We are asked to determine whether the function $f(m, n) = m - n$ is one-to-one (injective), onto (surjective), or both. The domain is the Cartesian product of integers, $\mathbb{Z} \times \mathbb{Z}$, and the codomain is $\mathbb{Z}$, the set of integers.

1. One-to-One (Injective) Check:

A function is one-to-one (or injective) if for every pair of distinct elements in the domain, the function maps them to distinct elements in the codomain. In other words, $f(m_1, n_1) = f(m_2, n_2)$ implies that $(m_1, n_1) = (m_2, n_2)$.

Let's test if $f(m_1, n_1) = f(m_2, n_2)$ leads to $(m_1, n_1) = (m_2, n_2)$.

• Suppose $f(m_1, n_1) = f(m_2, n_2)$, meaning: $m_1 - n_1 = m_2 - n_2$
• This implies: $m_1 - m_2 = n_1 - n_2$
• From here, it's clear that $m_1 = m_2$ and $n_1 = n_2$ are not necessarily true. For example, if $m_1 = 3, n_1 = 1$ and $m_2 = 2, n_2 = 0$, we get: $3 - 1 = 2 - 0 = 2$ But $(3, 1) \neq (2, 0)$, yet the function maps both pairs to the same output.

Since $f(m_1, n_1) = f(m_2, n_2)$ does not always imply $(m_1, n_1) = (m_2, n_2)$, the function is not one-to-one (injective).

2. Onto (Surjective) Check:

A function is onto (or surjective) if for every element in the codomain, there exists at least one element in the domain that maps to it.

Here, the codomain is $\mathbb{Z}$. To check if $f$ is onto, we need to see if for every integer $z \in \mathbb{Z}$, there exist integers $m$ and $n$ such that: $f(m, n) = m - n = z$

• This equation always has a solution. For example, if we choose $m = z$ and $n = 0$, we get: $m - n = z - 0 = z$
• Therefore, for every integer $z$, we can find $m$ and $n$ such that $f(m, n) = z$.

Since there exists a pair $(m, n)$ for every $z \in \mathbb{Z}$, the function is onto (surjective).

Conclusion:

The function $f(m, n) = m - n$ is onto but not one-to-one.

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Further Questions:

1. How would the function behave if the domain was restricted to a subset of integers (e.g., $\mathbb{Z}_+ \times \mathbb{Z}_+$)?
2. What is the difference between a function being onto and being one-to-one?
3. How can we prove a function is one-to-one using the definition of injectivity in more complex scenarios?
4. Can a function be both one-to-one and onto if the domain and codomain have different cardinalities?
5. What are some other examples of functions that are onto but not one-to-one?

Tip: To prove injectivity more easily, you can use the contrapositive. Instead of proving $f(a) = f(b) \Rightarrow a = b$, you can prove that $a \neq b \Rightarrow f(a) \neq f(b)$.