Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 5 - Sequences, Mathematical Induction, and Recursion - Exercise Set 5.3 - Page 266: 10

Answer

Proof by mathematical induction, Base Case: Show that P(0) is true: P(0) = 03 – 7*0 + 3 = 3 and 3 is divisible by 3. So, P(0) is true. Inductive Case: Show that for all integer k ≥ 0, if P(k) is true then P(k+1) is true: Let k be any integer such that k ≥ 0 and suppose P(k) is true. This means k3 - 7k + 3 is divisible by 3. Now we must show that P(k+1) is true. P(k+1) = (k+1)3 - 7(k+1) + 3 = k3 + 3k2 + 3k +1 -7k -7 + 3 = k3 + 3k2 -4k -3 = k3 – 7k + 3 +3k2 +3k – 6 [expressing above line in the form of P(k)] Now we see that k3 – 7k + 3 is equal to P(k) and P(k) is divisible by 3. Let (k3 – 7k + 3) / 3 = x. Where x is some integer Then, (k3 – 7k + 3) = 3x So, P(k+1) = 3x + 3k2 + 3k – 6 = 3(x+ k2 + k -2) which is divisible by 3. So, by definition of divisibility, n3 – 7n + 3 is divisible by 3

Work Step by Step

Steps: 1. Take any number k and show that the statement holds true for that value of k. 2. Show that if the statement is true for k, the statement holds true for the value k+1. This can be done by taking the statement where the value is k+1 and reducing it into the statement where the value is k.
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