Answer
No inverse
Work Step by Step
Given the function
\[
g : \mathbb{Z} \;\to\; \mathbb{Z}
\quad\text{by}\quad
g(n) = 4n - 5
\quad\text{for all integers } n.
\]
We need to determine whether \(g\) is one-to-one and whether it is onto, with domain \(\mathbb{Z}\) and codomain \(\mathbb{Z}\).
---
## 1. Is \(g\) one-to-one (injective)?
To check injectivity, suppose
\[
g(n_1) \;=\; g(n_2).
\]
That is,
\[
4n_1 - 5 \;=\; 4n_2 - 5.
\]
Adding 5 to both sides and dividing by 4, we get
\[
4n_1 \;=\; 4n_2
\quad\Longrightarrow\quad
n_1 \;=\; n_2.
\]
Hence, \(g\) is **injective**.
---
## 2. Is \(g\) onto (surjective)?
To check surjectivity, we ask: for an arbitrary integer \(m \in \mathbb{Z}\), can we solve
\[
g(n) = m \quad\Longleftrightarrow\quad 4n - 5 = m
\quad\Longleftrightarrow\quad 4n = m + 5
\quad\Longleftrightarrow\quad n = \frac{m + 5}{4}\,?
\]
Since \(n\) must be an integer, \(\tfrac{m+5}{4}\) must lie in \(\mathbb{Z}\). However, for many integers \(m\), \(m + 5\) is **not** divisible by 4. For instance:
- If \(m = 1\), then \(m+5 = 6\), and \(\tfrac{6}{4} = 1.5\notin\mathbb{Z}\).
Thus there is no integer \(n\) with \(g(n)=1\). More generally, \(g(n)\) only takes values that are congruent to \(3 \pmod{4}\) (since \(4n - 5 \equiv -5 \equiv 3 \pmod{4}\)).
Hence, **\(g\) is not onto** \(\mathbb{Z}\). It misses all integers that are not \(3 \pmod{4}\).
---
## 3. Conclusion
Because \(g\) is injective but **not** surjective, it is **not** a one-to-one correspondence \(\mathbb{Z}\to \mathbb{Z}\). Consequently, **there is no inverse function** \(g^{-1} : \mathbb{Z}\to \mathbb{Z}\).
