## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 5 - Number Theory and the Real Number System - 5.5 Real Numbers and Their Properties; Clock Addition - Exercise Set 5.5 - Page 309: 50

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#### Work Step by Step

(a) The set$\left\{ 0,1,2,3,4,5,6 \right\}$is closed under the operation of clock addition, because the entries of the table are all the elements of the set. Hence, the set $\left\{ 0,1,2,3,4,5,6 \right\}$is closed under the operation of clock addition. (b) Consider the expression of LHS. \begin{align} & \left( 3\oplus 5 \right)\oplus 6=1\oplus 6 \\ & =0 \end{align} Now, find RHS as: \begin{align} & 3\oplus \left( 5\oplus 6 \right)=3\oplus 4 \\ & =0 \end{align} Hence, $\left( 3\oplus 5 \right)\oplus 6=3\oplus \left( 5\oplus 6 \right)$. Associative property has been verified. (c) Identity element is that element, which does not change anything when clock addition is used. Now, as we can see that, $0\oplus 0=0,\ 0\oplus 1=1,\ 0\oplus 2=2,\ 0\oplus 3=3,\ 0\oplus 4=4,\ 0\oplus 5=5,\ 0\oplus 6=6$and, $0\oplus 0=0,\ 1\oplus 0=1,\ 2\oplus 0=2,\ 3\oplus 0=3,\ 4\oplus 0=4,\ 5\oplus 0=5,\ 6\oplus 0=6$ Thus, 0 is the identity element. Hence, 0 is the identity element in the 7-hour clock addition. (d) When an element is added to its inverse, the result is the identity element. The identity element is 0. Find inverse of each element as, Let $x$be the inverse of any element, such that $x\oplus 0=0$, and from the table $0\oplus 0=0$ Thus, 0 is the inverse of element 0. $x\oplus 1=0$, and from the table $6\oplus 1=0$ Thus, 6 is the inverse of element 1. $x\oplus 2=0$, and from the table $5\oplus 2=0$ Thus, 5 is the inverse of element 2. $x\oplus 3=0$, and from the table $4\oplus 3=0$ Thus, 4 is the inverse of element 3. $x\oplus 4=0$, and from the table $3\oplus 4=0$ Thus, 3 is the inverse of element 4. $x\oplus 5=0$, and from the table $2\oplus 5=0$ Thus, 2 is the inverse of element 5. $x\oplus 6=0$, and from the table $1\oplus 6=0$ Thus, 1 is the inverse of element 6. Hence, inverse of elements 0,1,2,3,4,5 and 6 are 0,6,5,4,3,2 and 1, respectively. (e) $4\oplus 5=2$ Now, find RHS as, $5\oplus 4=2$ And, $6\oplus 1=0$ Now, find RHS as, $1\oplus 6=0$ Both the equations follow commutative property. Hence,$4\oplus 5=5\oplus 4$and$6\oplus 1=1\oplus 6$follows commutative property and has been verified.

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