Question

Take the following statement as given: If \(p\) is a prime and \(a\) and \(b\) are integers such that \(p\) divides the product \(a b\), then \(p\) divides \(a\) or \(b\). (a) Prove: If \(p, p_{1}, \ldots, p_{k}\) are positive primes and \(p\) divides the product \(p_{1} \cdots p_{k}\), then \(p=p_{i}\) for some \(i\) in \(\\{1, \ldots, k\\}\). (b) Let \(n\) be an integer \(>1\). Show that the prime factorization of \(n\) found in Example 1.2 .7 is unique in the following sense: If $$ n=p_{1} \cdots p_{r} \quad \text { and } \quad n=q_{1} q_{2} \cdots q_{s} $$ where \(p_{1}, \ldots, p_{r}, q_{1}, \ldots, q_{s}\) are positive primes, then \(r=s\) and \(\left\\{q_{1}, \ldots, q_{r}\right\\}\) is a permutation of \(\left\\{p_{1}, \ldots, p_{r}\right\\}\)

Short answer

Question: Prove the statement: If p is a prime number and a and b are integers such that p divides the product ab, then p divides a or b. Also, show that this statement implies the uniqueness of the prime factorization of n > 1. Answer: To prove the given statement we consider two parts. (a) If p, p1, ..., pk are prime numbers and p divides the product p1...pk, then p = pi for some i in the range {1, ..., k}. We prove this by contradiction. Suppose p is not equal to any pi for i in {1, ..., k}. But if p divides the product p1...pk, then p must divide any of the prime factors pi for i in {1, ..., k}. This leads to a contradiction, and hence, p = pi for some i in {1, ..., k}. (b) To prove the uniqueness of the prime factorization of n > 1, let n have two different prime factorizations, n = p1...pr and n = q1...qs, where p1, ..., pr, q1, ..., qs are positive prime numbers. We show that r = s and {p1, ..., pr} is a permutation of {q1, ..., qs}. Using part (a), p1 must divide at least one factor qi in the range {1, ..., s}. Assuming p1 divides q1 and p1 = q1, we now consider the product p2...pr and q2...qs, which are still equal. Repeating this argument for all prime factors p_i, we find that p_i = q_i, and their corresponding products of remaining primes will be equal. At the end, we prove that r = s and {p1, ..., pr} is a permutation of {q1, ..., qs}. This implies the uniqueness of the prime factorization of n > 1.

Step-by-step solution

Part (a) - Proof of p equal to pi for some i

We will prove the given statement by contradiction. Let's suppose that p is not equal to any pi for i in {1, ..., k}. According to the given statement, if p divides the product p1...pk, then p must divide any of the prime factors pi for i in {1, ..., k}. But we assumed that p is not equal to any pi for i in {1, ..., k}. This leads to a contradiction. Hence, our assumption is wrong, and therefore, p = pi for some i in the range {1, ..., k}.

Part (b) - Proof of uniqueness of prime factorization

Let n have two different prime factorizations, n = p1...pr and n = q1...qs, where p1, ..., pr, q1, ..., qs are positive prime numbers. We will prove that r = s and {p1, ..., pr} is a permutation of {q1, ..., qs}. As p1 is a prime number, and p1 divides the product q1...qs (because n = p1...pr = q1...qs), according to the given statement, p1 must divide at least one factor qi in the range {1, ..., s}. Without loss of generality, assume p1 divides q1. Since q1 is a prime number, and p1 divides q1, then p1 = q1. Now, let's consider the product p2...pr and q2...qs. Since p1 = q1, we can divide both products by p1, and we still have two equal products of primes: (p2...pr) = (q2...qs). We can repeat this argument for all prime factors p_i. In each step, we will find that p_i = q_i, and their corresponding products of remaining primes will be equal. At the end, we will have proved that r = s and {p1, ..., pr} is a permutation of {q1, ..., qs}.

Key concepts

Proof by Contradiction

Understanding the concept of proof by contradiction is essential in grasping many mathematical theorems. Let's simplify the concept. Imagine asserting something is true, but for the sake of argument, you claim the opposite. Then, you proceed to show that this contrary claim leads to an impossible situation, a contradiction. This means your initial assertion must be true because assuming the opposite created an unreasonable outcome.

For example, let's say you claim that a number is the smallest positive integer, but to argue by contradiction, you assume that there is, in fact, a smaller positive integer. Through logical steps, you would find this smaller number contradicts the very definition of 'smallest positive integer', forcing you to conclude that your original number must indeed be the smallest. This simple illustration underpins many complex proofs in mathematics, including the proof that primes have a unique factorization, as showcased in the given exercise.

In the exercise, the assumption was that a prime number p does not equal any prime factors p_i in a product of primes. However, the nature of primes and divisibility leads to a logical impasse, proving by contradiction that p must, in fact, equal one of the p_i's.

Unique Prime Factorization Theorem

The unique prime factorization theorem, also known as the Fundamental Theorem of Arithmetic, is a cornerstone in the study of number theory. It states that every integer greater than 1 can be represented in exactly one way as a product of prime numbers, up to the order of the factors. This ensures that the prime factorization of a number is unique, making primes the basic building blocks of all integers.

Let's take an example. Consider the number 30. The prime factorization of 30 is 2 × 3 × 5. No matter how you attempt to factor 30 into primes, you'll always end up with these prime numbers, although their order may differ. In your exercise, two scenarios for the number n show different sets of prime factors. The exercise then demonstrates that these sets must be permutations of each other, upholding the tenets of the theorem.

Using fundamental logic and applying the principles of divisibility, we can conclude that other supposed factorizations would either repeat primes or introduce non-prime factors, which is not permissible. This theorem is not just a mathematical curiosity; it's a powerful tool used to solve various problems regarding divisibility, greatest common divisors, and more.

Divisibility in Prime Numbers

Delving into the concept of divisibility in prime numbers can illuminate why the unique prime factorization of numbers is possible. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. For instance, the numbers 2, 3, 5, and 7 are all prime because they can only be evenly divided by 1 and their own value.

In the context of our exercise, if a prime number p divides the product of a series of primes, this indicates that p must divide at least one of those prime numbers in the series. Since prime numbers don't have any divisors other than 1 and themselves, the only logical conclusion is that p must be one of the primes in that product.

This property of prime numbers being divisible only by 1 and themselves is what makes them such fundamental components of number theory, and is why they play a starring role in proving the uniqueness of prime factorization. The notion of divisibility within primes is highly intuitive once you grasp that primes are the 'atoms' of the number system, indivisible and fundamental.