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!)

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.

1

1n n

0k

kr r11

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.