Chapter 2.4, Problem 2.49PP

A.

Explanation of Solution

Formula for smallest positive integer:

• From the given question, a floating point format with an n-bit fraction.
  • So, the number “1”contain binary representation “1”, followed by “n” zeros, followed by “1” giving value ${\text{2}}^{\text{n}+\text{1}}$

B.

Explanation of Solution

Numeric value of given integer for single precision format:

• From the part A, the formula for smallest positive integer in single precision format is ${\text{2}}^{\text{n}+\text{1}}+\text{ 1}$.
• From the given question, the value of “n” is “23”. So, substitute “n = 23” in ${\text{2}}^{\text{n}+\text{1}}+\text{ 1}$.

$\begin{array}{l}{\text{2}}^{\text{n}+\text{1}}+\text{ 1}={\text{2}}^{\text{23}+\text{1}}+\text{ 1}\text{\hspace{0.17em}</}\end{array}$