Proof Composer

To provide a detailed example of a proof in number theory, let's consider a classic theorem: the Fundamental Theorem of Arithmetic. This theorem states that every integer greater than 1 is either a prime number or can be uniquely expressed as a product of prime numbers, ordered in a non-decreasing sequence. Theorem: Fundamental Theorem of Arithmetic. Statement: Every integer greater than 1 is either a prime number or can be expressed as a product of prime numbers, and this factorization is unique, apart from the order of the factors. Proof 1. Existence of Prime Factorization We first prove that every integer greater than 1 can be factored into primes. This is done using the principle of mathematical induction. - Base Case: The smallest integer greater than 1 is 2, which is a prime number. Hence, it trivially satisfies the theorem. - Inductive Step: Assume that every integer greater than 1 and less than \( n \) (where \( n > 2 \)) can be factored into primes. We need to show that \( n \) can also be factored into primes. - If \( n \) is prime, we are done. - If \( n \) is not prime, then \( n = ab \) where \( a, b < n \) and greater than 1. By the induction hypothesis, both \( a \) and \( b \) can be expressed as a product of primes. Therefore, \( n \) can also be expressed as a product of primes. 2. Uniqueness of Prime Factorization Next, we prove that this factorization is unique. We use proof by contradiction. - Assume that there are two distinct prime factorizations of \( n \). Let's call them \( p_1p_2...p_r \) and \( q_1q_2...q_s \), where \( p_i \) and \( q_j \) are prime numbers, and \( r \neq s \) or there exists some \( i \) such that \( p_i \neq q_i \). - Without loss of generality, assume that \( p_1 \) is not in the \( q \) sequence. Since \( p_1 \) divides \( n \) and \( n = q_1q_2...q_s \), \( p_1 \) must divide one of the \( q_i \)'s (by Euclid's lemma). But since \( q_i \)'s are also prime, this is only possible if \( p_1 = q_i \), which contradicts our assumption. - Therefore, the assumption that there are two distinct prime factorizations of \( n \) leads to a contradiction, implying that the factorization must be unique. This completes the proof of the Fundamental Theorem of Arithmetic. We have shown that every integer greater than 1 can be factored into primes (existence) and that this factorization is unique (uniqueness).