hw2sol

# hw2sol - CSE 20 Discrete Mathematics Fall 2010 Problem Set...

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CSE 20: Discrete Mathematics Fall 2010 Problem Set 2 Answers Instructor: Daniele Micciancio Due on: Wed. Oct. 13, 2010 Problem 1 (6 points) . 1. x Z . y Z .x < y (a) For every integer, there is an integer which is larger. (b) ¬ ( x Z . y Z .x < y ) ⇔ ∃ x Z . ¬ ( y Z .x < y ) ⇔ ∃ x Z . y Z . ¬ ( x < y ) ⇔ ∃ x Z . y Z .x y (c) The original statement is true, since for any integer x , there exists integer y = x + 1 such that x < x + 1 = y . 2. y Z . x Z .x < y (a) There is an integer which is larger than all integers. (b) ¬ ( y Z . x Z .x < y ) ⇔ ∀ y Z . ¬ ( x Z .x < y ) ⇔ ∀ y Z . x Z . ¬ ( x < y ) ⇔ ∀ y Z . x Z .x y (c) The negation of the original statement is true, since for any integer y , if we set x = y then x = y y , so such an integer always exists. Problem 2 (8 points)

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hw2sol - CSE 20 Discrete Mathematics Fall 2010 Problem Set...

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