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Magazine Chapter 4: Problem 20
Prove that if \(p\) is prime then \((p-1) ! \equiv-1 \bmod (p)\) (as in Wilson's Theorem, Corollary 4.5) by pairing off non-zero classes \(a, b \in \mathbb{Z}_{p}\) such that \(a b=1\) in \(\mathbb{Z}_{p}\).
Short Answer
Expert verified
We proved Wilson's Theorem: \((p-1)! \equiv -1 \pmod{p}\) for prime \(p\).
Step by step solution
01
Understanding the Problem
We need to prove Wilson's Theorem which states that for any prime number \(p\), the factorial \((p-1)!\) is congruent to \(-1\) mod \(p\). This means \((p-1)! \equiv -1 \pmod{p}\).
02
Identifying Elements in \(\mathbb{Z}_p\)
The set \(\mathbb{Z}_p\) contains the integers \(0, 1, 2, \ldots, p-1\). Our goal is to find a pairing among the non-zero integers from \(1\) to \(p-1\) such that each pair \((a, b)\) satisfies \(a \cdot b \equiv 1 \pmod{p}\).
03
Pairing Non-Zero Elements
Consider a non-zero class \(a\) in \(\mathbb{Z}_p\). We pair each element \(a\) with its multiplicative inverse \(b\equiv a^{-1} \pmod{p}\), satisfying the condition \(a \cdot b \equiv 1 \pmod{p}\). Since \(p\) is prime, every non-zero integer has a unique inverse.
04
Special Handling of Potential Fixed Points
The only elements that could potentially be paired with themselves (i.e., be their own inverse) are \(1\) and \(p-1\), because \(1 \cdot 1 \equiv 1 \pmod{p}\) and \((p-1) \, (p-1) \equiv 1 \pmod{p}\) since \(p-1 \equiv -1 \pmod{p}\).
05
Formulating \((p-1)!\) with Pairings
Since all elements except possibly \(1\) and \(p-1\) can be paired into these inverses that multiply to 1, their product is simplified. Therefore, the product of these pairings is 1 modulo \(p\). After multiplying all pairs, the result is: \((p-1)! = (1) \cdot (p-1) = -1 \equiv -1 \pmod{p}\).
06
Concluding the Proof
This completes the proof that \((p-1)! \equiv -1 \pmod{p}\) using Wilson's Theorem. The factorial of \(p-1\) modulo \(p\) results in \(-1\) due to the property of inverses in the field \(\mathbb{Z}_p\) and the special case handling of the elements 1 and \(p-1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factorial
In mathematics, the factorial of a non-negative integer n, denoted by \( n! \), is the product of all positive integers less than or equal to n.
For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
The factorial function grows very quickly and is commonly used in permutations, combinations, and other areas of mathematics.
For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
The factorial function grows very quickly and is commonly used in permutations, combinations, and other areas of mathematics.
- The factorial is defined as \( 0! = 1 \) by convention, even though zero is not a positive integer.
- Factorials are essential in various mathematical theorems, such as Wilson's Theorem.
Prime Numbers
Prime numbers are integers greater than 1 that have no divisors other than 1 and themselves. A prime number \( p \) has the unique property that any integer less than \( p \) is either relatively prime to \( p \) or is 1.
- Examples of prime numbers include 2, 3, 5, 7, and 11.
- Primes are the building blocks of the natural numbers due to their role in number factorization.
Modular Arithmetic
Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" a certain value, called the modulus.
The notation \( a \equiv b \pmod{m} \) means that \( a \) and \( b \) have the same remainder when divided by \( m \).
The notation \( a \equiv b \pmod{m} \) means that \( a \) and \( b \) have the same remainder when divided by \( m \).
- Often likened to a clock, where the hours reset to 1 after 12.
- It is extensively used in computer science, cryptography, and number theory.
Multiplicative Inverse
The multiplicative inverse of a number \( a \) in a given modular system is another number \( b \) such that \( a \cdot b \equiv 1 \pmod{m} \).
For a multiplicative inverse to exist, \( a \) and \( m \) must be coprime鈥攈aving no common factors other than 1.
For a multiplicative inverse to exist, \( a \) and \( m \) must be coprime鈥攈aving no common factors other than 1.
- In a field like \( \mathbb{Z}_p \), where \( p \) is prime, every non-zero element has a unique multiplicative inverse.
- This is because the Euclidean algorithm always finds inverses when dealing with integers that are coprime with the modulus.
