   Chapter 11.5, Problem 16ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
1 views

# Complete the proof of case 2 of the strong induction argument in Example 11.5.5. In other words, show that if k is an odd integer and w i = ⌊ log 2 i ⌋ + 1 for every integer i with 1 ≤ i ≤ k , then w k + 1 = ⌊ log 2 k + 1 ⌋ + 1 .

To determine

To prove that wk+1=log2(k+1)+1.

Explanation

Given:

k is an odd integer with k1 and wi=log2i+1 for each integer i with 1ik.

Proof:

Since k is odd, so k+1 is even.

wk+1=1+w( k+1)/2 by the definition of w1,w2,w3 …….

wk+1=1+(log2( ( k+1 )/2)+1) by inductive hypothesis

As, k is odd, k2 and so, 1(k+1)/2(k+1)/2<k

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 