91Ó°ÊÓ

In mathematics, the domain of a function is the set of all possible inputs for that function. For the function given in the exercise, denoted as \(f: \mathbb{Z} \rightarrow \mathbb{N} \cup \{0\}\), the domain is clearly specified as \(\mathbb{Z}\). This represents all integers, both positive and negative, including zero. This means that any integer you can think of can be put into the function \(f\) as an input.
Understanding the domain is crucial because it tells us what kind of numbers we can use with our function. In this case, because the function involves finding the remainder when any integer \(a\) is divided by \(n\), it makes sense for the function to accept any integer as input.

While the domain tells us what we can input into the function, the range tells us the possible outputs we can expect. For our function \(f\), where \(f(a) = r\) and \(0 \leq r < n\), the range is all the remainder values when any integer is divided by \(n\).

• The remainder \(r\) can be any non-negative integer less than \(n\).
• This means that possible values of \(r\) could be \(0, 1, 2, \ldots, n-1\).

So, the range of the function \(f\) is the set \(\{0, 1, 2, \ldots, n-1\}\). This provides a complete picture of what outputs can result from inputs into the function.

A one-to-one function, or injective function, is a type of function where each element of the domain maps to a unique element in the range. To determine whether the function \(f\) is one-to-one, we need to see if different inputs give different outputs.
For this specific function, \(f(a) = r\), several integers can produce the same remainder when divided by \(n\). For example, both the numbers \(0\) and \(n\) yield a remainder of \(0\). Therefore, this function is not one-to-one because the same remainder can be obtained from different integer inputs.

The concept of an onto function, also known as a surjective function, involves every element of the codomain being able to be mapped to from some element in the domain.
Our function \(f: \mathbb{Z} \rightarrow \mathbb{N} \cup \{0\}\) has an effective codomain of \(\{0, 1, 2, \ldots, n-1\}\), based on the range we determined earlier. For a function to be onto in its effective codomain, each element of \(\{0, 1, 2, \ldots, n-1\}\) must be an output for some input from the domain \(\mathbb{Z}\).

• Given our function design, indeed each number from \(0\) to \(n-1\) can be obtained as a remainder from some input in \(\mathbb{Z}\) when divided by \(n\).
• Thus, \(f\) is onto in the restricted codomain \(\{0, 1, \ldots, n-1\}\).

Therefore, while \(f\) is not onto the entire non-negative integers, it is onto its specific range, fulfilling the requirement of being surjective within its practical context.