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Chapter 3: Special Kinds of Theorems > Existence Theorems - Pg. 219

Solutions for the Exercises at the End of the Sections 219 34. a. The statement is true for k = 1, since 8 ­ 3 = 5. b. Assume it is true for a generic number n > 1. Thus, 8 n - 3 n = 5s with s an integer. c. Is the statement true for the next number, namely n + 1? Is 8 n+1 - 3 n+1 = 5t with t an integer? It is possible to rewrite the inductive hypothesis as 8 n = 3 n + 5s with s an integer. Using algebra and the inductive hypothesis we obtain 8 n+1 - 3 n+1 = 8 × 8 n - 3 n+1 = 8 × ð3 n + 5sÞ - 3 × 3 n = 8 × 3 n - 3 × 3 n + 5s = 3 n ð8 - 3Þ + 5s = 5ð3 n + sÞ: The number 3 n + s is an integer. Therefore the conclusion is true for all k 1 by the Principle of Mathematical Induction. It is possible to prove this statement using factorization techniques for differences of powers instead of mathematical induction. 35. a. Check the statement for n = 2. 1 0 1 0 1 0 2 = A = 1 1 1 1 2 1 1 0 . b. Assume it is true for a generic number n > 2; that is A n = n 1 c. Is the statement true for n + 1? Using the associative property of multiplication of matrices, we n+1 n