Let A = {n ∈ N | n = 6k−13, for some k ∈ N} and B = {m ∈ N | m = 5k + 14, for some k ∈ Z
+}.
Prove that A ̸= B.

To prove that the sets \( A \) and \( B \) are not equal, we need to show that there exists an element in one set that is not in the other.

### Definitions:
- \( A = \{n \in \mathbb{N} \mid n = 6k - 13 \text{ for some } k \in \mathbb{N}\} \)
- \( B = \{m \in \mathbb{N} \mid m = 5k + 14 \text{ for some } k \in \mathbb{Z}^+\} \)

Here, \( \mathbb{N} \) refers to the set of natural numbers \( \{1, 2, 3, \dots\} \) and \( \mathbb{Z}^+ \) refers to the set of positive integers.

### Elements of Set \( A \):
We start by identifying the first few elements of \( A \).

For \( k = 1 \):
\[
n = 6(1) - 13 = 6 - 13 = -7 \text{ (not a natural number, so not included in \( A \))}
\]

For \( k = 2 \):
\[
n = 6(2) - 13 = 12 - 13 = -1 \text{ (not a natural number, so not included in \( A \))}
\]

For \( k = 3 \):
\[
n = 6(3) - 13 = 18 - 13 = 5
\]

For \( k = 4 \):
\[
n = 6(4) - 13 = 24 - 13 = 11
\]

For \( k = 5 \):
\[
n = 6(5) - 13 = 30 - 13 = 17
\]

Thus, the first few elements of \( A \) are \( \{5, 11, 17, \dots\} \).

### Elements of Set \( B \):
Next, we identify the first few elements of \( B \).

For \( k = 1 \):
\[
m = 5(1) + 14 = 5 + 14 = 19
\]

For \( k = 2 \):
\[
m = 5(2) + 14 = 10 + 14 = 24
\]

For \( k = 3 \):
\[
m = 5(3) + 14 = 15 + 14 = 29
\]

For \( k = 4 \):
\[
m = 5(4) + 14 = 20 + 14 = 34
\]

Thus, the first few elements of \( B \) are \( \{19, 24, 29, 34, \dots\} \).

### Comparing Sets \( A \) and \( B \):
- The first few elements of \( A \) are \( \{5, 11, 17, \dots\} \).
- The first few elements of \( B \) are \( \{19, 24, 29, 34, \dots\} \).

Since the elements of \( A \) are not in \( B \) and vice versa (for example, \( 5 \in A \) but \( 5 \notin B \); \( 19 \in B \) but \( 19 \notin A \)), we can conclude that \( A \) and \( B \) are distinct sets.

### Conclusion:
The sets \( A \) and \( B \) are not equal, i.e., \( A \neq B \).

Would you like to go over the details, or do you have any other questions?

Here are 5 related questions to consider:

1. How do we determine the general form of elements in a set defined by an arithmetic sequence?
2. Can two arithmetic sequences ever produce identical sets for different starting values and increments?
3. What is the significance of natural numbers in defining sets like \( A \) and \( B \)?
4. How would the proof change if we altered the definitions of \( A \) and \( B \) slightly?
5. What are some other methods to prove that two sets are unequal?

**Tip:** When comparing sets, always start by identifying elements explicitly to see if one set contains an element that the other does not. This can provide a straightforward way to determine if the sets are different.

Let A = {n ∈ N | n = 6k−13, for some k ∈ N} and B = {m ∈ N | m = 5k + 14, for some k ∈ Z+}. Prove that A ̸= B.

Explore how to prove that the sets A = {n ∈ N | n = 6k−13} and B = {m ∈ N | m = 5k + 14} are not equal. This problem involves set theory and arithmetic sequences, making it suitable for students in Grades 10-12.

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