## 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: 64

#### Answer

The binary code for $p$ is: $1110000$

#### Work Step by Step

The essay states that the lowercase letter $p$ is assigned the number $112$. To find the binary representation for this, use division. The place values in base two are $..., 2^7, 2^6, 2^5, 2^4, 2^3, 2^2, 2^1, 1$ The place values that are less than or equal to $112$ are $2^6, 2^5, 2^4, 2^3, 2^2, 2^1$ and $1$. Divide $112$ by $2^6$ or $64$: $112 \div 64 = 1$ remainder $48$ Divide $48$ by $2^5$ or $32$ $48 \div 32 = 1$ remainder $16$ Divide $16$ by $2^4$ or $16$: $16 \div 16 = 1$ remainder $0$ Divide $0$ by $2^3$ or $8$: $0 \div 8 = 0$ remainder $0$ Divide $0$ by $2^2$ or $4$: $0 \div 4=0$ remainder $0$ Divide $0$ by $2^1$ or $2$: $0 \div 2 = 0$ remainder $0$ Divide $0$ by $1$: $0 \div 1 = 0$ Thus, $112 = (1 \times 2^6) + (1 \times 2^5) + (1 \times 2^4)+(0 \times 2^3)+(0\times 2^2) + (0 \times 2^1)+(0 \times 1) \\112=1110000_{\text{two}}$

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