Use the Fundamental Theorem of Arithmetic and the Division Algorithm to prove that there are infinitely many prime numbers. Note: we will offer two different proofs. Proof #1 (in the style of Euclid, c. 200 b.c.e.). First, we will rephrase the claim: Prove that for every natural number n, there are more than n prime numbers. Proof Suppose we have any collection of n prime numbers. We will prove that there is at least one more prime number that isn’t in our collection. Let our n prime numbers be enumerated: p1, p2, p3, …, pn. Let K = (p1 × p2 × p3 ×… × pn) + 1. The natural number K is either prime or composite. Suppose K is prime. Since K is clearly not equal to any of our primes p1, p2, … , pn, (it is greater than each of them) we have found another prime that wasn’t in our collection of n primes, so the number of primes is greater than n. On the other hand, suppose K is composite. According to the Fundamental Theorem of Arithmetic, K must have at least one prime factor. Note that none of our primes p1, p2, …, pn is a factor of K, because, according to the Division Algorithm, each of p1, p2, ..,pn leaves a remainder of 1* when divided into K. Since K must have a prime factor, this means that there exists at least one prime that wasn’t in our collection of n primes, therefore the number of primes is greater than n. Since we have proven that the number of primes is greater than every natural number n, we have proven that there are infinitely many primes. *For example, K = p1q + 1 where q = p2 × p3 ×… × pn. Second Proof (proof by Contradiction). Assume that there are only finitely many prime numbers. Enumerate them and let the complete P set of prime numbers be P = {p1, p2, p3, …, pn} Let K = (p1 × p2 × p3 ×… × pn) + 1. The natural number K is either prime or composite. Case 1: Suppose K is prime. Since K is clearly greater than any of the primes in set P, we have found a prime that wasn’t included in P. This contradicts the assumption that P contains every prime. Case 2: Suppose K is composite. According to the Fundamental Theorem of Arithmetic, K must have at least one prime factor. Note that none of the primes in P is a factor of K, because, according to the Division Algorithm, each of p1, p2, ..,pn leaves a remainder of 1 when divided into K. Since K must have a prime factor, this means that there exists at least one prime that wasn’t in set P, which contradicts the assumption that P contains every prime.