## Invitation to Computer Science 8th Edition

Published by Cengage Learning

# Chapter 4 - 4.2 - The Binary Numbering System - Practice Problems - Page 166: 1

#### Answer

\begin{aligned} \text { a. } 10101000 &=\left(1 \times 2^{3}\right)+\left(1 \times 2^{5}\right)+\left(1 \times 2^{7}\right) \\ &=8+32+128 \\ &=168 \text { as an unsigned integer value } \end{aligned} \begin{aligned} \text { b. } 10101000 &=\left(1 \times 2^{3}\right)+\left(1 \times 2^{5}\right) \\ &=8+32 \\ &=40 \end{aligned} This is the value of the magnitude portion of the number. The left-most bit represents the sign bit. In this example, it is a $1,$ which is a negative sign. $$=-40 \text { as a signed integer value }$$

#### Work Step by Step

\begin{aligned} \text { a. } 10101000 &=\left(1 \times 2^{3}\right)+\left(1 \times 2^{5}\right)+\left(1 \times 2^{7}\right) \\ &=8+32+128 \\ &=168 \text { as an unsigned integer value } \end{aligned} \begin{aligned} \text { b. } 10101000 &=\left(1 \times 2^{3}\right)+\left(1 \times 2^{5}\right) \\ &=8+32 \\ &=40 \end{aligned} This is the value of the magnitude portion of the number. The left-most bit represents the sign bit. In this example, it is a $1,$ which is a negative sign. $$=-40 \text { as a signed integer value }$$

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