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 sentencesBy the definition of odd, n = 2k + 1 for someinteger k.By the definition of even, n = 2k for some integerk.It follows thatn² + 3n+ 5 = (2k)² + 3(2k) + 5= 4k² + 6k+5= 2(2k² + 3k + 2) + 1Hence, 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 the following sentences so that they form a direct proof of the statement: If x and y are rational numbers, then 3x + 2y is also a rational number.Choose from this list of sentencesSince b, d are integers, bd is an integer.By the definition of rational numbers, x = a/bwhere a and b are integers and b# 0, andy = c/d where c and d are integers and d 0.Since 3x + 2y can be expressed as a ratio of twointegers where the denominator is not zero,3x + 2y is a rational number.Since a, b, c, d are integers, 3ad + 2bc is aninteger.Let x and y be rational numbers.Since b 0 and d # 0, bd # 0.It follows that 3x + 2y = 3a/b + 2c/d= (3ad + 2bc)/(bd)Direct proof of the statement (in order):

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Order 7 of the following sentences so that they form a direct proof of the statement. Choose from this list of sentences Direct proof of the statement (in order): Since k is an integer, 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. Choose from this list of sentences Direct proof of the statement (in order): Since k is an integer, 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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