Answer
Student A is correct that \(R\) *is* well-defined, because in \(\mathbb{Z}_5\) (the integers mod 5), every nonzero element has a *unique* multiplicative inverse.
Work Step by Step
## Reasoning
1. **The set \(\mathbb{Z}_5 - \{0\}\) is \(\{1,2,3,4\}\).**
Since 5 is prime, \(\mathbb{Z}_5\) is a field. Consequently, *every* nonzero element \(x\in\{1,2,3,4\}\) has exactly one inverse \(y\) in \(\{1,2,3,4\}\) such that
\[
x \cdot y \equiv 1 \pmod{5}.
\]
2. **Uniqueness of \(y\).**
Because \(\mathbb{Z}_5\) is a field, the inverse of each nonzero element is unique. In other words, there is *exactly one* \(y\) that makes \(x \cdot y \equiv 1 \pmod{5}\). Hence the rule
\[
R(x) = \text{the unique \(y\) such that } x \cdot y \equiv 1 \pmod{5}
\]
always yields a single well-defined value of \(y\) for each \(x\).
3. **Examples.**
- \(1 \cdot 1 \equiv 1 \pmod{5}\), so \(R(1)=1\).
- \(2 \cdot 3 \equiv 6 \equiv 1 \pmod{5}\), so \(R(2)=3\).
- \(3 \cdot 2 \equiv 6 \equiv 1 \pmod{5}\), so \(R(3)=2\).
- \(4 \cdot 4 \equiv 16 \equiv 1 \pmod{5}\), so \(R(4)=4\).
Since for *each* \(x\) there is *exactly one* suitable \(y\), \(R\) is indeed a well-defined function. Thus, **Student A** is correct.
