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: 26

Answer

$110_{\text{two}}$

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 $6$ are $2^2, 2^1$ and $1$ Divide $6$ by $2^2$ or $4$ to obtain: $6 \div 4 = 1$ remainder $2$ Divide the remainder $2$ by $2^1$ or $2$ to obtain: $2 \div 2 = 1$ remainder $0$ Divide $0$ by $1$ to obtain $0 \div 1= 0$ Thus, $6=(1 \times 2^2) + (1 \times 2) + (0 \times 1) \\5=110_{\text{two}}$
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