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Click and drag the steps in the correct order to show that if n is a positive integer, then n is even if and only if 7n +4 is even. Suppose that n is even. Suppose that n is not even. Then, n can be written as 2k +1 for some integer k. Thus, 7n+ 4 = 14k +10= 2(7k+5). Then, 7n + 4 = 14k+ 2 = 2(7k+ 1). Thus, 7n+ 4 = 14k+11 = 2(7k+5)+1. Then, 7n + 4 = 14k+ 4 = 2(7k+ 2). Since n is even, it can be written as 2k for some integer k. This is 1 more than 2 times an integer, so it is odd. This is 2 times an integer; so it is even as desired.

Click and drag the steps in the correct order to show that if n is a positive integer, then n is even if and only if 7n +4 is even. Suppose that n is even. Suppose that n is not even. Then, n can be written as 2k +1 for some integer k. Thus, 7n+ 4 = 14k +10= 2(7k+5). Then, 7n + 4 = 14k+ 2 = 2(7k+ 1). Thus, 7n+ 4 = 14k+11 = 2(7k+5)+1. Then, 7n + 4 = 14k+ 4 = 2(7k+ 2). Since n is even, it can be written as 2k for some integer k. This is 1 more than 2 times an integer, so it is odd. This is 2 times an integer; so it is even as desired.

Database System Concepts
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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Click and drag the steps in the correct order to show that if n is a positive integer, then n is even if and only if 7n + 4 is even. Suppose that n is even. Suppose that n is not even. Then, n can be written as 2k +1 for some integer k. Thus, 7n + 4 = 14k +10 = 2(7k+5). Then, 7n + 4 = 14k+ 2 = 2(7k+ 1). Thus, 7n + 4 = 14k +11 = = 2(7k+5)+1. Then, 7n + 4 = 14k + 4 = 2(7k + 2). Since n is even, it can be written as 2k for some integer k. This is 1 more than 2 times an integer, so it is odd. This is 2 times an integer;: so it is even as desired.
Transcribed Image Text:Click and drag the steps in the correct order to show that if n is a positive integer, then n is even if and only if 7n + 4 is even. Suppose that n is even. Suppose that n is not even. Then, n can be written as 2k +1 for some integer k. Thus, 7n + 4 = 14k +10 = 2(7k+5). Then, 7n + 4 = 14k+ 2 = 2(7k+ 1). Thus, 7n + 4 = 14k +11 = = 2(7k+5)+1. Then, 7n + 4 = 14k + 4 = 2(7k + 2). Since n is even, it can be written as 2k for some integer k. This is 1 more than 2 times an integer, so it is odd. This is 2 times an integer;: so it is even as desired.
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