Show that Z + × Z + is in bijection with Z + × Z + × Z + .

To show that \(\mathbb{Z}^+ \times \mathbb{Z}^+\) is in bijection with \(\mathbb{Z}^+ \times \mathbb{Z}^+ \times \mathbb{Z}^+\), we need to find a bijective function \( f: \mathbb{Z}^+ \times \mathbb{Z}^+ \to \mathbb{Z}^+ \times \mathbb{Z}^+ \times \mathbb{Z}^+ \) and its inverse.

### Step 1: Finding a Bijection

To establish a bijection, consider a pair \((a, b) \in \mathbb{Z}^+ \times \mathbb{Z}^+\) and map it to a triplet \((x, y, z) \in \mathbb{Z}^+ \times \mathbb{Z}^+ \times \mathbb{Z}^+\). One way to construct such a mapping is to split the pair into three positive integers using a systematic method.

**One possible mapping:**
Given \( (a, b) \in \mathbb{Z}^+ \times \mathbb{Z}^+ \), define:
1. \( x = 2a - 1 \) (odd integer)
2. \( y = 2b - 1 \) (odd integer)
3. \( z = a + b \)

Then, we have \( f: \mathbb{Z}^+ \times \mathbb{Z}^+ \to \mathbb{Z}^+ \times \mathbb{Z}^+ \times \mathbb{Z}^+ \) given by:
\[
f(a, b) = (2a - 1, 2b - 1, a + b).
\]

### Step 2: Finding the Inverse

Now, we need to find the inverse function \( f^{-1}: \mathbb{Z}^+ \times \mathbb{Z}^+ \times \mathbb{Z}^+ \to \mathbb{Z}^+ \times \mathbb{Z}^+ \). Given a triplet \( (x, y, z) \):
1. Since \( x = 2a - 1 \), solve for \( a \):
   \[
   a = \frac{x + 1}{2}.
   \]
   Since \( x \) is odd, \( a \) is guaranteed to be a positive integer.
2. Similarly, since \( y = 2b - 1 \), solve for \( b \):
   \[
   b = \frac{y + 1}{2}.
   \]
   Since \( y \) is odd, \( b \) is also a positive integer.
3. We also know that \( z = a + b \). Substitute the expressions for \( a \) and \( b \):
   \[
   z = \frac{x + 1}{2} + \frac{y + 1}{2}.
   \]

Now, reconstruct the pair \( (a, b) \) from \( (x, y, z) \):
\[
f^{-1}(x, y, z) = \left(\frac{x + 1}{2}, \frac{y + 1}{2}\right).
\]

### Step 3: Proving Bijectiveness

1. **Injectivity:** Suppose \( f(a_1, b_1) = f(a_2, b_2) \). This means
   \[
   (2a_1 - 1, 2b_1 - 1, a_1 + b_1) = (2a_2 - 1, 2b_2 - 1, a_2 + b_2).
   \]
   From this, \( 2a_1 - 1 = 2a_2 - 1 \) implies \( a_1 = a_2 \) and \( 2b_1 - 1 = 2b_2 - 1 \) implies \( b_1 = b_2 \). Therefore, \( f \) is injective.

2. **Surjectivity:** For any \( (x, y, z) \in \mathbb{Z}^+ \times \mathbb{Z}^+ \times \mathbb{Z}^+ \), we can find \( (a, b) \in \mathbb{Z}^+ \times \mathbb{Z}^+ \) such that \( f(a, b) = (x, y, z) \). We set
   \[
   a = \frac{x + 1}{2} \quad \text{and} \quad b = \frac{y + 1}{2}.
   \]
   Thus, every element in \( \mathbb{Z}^+ \times \mathbb{Z}^+ \times \mathbb{Z}^+ \) has a preimage in \( \mathbb{Z}^+ \times \mathbb{Z}^+ \), so \( f \) is surjective.

Since \( f \) is both injective and surjective, it is a bijection.

### Conclusion

Therefore, there exists a bijection between \( \mathbb{Z}^+ \times \mathbb{Z}^+ \) and \( \mathbb{Z}^+ \times \mathbb{Z}^+ \times \mathbb{Z}^+ \).

Do you want more details or have any questions about the solution?

Here are 5 related questions:
1. How can we construct a bijection between \( \mathbb{Z}^+ \) and \( \mathbb{Z}^+ \times \mathbb{Z}^+ \)?
2. What is a general method to find a bijection between \( \mathbb{Z}^+^n \) and \( \mathbb{Z}^+^{n+1} \)?
3. How do we prove the countability of \( \mathbb{Z}^+ \times \mathbb{Z}^+ \)?
4. Can you find a bijection between \( \mathbb{N} \) and \( \mathbb{N} \times \mathbb{N} \)?
5. How does the Cantor pairing function provide a bijection between \( \mathbb{Z} \) and \( \mathbb{Z} \times \mathbb{Z} \)?

**Tip:** Bijections are crucial in mathematics for showing that two sets have the same cardinality, especially in set theory and the study of infinite sets.

Show that Z + × Z + is in bijection with Z + × Z + × Z +.

Explore the proof that demonstrates the bijection between Z+ × Z+ and Z+ × Z+ × Z+. This detailed explanation covers key set theory concepts, including Cartesian products, bijective functions, and the use of injectivity and surjectivity in establishing a bijection. Suitable for undergraduate mathematics students.

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