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Unformatted text preview: Homework 2  For 10/15/2009 Exercise 1 p q r p ∨ q p → r q → r θ :[( p ∨ q ) ∧ ( p → r ) ∧ ( q → r )] θ → 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 θ :( p ∨ q ) ∧ ( ¬ p ∨ r ) q ∨ r θ → ( 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 = π 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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This note was uploaded on 11/10/2009 for the course ECS 12 taught by Professor Khoel during the Spring '09 term at UC Davis.
 Spring '09
 Khoel

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