CS0441 Discrete Structures Recitation 3 Xiang Xiao
Section 1.5 Q10 Let F(x, y) be the statement x can fool y, where the domain consists of all people in the world. Use quantifiers to express each of these statements. g. Nancy can fool exactly two people. x y( F( Nancy, x) F( Nancy, y) ( x y) z( F( Nancy, z) ( z x) ( z y))) h. There is exactly one person whom everybody can fool. x y( F( x, y) z( F( x, z) ( z y))) j. There is someone who can fool can fool exactly one person besides himself or herself x y( F( x, y) ( x y) z( F( x, z) ( z y)))
Section 1.5 Q14 Use quantifiers and predicates with more than one variable to express these statements e. There is a student in this class who has taken every course offered by one of the departments in this school. T(x, y): student x has taken course y; O(y, z): course y is offered by department z. x y z( O( y, z) T ( x, y))
Section 1.5 Q14 Use quantifiers and predicates with more than one variable to express these statements f. Some student in this class grew up in the same town as exactly one other student in this class Domain: student in this class S(x, y): x grew up in the same town as y. x y( S( x, y) ( x y) z( S( x, z) ( z y)))
Section 1.5 Q32 Express the negations of each of these statements so that all negation symbols immediately precede predicates. c. x y( Q( x, y) Q( y, x) Step 1: x y( Q( x, y) Q( y, x) x y ( Q( x, y) Q( y, x) Step 2: ( Q( x, y) Q( y, x)) (( Q( x, y) Q( y, x)) ( Q( y, x) Q( x, y))) (( Q( x, y) Q( y, x)) ( Q( y, x) Q( x, y))) ( ( Q( x, y) Q( y, x)) ( ( Q( y, x) Q( x, y))) ( Q( x, y) Q( y, x)) ( Q( y, x) Q( x, y))) Step 3: x y( Q( x, y) Q( y, x) x y( Q( x, y) Q( y, x)) ( Q( y, x) Q( x, y)))
Section 1.6 Q6 Use rules of inference to show that the hypotheses If it does not rain or if it is not foggy, then the sailing race will be held and the lifesaving demonstration will go on, If the sailing race is held, then the trophy will be awarded, and The trophy was not awarded imply the conclusion It rained. p: It rains; q: It is foggy; r: the sailing race will is held; s: the lifesaving demonstration goes on t: the trophy is awarded Step Reason
Section 1.6 Q6 Use rules of inference to show that the hypotheses If it does not rain or if it is not foggy, then the sailing race will be held and the lifesaving demonstration will go on, If the sailing race is held, then the trophy will be awarded, and The trophy was not awarded imply the conclusion It rained. Step Reason 1. t Premise 2. r t Premise 3. r Modus tollens from (1) and (2) 4. p q r s Premise 5. r s r Simplification 6. p q r Hypothetical syllogism from (4) and (5) 7. ( p q) Modus tollens from (3) and (6) 8. p q De Morgan's law, Double negation 9. p Simplification from (8)
Section 1.6 Question 14 For each of these arguments, explain which rules of inference are used for each step. a. Linda, a student in this class, owns a red convertible. Everyone who owns a red convertible has gotten at least one speeding ticked. Therefore, some one in this class has gotten a speeding ticket. C(x): x is in the class; O(x): x owns a red convertible; S(x): x has gotten a speeding ticket. Step Reason 1. x( R( x) S( x)) Premise 2. R( Linda) S( Linda) Universal instantiation from (1) 3. R( Linda) Premise 4. S( Linda) Modus pones from (2) and (3) 5. C( Linda) Premise 6. C( Linda) S( Linda) Conjuction from (4) and (5) 7. x( C( x) S( x)) Existential generalization from (6)
Section 1.6 Question 16 Prove that if n is a perfect square, then n + 2 is not a perfect square. Solution: direct proof Assume n is a perfect square, prove that n+2 is not a perfect square Assume n is a perfect square, n = k 2 (k >= 0) The next largest perfect square after n should be (k+1) 2 = k 2 +2k+1 = n+1+2k When k = 0, n = 0, n+2 = 2 is not a perfect square When k >= 1, (k+1) 2 = n+1+2k >= n + 3 > n+2, n+ 2 is not a perfect square Therefore, n + 2 is not a perfect square.
Section 1.6 Question 16 Prove that if n is a perfect square, then n + 2 is not a perfect square. Solution: proof by contraposition Proof that when n+2 is a perfect square, n cannot be a perfect square. Assume n+2 is a perfect square, n+2 = k 2 (k >= 0), then n = k 2-2. When k = 0; n = -2, which is not a perfect square. Let s discuss situations when k >= 1: The perfect square right before (n+2) should be (k-1) 2 = k 2-2k+1 = n+1-2k. Because k >= 1, (k-1) 2 = n+1-2k <= n - 1 < n, n is not a perfect square Therefore, when n+2 is a perfect square, n cannot be a perfect square.
Section 1.6 Question 16 Prove that if n is a perfect square, then n + 2 is not a perfect square. Solution: proof by contradiction Proof that when n and n+2 are both perfect squares, there are some contradictions Assume n+2 is a perfect square, n + 2 = a 2 (a >= 0) Given the condition n is a perfect square, n = b 2 (b >= 0). (n + 2) n = a 2 - b 2 = (a b)(a + b) = 2 a and b are both integers larger all equal to zero. Therefore, (a - b) and (a + b) are all integers. Therefore, we have a b = 1 and a + b = 2. As a result, a = 1.5, b = 0.5. This leads to the contradiction a and b are all integers. Therefore, the assumption does not stand, n+2 is not a perfect square.
Section 1.6 Question 28 Prove that m 2 = n 2 if and only if m = n or m = -n. You need to prove: m 2 = n 2 m = n or m = -n and m = n or m = -n m 2 = n 2