91Ó°ÊÓ

Question: Prove Wilson's Theorem using the concept of modular arithmetic and multiplicative inverses for a given prime number \(p\). Answer: To prove Wilson's Theorem, which states that \((p-1)! \equiv -1 \pmod{p}\) for a prime number \(p\), we can follow these steps: 1. Define a modular inverse: For an integer \(a\) modulo \(p\), if it has a multiplicative inverse \(b\), write it as \(a^{-1} \equiv b \pmod{p}\). 2. Break down the factorial: \((p-1)! = 1 \cdot 2 \cdot 3 \cdots (p-1)\). 3. Pair up the numbers with their inverses: For every number in the range \(1\) to \(p-1\), pair it with its unique multiplicative inverse modulo \(p\). 4. Recognize that the product of each pair is congruent to 1: For each pair \((a, a^{-1})\), their product is congruent to 1 modulo \(p\): \(a \cdot a^{-1} \equiv 1 \pmod{p}\). 5. Multiply all pairs: \((1) \cdot (2 \cdot 2^{-1}) \cdot (3 \cdot 3^{-1}) \cdots ((p-1) \cdot (p -1)^{-1}) \equiv 1^{p-2} \cdot (p-1) \pmod{p}\). 6. Simplify the expression: Since \(1^{p-2} = 1\), the expression simplifies to \((p-1)! \equiv 1 \cdot (p-1) \pmod{p}\). 7. Prove Wilson's Theorem: Since \((p-1)! \equiv (p-1) \pmod{p}\), it follows that \((p-1)! \equiv -1 \pmod{p}\), proving Wilson's Theorem.