i. Which of the following relations f: Q → Q define a mapping? In each case, supply a reason why f is or is not a mapping. p+1 3p a. f(p/q): = b. f(p/q) = = p-2 3q p+q Р c. f(p/q): d. f(p/q) = 32²2-22 q² 7q² =
i. Which of the following relations f: Q → Q define a mapping? In each case, supply a reason why f is or is not a mapping. p+1 3p a. f(p/q): = b. f(p/q) = = p-2 3q p+q Р c. f(p/q): d. f(p/q) = 32²2-22 q² 7q² =
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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Help with d please
![### Identifying Mappings in Relations from Rational Numbers to Rational Numbers
This exercise focuses on evaluating whether given relations \( f : \mathbb{Q} \to \mathbb{Q} \) define a mapping. In each case, we are required to justify whether \( f \) is or is not a mapping.
#### Definitions and Relations
Consider the following functions \( f \), where \( p \) and \( q \) are rational numbers (\( q \neq 0 \)):
a. \( f(p/q) = \frac{p+1}{p-2} \)
b. \( f(p/q) = \frac{3p}{3q} \)
c. \( f(p/q) = \frac{p+q}{q^2} \)
d. \( f(p/q) = \frac{3p^2}{7q^2} - \frac{p}{q} \)
#### Analysis
##### Option a
**Function:**
\[ f(p/q) = \frac{p+1}{p-2} \]
**Reasoning:**
- To determine if it is a valid function, consider the denominator \( p-2 = 0 \), which implies \( p = 2 \). At \( p = 2 \), the function is undefined because division by zero is not allowed.
- Therefore, \( f(p/q) \) is not defined for all rational numbers, making it not a proper mapping over the domain of all rational numbers \( \mathbb{Q} \).
##### Option b
**Function:**
\[ f(p/q) = \frac{3p}{3q} \]
**Reasoning:**
- Simplify the function: \( \frac{3p}{3q} = \frac{p}{q} \).
- The function essentially maps \( p/q \) to itself, which is well-defined for all rational \( p \) and \( q \) ( \( q \neq 0 \)).
- Hence, \( f(p/q) \) is a valid mapping.
##### Option c
**Function:**
\[ f(p/q) = \frac{p+q}{q^2} \]
**Reasoning:**
- The expression \( \frac{p+q}{q^2} \) is defined for all rational \( p \) and \( q \) ( \( q \neq 0 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fed14a3ea-da26-4be7-a143-8b845df95e91%2Fcfd4ad00-5ff2-45c6-b2f3-62ae3cb22677%2Fh6hyq5p_processed.png&w=3840&q=75)
Transcribed Image Text:### Identifying Mappings in Relations from Rational Numbers to Rational Numbers
This exercise focuses on evaluating whether given relations \( f : \mathbb{Q} \to \mathbb{Q} \) define a mapping. In each case, we are required to justify whether \( f \) is or is not a mapping.
#### Definitions and Relations
Consider the following functions \( f \), where \( p \) and \( q \) are rational numbers (\( q \neq 0 \)):
a. \( f(p/q) = \frac{p+1}{p-2} \)
b. \( f(p/q) = \frac{3p}{3q} \)
c. \( f(p/q) = \frac{p+q}{q^2} \)
d. \( f(p/q) = \frac{3p^2}{7q^2} - \frac{p}{q} \)
#### Analysis
##### Option a
**Function:**
\[ f(p/q) = \frac{p+1}{p-2} \]
**Reasoning:**
- To determine if it is a valid function, consider the denominator \( p-2 = 0 \), which implies \( p = 2 \). At \( p = 2 \), the function is undefined because division by zero is not allowed.
- Therefore, \( f(p/q) \) is not defined for all rational numbers, making it not a proper mapping over the domain of all rational numbers \( \mathbb{Q} \).
##### Option b
**Function:**
\[ f(p/q) = \frac{3p}{3q} \]
**Reasoning:**
- Simplify the function: \( \frac{3p}{3q} = \frac{p}{q} \).
- The function essentially maps \( p/q \) to itself, which is well-defined for all rational \( p \) and \( q \) ( \( q \neq 0 \)).
- Hence, \( f(p/q) \) is a valid mapping.
##### Option c
**Function:**
\[ f(p/q) = \frac{p+q}{q^2} \]
**Reasoning:**
- The expression \( \frac{p+q}{q^2} \) is defined for all rational \( p \) and \( q \) ( \( q \neq 0 \
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