Exercise 17.2.17. (a) Give a counterexample that proves that the binary relation C in Fig- ure 17.2.3 is not transitive. (*Hint*) (b) Explain why the binary relation R = {(1,4), (1, 1), (4,1)} is not transitive. (*Hint*) (c) Explain why the binary relation is transitive. (*Hint*) R= {(1,2), (1,3), (1,4)}
Exercise 17.2.17. (a) Give a counterexample that proves that the binary relation C in Fig- ure 17.2.3 is not transitive. (*Hint*) (b) Explain why the binary relation R = {(1,4), (1, 1), (4,1)} is not transitive. (*Hint*) (c) Explain why the binary relation is transitive. (*Hint*) R= {(1,2), (1,3), (1,4)}
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 do Exercise 17.2.17 part A,B, and C and please show step by step and explain
Hint for A: There are actually three counterexamples, you only need to find one.
Hint for B: Give a specific example where a ∼ b and b ∼ c but a (/∼) c. In other words, (a, b) and (b, c) are elements of R∼, but (a, c) is not in R∼. It is not necessary for a, b, and c to be distinct.
Hint for C: Explain why it is impossible to find a counterexample.
![**Exercise 17.2.17.**
(a) Give a counterexample that proves that the binary relation \( C \) in Figure 17.2.3 is *not* transitive. (*Hint*)
(b) Explain why the binary relation
\[ R_{\sim} = \{(1, 4), (1, 1), (4, 1)\} \]
is *not* transitive. (*Hint*)
(c) Explain why the binary relation
\[ R_{\sim} = \{(1, 2), (1, 3), (1, 4)\} \]
is transitive. (*Hint*)
\( \diamond \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F892e817a-9b32-4eeb-b8fc-5dd7ffde6479%2Fc3e6d078-e9f6-49e0-a6cc-2ae2baa5b376%2Fra5jnk_processed.png&w=3840&q=75)
Transcribed Image Text:**Exercise 17.2.17.**
(a) Give a counterexample that proves that the binary relation \( C \) in Figure 17.2.3 is *not* transitive. (*Hint*)
(b) Explain why the binary relation
\[ R_{\sim} = \{(1, 4), (1, 1), (4, 1)\} \]
is *not* transitive. (*Hint*)
(c) Explain why the binary relation
\[ R_{\sim} = \{(1, 2), (1, 3), (1, 4)\} \]
is transitive. (*Hint*)
\( \diamond \)
![**Example 17.2.16**
Let's think about binary relations on the set \(\{1, 2, 3\}\) as seen in Figure 17.2.3. Which of the binary relations, A, B, or C, are transitive? Why or why not?
---
**Figure 17.2.3**: Digraphs corresponding to Example 17.2.16
- **Diagram A**:
- There are directed edges from 1 to 2, 2 to 3, and 1 to 3, forming a complete directed triangle among the three nodes.
- **Diagram B**:
- There is a directed edge from 1 to 3 and from 1 to 2. There is no direct edge from 2 to 3.
- **Diagram C**:
- There are directed edges from 1 to 3, 3 to 2, and 2 to 1, forming a cycle.
---
**Is the relation in A transitive?**
Let's revisit the concept of transitivity: if \(a \sim b\) and \(b \sim c\) then \(a \sim c\). More simply, if there's an arrow from \(a\) to \(b\) and another from \(b\) to \(c\), then there's also an arrow directly from \(a\) to \(c\).
Consider \(a\), \(b\), and \(c\) as airports, and the arrows as flights. Transitivity implies that whenever there's a multi-stop route between airports, a direct route must also exist.
**In Diagram A**, there's an indirect route from 3 to 2 via 1. However, there is also a direct route from 3 to 2 (3 to 1, 1 to 2, and 3 to 2). Hence, this relation is transitive, as these conditions satisfy all possible indirect to direct routes.
If the arrow from 3 to 2 is removed, the relation loses transitivity, as the indirect path still remains (3 to 1 to 2), but there would be no direct connection from 3 to 2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F892e817a-9b32-4eeb-b8fc-5dd7ffde6479%2Fc3e6d078-e9f6-49e0-a6cc-2ae2baa5b376%2Fehy4vydj_processed.png&w=3840&q=75)
Transcribed Image Text:**Example 17.2.16**
Let's think about binary relations on the set \(\{1, 2, 3\}\) as seen in Figure 17.2.3. Which of the binary relations, A, B, or C, are transitive? Why or why not?
---
**Figure 17.2.3**: Digraphs corresponding to Example 17.2.16
- **Diagram A**:
- There are directed edges from 1 to 2, 2 to 3, and 1 to 3, forming a complete directed triangle among the three nodes.
- **Diagram B**:
- There is a directed edge from 1 to 3 and from 1 to 2. There is no direct edge from 2 to 3.
- **Diagram C**:
- There are directed edges from 1 to 3, 3 to 2, and 2 to 1, forming a cycle.
---
**Is the relation in A transitive?**
Let's revisit the concept of transitivity: if \(a \sim b\) and \(b \sim c\) then \(a \sim c\). More simply, if there's an arrow from \(a\) to \(b\) and another from \(b\) to \(c\), then there's also an arrow directly from \(a\) to \(c\).
Consider \(a\), \(b\), and \(c\) as airports, and the arrows as flights. Transitivity implies that whenever there's a multi-stop route between airports, a direct route must also exist.
**In Diagram A**, there's an indirect route from 3 to 2 via 1. However, there is also a direct route from 3 to 2 (3 to 1, 1 to 2, and 3 to 2). Hence, this relation is transitive, as these conditions satisfy all possible indirect to direct routes.
If the arrow from 3 to 2 is removed, the relation loses transitivity, as the indirect path still remains (3 to 1 to 2), but there would be no direct connection from 3 to 2.
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