Answer The given equation illustrates the “Identity Property of the Addition.” Explanation Given: The given equation is j k + 0 = j k . Calculation: The following table shows the real numbers with reference to any three variables a , b , and c . Property Addition Multiplication Commutative a + b = b + a a ⋅ b = b ⋅ a Associative ( a + b ) + c = a + ( b + c ) ( a ⋅ b ) ⋅ c = a ⋅ ( b ⋅ c ) Identity a + 0 = a = 0 + a a ⋅ 1 = a = 1 ⋅ a Inverse a + ( − a ) = 0 = ( − a ) + a a ⋅ 1 a = 1 = 1 a ⋅ a and a ≠ 0 Distributive a ( b + c ) = a b + a c and ( b + c ) a = b a + c a a ( b + c ) = a b + a c and ( b + c ) a = b a + c a By comparing the equation j k + 0 = j k with the above Properties of the addition, it is observed that the equation j k + 0 = j k illustrate the Identity Property of the Addition. Therefore, the given equation illustrates the “Identity Property of the Addition.”

1st Edition

ISBN: 9780078884825

Author: McGraw-Hill/Glencoe

Publisher: Glencoe/McGraw-Hill School Pub Co

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Question

Chapter EP, Problem 1.2.11EP

To determine

To Name: The property illustrated by the given equation.

Expert Solution & Answer

Answer to Problem 1.2.11EP

The given equation illustrates the “Identity Property of the Addition.”

Explanation of Solution

Given:

The given equation is $jk+0=jk$ .

Calculation:

The following table shows the real numbers with reference to any three variables a , b , and c .

Property	Addition	Multiplication
Commutative	$a+b=b+a$	$a⋅b=b⋅a$
Associative	$(a+b)+c=a+(b+c)$	$(a⋅b)⋅c=a⋅(b⋅c)$
Identity	$a+0=a=0+a$	$a⋅1=a=1⋅a$
Inverse	$a+(−a)=0=(−a)+a$	$a⋅1a=1=1a⋅a$ and $a≠0$
Distributive	$a(b+c)=ab+ac$ and $(b+c)a=ba+ca$	$a(b+c)=ab+ac$ and $(b+c)a=ba+ca$

By comparing the equation $jk+0=jk$ with the above Properties of the addition, it is observed that the equation $jk+0=jk$ illustrate the Identity Property of the Addition.

Therefore, the given equation illustrates the “Identity Property of the Addition.”

Chapter EP, Problem 1.2.10EP

Chapter EP, Problem 1.2.12EP

Chapter EP Solutions

Algebra 2

Ch. EP - Prob. 1.1.1EP Ch. EP - Prob. 1.1.2EP Ch. EP - Prob. 1.1.3EP Ch. EP - Prob. 1.1.4EP Ch. EP - Prob. 1.1.5EP Ch. EP - Prob. 1.1.6EP Ch. EP - Prob. 1.1.7EP Ch. EP - Prob. 1.1.8EP Ch. EP - Prob. 1.1.9EP Ch. EP - Prob. 1.1.10EP

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