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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.
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