1.6.2 (1) [Used in Example 1.6. 3] Let T be the set defined in Equation 1. 6. 1. (1)Prove that Tis a Dedekind cut.
1.6.2 (1) [Used in Example 1.6. 3] Let T be the set defined in Equation 1. 6. 1. (1)Prove that Tis a Dedekind cut.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please include reasons for each step.
![**1.6.2 (1)**
*[Used in Example 1.6.3] Let T be the set defined in Equation 1.6.1 (1). Prove that T is a Dedekind cut.*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8975dae8-fbcb-4b49-8e3a-715e0e53990f%2Fa6eced6f-d3c5-4e9c-a3d5-98b4dea3c6f6%2Fwmky8ei_processed.png&w=3840&q=75)
Transcribed Image Text:**1.6.2 (1)**
*[Used in Example 1.6.3] Let T be the set defined in Equation 1.6.1 (1). Prove that T is a Dedekind cut.*
![**Example 1.6.3**
To find a Dedekind cut that is not of the form given in Lemma 1.6.2, the simplest idea is to look at the set of all rational numbers greater than a real number that is not rational; the problem is how to describe such a set without making use of real numbers. In the case of the number \( x = \sqrt{2} \), there turns out to be a simple solution to this problem, as we will now see. We note first, however, that we have not yet formally defined what \( \sqrt{2} \) means, nor proved that there is such a real number, though we will do so in Theorem 2.6.9 and Definition 2.6.10. We have not yet proved that \( \sqrt{2} \) is not rational, a fact with which the reader is, at least informally, familiar; we will see a proof of this fact in Theorem 2.6.11. More precisely, it will be seen in that example that there is no rational number \( x \) such that \( x^2 = 2 \), and this last statement makes use only of rational numbers, so it is suited to our purpose at present. Nothing in our subsequent treatment of \( \sqrt{2} \) in Section 2.6 makes use of the current example, so it will not be circular reasoning for us to make use of these subsequently proved facts here.
Let
\[ T = \{ x \in \mathbb{Q} \mid x > 0 \text{ and } x^2 < 2 \}. \]
(1.6.1)
It is seen by Exercise 1.6.2 (1) that \( T \) is a Dedekind cut, and by Part (2) of that exercise it is seen that if \( T \) has the form \( \{ x \in \mathbb{Q} \mid x > r \} \) for some \( r \in \mathbb{Q} \), then \( r^2 = 2 \). By Theorem 2.6.11 we know that there is no rational number \( x \) such that \( x^2 = 2 \), and this shows that \( T \) is a Dedekind cut that is not of the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8975dae8-fbcb-4b49-8e3a-715e0e53990f%2Fa6eced6f-d3c5-4e9c-a3d5-98b4dea3c6f6%2Fhpgx9if_processed.png&w=3840&q=75)
Transcribed Image Text:**Example 1.6.3**
To find a Dedekind cut that is not of the form given in Lemma 1.6.2, the simplest idea is to look at the set of all rational numbers greater than a real number that is not rational; the problem is how to describe such a set without making use of real numbers. In the case of the number \( x = \sqrt{2} \), there turns out to be a simple solution to this problem, as we will now see. We note first, however, that we have not yet formally defined what \( \sqrt{2} \) means, nor proved that there is such a real number, though we will do so in Theorem 2.6.9 and Definition 2.6.10. We have not yet proved that \( \sqrt{2} \) is not rational, a fact with which the reader is, at least informally, familiar; we will see a proof of this fact in Theorem 2.6.11. More precisely, it will be seen in that example that there is no rational number \( x \) such that \( x^2 = 2 \), and this last statement makes use only of rational numbers, so it is suited to our purpose at present. Nothing in our subsequent treatment of \( \sqrt{2} \) in Section 2.6 makes use of the current example, so it will not be circular reasoning for us to make use of these subsequently proved facts here.
Let
\[ T = \{ x \in \mathbb{Q} \mid x > 0 \text{ and } x^2 < 2 \}. \]
(1.6.1)
It is seen by Exercise 1.6.2 (1) that \( T \) is a Dedekind cut, and by Part (2) of that exercise it is seen that if \( T \) has the form \( \{ x \in \mathbb{Q} \mid x > r \} \) for some \( r \in \mathbb{Q} \), then \( r^2 = 2 \). By Theorem 2.6.11 we know that there is no rational number \( x \) such that \( x^2 = 2 \), and this shows that \( T \) is a Dedekind cut that is not of the
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
