DS-PartA-May12-Sol

# Thus the inequality holds for n 1 2 pts inductive

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Unformatted text preview: nequality holds for n = 1. (2 pts) Inductive hypothesis: Assume for an arbitrary positive integer n = k that (2 pts) Inductive step: Prove for n = k + 1 that (2 pts) (2 pts) , using I. H. (3 pts) (0 pts) , since using the inequality given it follows that (4 pts) This completes proving the inductive step. It follows that the original assertion holds for all positive integers n. Page 2 of 4 Spring 2012 Discrete Structures Exam, Part A 2) (15 pts) PRF (Logic) Prove that the following logical expression is a tautology using the laws of logic equivalence and the definition of the conditional statement only. Show each step and state which rule is being used. Note: You may combine both associative and commutative laws in a single step, s o long as you do this properly. Solution ) ( T T defn of implication Double Negation DeMorgan’s Law Commutative Law Inverse Law (using Domination Law (3 pts) (2 pts) (3 pts) (2 pts) ) (3 pts) (2 pts) Grading: Give positive points for steps in the correct direction, roughly the steps are outlined above. Students may use mo...
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