solutions 2

# Discrete Mathematics and Its Applications with MathZone

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ICS 6D ( Discrete Mathematic): Solution to Homework #2 Section 1.5 4. (a) simplification (b) disjunctive syllogism (c) modus ponens (d) addition (e) hypothetical syllogism 6. Let p be ”It rains”, q be ”It is foggy”, r be ”Sailing race is held”, s be ”Lifesaving demonstration goes on” and t be ”Trophy is awarded”. Our premises are ( ¬ p ∨ ¬ q ) ( r s ), r t and ¬ t . And the conclusion is p . The following argument shows how to derive the truth of p . r t ¬ t : ¬ r ¬ r : ¬ r ∨ ¬ s ( ¬ p ∨ ¬ q ) ( r s ) ¬ ( r s ) : ¬ ( ¬ p ∨ ¬ q ) p q : p (Above we also used the DeMorgan Law twice: First that ¬ r ∨ ¬ s is equivalent to ¬ ( r s ), and then the same for p, q instead of r, s , which implies in particular that ¬ ( ¬ p ∨ ¬ q ) is equivalent to p q . 24. The error occurs in step 2. We cannot say that p ( c ) is true when p ( c ) q ( c ) is true. In fact this might look like simplification rule but note that simplification only applies for conjunction. Section 2.1 4. There are 4 × 4 different combination for subsets from which the following statements are true. A A , B A , B B , C A , C C , C D , D D . 8. (a) True, (b) True, (c) False, (d) True, (e) True, (f) True, (g) True. 22. (a) It is not power set, (b) Power set of { a } , (c) It is not power set, (d) Power set of { a, b } 34. (a) There exists a real number x , such that x 3 = 1. ”True” (b) There exists an integer x , such that x + 1 > x . ”True” (c) For every integer x , x 1 is an integer. ”True” (d) For every integer x , x 2 is an integer. ”True” Section 2.2 2. (a) A B , (b) A B , (c) A B , (d) ¯ A ¯ B 4.

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