Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 4 - Number Representation and Calculation - 4.2 Number Bases in Positional Systems - Exercise Set 4.2: 27

Answer

$1000_{\text{two}}$

Work Step by Step

The place values of base two are: $,,,, 2^3, 2^2, 2^1, 1$. The place value that are less then $8$ are $2^3, 2^2, 2^1$ and $1$ Divide $8$ by $2^3$ or $8$ to obtain: $8 \div 8 = 1$ remainder $0$ Divide the remainder $0$ by $2^1$ or $2$ to obtain: $0 \div 2 = 0$ remainder $0$ Divide $0$ by $1$ to obtain $0 \div 1= 0$ Thus, $8=(1 \times 2^3) + (0 \times 2^2) + (0 \times 2) + (0 \times 1) \\8=1000_{\text{two}}$
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