Answer
No, \(f(n) = n^2\) is *not* well-defined as a function \(f : S \to S\) if we interpret “\(n^2\)” in the usual mathematical sense. For some large integers \(n \in S\), the value \(n^2\) lies outside of \(S\). Consequently, it does not yield an output in \(S\) for every input in \(S\).
Work Step by Step
## Detailed Explanation
1. **Domain and Codomain**
Let
\[
S = \{-2{,}147{,}483{,}648,\,-2{,}147{,}483{,}647,\dots,\,2{,}147{,}483{,}647\}.
\]
We attempt to define a function
\[
f : S \;\longrightarrow\; S,
\quad
n \;\mapsto\; n^2.
\]
2. **Problem: Large Squares Exceed the Range**
- The maximum absolute value in \(S\) is \(2{,}147{,}483{,}648\).
- Squaring such a number in the usual mathematical sense gives \(\bigl(2{,}147{,}483{,}648\bigr)^2 \approx 4.611686 \times 10^{18}\), which is *far* larger than \(2{,}147{,}483{,}647\).
- Hence, for large \(\lvert n\rvert\), the value \(n^2\not\in S\).
3. **Definition of “Well-Defined”**
In mathematics, saying a rule “defines a function” \(f : A \to B\) requires that:
1. Every element \(a \in A\) has a *single* unambiguous output \(f(a)\).
2. This output *must lie in* \(B\).
If an alleged “function” produces values outside \(B\) for some inputs in \(A\), it is *not* a well-defined function \(A \to B\).
4. **Conclusion**
Since \(n^2\) for large \(\lvert n\rvert\) in \(S\) does *not* stay within \(S\), the rule \(f(n) = n^2\) does *not* define a function \(S \to S\) in the standard mathematical sense. It fails the requirement that outputs lie in \(S\) for *all* inputs \(n \in S\).
In practice, a computer might store only the low-order 32 bits of \(n^2\) (causing overflow), but that is *not* the same as the usual mathematical operation “square,” and thus also fails to be “\(n^2\)” in the strict sense.
