Problem 13 - Discrete Mathematics with Applications

Question

Let \(a\) be a positive integer and \(p\) a prime number such that \(p | a^{n} .\) Then \(p | a,\) where \(n \geq 1.\) (Hint: Use Exercise 37 in Section 4.2.)

Step-by-Step Solution

Answer

Using mathematical induction and Exercise 37 as a hint, we can prove that if a prime number \(p\) divides \(a^n\) (where \(a\) is a positive integer and \(n \geq 1\)), then \(p\) also divides \(a\). Base case: For \(n=1\), we have \(p | a^1\), so \(p | a\). Inductive hypothesis: Assume that \(p | a^n\), meaning that there exists an integer \(k\) such that \(a^n = pk\). Inductive step: We must prove that \(p | a^{(n+1)}\). Using the inductive hypothesis, we get \(a^{(n+1)} = (pk)a\) which results in \(a^{(n+1)} = p(ka)\). Applying Exercise 37, we conclude that \(p | a\). Thus, if a prime number \(p\) divides \(a^n\), then \(p\) must also divide \(a\).

Step 1: Recap Exercise 37 and prove the given problem

First, let's recall Exercise 37: It says that if p is a prime number and p | ab, then p | a or p | b. Now, let's prove the given problem: We are given that p | a^n. We will prove that p | a using mathematical induction on the power n.

Step 2: Base case (n=1)

For n = 1, we have that p | a^1. This means p | a, which proves the base case.

Step 3: Inductive hypothesis

Now, let's assume that p | a^n for some integer n. This means that there exists an integer k such that a^n = pk.

Step 4: Inductive step

We need to prove that p | a^(n+1). Now let's calculate a^(n+1): \[a^{(n+1)} = a^n \cdot a\] Using the inductive hypothesis, we know that a^n = pk. Therefore, \[a^{(n+1)} = (pk) \cdot a\] \[a^{(n+1)} = p(ka)\] Now, applying Exercise 37, since p is a prime and p | p(ka), then p must divide one of the factors of the product. In this case, since p cannot divide itself, it must divide the other factor. Hence: \[p | a\]

Step 5: Conclusion

By using mathematical induction and applying Exercise 37, we have proven that if p | a^n, then p | a. This holds for any positive integer a and any prime number p, with n greater than or equal to 1.

Key Concepts

Prime Numbers

A prime number is a special type of number that is greater than 1 and only divisible by 1 and itself.
Prime numbers play a fundamental role in number theory because they are the building blocks of whole numbers.

• Prime numbers include numbers like 2, 3, 5, 7, and 11.
• Each prime number has exactly two distinct positive divisors.

Primes are key for various mathematical propositions. For example, in Exercise 37, a prime number like \( p \) has unique properties that allow us to deduce divisibility rules when \( p \) divides products of numbers.

Divisibility

Divisibility is the concept where one integer can be evenly divided by another integer without leaving a remainder. In the context of the original exercise, divisibility is crucial to understanding how primes affect other numbers.When we say \( p | a^n \), it means that the prime \( p \) divides the integer \( a^n \) without remainder.
This divisibility relationship can be extended further using mathematical induction.

• For example, if \( p | a^n \), we are investigating whether \( p | a \).
• This means checking if there is an integer \( k \) such that \( a^n = pk \).

Divisibility by a prime number imposes special conditions on the numbers it divides, making it a powerful tool in proving mathematical statements.

Inductive Step

The inductive step is a crucial part of mathematical induction. It involves taking a general case and showing it leads logically to the next case in the sequence.In our original exercise, after verifying the base case, we assume the inductive hypothesis: \( p | a^n \) for some \( n \). This step prepares us to prove \( p | a^{n+1} \).The transitional process can be broken down as follows:

• First, express \( a^{n+1} = a^n \cdot a \).
• Substitute the inductive hypothesis \( a^n = pk \), leading to \( a^{n+1} = p(ka) \).
• Using what we know about primaries from Exercise 37, we establish that \( p | a \).

This logical transition demonstrates how the properties are preserved and extended from \( n \) to \( n+1 \).

Base Case

The base case is the first step in the process of mathematical induction. It sets the standard or starting point for the inductive reasoning method.In our scenario, the base case checks whether the statement holds true for \( n = 1 \). Specifically, we examine if \( p | a \) when \( p | a^1 \).

• For \( n = 1 \), \( a^1 = a \), so if \( p | a^1 \), it directly implies \( p | a \).
• This initial verification guarantees that our statement begins correctly.

The base case confirms that the hypothesis stands when \( n \) is at its smallest. Successfully proving the base case is crucial as it forms the foundation for all subsequent steps in induction.