## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 4 - Number Representation and Calculation - 4.3 Computation in Positional Systems - Exercise Set 4.3 - Page 234: 23

#### Answer

$111_{three}$

#### Work Step by Step

Start with right-most column. 0 is less than 2, we need to borrow 1 three... we can't borrow from zero, so we borrow 1 from $20_{three}$. This leaves 12 instead of 20 in $1200_{three}$in the top row... After borrowing: In base 10, $(0+3)-2=1$, which we record. Also, adjust the top row to reflect the borrowing. $\begin{array}{lllll} & 1 & *1 & *2 & 0_{three}\\ - & 1 & 0 & 1 & 2_{three}\\ -- & -- & -- & -- & --\\ & & & & 1_{three} \end{array}$ Next column: $2-1=1.$ Record the 1. $\begin{array}{lllll} & 1 & *1 & *2 & 0_{three}\\ - & 1 & 0 & 1 & 2_{three}\\ -- & -- & -- & -- & --\\ & & & 1 & 1_{three} \end{array}$ Next column: $1-0=1,$ Record the 1. $\begin{array}{lllll} & 1 & *1 & *2 & 0_{three}\\ - & 1 & 0 & 1 & 2_{three}\\ -- & -- & -- & -- & --\\ & & 1 & 1 & 1_{three} \end{array}$ Next column:$1-1=0$ Result: $111_{three}$

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