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]
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fed14a3ea-da26-4be7-a143-8b845df95e91%2F61568a93-5fc6-44ab-8a03-768271fc5c1a%2Fkmusyw6_processed.png&w=3840&q=75)
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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