{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 4 Solutions

# Homework 4 Solutions - k where k ≥ 1 We want to show that...

This preview shows page 1. Sign up to view the full content.

HOMEWORK 4: COMMENTS 1. Non-book problems (1) We proceed by strong induction on n . We need two base cases: a 1 = 2 1 +1, a 2 = 2 2 +1. Assume, as inductive hypothesis that a i = 2 i + 1 for integers i in the range 1 , . . . , k , where k 2. We want to show that a k +1 = 2 k +1 + 1. Note that a k +1 = 3 a k - 2 a k - 1 because k 2 , = 3(2 k + 1) - 2(2 k - 1 + 1) by inductive hypothesis = 3 · 2 k + 3 - 2 k - 2 = 2 · 2 k + 1 = 2 k +1 + 1 . Hence a n = 2 n + 1 for all n N . Book Problems. Problem 21 Solution. We proceed by strong induction on n . The base case is true by by hypothesis , i.e., x + 1 /x is an integer because we are assuming so. Assume, as inductive hypothesis that x i + 1 /x i is an integer for integers i in the range 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , . . . , k , where k ≥ 1. We want to show that x k +1 + 1 /x k +1 is an integer. We have ( x k +1 + 1 /x k +1 ) = ( x k + 1 /x k )( x + 1 /x )-( x k-1 + 1 /x k-1 ) Now, the right hand side is an integer since ( x k + 1 /x k ) , ( x k-1 + 1 /x k-1 ), and ( x + 1 /x ) are all integers, the ﬁrst two by inductive hypothesis, the later by the original problem’s hypothesis. Hence the left hand side is an integer. By induction, x n + 1 /x n is an integer for all n ∈ N . ± Date : February 21th, 2008. 1...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online