At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Let's solve the given problem carefully, step by step.
We are given:
1. A set [tex]\( S = \{ a \mid a \in \mathbb{N}, a \leq 36 \} \)[/tex], which contains all the natural numbers up to and including 36.
2. Two relations, [tex]\( R_1 \)[/tex] and [tex]\( R_2 \)[/tex], from [tex]\( S \)[/tex] to [tex]\( S \)[/tex].
Without copying or referring to any code, we can address the problem as follows:
Consider the relations defined as:
[tex]\[ R_1 = \{(x, y) \mid x, y \in S, y = f(x)\} \][/tex]
[tex]\[ R_2 = \{(x, y) \mid x, y \in S, y = g(x)\} \][/tex]
The question specifies finding a new set which is the difference of [tex]\( R_1 \)[/tex] and the intersection of [tex]\( R_1 \)[/tex] and [tex]\( R_2 \)[/tex]:
[tex]\[ R_1 \backslash (R_1 \cap R_2) \][/tex]
To proceed, we need to understand what this means in simple terms:
- [tex]\( R_1 \cap R_2 \)[/tex] is the set of all ordered pairs that are common to both [tex]\( R_1 \)[/tex] and [tex]\( R_2 \)[/tex].
- [tex]\( R_1 \backslash (R_1 \cap R_2) \)[/tex] is the set of all ordered pairs in [tex]\( R_1 \)[/tex] that are not in [tex]\( R_1 \cap R_2 \)[/tex].
Step-by-Step Solution:
1. Identify the Set [tex]\( S \)[/tex]:
[tex]\[ S = \{1, 2, 3, \ldots, 36\} \][/tex]
2. Define [tex]\( R_1 \)[/tex] and [tex]\( R_2 \)[/tex]:
Without specific definitions for functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex], we can assume they follow specific, defined rules which map x to y within the range specified by [tex]\( S \)[/tex].
3. Find [tex]\( R_1 \cap R_2 \)[/tex]:
This includes all pairs that satisfy both relations. For example, if [tex]\( y = f(x) \)[/tex] for [tex]\( R_1 \)[/tex] and [tex]\( y = g(x) \)[/tex] for [tex]\( R_2 \)[/tex], then:
[tex]\[ R_1 \cap R_2 = \{(x, y) \mid y = f(x) \text{ and } y = g(x)\} \][/tex]
4. Calculate [tex]\( R_1 \backslash (R_1 \cap R_2) \)[/tex]:
These are pairs in [tex]\( R_1 \)[/tex] but not in their intersection:
[tex]\[ R_1 \backslash (R_1 \cap R_2) = \{(x, y) \mid y = f(x) \text{ and } y \neq g(x) \} \][/tex]
By isolating the concept of intersections and differences in sets and relations, we have created a step-by-step outline to handle this mathematical query methodically. This process leverages basic set operations and functions mapping to find solutions intricately.
We are given:
1. A set [tex]\( S = \{ a \mid a \in \mathbb{N}, a \leq 36 \} \)[/tex], which contains all the natural numbers up to and including 36.
2. Two relations, [tex]\( R_1 \)[/tex] and [tex]\( R_2 \)[/tex], from [tex]\( S \)[/tex] to [tex]\( S \)[/tex].
Without copying or referring to any code, we can address the problem as follows:
Consider the relations defined as:
[tex]\[ R_1 = \{(x, y) \mid x, y \in S, y = f(x)\} \][/tex]
[tex]\[ R_2 = \{(x, y) \mid x, y \in S, y = g(x)\} \][/tex]
The question specifies finding a new set which is the difference of [tex]\( R_1 \)[/tex] and the intersection of [tex]\( R_1 \)[/tex] and [tex]\( R_2 \)[/tex]:
[tex]\[ R_1 \backslash (R_1 \cap R_2) \][/tex]
To proceed, we need to understand what this means in simple terms:
- [tex]\( R_1 \cap R_2 \)[/tex] is the set of all ordered pairs that are common to both [tex]\( R_1 \)[/tex] and [tex]\( R_2 \)[/tex].
- [tex]\( R_1 \backslash (R_1 \cap R_2) \)[/tex] is the set of all ordered pairs in [tex]\( R_1 \)[/tex] that are not in [tex]\( R_1 \cap R_2 \)[/tex].
Step-by-Step Solution:
1. Identify the Set [tex]\( S \)[/tex]:
[tex]\[ S = \{1, 2, 3, \ldots, 36\} \][/tex]
2. Define [tex]\( R_1 \)[/tex] and [tex]\( R_2 \)[/tex]:
Without specific definitions for functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex], we can assume they follow specific, defined rules which map x to y within the range specified by [tex]\( S \)[/tex].
3. Find [tex]\( R_1 \cap R_2 \)[/tex]:
This includes all pairs that satisfy both relations. For example, if [tex]\( y = f(x) \)[/tex] for [tex]\( R_1 \)[/tex] and [tex]\( y = g(x) \)[/tex] for [tex]\( R_2 \)[/tex], then:
[tex]\[ R_1 \cap R_2 = \{(x, y) \mid y = f(x) \text{ and } y = g(x)\} \][/tex]
4. Calculate [tex]\( R_1 \backslash (R_1 \cap R_2) \)[/tex]:
These are pairs in [tex]\( R_1 \)[/tex] but not in their intersection:
[tex]\[ R_1 \backslash (R_1 \cap R_2) = \{(x, y) \mid y = f(x) \text{ and } y \neq g(x) \} \][/tex]
By isolating the concept of intersections and differences in sets and relations, we have created a step-by-step outline to handle this mathematical query methodically. This process leverages basic set operations and functions mapping to find solutions intricately.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.