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): Choose from this list of sentences By the definition of odd, n = 2k + 1 for some integer k. Let n be odd. 2k² + 3k+2 is even. Let n be even. By the definition of even, n = 2k for some integer k. Since k is an integer, It follows that n² + 3n+ 5 = (2k)² + 3(2k) +5 = 4k² + 6k+5 = 2(2k2 + 3k + 2) + 1 Thus, by the definition of odd, 2(2k² + 3k + 2) + 1 is odd. 2k² + 3k + 2 is an integer. Hence, n² + 3n+ 5 is odd.

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
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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):
Choose from this list of sentences
By the definition of odd, n = 2k + 1 for some
integer k.
Let n be odd.
2k2 + 3k + 2 is even.
Let n be even.
By the definition of even, n = 2k for some integer
k.
Since k is an integer,
It follows that
n² + 3n+ 5 = (2k)² + 3(2k) +5
= 4k²2 +6k+5
= 2(2k² + 3k + 2) + 1
Thus, by the definition of odd,
2(2k² + 3k + 2) + 1 is odd.
2k2 + 3k + 2 is an integer.
Hence, n² + 3n+ 5 is odd.
Transcribed Image Text: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): Choose from this list of sentences By the definition of odd, n = 2k + 1 for some integer k. Let n be odd. 2k2 + 3k + 2 is even. Let n be even. By the definition of even, n = 2k for some integer k. Since k is an integer, It follows that n² + 3n+ 5 = (2k)² + 3(2k) +5 = 4k²2 +6k+5 = 2(2k² + 3k + 2) + 1 Thus, by the definition of odd, 2(2k² + 3k + 2) + 1 is odd. 2k2 + 3k + 2 is an integer. Hence, n² + 3n+ 5 is odd.
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