Answer
**Analysis of the Mistake**
The “proof” given attempts to show that \(f(n) = 4n + 3\) is one-to-one (injective) by arguing that “for each integer \(n\), there is exactly one value of \(f(n)\).” This statement, however, merely shows that \(f\) is *well-defined*—that is, each input \(n\) indeed has a *unique* output. **It does not** prove that *different* inputs must yield *different* outputs.
Work Step by Step
### What “One-to-One” Really Requires
A function \(f\colon X \to Y\) is **one-to-one (injective)** if:
> **Whenever** \(f(x_1) = f(x_2)\), we must have \(x_1 = x_2.\)
Equivalently, an injective function *never* takes two distinct inputs to the same output.
### The Actual Error
1. The proof states:
> “Given any integer \(n\), there is only one possible value for \(f(n)\). Hence \(f\) is one-to-one.”
2. That claim shows only that \(f\) *assigns* a single value to each \(n\)—which is true for *every* function (that’s the definition of a function being well-defined).
3. To prove injectivity properly, one must show:
\[
\text{If } 4n_1 + 3 = 4n_2 + 3, \text{ then } n_1 = n_2.
\] This indeed holds for \(f(n) = 4n + 3\), but the proof as written does *not* address that requirement.
**In short,** the proof confuses *being a function* (which guarantees one output per input) with *being injective* (which requires distinct inputs to produce distinct outputs).
