Order 7 of the following sentences so that they form a direct proof of the statement: If n is even, then n² + 3n+ 5 is odd. Direct proof of the statement (in order): Since k is an integer, Choose from this list of sentences By the definition of odd, n = 2k + 1 for some integer k. By the definition of even, n = 2k for some integer k. It follows that n² + 3n+ 5 = (2k)² + 3(2k) + 5 = 4k² + 6k+5 = 2(2k² + 3k + 2) + 1 Hence, n² + 3n+ 5 is odd. Let n be odd. Thus, by the definition of odd, . 2(2k² + 3k + 2) + 1 is odd. 2k² + 3k + 2 is even. 2k² + 3k + 2 is an integer. Let n be even.
Order 7 of the following sentences so that they form a direct proof of the statement: If n is even, then n² + 3n+ 5 is odd. Direct proof of the statement (in order): Since k is an integer, Choose from this list of sentences By the definition of odd, n = 2k + 1 for some integer k. By the definition of even, n = 2k for some integer k. It follows that n² + 3n+ 5 = (2k)² + 3(2k) + 5 = 4k² + 6k+5 = 2(2k² + 3k + 2) + 1 Hence, n² + 3n+ 5 is odd. Let n be odd. Thus, by the definition of odd, . 2(2k² + 3k + 2) + 1 is odd. 2k² + 3k + 2 is even. 2k² + 3k + 2 is an integer. Let n be even.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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