Let A = {n ∈ Z | n = 4 − k, for some k ∈ N and 5|k} and C = {m ∈ Z | m = 24 −
5k for some k ∈ N and k ≥ 4}. Prove that A = C.

To prove that \( A = C \), we need to show that every element of set \( A \) is in set \( C \), and vice versa.

### Definitions of Sets \( A \) and \( C \)

1. **Set \( A \):**
   \[
   A = \{n \in \mathbb{Z} \mid n = 4 - k, \text{ for some } k \in \mathbb{N} \text{ and } 5 \mid k \}
   \]
   This means \( n \) is of the form \( 4 - k \) where \( k \) is a natural number and a multiple of 5, i.e., \( k = 5m \) for some \( m \in \mathbb{N} \).

2. **Set \( C \):**
   \[
   C = \{m \in \mathbb{Z} \mid m = 24 - 5k, \text{ for some } k \in \mathbb{N} \text{ and } k \geq 4 \}
   \]
   This means \( m \) is of the form \( 24 - 5k \) where \( k \) is a natural number and \( k \geq 4 \).

### Proof that \( A \subseteq C \)

1. Let \( n \in A \). By definition, \( n = 4 - k \) where \( k = 5m \) for some \( m \in \mathbb{N} \).
   
2. Substituting \( k = 5m \):
   \[
   n = 4 - 5m
   \]

3. To express \( n \) in the form required for set \( C \):
   \[
   n = 24 - 5(k')
   \]
   where \( k' \) needs to be determined.

4. Set \( 4 - 5m = 24 - 5k' \). Solving for \( k' \):
   \[
   24 - 5k' = 4 - 5m
   \]
   \[
   20 = 5k' - 5m
   \]
   \[
   4 = k' - m
   \]
   \[
   k' = m + 4
   \]

5. Since \( m \in \mathbb{N} \), \( k' = m + 4 \geq 4 \) because \( m \geq 1 \). Therefore, \( k' \in \mathbb{N} \) and \( k' \geq 4 \).

6. Thus, \( n = 4 - 5m = 24 - 5k' \) with \( k' \geq 4 \), implying \( n \in C \). Therefore, \( A \subseteq C \).

### Proof that \( C \subseteq A \)

1. Let \( m \in C \). By definition, \( m = 24 - 5k \) where \( k \geq 4 \) and \( k \in \mathbb{N} \).

2. Set \( k = 4 + m' \) where \( m' \in \mathbb{N} \) to satisfy \( k \geq 4 \).

3. Substituting \( k = 4 + m' \):
   \[
   m = 24 - 5(4 + m')
   \]
   \[
   m = 24 - 20 - 5m'
   \]
   \[
   m = 4 - 5m'
   \]

4. Since \( m' \in \mathbb{N} \), let \( m' = m \) (to maintain the natural number property). Thus, \( m = 4 - 5m' = 4 - 5m \) with \( 5m \in \mathbb{N} \) as required by \( A \).

5. Therefore, \( m \in A \). This implies \( C \subseteq A \).

### Conclusion

Since we have shown that \( A \subseteq C \) and \( C \subseteq A \), it follows that:

\[
A = C
\]

Do you have any questions or would you like more details?

### Related Questions:
1. How would this proof change if we altered the conditions on \( k \)?
2. Can we use a different method, such as set notation manipulation, to prove \( A = C \)?
3. How can we visualize sets \( A \) and \( C \) on a number line?
4. How does changing the multiple constraint from \( 5 \) to another number affect the sets?
5. What happens to the sets if we allow \( k \) to be in \( \mathbb{Z} \) instead of \( \mathbb{N} \)?

**Tip:** Always clearly define the elements of each set before attempting to prove equality; this ensures you accurately capture the conditions needed for membership in each set.

Let A = {n ∈ Z | n = 4 − k, for some k ∈ N and 5|k} and C = {m ∈ Z | m = 24 − 5k for some k ∈ N and k ≥ 4}. Prove that A = C.

Explore the detailed proof that shows the equality of two sets A and C, defined with specific algebraic conditions. The problem covers key concepts in set theory and algebra, suitable for students in Grades 10-12.

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