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: 8

Answer

$Answer\ of\ a:$ $\begin{aligned} \text +0.25 &=0.01 \text { in binary } \\ &=0.1 \times 2^{-1} \text { in scientific notation } \end{aligned}$ so the base is $+0.1$ and the exponent is $-1 .$ $\ \ \ \ \ =0\ \underbrace{100000000}_{\text { mantissa }} \qquad \underbrace{1\ 00001}_{\text { exponent }}$ --- $Answer\ of\ b:$ $\begin{aligned}-32\ 1 / 16 &=-100000.0001 \\ &=-0.1000000001 \times 2^{6} \\ &= 1\ \underbrace{100000000}_{\text { mantissa }} \qquad \underbrace{0\ 00110}_{\text { exponent }} \end{aligned}$ Note that the last 1 in the mantissa was not stored because there was not enough room. The loss of accuracy that results from limiting the number of digits available is called a truncation error.
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Work Step by Step

$Answer\ of\ a:$ $\begin{aligned} \text +0.25 &=0.01 \text { in binary } \\ &=0.1 \times 2^{-1} \text { in scientific notation } \end{aligned}$ so the base is $+0.1$ and the exponent is $-1 .$ $\ \ \ \ \ =0\ \underbrace{100000000}_{\text { mantissa }} \qquad \underbrace{1\ 00001}_{\text { exponent }}$ --- $Answer\ of\ b:$ $\begin{aligned}-32\ 1 / 16 &=-100000.0001 \\ &=-0.1000000001 \times 2^{6} \\ &= 1\ \underbrace{100000000}_{\text { mantissa }} \qquad \underbrace{0\ 00110}_{\text { exponent }} \end{aligned}$ Note that the last 1 in the mantissa was not stored because there was not enough room. The loss of accuracy that results from limiting the number of digits available is called a truncation error.
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