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 two things:

1. \( A \subseteq C \): Every element of \( A \) is also an element of \( C \).
2. \( C \subseteq A \): Every element of \( C \) is also an element of \( A \).

### Step 1: Show \( A \subseteq C \)

Let \( n \in A \). By the definition of \( A \), this means there exists some \( k_1 \in \mathbb{N} \) such that \( k_1 \) is a multiple of 5 (i.e., \( k_1 = 5m_1 \) for some \( m_1 \in \mathbb{N} \)) and \( n = 4 - k_1 \).

So, we can write:
\[
n = 4 - 5m_1
\]

Now, we need to show that \( n \) belongs to \( C \). For \( n \) to belong to \( C \), there must exist some \( k_2 \in \mathbb{N} \) with \( k_2 \geq 4 \) such that:
\[
n = 24 - 5k_2
\]

Comparing the two expressions for \( n \), we get:
\[
4 - 5m_1 = 24 - 5k_2
\]

Rearranging this equation:
\[
20 = 5(k_2 - m_1)
\]

Simplifying:
\[
k_2 - m_1 = 4
\]

Thus, \( k_2 = m_1 + 4 \). Since \( m_1 \in \mathbb{N} \), and adding 4 to a natural number is still a natural number, \( k_2 \) is also a natural number and \( k_2 \geq 4 \). This shows that \( n \in C \), so \( A \subseteq C \).

### Step 2: Show \( C \subseteq A \)

Let \( n \in C \). By the definition of \( C \), this means there exists some \( k_2 \in \mathbb{N} \) with \( k_2 \geq 4 \) such that:
\[
n = 24 - 5k_2
\]

Now, we need to show that \( n \) belongs to \( A \). For \( n \) to belong to \( A \), there must exist some \( k_1 \in \mathbb{N} \) such that \( k_1 = 5m_1 \) for some \( m_1 \in \mathbb{N} \) and:
\[
n = 4 - k_1
\]

Again, comparing the two expressions for \( n \), we get:
\[
24 - 5k_2 = 4 - k_1
\]

Rearranging:
\[
20 = k_1 - 5k_2
\]

Since \( k_1 \) is a multiple of 5, say \( k_1 = 5m_1 \) for some \( m_1 \in \mathbb{N} \), we substitute:
\[
20 = 5m_1 - 5k_2
\]

Simplifying:
\[
4 = m_1 - k_2
\]

So, \( m_1 = k_2 + 4 \). Since \( k_2 \geq 4 \) and \( k_2 \in \mathbb{N} \), it follows that \( m_1 \in \mathbb{N} \) as well. This shows that \( k_1 = 5m_1 \) and \( n = 4 - k_1 \), which implies \( n \in A \). Hence, \( C \subseteq A \).

### Conclusion

Since we've shown both \( A \subseteq C \) and \( C \subseteq A \), we conclude that \( A = C \).

---

Do you want details or have any questions?

Here are five related questions for further exploration:

1. How would the sets \( A \) and \( C \) change if the definitions of \( k \) were altered?
2. Can you prove the equality of sets \( A \) and \( C \) using a different method, such as by direct enumeration of elements?
3. What if \( k \) had to satisfy additional conditions in either set, such as \( k \) being even?
4. How does the nature of the sets \( A \) and \( C \) change if \( k \) is required to be a non-negative integer?
5. Could the approach used here be applied to other sets defined by similar conditions? How?

**Tip:** When dealing with set equality, it's often helpful to use the technique of proving mutual inclusion (\( A \subseteq B \) and \( B \subseteq A \)) to establish that \( A = B \).

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 proof of set equality between 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}. This problem covers key concepts in set theory and number theory, suitable for undergraduate students.

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