Answer
No inverse
Work Step by Step
Given the function
\[
F : \mathbb{Z} \;\to\; \mathbb{Z}
\quad\text{by}\quad
F(n) \;=\; 2 \;-\; 3n,
\]
for all integers \(n\). We need to determine whether \(F\) is one-to-one (injective) and onto (surjective) when considered as a map \(\mathbb{Z}\to \mathbb{Z}\).
---
## 1. Is \(F\) One-to-One (Injective)?
Suppose
\[
F(n_1) = F(n_2).
\]
That is,
\[
2 - 3n_1 \;=\; 2 - 3n_2.
\]
Subtracting 2 from both sides gives
\[
-3n_1 = -3n_2
\quad\Longrightarrow\quad
3n_1 = 3n_2
\quad\Longrightarrow\quad
n_1 = n_2.
\]
Hence, \(F\) is **injective**.
---
## 2. Is \(F\) Onto (Surjective)?
We ask whether every integer \(m\) in the codomain \(\mathbb{Z}\) has a preimage \(n\in \mathbb{Z}\) such that
\[
F(n) = m \quad\Longleftrightarrow\quad 2 - 3n = m.
\]
Rearranging gives
\[
-3n = m - 2
\quad\Longleftrightarrow\quad
n = \frac{2 - m}{3}.
\]
For \(n\) to be an integer, \((2 - m)\) must be divisible by 3. Concretely, \(m\) must be congruent to \(2 \pmod{3}\). Therefore:
- If \(m \equiv 2 \pmod{3}\), then \(n = \frac{2 - m}{3}\) is indeed an integer, and \(m\) is in the image of \(F\).
- If \(m\) is **not** of the form \(3k + 2\), there is no integer \(n\) with \(F(n) = m\).
Thus, \(F\) **does not** map onto all of \(\mathbb{Z}\). It misses any integer \(m\) that is not congruent to \(2\) modulo \(3\).
---
## 3. Conclusion
Because \(F\) is injective but **not** surjective, it is **not** a one-to-one correspondence \(\mathbb{Z}\to \mathbb{Z}\). Consequently, there is **no inverse function** \(F^{-1}: \mathbb{Z}\to \mathbb{Z}\).
