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Prove that G = {a+b√2: a, b € Q and a and b are not both zero} is a subgroup of R* under the group operation of multiplication. [#41, 3.5]

Prove that G = {a+b√2: a, b € Q and a and b are not both zero} is a subgroup of R* under the group operation of multiplication. [#41, 3.5]

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter7: Real And Complex Numbers
Section7.3: De Moivre’s Theorem And Roots Of Complex Numbers
Problem 23E: Prove that the set of all complex numbers that have absolute value forms a group with respect to...
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## Problem Statement: **Prove that \( G = \{ a + b\sqrt{2} : a, b \in \mathbb{Q} \text{ and } a \text{ and } b \text{ are not both zero} \} \) is a subgroup of \( \mathbb{R}^* \) under the group operation of multiplication.** \[ #[41, 3.5] \] ### Explanation: The problem asks you to prove that the set \( G \) defined as follows: \[ G = \{ a + b\sqrt{2} : a, b \in \mathbb{Q} \text{ and } a \text{ and } b \text{ are not both zero} \} \] is a subgroup of the multiplicative group of nonzero real numbers, \( \mathbb{R}^* \). ### Key Concepts: **Subgroup:** To show that \( G \) is a subgroup of \( \mathbb{R}^* \), you need to verify the following subgroup criteria for \( G \): 1. The set \( G \) is closed under multiplication. 2. The set \( G \) contains the multiplicative identity. 3. The set \( G \) is closed under taking inverses. ### Steps for Proof: 1. **Closure under Multiplication:** - Take any two elements \( x, y \in G \). - Show that the product \( xy \) is also in \( G \). 2. **Identity Element:** - Identify the multiplicative identity in \( \mathbb{R}^* \) and show that it lies in \( G \). 3. **Closure under Taking Inverses:** - Take any element \( x \in G \). - Show that its inverse \( x^{-1} \) is also in \( G \). This proof shows whether or not the set \( G \) forms a well-defined structure under the operation of multiplication which is significant in higher mathematics, particularly in the context of group theory.
Transcribed Image Text:## Problem Statement: **Prove that \( G = \{ a + b\sqrt{2} : a, b \in \mathbb{Q} \text{ and } a \text{ and } b \text{ are not both zero} \} \) is a subgroup of \( \mathbb{R}^* \) under the group operation of multiplication.** \[ #[41, 3.5] \] ### Explanation: The problem asks you to prove that the set \( G \) defined as follows: \[ G = \{ a + b\sqrt{2} : a, b \in \mathbb{Q} \text{ and } a \text{ and } b \text{ are not both zero} \} \] is a subgroup of the multiplicative group of nonzero real numbers, \( \mathbb{R}^* \). ### Key Concepts: **Subgroup:** To show that \( G \) is a subgroup of \( \mathbb{R}^* \), you need to verify the following subgroup criteria for \( G \): 1. The set \( G \) is closed under multiplication. 2. The set \( G \) contains the multiplicative identity. 3. The set \( G \) is closed under taking inverses. ### Steps for Proof: 1. **Closure under Multiplication:** - Take any two elements \( x, y \in G \). - Show that the product \( xy \) is also in \( G \). 2. **Identity Element:** - Identify the multiplicative identity in \( \mathbb{R}^* \) and show that it lies in \( G \). 3. **Closure under Taking Inverses:** - Take any element \( x \in G \). - Show that its inverse \( x^{-1} \) is also in \( G \). This proof shows whether or not the set \( G \) forms a well-defined structure under the operation of multiplication which is significant in higher mathematics, particularly in the context of group theory.
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