Download As discussed earlier: Definition An integer p > 1 is

Prime numbers
As discussed earlier:
Definition
An integer p > 1 is called prime if the only positive divisors of p
are 1 and p. An integer n > 1 is called composite if it is not prime.
The integer n = 1 is neither prime nor composite; it is a unit.
Prime numbers have been studied for millenia, starting with the
ancient Greeks, and continuing today. Finding lists of primes,
determining whether a number is prime, finding patterns in the
primes, are all questions with both ancient and modern interest.
Can find primes using the Sieve of Eratosthenes. See the
book.
The Prime number theorem answers the question, if n is a
large integer, how many of the first n integers are primes?
Factoring large numbers is relevant in cryptography.
Prime numbers
Theorem (Fundamental theorem of arithmetic)
Every positive integer greater than 1 can be written as a product
of prime numbers. This expression is unique, up to a reordering of
the factors.
Example
60 = 2 × 2 × 3 × 5 = 3 × 2 × 5 × 2.
Proof.
First we deal with existence of prime factorizations. This part of
the proof is by strong induction. The base case: any prime number
p satisfies the statement: p is a product of a single prime.
Prime numbers
Proof, continued.
For the inductive step, suppose that for some integer k ≥ 2, for all
n with 2 ≤ n ≤ k, the integer n can be written as a product of
primes. We want to show that k + 1 can be written as a product
of primes.
There are two cases: if k + 1 is a prime, then we’re done: it’s a
product of a single prime. Otherwise k + 1 is composite: k + 1 can
be factored as k + 1 = a × b with 2 ≤ a ≤ b ≤ k. By the inductive
hypothesis, both a and b can be written as products of primes.
Therefore their product can be so written.
This finishes the “existence” part of the proof: every integer
greater than 1 can be written as a product of primes.
Prime numbers
To prove uniqueness of factorizations, we need a lemma.
Lemma
Suppose that p is a prime number and p|a1 a2 . . . am . Then p|ai for
some i.
Proof.
The proof is by induction on m. If m = 1, it’s clear.
Suppose that for some k ≥ 1, the result is true when m = k. We
want to prove it when m = k + 1, so suppose that p|a1 a2 . . . ak+1 .
Then p|(a1 a2 . . . ak )ak+1 , so by an earlier result, either
p|(a1 a2 . . . ak ) or p|ak+1 . In the first case, the inductive hypothesis
tells us that p|ai for some i ≤ k. In the second case, p|ak+1 . In
either case, the inductive step is completed, and this finishes the
proof.
Prime numbers
Proof of the fundamental theorem of arithmetic, continued.
Now we prove uniqueness. Suppose that
k = p1 p2 . . . pm = q1 q2 . . . qn ,
where each pi and each qj is prime, and suppose that
p1 ≤ p2 ≤ · · · ≤ pm ,
q1 ≤ q2 ≤ · · · ≤ qn .
Then p1 divides k, so p1 |qi for some i. Since p1 and qi are prime,
this means that p1 = qi , so cancel them. Induction on m
completes the proof.
Prime numbers
Proposition
Suppose that a = p1r1 p2r2 . . . pnrn , where the pi ’s are distinct primes,
and each ri is a positive integer. Then b is a divisor of a if and
only if b = p1s1 p2s2 . . . pnsn , where 0 ≤ si ≤ ri for each i.
Proof.
If a and b are as in the proposition and if 0 ≤ si ≤ ri for each i,
then certainly b is a divisor of a:
a = b p1r1 −s1 p2r2 −s2 . . . pnrn −sn .
Since ri ≥ si for each i, each exponent here is non-negative, so the
term in parentheses is an integer.
Conversely, suppose that a is as in the proposition and a = bq for
some q. Factor b and q into prime factors; their factorization must
combine to give the factorization for a, so the only prime factors
which can appear in b (with possibly smaller exponents) are those
which appear in a.
Prime numbers
Corollary
Suppose that a and b are positive integers, and
a = p1r1 p2r2 . . . pnrn ,
b = p1s1 p2s2 . . . pnsn ,
with each ri and si non-negative. Then c is a common divisor of a
and b if and only if
c = p1m1 p2m2 . . . pnmn ,
where mi ≤ min(ri , si ).
In particular,
min(r1 ,s1 ) min(r2 ,s2 )
min(r ,s )
p2
. . . pn n n .
gcd(a, b) = p1
Prime numbers
Example
17238 = 2 × 3 × 132 × 17 = 21 × 31 × 50 × 132 × 171 ,
3315 = 3 × 5 × 13 × 17 = 20 × 31 × 51 × 131 × 171 ,
so
gcd(17238, 3315) = 20 × 31 × 50 × 131 × 171 = 3 × 13 × 17 = 663.
The previous corollary is not actually very practical most of the
time, because, as we’ve discussed, factoring integers into primes
cannot be done very efficiently.
Prime numbers
Theorem
√
2 is not a rational number: there is no rational number whose
square is 2.
Of course we’ve already proved this, but we can use the
fundamental theorem of arithmetic to give another proof.
Prime numbers
Proof.
Suppose, for the sake of contradiction, that there is a rational
number q so that q 2 = 2. Let q = a/b where a and b are integers,
and factor a and b into primes:
a = 2m1 p2m2 . . . prmr ,
b = 2n1 p2n2 . . . prnr .
We are assuming that (a/b)2 = 2, so a2 = 2b 2 . On the left side of
this equation, the exponent of 2 is 2m1 . On the right side, the
exponent of 2 is 2n1 + 1. By the uniqueness clause of the
fundamental theorem of arithmetic, these exponents must agree:
2m1 = 2n1 + 1. This can’t happen: an even number cannot equal
an
√ odd number. This is a contradiction, so our assumption that
2 ∈ Q must be false.
Prime numbers
You should now be able to supply at least two proofs of the
following.
Theorem
Let p be a prime number. Then there is no rational number whose
square is p.
(Good practice for the final exam. . . )
Prime numbers
Theorem
There are infinitely many prime numbers.
This proof is usually credited to Euclid (about 2300 years ago). It
is a classic example of proof by contradiction.
Proof.
Suppose that there were only finitely many primes, say
p1 , p2 , p3 , . . . , pn
for some integer n. Let m = p1 p2 p3 . . . pn + 1. When we divide m
by any pi , we get a remainder of 1: m is not divisible by any prime
pi . By the fundamental theorem of arithmetic, m is either prime
(and is not on our list) or factors as a product of primes (none of
which is on our list). In either case, our list of primes is not
complete, and this is a contradiction. This finishes the proof.
Prime numbers
Twin primes are primes whose difference is 2: 3 and 5, 5 and 7, 11
and 13, 59 and 61, . . .
Also dating from Euclid’s time:
The Twin Prime Conjecture
There are infinitely many pairs of twin primes.
Researchers today are still trying to prove this. About a week ago,
Yitang Zhang announced some progress on this problem. . .