2. Let d € Z – {0} and assume that √d & Q (we will eventually prove that this is true if and only if d is not the square of an integer, i.e., if and only if √d Z). In this case, the elements of Z[√d] are called quadratic integers. = a+b√d. (a) Prove that, if z € Z[√d], then there are unique integers a and b such that z = (Hint: Such integers exist by the definition of Z[√d]. Assuming that there are integers a₁, a2, b₁, and b2 such that z = a₁ + b₁√d and z = a₂ + b₂√d, we need to show that a₁ = a2 and b₁ = b₂. Ignoring the first equality, suppose for the sake of a contradiction that b₁ b₂. Use the irrationality of √d to reach a contradiction. How does the desired result follow?) (b) We would like to define a function oa : Z[√d] → Z[√d] by oa(z) = a - b√d, where a, b € Z satisfy z = a +b√d. Why is there a potential issue with trying to define a function using this formula? Explain how (a) resolves this issue, so that the displayed formula does indeed give a well-defined function, which we call conjugation. (Hint: compare with the first problem of Homework 1.) (c) Prove that the function od defined in (b) is bijective.
2. Let d € Z – {0} and assume that √d & Q (we will eventually prove that this is true if and only if d is not the square of an integer, i.e., if and only if √d Z). In this case, the elements of Z[√d] are called quadratic integers. = a+b√d. (a) Prove that, if z € Z[√d], then there are unique integers a and b such that z = (Hint: Such integers exist by the definition of Z[√d]. Assuming that there are integers a₁, a2, b₁, and b2 such that z = a₁ + b₁√d and z = a₂ + b₂√d, we need to show that a₁ = a2 and b₁ = b₂. Ignoring the first equality, suppose for the sake of a contradiction that b₁ b₂. Use the irrationality of √d to reach a contradiction. How does the desired result follow?) (b) We would like to define a function oa : Z[√d] → Z[√d] by oa(z) = a - b√d, where a, b € Z satisfy z = a +b√d. Why is there a potential issue with trying to define a function using this formula? Explain how (a) resolves this issue, so that the displayed formula does indeed give a well-defined function, which we call conjugation. (Hint: compare with the first problem of Homework 1.) (c) Prove that the function od defined in (b) is bijective.
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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![The following transcription and explanation are for an educational website:
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### Quadratic Integers and Their Properties
Consider \( d \in \mathbb{Z} - \{0\} \) and assume that \( \sqrt{d} \notin \mathbb{Q} \) (we will eventually prove that this is true if and only if \( d \) is not the square of an integer, i.e., if and only if \( \sqrt{d} \notin \mathbb{Z} \)). In this case, the elements of \( \mathbb{Z}[\sqrt{d}] \) are called **quadratic integers**.
#### Part (a)
**Problem Statement:**
Prove that, if \( z \in \mathbb{Z}[\sqrt{d}] \), then there are unique integers \( a \) and \( b \) such that \( z = a + b\sqrt{d} \).
**Hint:** Such integers **exist** by the definition of \( \mathbb{Z}[\sqrt{d}] \). Assuming that there are integers \( a_1, a_2, b_1, \) and \( b_2 \) such that \( z = a_1 + b_1\sqrt{d} \) and \( z = a_2 + b_2\sqrt{d} \), we need to show that \( a_1 = a_2 \) and \( b_1 = b_2 \). Ignoring the first equality, suppose for the sake of a contradiction that \( b_1 \neq b_2 \). Use the irrationality of \( \sqrt{d} \) to reach a contradiction. How does the desired result follow?
#### Part (b)
**Problem Statement:**
We would like to define a function \( \sigma_d: \mathbb{Z}[\sqrt{d}] \rightarrow \mathbb{Z}[\sqrt{d}] \) by
\[ \sigma_d(z) = a - b\sqrt{d}, \]
where \( a, b \in \mathbb{Z} \) satisfy \( z = a + b\sqrt{d} \).
**Question:** Why is there a **potential** issue with trying to define a function using this formula? Explain how (a) resolves this issue so that the displayed formula does indeed give a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Febb913e1-4986-4d74-b6ce-ad576ddf43d3%2F98d79405-8675-4d06-8022-b5a492d66b64%2Fjzcxsn_processed.png&w=3840&q=75)
Transcribed Image Text:The following transcription and explanation are for an educational website:
---
### Quadratic Integers and Their Properties
Consider \( d \in \mathbb{Z} - \{0\} \) and assume that \( \sqrt{d} \notin \mathbb{Q} \) (we will eventually prove that this is true if and only if \( d \) is not the square of an integer, i.e., if and only if \( \sqrt{d} \notin \mathbb{Z} \)). In this case, the elements of \( \mathbb{Z}[\sqrt{d}] \) are called **quadratic integers**.
#### Part (a)
**Problem Statement:**
Prove that, if \( z \in \mathbb{Z}[\sqrt{d}] \), then there are unique integers \( a \) and \( b \) such that \( z = a + b\sqrt{d} \).
**Hint:** Such integers **exist** by the definition of \( \mathbb{Z}[\sqrt{d}] \). Assuming that there are integers \( a_1, a_2, b_1, \) and \( b_2 \) such that \( z = a_1 + b_1\sqrt{d} \) and \( z = a_2 + b_2\sqrt{d} \), we need to show that \( a_1 = a_2 \) and \( b_1 = b_2 \). Ignoring the first equality, suppose for the sake of a contradiction that \( b_1 \neq b_2 \). Use the irrationality of \( \sqrt{d} \) to reach a contradiction. How does the desired result follow?
#### Part (b)
**Problem Statement:**
We would like to define a function \( \sigma_d: \mathbb{Z}[\sqrt{d}] \rightarrow \mathbb{Z}[\sqrt{d}] \) by
\[ \sigma_d(z) = a - b\sqrt{d}, \]
where \( a, b \in \mathbb{Z} \) satisfy \( z = a + b\sqrt{d} \).
**Question:** Why is there a **potential** issue with trying to define a function using this formula? Explain how (a) resolves this issue so that the displayed formula does indeed give a
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