, b), (4, 1), (a, c), (b, 2), (c, 3)} d. {(1, a), (2, b), (3, 4), (4, 3), (5, c), (a, a), (b, 1), (c, 2)} Question 3 Complete Marked out of 2.00 Question 4 Complete Marked out of 2.00 Question 5 Complete Marked out of 2.00 Let G and L be relations on A = {1, 2, 3, 4} with G = {(1, 2), (2, 3), (4, 3)} and L = {(2, 2), (1, 3), (3, 4)}. Which one of the following alternatives represents the relation L ￮ G = G; L? a. {(2, 3), (3, 3)} b. {(1, 2), (2, 4), (4, 4)} c. {(1, 2), (2, 1), (3, 3), (4, 4)} d. {(2, 4), (4, 4)} Let g be a function from Z (the set of positive integers) to Q (the set of rational numbers) defi ned by (x, y) ∈ g iff y = (g ⊆ Z x Q) and let f be a function on Z defi ned by (x, y) ∈ f iff y = 5x + 2x – 3 (f ⊆ Z x Z ). Consider the function f on Z . For which values of x is it the case that 5x + 2x – 3 0? Hint: Solve 5x + 2x – 3 0 and keep in mind that x ∈ Z . a. x 5, x ∈ Z b. x 1, x ∈ Z c. x ≥ 1, x ∈ Z d. x 1, x ∈ Z + 4x − 3/7 + + 2 + + + 2 2 + + 3/5 + + + Let g be a function from Z (the set of positive integers) to Q (the set of rational numbers) defi ned by (x, y) ∈ g iff y = (g ⊆ Z x Q) and let f be a function on Z defi ned by (x, y) ∈ f iff y = 5x + 2x – 3 (f ⊆ Z x Z ). Which one of the following is an ordered pair belonging to f? a. (–1, 0) b. (2, 21) c. (1, 5) d. (3, 44) + 4x − 3/7 + + 2 + + Question 6 Complete Marked out of 2.00 Question 7 Complete Marked out of 2.00 Question 8 Complete Marked out of 2.00 Let g be a function from Z (the set of positive integers) to Q (the set of rational numbers) defi ned by (x, y) ∈ g iff y = (g ⊆ Z x Q) and let f be a function on Z defi ned by (x, y) ∈ f iff y = 5x + 2x – 3 (f ⊆ Z x Z ). Which one of the following alternatives represents the image of x under g ￮ f (ie g ￮ f(x)))? a. 20x + 8x – 12 b. 80x + 4 x – c. 20x + 8x + 3 d. 80x + 4 x – 3 + 4x − 3/7 + + 2 + + 2 3/7 2 4/7 180/49 2 3/7 2 4/7 Let g be a function from Z (the set of positive integers) to Q (the set of rational numbers) defi ned by (x, y) ∈ g iff y = (g ⊆ Z x Q), and let f be a function on Z defi ned by (x, y) ∈ f iff y = 5x + 2x – 3 (f ⊆ Z x Z ). Which one of the following statements regarding the function g is TRUE? (Remember, g ⊆ Z x Q.) a. g can be presented as a straight line graph. b. g is injective. c. g is surjective. d. g is bijective. + 4x − 3/7 + + 2 + + + Which one of the following alternatives gives the format for the list notation of a. (((a, a), a), ((a, b), d), ((a, c), c), ((a, d), b), ((b, a), b), ((b, b), a), ((b, c), d), ((b, d), a), ((c, a), c), ((c, b), b), ((c, c), a), ((c, d),c), ((d, a), d), ((d, b), b), ((d, c), c), ((d, d), a)) b. {((a, a), a), ((a, b), d), ((a, c), c), ((a, d), b), ((b, a), b), ((b, b), a), ((b, c), d), ((b, d), a), ((c, a), c), ((c, b), b), ((c, c), a), ((c, d),c), ((d, a), d), ((d, b), b), ((d, c), c), ((d, d), a)} c. {{(a, a), a}, {(a, b), d}, {(a, c), c}, {(a, d), b}, {(b, a), b}, {(b, b), a}, {(b, c), d}, {(b, d), {(c, a), c}, {(c, b), b}, {(c, c), a}, {(c, d), c},{(d, a), d}, {(d, b), b}, {(d, c), c}, {(d, d), a}} d. {({a, a}, a), ({a, b}, d), ({a, c}, c), ({a, d}, b), ({b, a}, b), ({b, b}, a), ({b, c}, d), ({b, d}, a), ({c, a}, c), ({c, b}, b), ({c, c}, a), ({c, d},c), ({d, a}, d), ({d, b}, b), ({d, c}, c), ({d, d}, a)} Question 9 Complete Marked out of 2.00 Question 10 Complete Marked out of 2.00 Which one of the following options regarding the binary operation * is FALSE? a. (a * b) * (c * d) = (a * (b * d)) * (d * c) b. (a * b) (b * a) can be used as a counterexample to prove that the binary operation * is not commutative. c. (a * b) * d = a * (b * d) proves that the binary operation * is associative. d. The binary operation * does not have an identity element. ≠ Let A = {□, ◊, ☼, ⌂} and let # be a binary operation from A X A to A presented by the following table: Which one of the following statements pertaining to the binary operation # is TRUE? a. is the identity element for #. b. # is symmetric (commutative). c. # is associative. d. [(⌂ # ◊) # ☼] = [⌂ # (◊ # ☼)] Question 11 Complete Marked out of 2.00 Question 12 Complete Marked out of 2.00 Consider the representation of the binary operation # below: # can be written in list notation. Which one of the following ordered pairs is an element of the list notation set representing #? a. ((□, ◊), ⌂) b. ((⌂, ☼), ◊) c. ((☼, ◊), ◊) d. ((⌂, ◊), ◊) Perform the following matrix multiplication operation: Which one of the following alternatives represents the correct answer to the above operation? a. The operation is not possible. b. c. d. Question 13 Complete Marked out of 2.00 Consider the following matrices: Which one of the following statements is FALSE? a. The result of B C is b. The result of A C is c. B C = C B d. The operation (C B) A is not possible. ∙ ∙ ∙ ∙ ∙ ∙ Question 14 Complete Marked out of 2.00 Consider the truth table for the connective ‘ ’ with two simple declarative statements p and q. (i) Which one of the given alternatives represents ‘‘ ’ as a binary operation on the set of truth values {T, F}? (ii) Does the binary operation ‘ ’ have an identity element? For each alternative, please look at (i) and (ii). a. (i) (ii) The binary operation ‘ ’ does not have an identity element. b. (i) (ii) The binary operation ‘ ’ has an identity element. c. (i) (ii) The binary operation ‘ ’ does not have an identity element. d. (i) (ii) The binary operation ‘ ’ has an identity element. ↔ ↔ ↔ ↔ ↔ ↔ ↔ Question 15 Complete Marked out of 2.00 Let p, q and r be simple declarative statements. Which alternative provides the truth values for the biconditional ‘ ’ of thecompound statement provided in the given table? Hint: Determine the truth values of p r, q ∨ r, (p r) ⋀ (q ∨ r), q p, ¬(q p) and ¬(q p) ⋀ r in separate columnsbefore determining the truth values of [(p r) ⋀ (q ∨ r}] [¬(q p) ⋀ r]. a. b. c. d. ↔ → → → → → → ↔ → Question 16 Complete Marked out of 2.00 Question 17 Complete Marked out of 2.00 Consider the following quantifi ed statement: ∀x ∈ Z [(x ≥ 0) ∨ (x + 2x – 8＞0)] Which one of the alternatives provides a true statement regarding the given statement or its negation? a. The negation ∃x ∈ Z [(x 0) ∨ (x + 2x – 8 ≤ 0)] is not true. b. x = – 3 would be a counterexample to prove that the negation is not true. c. x = – 6 would be a counterexample to prove that the statement is not true. d. The negation ∃x ∈ Z [(x 0) ∧ (x + 2x – 8 ≤ 0)] is true. 2 2 2 2 2 2 Consider the following proposition: "For any predicates P(x) and Q(x) over a domain D, the negation of the statement ∃x ∈ D, P(x) ∧ Q(x)" is the statement "∀x ∈ D, P(x) ¬Q(x)". We can use this truth to write the negation of the following statement: "There exist integers a and d such that a and d are negative and a/d = 1 + d/a". Which one of the alternatives provides the negation of this statement? a. There exist integers a and d such that a and d are positive and a/d = 1 + d/a. b. For all integers a and d, if a and d are positive then a/d 1 + d/a. c. For all integers a and d, if a and d are negative then a/d 1 + d/a. d. For all integers a and d, a and d are positive and a/d 1 + d/a. → ≠ ≠ ≠ Question 18 Complete Marked out of 2.00 Question 19 Complete Marked out of 2.00 Question 20 Complete Marked out of 2.00 Which one of the alternatives is a proof by contrapositive of the statement If x – x + 4 is not divisible by 4, then x even. a. Required to prove: If x – x + 4 is not divisible by 4 then x even. Proof: Suppose x is odd. Let x = 2k + 1, then we have to prove that x – x + 4 is divisible by 4. x – x + 4 = (2k + 1) – (2k + 1) + 4 = (2k + 1)(4k + 4k +1) – 2k – 1 + 4 = 8k + 8k + 2k + 4k + 4k +1 – 2k – 1 + 4 = 8k + 12k + 4k + 4 = 4(2k + 3k + k + 1), which is divisible by 4. (4 multiplied by any integer is divisible by 4) b. Required to prove: If x – x + 4 is not divisible by 4, then x even. Proof: Assume that x – x + 4 is not divisible by 4. Then x can be even or odd. We assume that x is odd. Let x = 2k + 1, then x – x + 4 = (2k+1) – (2k + 1) + 4 = (2k + 1)(4k + 4k +1) – 2k – 1 + 4 = 8k + 8k + 2k + 4k + 4k +1 – 2k – 1 + 4 = 8k + 12k + 4k + 4 = 4(2k + 3k + k + 1), which is divisible by 4. (4 multiplied by any integer is divisible by 4) But this is a contradiction to our original assumption. Therefore x must be even if x – x + 4 is not divisible by 4. c. Required to prove: If x – x + 4 is not divisible by 4, then x even. Proof: Let x = 4 be an even element of Z. We can replace x with 4 in the expression x – x + 4. x – x + 4 = 64 – 4 + 4 = 64 which is divisible by 4. d. Required to prove: If x – x + 4 is not divisible by 4, then x even. Proof: Assume that x is even, i.e. x = 4k, then x – x + 4 = (4k) – (4k) + 4 = 64k – 4k + 4 = 4(16k – k + 1), which is divisible by 4. 3 3 3 3 3 2 3 2 2 3 2 3 2 3 3 3 3 2 3 2 2 3 2 3 2 3 3 3 3 3 3 3 3 3 By using logical equivalences and de Morgan’s rules, we can show that the statements ¬p ⋁ q and (p ⋀ ¬q) (¬p ⋁ q) are equivalent. a. True b. False → The statement [p ⋀ (r q)] [(r ⋁ q) ⋀ (p q)] is a contradiction. a. True b. False → ↔ → Question 21 Complete Marked out of 2.00 Question 22 Complete Marked out of 2.00 Question 23 Complete Marked out of 2.00 Question 24 Complete Marked out of 2.00 Consider the two statements below: Statement 1: ∀x Z , [(2x + 1 3) ⋁ (2x – 1 1)] Statement 2: ∃x Z, [(x – 1 0) ∧ (2x - 2 0)] It is true that both statements 1 and 2 are false? a. True b. False ∈ + 2 ≥ ∈ 2 ≥ Consider the following statement: ∀x Z, [(2x + 4 0) ⋁ (4 - x ≤ 0)] The negation of the above statement is: ¬[∀x Z, [(2x + 4 0) ⋁ (4 - x ≤ 0)]] ≡ ∃x Z, ¬[(2x + 4 0) ⋁ (4 - x ≤ 0)] ≡ ∃x Z, [¬(2x + 4 0) ∧ ¬(4 - x ≤ 0)] ≡ ∃x Z, [(2x + 4 ≤ 0) ∧ (4 - x 0)] a. True b. False ∈ 2 ∈ 2 ∈ 2 ∈ 2 ∈ 2 Consider the statement If n is even, then 4n - 3 is odd. The contrapositive of the given statement is: If 4n - 3 is odd, then n is even. a. True b. False 2 2 Consider the statement If n is a multiple of 3, then 2n + 2 is not a multiple of 3. The converse of the given statement is: If n is not a multiple of 3, then 2n + 2 is a multiple of 3. a. True b. False Question 25 Complete Marked out of 2.00 Consider the following statement, for all x Z: If x + 1 is even, then 3x - 4 is odd. The correct way to start a direct proof to determine if the statement is true is as follows: Assume x is even, then x = 2k for some k Z, then 3x – 4 ie 3(2k) - 4 i.e. ……….. a. True b. False