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. Thus, 7n + 4 = 14k +10= 2(7k+5). Then, 7n + 4 = 14k+ 2 = 2(7k + 1). Since n is even, it can be written as 2k for some integer k. This is 2 times an integer; so it is even as desired. Then, n can be written as 2k +1 for some integer k. Thus, 7n + 4 = 14k +11 = 2(7k+5)+1. Then, 7n + 4 = 14k + 4 = 2(7k+ 2). This is 1 more than 2 times an integer, so it is odd. Reset
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. Thus, 7n + 4 = 14k +10= 2(7k+5). Then, 7n + 4 = 14k+ 2 = 2(7k + 1). Since n is even, it can be written as 2k for some integer k. This is 2 times an integer; so it is even as desired. Then, n can be written as 2k +1 for some integer k. Thus, 7n + 4 = 14k +11 = 2(7k+5)+1. Then, 7n + 4 = 14k + 4 = 2(7k+ 2). This is 1 more than 2 times an integer, so it is odd. Reset
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
Consider these statements about the real number:
(i) x is rational.
(ii) x/2 is rational.
(iii) 3x - 1 is rational.
Put the steps in correct order to prove that (i) implies (ii).
(You must provide an answer before moving to the next part.)
Rank the options below.
e to search
Then, x/2 = p/(2q). Since q#0. 2q # 0.
Hence, x/2 is rational.
Suppose x is rational.
So, x = p/q where p and q are integers with q‡ 0.
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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.
Thus, 7n + 4 = 14k +10= 2(7k+5).
Then, 7n + 4 = 14k + 2 = 2(7k + 1).
Since n is even, it can be written as 2k for some
integer k.
This is 2 times an integer; so it is even as desired.
Then, n can be written as 2k +1 for some integer
k.
Thus, 7n + 4 = 14k +11 = 2(7k+5)+1.
Then, 7n + 4 = 14k + 4 = 2(7k+ 2).
This is 1 more than 2 times an integer, so it is odd.
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