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 naturalnumber 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.