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.