Answer :
Sure, let's solve each part step by step.
### Part A: Cartesian Products of Sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex]
Step 1: Define Set [tex]\( A \)[/tex]
Given:
[tex]\[ A = \{ x \leq 4 \mid x \in \mathbb{N} \} \][/tex]
This means that [tex]\( A \)[/tex] contains all natural numbers [tex]\( x \)[/tex] such that [tex]\( x \leq 4 \)[/tex]. Thus:
[tex]\[ A = \{ 0, 1, 2, 3, 4 \} \][/tex]
Step 2: Define Set [tex]\( B \)[/tex]
Given:
[tex]\[ B = \{ x \mid x^2 - 5x - 6 = 0 \} \][/tex]
To solve for [tex]\( B \)[/tex], we must find the roots of the quadratic equation [tex]\( x^2 - 5x - 6 = 0 \)[/tex]. Factoring the quadratic equation, we get:
[tex]\[ (x - 6)(x + 1) = 0 \][/tex]
Hence, the solutions are:
[tex]\[ x = 3 \text{ and } x = -2 \][/tex]
So,
[tex]\[ B = \{ 3, -2 \} \][/tex]
Step 3: Find [tex]\( A \times B \)[/tex]
The Cartesian product [tex]\( A \times B \)[/tex] is the set of all possible ordered pairs [tex]\( (a, b) \)[/tex] where [tex]\( a \in A \)[/tex] and [tex]\( b \in B \)[/tex]. Thus:
[tex]\[ A \times B = \{ (a, b) \mid a \in A \text{ and } b \in B \} \][/tex]
Calculating the pairs:
[tex]\[ A \times B = \{ (0, 3), (0, -2), (1, 3), (1, -2), (2, 3), (2, -2), (3, 3), (3, -2), (4, 3), (4, -2) \} \][/tex]
Step 4: Find [tex]\( B \times A \)[/tex]
The Cartesian product [tex]\( B \times A \)[/tex] is the set of all possible ordered pairs [tex]\( (b, a) \)[/tex] where [tex]\( b \in B \)[/tex] and [tex]\( a \in A \)[/tex]. Thus:
[tex]\[ B \times A = \{ (b, a) \mid b \in B \text{ and } a \in A \} \][/tex]
Calculating the pairs:
[tex]\[ B \times A = \{ (3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (-2, 0), (-2, 1), (-2, 2), (-2, 3), (-2, 4) \} \][/tex]
### Part B: Cartesian Products of Sets [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]
Step 1: Define Set [tex]\( P \)[/tex]
Given:
[tex]\[ P = \{ x \mid 2 < x < 7 \text{ and } x \in \mathbb{N} \} \][/tex]
This means that [tex]\( P \)[/tex] contains all natural numbers [tex]\( x \)[/tex] such that [tex]\( 2 < x < 7 \)[/tex]. Thus:
[tex]\[ P = \{ 3, 4, 5, 6 \} \][/tex]
Step 2: Define Set [tex]\( Q \)[/tex]
Given:
[tex]\[ Q = \{ x \mid x^2 = 3x \} \][/tex]
To solve for [tex]\( Q \)[/tex], we need to find the roots of the equation [tex]\( x^2 - 3x = 0 \)[/tex]. Factoring the equation we get:
[tex]\[ x(x - 3) = 0 \][/tex]
Hence, the solutions are:
[tex]\[ x = 0 \text{ and } x = 3 \][/tex]
So,
[tex]\[ Q = \{ 0, 3 \} \][/tex]
Step 3: Find [tex]\( P \times Q \)[/tex]
The Cartesian product [tex]\( P \times Q \)[/tex] is the set of all possible ordered pairs [tex]\( (p, q) \)[/tex] where [tex]\( p \in P \)[/tex] and [tex]\( q \in Q \)[/tex]. Thus:
[tex]\[ P \times Q = \{ (p, q) \mid p \in P \text{ and } q \in Q \} \][/tex]
Calculating the pairs:
[tex]\[ P \times Q = \{ (3, 0), (3, 3), (4, 0), (4, 3), (5, 0), (5, 3), (6, 0), (6, 3) \} \][/tex]
Step 4: Find [tex]\( Q \times P \)[/tex]
The Cartesian product [tex]\( Q \times P \)[/tex] is the set of all possible ordered pairs [tex]\( (q, p) \)[/tex] where [tex]\( q \in Q \)[/tex] and [tex]\( p \in P \)[/tex]. Thus:
[tex]\[ Q \times P = \{ (q, p) \mid q \in Q \text{ and } p \in P \} \][/tex]
Calculating the pairs:
[tex]\[ Q \times P = \{ (0, 3), (0, 4), (0, 5), (0, 6), (3, 3), (3, 4), (3, 5), (3, 6) \} \][/tex]
### Summary of Results
- [tex]\( A \times B = \{ (0, 3), (0, -2), (1, 3), (1, -2), (2, 3), (2, -2), (3, 3), (3, -2), (4, 3), (4, -2) \} \)[/tex]
- [tex]\( B \times A = \{ (3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (-2, 0), (-2, 1), (-2, 2), (-2, 3), (-2, 4) \} \)[/tex]
- [tex]\( P \times Q = \{ (3, 0), (3, 3), (4, 0), (4, 3), (5, 0), (5, 3), (6, 0), (6, 3) \} \)[/tex]
- [tex]\( Q \times P = \{ (0, 3), (0, 4), (0, 5), (0, 6), (3, 3), (3, 4), (3, 5), (3, 6) \} \)[/tex]
### Part A: Cartesian Products of Sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex]
Step 1: Define Set [tex]\( A \)[/tex]
Given:
[tex]\[ A = \{ x \leq 4 \mid x \in \mathbb{N} \} \][/tex]
This means that [tex]\( A \)[/tex] contains all natural numbers [tex]\( x \)[/tex] such that [tex]\( x \leq 4 \)[/tex]. Thus:
[tex]\[ A = \{ 0, 1, 2, 3, 4 \} \][/tex]
Step 2: Define Set [tex]\( B \)[/tex]
Given:
[tex]\[ B = \{ x \mid x^2 - 5x - 6 = 0 \} \][/tex]
To solve for [tex]\( B \)[/tex], we must find the roots of the quadratic equation [tex]\( x^2 - 5x - 6 = 0 \)[/tex]. Factoring the quadratic equation, we get:
[tex]\[ (x - 6)(x + 1) = 0 \][/tex]
Hence, the solutions are:
[tex]\[ x = 3 \text{ and } x = -2 \][/tex]
So,
[tex]\[ B = \{ 3, -2 \} \][/tex]
Step 3: Find [tex]\( A \times B \)[/tex]
The Cartesian product [tex]\( A \times B \)[/tex] is the set of all possible ordered pairs [tex]\( (a, b) \)[/tex] where [tex]\( a \in A \)[/tex] and [tex]\( b \in B \)[/tex]. Thus:
[tex]\[ A \times B = \{ (a, b) \mid a \in A \text{ and } b \in B \} \][/tex]
Calculating the pairs:
[tex]\[ A \times B = \{ (0, 3), (0, -2), (1, 3), (1, -2), (2, 3), (2, -2), (3, 3), (3, -2), (4, 3), (4, -2) \} \][/tex]
Step 4: Find [tex]\( B \times A \)[/tex]
The Cartesian product [tex]\( B \times A \)[/tex] is the set of all possible ordered pairs [tex]\( (b, a) \)[/tex] where [tex]\( b \in B \)[/tex] and [tex]\( a \in A \)[/tex]. Thus:
[tex]\[ B \times A = \{ (b, a) \mid b \in B \text{ and } a \in A \} \][/tex]
Calculating the pairs:
[tex]\[ B \times A = \{ (3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (-2, 0), (-2, 1), (-2, 2), (-2, 3), (-2, 4) \} \][/tex]
### Part B: Cartesian Products of Sets [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]
Step 1: Define Set [tex]\( P \)[/tex]
Given:
[tex]\[ P = \{ x \mid 2 < x < 7 \text{ and } x \in \mathbb{N} \} \][/tex]
This means that [tex]\( P \)[/tex] contains all natural numbers [tex]\( x \)[/tex] such that [tex]\( 2 < x < 7 \)[/tex]. Thus:
[tex]\[ P = \{ 3, 4, 5, 6 \} \][/tex]
Step 2: Define Set [tex]\( Q \)[/tex]
Given:
[tex]\[ Q = \{ x \mid x^2 = 3x \} \][/tex]
To solve for [tex]\( Q \)[/tex], we need to find the roots of the equation [tex]\( x^2 - 3x = 0 \)[/tex]. Factoring the equation we get:
[tex]\[ x(x - 3) = 0 \][/tex]
Hence, the solutions are:
[tex]\[ x = 0 \text{ and } x = 3 \][/tex]
So,
[tex]\[ Q = \{ 0, 3 \} \][/tex]
Step 3: Find [tex]\( P \times Q \)[/tex]
The Cartesian product [tex]\( P \times Q \)[/tex] is the set of all possible ordered pairs [tex]\( (p, q) \)[/tex] where [tex]\( p \in P \)[/tex] and [tex]\( q \in Q \)[/tex]. Thus:
[tex]\[ P \times Q = \{ (p, q) \mid p \in P \text{ and } q \in Q \} \][/tex]
Calculating the pairs:
[tex]\[ P \times Q = \{ (3, 0), (3, 3), (4, 0), (4, 3), (5, 0), (5, 3), (6, 0), (6, 3) \} \][/tex]
Step 4: Find [tex]\( Q \times P \)[/tex]
The Cartesian product [tex]\( Q \times P \)[/tex] is the set of all possible ordered pairs [tex]\( (q, p) \)[/tex] where [tex]\( q \in Q \)[/tex] and [tex]\( p \in P \)[/tex]. Thus:
[tex]\[ Q \times P = \{ (q, p) \mid q \in Q \text{ and } p \in P \} \][/tex]
Calculating the pairs:
[tex]\[ Q \times P = \{ (0, 3), (0, 4), (0, 5), (0, 6), (3, 3), (3, 4), (3, 5), (3, 6) \} \][/tex]
### Summary of Results
- [tex]\( A \times B = \{ (0, 3), (0, -2), (1, 3), (1, -2), (2, 3), (2, -2), (3, 3), (3, -2), (4, 3), (4, -2) \} \)[/tex]
- [tex]\( B \times A = \{ (3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (-2, 0), (-2, 1), (-2, 2), (-2, 3), (-2, 4) \} \)[/tex]
- [tex]\( P \times Q = \{ (3, 0), (3, 3), (4, 0), (4, 3), (5, 0), (5, 3), (6, 0), (6, 3) \} \)[/tex]
- [tex]\( Q \times P = \{ (0, 3), (0, 4), (0, 5), (0, 6), (3, 3), (3, 4), (3, 5), (3, 6) \} \)[/tex]
