Answer
No inverse
Work Step by Step
Given the function
\[
f: \mathbb{Z} \;\to\; \mathbb{Z}
\quad\text{by}\quad
f(n) \;=\; 2n.
\]
We are to determine whether \(f\) is a one-to-one correspondence (i.e., bijective) from \(\mathbb{Z}\) to \(\mathbb{Z}\). If it is, we must find its inverse.
---
### 1. Is \(f\) one-to-one (injective)?
To check injectivity, suppose \(f(n_1) = f(n_2)\). That is:
\[
2n_1 \;=\; 2n_2.
\]
Dividing both sides by 2 gives \(n_1 = n_2\). Therefore, \(f\) is injective.
---
### 2. Is \(f\) onto (surjective)?
To check surjectivity, we ask: for an arbitrary integer \(m \in \mathbb{Z}\), does there exist an integer \(n \in \mathbb{Z}\) such that \(f(n) = m\)? That would require
\[
2n \;=\; m
\quad\Longrightarrow\quad
n \;=\; \frac{m}{2}.
\]
- If \(m\) is even, say \(m = 2k\), then \(n = k\) is an integer, so \(m\) is in the image of \(f\).
- If \(m\) is odd, say \(m = 2k + 1\), then \(n = \frac{2k+1}{2} = k + \tfrac12\) is not an integer. So there is no integer \(n\) that maps to an odd \(m\).
Hence, **\(f\) is not onto** \(\mathbb{Z}\), because it misses all odd integers in the codomain.
---
### 3. Conclusion
Since \(f\) is **not** onto \(\mathbb{Z}\), it fails to be a bijection. Therefore, **it is not a one-to-one correspondence** and does not have an inverse function \(f^{-1}: \mathbb{Z} \to \mathbb{Z}\).
\(f(n) = 2n\) is injective but not surjective as a map \(\mathbb{Z}\to \mathbb{Z}\). Consequently, it is not a one-to-one correspondence, and no inverse function exists over all of \(\mathbb{Z}\).
