Consider the set [tex]\( S = \{a \mid a \in \mathbb{N}, a \leq 36\} \)[/tex].

Let [tex]\( R_1 \)[/tex] and [tex]\( R_2 \)[/tex] be relations from [tex]\( S \)[/tex] to [tex]\( S \)[/tex] defined as [tex]\( R_1 = \{(x, y) \mid x, y \in S \} \)[/tex].

Find [tex]\( R_1 \setminus (R_1 \cap R_2) \)[/tex].



Answer :

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.