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Chapter 3: Problem 12

Determine if each function from \(\mathbb{R}\) to \(\mathbf{Z}\) is surjective. $$g(x)=\lfloor x\rfloor$$

Short Answer

Expert verified
The function \(g(x)=\lfloor x\rfloor\) is surjective, as for every integer \(y\in\mathbb{Z}\), we have \(g(y)=\lfloor y\rfloor = y\), indicating that every element in the codomain \(\mathbb{Z}\) is mapped to at least one element in the domain \(\mathbb{R}\).

Step by step solution

01

Definition of Surjective Function

To show that a function is surjective, we need to show that for every element \(y\) in the codomain, \(\mathbb{Z}\), there exists at least one element \(x\) in the domain, \(\mathbb{R}\), such that \(g(x) = y\).
02

Determine the value of \(g(x)\) using the definition

We are given that \(g(x)=\lfloor x\rfloor\). As a reminder, \(\lfloor x \rfloor\) denotes the greatest integer less than or equal to \(x\).
03

Let \(y \in \mathbb{Z}\)

Since we want to show that the function \(g(x)\) is surjective, we should first suppose that \(y\) is an integer contained in the codomain \(\mathbb{Z}\).
04

Find x such that \(g(x) = y\)

We are trying to find at least one value of \(x\) that satisfies \(g(x) = y\). If we look at the definition of \(g(x)\), we notice that if \(x = y\), the function would be evaluated as \(g(y) = \lfloor y \rfloor\). Since \(y\) is already an integer, the greatest integer less than or equal to \(y\) would be \(y\) itself. Therefore, \(g(y) = y\).
05

Conclude that \(g(x)\) is surjective

We showed that for any integer \(y \in \mathbb{Z}\), there exists at least one value of \(x\) in \(\mathbb{R}\) (which we showed is just \(x = y\)) such that \(g(x) = y\). Therefore, the function \(g(x) = \lfloor x \rfloor\) is surjective, as it maps every element in the codomain \(\mathbb{Z}\) to at least one element in the domain \(\mathbb{R}\).

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