## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 4 - Number Representation and Calculation - 4.2 Number Bases in Positional Systems - Exercise Set 4.2 - Page 226: 28

#### Answer

$1001_{\text{two}}$

#### Work Step by Step

The place values of base two are: $,,,, 2^3, 2^2, 2^1, 1$. The place value that are less than or equal to $9$ are $2^3, 2^2, 2^1$ and $1$ Divide $9$ by $2^3$ or $8$ to obtain: $9 \div 8 = 1$ remainder $1$ Divide the remainder $1$ by $2^2$ or $4$ to obtain: $1 \div 4 = 0$ remainder $1$ Divide the remainder $1$ by $2^1$ or $2$ to obtain $1 \div 2= 0$ remainder $1$ Divide the remainder $1$ by $1$ to obtain: $1 \div 1 =$1$remainder$0$Thus,$9=(1 \times 2^3) + (0 \times 2^2) + (0 \times 2) + (1 \times 1) \\8=1001_{\text{two}}\$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.