Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 5 - Section 5.1 - Mathematical Induction - Exercises - Page 331: 55

Answer

There are six base cases, for the cases when i + j ≤ 2. The knight is already at (0, 0) to start, so the empty sequence of moves reaches that square.

Work Step by Step

-We use the notation (i, j ) to mean the square in row i and column j and use induction on i +j to show that every square can be reached by the knight. --Basis step: There are six base cases, for the cases when i + j ≤ 2. The knight is already at (0, 0) to start, so the empty sequence of moves reaches that square. To reach (1, 0), the knight moves (0, 0)→(2, 1)→(0, 2)→(1, 0). Similarly, to reach (0, 1),the knight moves (0, 0) → (1, 2) → (2, 0) → (0, 1). -Note that the knight has reached (2, 0) and (0, 2) in the process. For the last basis step there is (0, 0) → (1, 2) → (2, 0) →(0, 1) → (2, 2) → (0, 3) → (1, 1). -- Inductive step: Assume the inductive hypothesis, that the knight can reach any square (i, j ) for which i + j = k, where k is an integer greater than 1. We must show how the knight can reach each square (i, j )when i+j=k+ 1. Because k + 1 ≥ 3, at least one of i and j is at least 2. If i ≥ 2, then by the inductive hypothesis, there is a sequence of moves ending at (i − 2, j + 1), because i − 2 + j + 1 = i + j − 1 = k; from there it is just one step to (i, j ); similarly, if j ≥ 2.
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