   Chapter 11.4, 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
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# If, n is an odd integer and n > 1 , is ⌈ log 2 ( n + 1 ) ⌉ = ⌈ log 2 ( n ) ⌉ ? Justify your answer.

To determine

Explanation

Given information:

If n is an odd integer and n > 1.

Calculation:

The given statement is true, which we will justify with a proof.

Let n be an odd integer and let n > 1.

There exists an integer k such that:

2k1<n2k

We note that k needs to be a positive integer since n > 1 = 20 (thus k > 0).

However, since n is an odd integer, there cannot exists a positive integer k such that n = 2k (as 2k is even for all positive integers k ).

2k1<n<2k

Since the logarithm with base 2 is a increasing function, we can take the logarithm with base 2 of each side of the equation (and thus won’t affect the direction of the inequality signs).

log22k1<log2n<log22k

However, log22k1=k1 and 2k=k

k1<log2n<k

By definition of the ceil function:

log2n=k

We previously derived 2k1<n<2k. since n is odd, n + 1 is even and thus n2k (as 2k is some even integer larger than n ).

2k1<n+12k

Since the logarithm with base 2 is a strictly increasing function, we can take the logarithm with base 2 of each side of the equation (and thus won’t affect the direction of the inequality signs)

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