Let 67 = {n € Z: there exists m € Is.t.n = 6m}. Let
3Z = {n EZ: there exists mE Zs.t. n = 3m}. Prove that 67 C 37.

The problem is asking to prove that the set \( 6\mathbb{Z} \) (multiples of 6) is a subset of \( 3\mathbb{Z} \) (multiples of 3). Here's how we can prove it formally.

### Definitions:
- \( 6\mathbb{Z} = \{ n \in \mathbb{Z} \mid \exists m \in \mathbb{Z}, n = 6m \} \)
- \( 3\mathbb{Z} = \{ n \in \mathbb{Z} \mid \exists m \in \mathbb{Z}, n = 3m \} \)

### Goal:
We want to prove that \( 6\mathbb{Z} \subseteq 3\mathbb{Z} \), i.e., every element in \( 6\mathbb{Z} \) is also in \( 3\mathbb{Z} \).

### Proof:
1. **Take any arbitrary element from \( 6\mathbb{Z} \):**
   Let \( n \in 6\mathbb{Z} \), so by definition of \( 6\mathbb{Z} \), there exists some integer \( m \in \mathbb{Z} \) such that:
   \[
   n = 6m
   \]
   
2. **Show that \( n \in 3\mathbb{Z} \):**
   Now, we know that \( n = 6m \). We can factor \( 6m \) as:
   \[
   n = 3(2m)
   \]
   Since \( m \in \mathbb{Z} \), the product \( 2m \) is also an integer (because the set of integers is closed under multiplication). Therefore, there exists some integer \( k = 2m \in \mathbb{Z} \) such that:
   \[
   n = 3k
   \]
   Thus, \( n \in 3\mathbb{Z} \).

### Conclusion:
Since \( n \in 6\mathbb{Z} \) implies \( n \in 3\mathbb{Z} \), we have shown that \( 6\mathbb{Z} \subseteq 3\mathbb{Z} \). Therefore, \( 6\mathbb{Z} \subseteq 3\mathbb{Z} \) is true.

---

Would you like more details or have any questions?

Here are five related questions to expand your understanding:

1. What is the difference between subset and proper subset?
2. Can you prove that \( 3\mathbb{Z} \subseteq 6\mathbb{Z} \)?
3. How would you represent the set of multiples of 4 in a similar form?
4. What are the properties of the set of integers \( \mathbb{Z} \)?
5. Can you provide a counterexample to show that \( 3\mathbb{Z} \) is not a subset of \( 6\mathbb{Z} \)?

**Tip:** When proving subsets, try to express elements of one set in terms of elements from another set, using the definitions of the sets carefully.

Let 67 = {n € Z: there exists m € Z s.t. n = 6m}. Let 3Z = {n € Z: there exists m € Z s.t. n = 3m}. Prove that 67 C 37.

This problem explores the proof that the set of multiples of 6 (6Z) is a subset of the set of multiples of 3 (3Z). Using key set theory and number theory concepts, the proof demonstrates how any element of 6Z can be expressed as an element of 3Z. Suitable for students in Grades 9-12.

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