Let's analyze the function $f(m, n) = m^2 + n^2$, where the domain is $\mathbb{Z} \times \mathbb{Z}$ (the set of ordered pairs of integers) and the codomain is $\mathbb{Z}$ (the set of integers).

1. One-to-One (Injective):

A function is one-to-one (or injective) if different inputs always map to different outputs. In other words, for any two distinct pairs $(m_1, n_1)$ and $(m_2, n_2)$, if $f(m_1, n_1) = f(m_2, n_2)$, then it must follow that $(m_1, n_1) = (m_2, n_2)$.

Let's test whether this function is injective. Suppose that: $f(m_1, n_1) = f(m_2, n_2)$ This means: $m_1^2 + n_1^2 = m_2^2 + n_2^2$ However, this does not necessarily imply that $(m_1, n_1) = (m_2, n_2)$. For example, let's check the following:

• $f(1, 1) = 1^2 + 1^2 = 2$
• $f(-1, -1) = (-1)^2 + (-1)^2 = 2$

Here, $f(1, 1) = f(-1, -1)$, but clearly $(1, 1) \neq (-1, -1)$.

This example shows that different pairs $(m, n)$ can give the same output. Therefore, the function is not one-to-one (not injective).

2. Onto (Surjective):

A function is onto (or surjective) if every element in the codomain has a preimage in the domain. In other words, for every integer $y$ in the codomain $\mathbb{Z}$, there must be some pair $(m, n) \in \mathbb{Z} \times \mathbb{Z}$ such that $f(m, n) = y$.

We need to check if for every integer $y \in \mathbb{Z}$, there are integers $m$ and $n$ such that: $m^2 + n^2 = y$ Consider some values of $y$:

• If $y = -1$, then we need to solve $m^2 + n^2 = -1$. But the sum of two squares is always non-negative, meaning there are no integer solutions for $m$ and $n$ in this case.
• For other negative numbers, the sum of squares $m^2 + n^2$ cannot produce a negative value. Thus, there are no integer pairs that map to negative numbers.

This shows that there are integers in the codomain that are not achievable by the function. Therefore, the function is not onto (not surjective).

Conclusion:

The function $f(m, n) = m^2 + n^2$ is neither one-to-one nor onto.

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Would you like further details or have any questions about this analysis?

Here are 5 related questions to explore:

1. What would happen if the codomain were restricted to non-negative integers?
2. How can we prove whether a function is injective using counterexamples?
3. What are the conditions for a function to be surjective?
4. Can a function be injective but not surjective? Give an example.
5. How does the Pythagorean theorem relate to the sum of squares of integers?

Tip: When checking for injectivity, always look for counterexamples where different inputs yield the same output. This is a quick way to disprove injectivity.