https://file.aiquickdraw.com/mathbot/file/1726970136354j47ewcpc.jpg

Let's break down the problem presented in the image and solve it step by step. We are given the following sets:

- \( X = \{ 0 \} \)
- \( A = \{ -1, 0, 1 \} \)
- \( B = \{ 4, 5 \} \)
- \( C = \left\{ \frac{p}{q+1} \in \mathbb{Q} \mid p, q \in \mathbb{Z}, 5 \leq p \leq 6 \text{ and } -1 \leq q \leq 1 \right\} \)
- \( D = \left\{ \frac{4}{5}, \frac{9}{6}, \frac{16}{7}, \frac{25}{8}, \frac{36}{9} \right\} \)

Now, let's solve each part.

### 1. List all elements of \( C \)
Set \( C \) is defined by a rule involving integers \( p \) and \( q \). The conditions are:

- \( p \in \mathbb{Z} \), where \( 5 \leq p \leq 6 \)
- \( q \in \mathbb{Z} \), where \( -1 \leq q \leq 1 \)

Let's calculate all possible values of \( \frac{p}{q+1} \):

For \( p = 5 \):
- When \( q = -1 \), \( \frac{5}{q+1} = \frac{5}{0} \), which is undefined.
- When \( q = 0 \), \( \frac{5}{q+1} = \frac{5}{1} = 5 \).
- When \( q = 1 \), \( \frac{5}{q+1} = \frac{5}{2} \).

For \( p = 6 \):
- When \( q = -1 \), \( \frac{6}{q+1} = \frac{6}{0} \), which is undefined.
- When \( q = 0 \), \( \frac{6}{q+1} = \frac{6}{1} = 6 \).
- When \( q = 1 \), \( \frac{6}{q+1} = \frac{6}{2} = 3 \).

Thus, \( C = \left\{ 5, \frac{5}{2}, 6, 3 \right\} \).

### 2. \( A \times B \) 
This represents the Cartesian product of sets \( A \) and \( B \), meaning we take all possible pairs \( (a, b) \) where \( a \in A \) and \( b \in B \). The pairs are:

\[
A \times B = \{ (-1, 4), (-1, 5), (0, 4), (0, 5), (1, 4), (1, 5) \}
\]

### 3. \( A \times A \) 
Similarly, we calculate the Cartesian product of \( A \) with itself:

\[
A \times A = \{ (-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1) \}
\]

### 4. \( 2^{X \times A} \) 
This refers to the power set of the Cartesian product \( X \times A \). First, we calculate \( X \times A \):

\[
X \times A = \{ (0, -1), (0, 0), (0, 1) \}
\]

The power set of \( X \times A \) is the set of all subsets of \( X \times A \). So, the power set is:

\[
2^{X \times A} = \{ \emptyset, \{(0, -1)\}, \{(0, 0)\}, \{(0, 1)\}, \{(0, -1), (0, 0)\}, \{(0, -1), (0, 1)\}, \{(0, 0), (0, 1)\}, \{(0, -1), (0, 0), (0, 1)\} \}
\]

### 5. Describe \( A \) and \( D \) using set-builder notation

- \( A = \{ x \in \mathbb{Z} \mid -1 \leq x \leq 1 \} \)
- \( D = \left\{ \frac{n^2}{n+4} \mid n \in \mathbb{Z}, 2 \leq n \leq 6 \right\} \)

These notations describe \( A \) and \( D \) using conditions on \( x \) and \( n \).

---

Let me know if you'd like further details or explanations. 

Here are five related questions to explore further:

1. What is the cardinality of set \( C \)?
2. How many elements are in the power set \( 2^{A \times B} \)?
3. Can \( C \) be expressed using a different interval for \( p \) and \( q \)?
4. How does the Cartesian product change if \( B \) had more elements?
5. What is the complement of set \( D \) with respect to the rational numbers?

**Tip:** When working with Cartesian products, remember that the order of elements matters in pairs like \( (a, b) \) and \( (b, a) \).

Consider the sets X = {0}, A = {-1, 0, 1}, B = {4, 5}, C = {p / (q + 1) ∈ Q | p, q ∈ Z, 5 ≤ p ≤ 6 and -1 ≤ q ≤ 1}, and D = {4/5, 9/6, 16/7, 25/8, 36/9}. List the elements of C, A × B, A × A, 2^{X × A}, and describe A and D using set builder notation.

Explore set operations involving Cartesian products, power sets, and rational numbers. This problem covers set theory concepts, including listing elements of sets, performing Cartesian products, and describing sets using set-builder notation. Suitable for students in Grades 9-12.

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