Primes - ECE 3250 PRIME NUMBERS AND FACTORIZATION Fall 2008...

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ECE 3250 PRIME NUMBERS AND FACTORIZATION Fall 2008 I think you all know what a prime number is, but let’s start from the beginning. Given two natural numbers a and b , we say that a is a divisor of b if and only if b = am for some natural number m . The standard notation for “ a is a divisor of b ” is a | b . Often we just say “ a divides b ” for short. Observe that every a N divides 0 and that 1 divides every a N . Note also that if a | b and b | c , then a | c . A natural number p is prime if and only if the only natural-number divisors of p are 1 and p itself. By convention, 1 is not a prime number even though it satisfies the technical definition. The first few prime numbers are 2, 3, 5, 7, 11, and 13. Note that 2 is the only even prime number. Can you see why? In what follows, I’ll be using induction a fair amount to prove things. It takes some practice to get the hang of using inductive arguments, but they’re extremely useful tools for proving a lot of facts about the natural numbers. Here’s an example of a typical such argument. I’ll prove that every a N , a > 1, has at least one prime divisor. You start with the base case a = 2: clearly, 2 has a prime divisor (2 itself). Then you move on to the induction step: suppose we have shown that every a n has at least one prime divisor. Consider a = n + 1. If n + 1 is prime, we’re done, since n + 1 is then a prime divisor of itself. If n + 1 is not prime, then we can write n + 1 = ab for some natural numbers a and b with 1 < a, b n . But since we’ve shown already that every such a and b has at least one prime divisor, and since a and b here are both divisors of n + 1, we conclude that n + 1 must have at least one prime divisor. I hope you see how the induction works. We know that the theorem is true for n = 2 by the base case. What about n = 3? It’s true for n = 2, and the induction step shows that if it’s true for n = 2, then it’s true for n = 3; hence it’s also true for n = 3. What about n = 4? We know now that it’s true for n = 2 and n = 3, and the induction step enables us to conclude that it’s also true for n = 4. And so the dominoes fall. Euclid used the result we just proved to demonstrate that there are infinitely many prime numbers. His argument proceeds as follows. Suppose that we have a list of K primes. Index them as p 1 , p 2 , . . . , p K . Consider the number R = 1 + p 1 p 2 p 3 · · · p K . Our “theorem” above guarantees that R has at least one prime divisor p , and p could not possibly be among the p j on our list. If it were on the list, then p | R would imply that p | ( R - p 1 p 2 p 3 · · · p K ) i . e . p | 1 , which is impossible. So p is not on our list. In particular, our list doesn’t contain every prime. More trenchantly, nothing about our list is special, so no finite list of primes can be exhaustive. In other words, infinitely many primes exist.
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