Prime Numbers (3/17 )

1. Prime Numbers (3/17 ) • We all know what a prime number is. • Theorem (Euclid). There are infinitely many primes. • Euclid’s original proof idea can be stated in various different ways. Our text’s way is fine. It shows that given any finite collection of primes, we can always produce a new one. Hence there are infinitely many. • So let {p1, p2, p3,..., pk} be any collection of primes and let A = p1p2p3...pk+ 1. If A is prime we are done since it’s clearly bigger that the ones in our list. If A is composite, than it’s divisible by a prime q. But q cannot be from our list, since if it were, it would divide both A and p1p2p3...pk, so it would divide their difference, which is 1. (Crossed swords!)

2. A Calculus Proof of This • Recall from Calculus II: • The harmonic series diverges(discuss). • If |r | < 1, the geometric series has value (discuss). • Now assume there are only finitely may primes{2, 3, 5, ..., pn}. Use the two facts above to reach a contradiction.

3. Primes in Arithmetic Progression • Question: Given a modulus m and a remainder b, are there infinitely many primes which are congruent to b (mod m). That is, are there infinitely many primes of the form mk+ b? • Example: Prove that there are infinitely many primes of the form 4k + 3 (i.e., the case m = 4 and b = 3 is okay). • It turns out to be quite a bit harder to proof the case 4k + 1. • Note that Euclid’s Theorem is a special case of this question. For what m and what b ? • Dirichlet’s Theorem. If GCD(m, b) = 1, then there are infinitely many primes congruent to b (mod m). • Look over test, read Chapter 12 and do Ex 12.1 & 12.2.