Invitation to Computer Science 8th Edition

Published by Cengage Learning
ISBN 10: 1337561916
ISBN 13: 978-1-33756-191-4

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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