2_logic.proofs

2_logic.proofs - Homework 2 - For 10/15/2009 Exercise 1 p q...

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Unformatted text preview: Homework 2 - For 10/15/2009 Exercise 1 p q r p q p r q r :[( p q ) ( p r ) ( q r )] r T T T T T T T T T T F T F F F T T F T T T T T T T F F T F T F T F T T T T T T T F T F T T F F T F F T F T T F T F F F F T T F T Exercise 2 p q r p q p r :( p q ) ( p r ) q r ( q r ) T T T T T T T T T T F T F F T T T F T T T T T T T F F T F F F T F T T T T T T T F T F T T T T T F F T F T F T T F F F F T F F T Exercise 3 a . It is not a valid argument. The two statements can be simplified as p q , therefore q p . The second statement is the converse of the first statement: they are not equivalent. b . It is a valid argument: we are just applying the result to the specific case x = Exercise 4 Let n bean integer; there exists two integers k and l such that n = 10 k + l where 0 l 9. We get: n 2 = (10 k + l ) 2 (1) = 100 k + 20 kl + l 2 (2) = k * 100 + 2 kl * 10 + l 2 (3) k * 100 and 2 kl * 10 are multiples of 10. Therefore, n 2 ends as l 2 . In the following table, we show that l 2 must end with a 0, 1, 4, 5, 6, or 9. 1 l l 2 end 1 1 1 2 4 4 3 9 9 4 16 6 5 25 5 6 36 6 7 49 9 8 64 4 9 81 1 Exercise 5 Let p be the proposition n is even and q be the proposition 7 n +4 is even. We want to show that p q , which is logically equivalent to show that p q and q p . i) Let us show p q : Hypothesis: p is true, i.e. n is even. If n is even, then let n = 2 m , where m is also an integer. Therefore, we get: 7 n + 4 = 7(2 m ) + 4 = 14 m + 4 = 2 * (7 m + 2) Thus, 7 n + 4 is a multiple of 2: it is even. ii) Let us show q p : Hypothesis: q is true, i.e. 7 n +4 is even. If 7 n + 4 is even, then let 7 n + 4 = 2 m , where m is also an integer. Therefore, we get: 7 n = 2 m- 4 n = 2 m- 4- 6 n n = 2 * ( m- 2- 3 n ) Thus, n is a multiple of 2: it is even.is a multiple of 2: it is even....
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2_logic.proofs - Homework 2 - For 10/15/2009 Exercise 1 p q...

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