Answer
Student B is correct because the recursion leads to a *loop* for certain inputs. For example, when \(n=3\), the definition forces
\[
F(3)\;=\;1 + F(6)\quad\text{and}\quad F(6)\;=\;F(3).
\]
Putting these together yields \(F(3) = 1 + F(3)\), a contradiction. Thus \(F\) cannot be assigned a consistent value at \(n=3\), so the “definition” fails to be well‐defined.
Work Step by Step
## Why the Definition Breaks Down
The function \(F\colon \mathbb{Z}^+ \to \mathbb{Z}\) is *claimed* to be defined by:
\[
F(n) \;=\;
\begin{cases}
1, & \text{if }n=1,\\
F\!\bigl(\tfrac{n}{2}\bigr), & \text{if }n\text{ is even},\\
1 + F(5n - 9), & \text{if }n\text{ is odd and }n>1.
\end{cases}
\]
On the surface, this resembles Collatz‐style recursions, but there is a fatal problem:
1. **Check \(n=3\):**
- \(3\) is odd and \(>1\), so
\[
F(3) \;=\; 1 + F(5\cdot3 - 9) \;=\; 1 + F(6).
\]
- \(6\) is even, so
\[
F(6) \;=\; F\!\bigl(\tfrac{6}{2}\bigr) \;=\; F(3).
\]
Hence from these two lines,
\[
F(3) \;=\; 1 + F(6) \;=\; 1 + F(3),
\]
implying \(0 = 1\). This is a contradiction.
2. **No Value Can Be Assigned:**
Because the definition of \(F(3)\) immediately depends on \(F(6)\), and \(F(6)\) depends *back* on \(F(3)\), we get a cycle of dependencies with no way to “bottom out” at the base case \(F(1)=1\).
In other words, the proposed rule never yields a consistent value for \(F(3)\) (and hence for all inputs leading to \(3\) in the recursion). A well‐defined function must assign *exactly one* output to *each* input in its domain. Here, \(n=3\) admits *no* consistent assignment, so \(F\) is *not* well‐defined.
---
## Key Takeaway
- A recursive definition must always reduce in a way that eventually reaches a base case without contradiction.
- Because \(n=3\) leads to a *direct cycle* \(F(3)\leftrightarrow F(6)\), the recursion never resolves.
- Therefore, Student B’s claim is justified: **the function \(F\) is not well‐defined** on all positive integers.
