Chapter 3, Problem 4RQ

Explanation of Solution

Range of a 16 bit twos complement value:

• A numeric range of twos complement is “$-\left({\text{2}}^{\text{n}-\text{1}}\right)$” to “$\left({\text{2}}^{\text{n}-\text{1}}\right)-1$” where “n” is the number of bits.
• Substitute the value “16” for “n” to find the range of 16 bit twos complement value.

$\begin{array}{c}-\left({\text{2}}^{\text{n}-\text{1}}\right)=-\left(2^{16-1}\right)\\ =-(\end{array}$

Explanation of Solution

Range of a 16 bit excess notation:

• A numeric range of excess notation is “$-\left({\text{2}}^{\text{n}-\text{1}}\right)$” to “${\text{2}}^{\left(\text{n}-\text{1}\right)}-\text{1}$” where “n” is the number of bits.
• Substitute the value “16” for “n” to find the range of 16 bit twos complement value.

$\begin{array}{c}-\left({\text{2}}^{\text{n}-\text{1}}\right)=-\left(2^{16-1}\right)\\ =-(\end{array}$

Explanation of Solution

Range of a 16 bit unsigned integer value:

• A numeric range of unsigned integer is “0” to “${\text{2}}^{\text{n}}-\text{1}$” where “n” is the number of bits.
• Substitute the value “16” for “n” to find the range of 16 bit unsigned value