Explanation Given information: ℂ is the set of all ordered pairs ( a , b ) of real numbers a and b . Equality, addition, and multiplication are defined in ℂ by ( a , b ) = ( c , d ) if and only if a = c and b = d ( a , b ) + ( c , d ) = ( a + c , b + d ) ( a , b ) ( c , d ) = ( a c − b d , a d + b c ) The elements of ℂ are called complex numbers. Formula used: i) S is a subset of a field F that contains at least two elements and satisfies both of the following conditions: x ∈ S and y ∈ S imply, x − y ∈ S , and x ∈ S and y ≠ 0 ∈ S imply, x y − 1 ∈ S . This S is called a subfield of F . ii) ( a , b ) − 1 = ( a a 2 + b 2 , − b a 2 + b 2 ) Proof: Given that, R ' = { ( a , 0 ) | a ∈ ℝ } Let ℂ be the set of all ordered pairs ( a , b ) of real numbers a and b . In particular if b = 0 implies that ( a , 0 ) ∈ ℂ for all real numbers a . Thus, R ' is a subset of ℂ that contains at least two elements. Let ( x , 0 ) ∈ R ' and ( y , 0 ) ∈ R ' with x , y ∈ ℝ Consider ( x , 0 ) − ( y , 0 ) = ( x , 0 ) + ( − ( y , 0 ) ) By using definition of addition in complex number, ⇒ ( x , 0 ) − ( y , 0 ) = ( x − y , 0 ) By using, ℝ is a field

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Author: Gilbert

Publisher: Cengage

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Chapter 7.2, Problem 17E

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To prove: R'={(a, 0)|a∈ℝ} is a subfield of ℂ.

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Chapter 7.2, Problem 16E

Chapter 7.2, Problem 18E

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Chapter 7 Solutions

Elements Of Modern Algebra

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To prove: R ' = { ( a , 0 ) | a ∈ ℝ } is a subfield of ℂ .

Solution Summary: The author explains that C is the set of all ordered pairs (a,b) of real numbers a and b.

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To prove: R'={(a, 0)|a∈ℝ} is a subfield of ℂ.

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Chapter 7 Solutions