The Fundamental Theorem of Arithmetic Every
integer n!> 1 is uniquely representable as a product of
primes. That is, for every n > 1, there is a unique set of
primes p1, p2 ,K, pk and a unique collection of positive
integers e1,e2 ,K,ek so that n =
p1e1 p 2e2 L p kek
k
= ’ piei .
i=1
†
Proof First we show the existence of a factorization for
any
† integer greater than 1. For if not, there must be a
† m. Necessarily, m is not prime
least counterexample
(as such a number certainly has a prime factorization),
so there exist integers a, b for which m = ab and
1 < a,b < m. But both a and b, being less than m, must
have prime factorizations. As their product provides a
prime factorization of m, we have a contradicton.
†
†
Next we show the uniqueness of the factorization of any
such n. Suppose that n has two prime factorizations:
k
n=’
i=1
piei
l
d
= ’qjj ,
j=1
and is the smallest integer with this property. Then all
the primes that appear in both factorizations (the p’s
and q’s) are†distinct, else we could cancel common
factors and obtain a smaller positive integer having two
prime factorizations. Choose the notation so that p1
and q1 are the smallest primes in their respective
factorizations, and assume that p1 < q1. Then it is easy
to see that neither k nor l can equal 1 (for†then we find
†
†
†
†
that a single prime is divisible by, hence greater than,
some other primes, which contradicts the definition of
prime). Thus, p1 and q1 are nontrivial factors of n,
whence p1,q1 £ n . So p1q1 £ n . It follows that both p1
and q1 are factors of m = n - p1q1 , hence we can write
m = ap†
1 for some
† positive integer a. Since q1 | ap1 , and q1
†does not divide†p1, we must have q1 | a. If we
† write
chosen integer b, then m = bp1q1 ,
† a = bq1 for suitably
†
whence n = (b +1) p1q1 and p1q1 | n =†p1e1 p2e2 L pkek†. But
then we
| p1e1 p2e2 L p ekk . Again, as q1 does
† conclude that q1 †
not divide p1, this reduces to q1 | p†2e2 L pkek . But q1 is
†different from all†the other p’s, hence it divides none of
them, contradicting
the fact that if a prime
† divides a
†
product, it must divide one of the factors. //
†
†
†
Corollary
2 is irrational. //
k
Corollary If m | n and n has prime factorization ’ piei ,
i=1
†
k
then m must have prime factorization of the form ’ pidi
where†0 £ di £ ei for each i = 1, 2, …, k. //
i=1
†
k
Corollary If n has prime factorization ’†piei , then the
†
i=1
k
number of positive divisors of n is n (n) = ’ (ei +1) . //
i=1
†
†
†
Proposition The number of postive multiples of m
which are less than or equal to n ( m,n Œ N) is Îmn ˚.
n . Since k is an integer,
Proof If km £ n, then k £ m
k £ Îmn ˚. This is true for all†values of k†up to the largest
multiple of m less than or equal to n. But this counts
precisely
how †
many postive multiples of m there are in
†
this range. //
k
Corollary If n! has prime factorization ’ piei , then
i=1
•
Í ˙
ei = Â Í nk ˙.
k=1Î pi ˚
†
Proof There are Î pni ˚ multiples of pi which are ≤ n, each
of which contributes
a factor to the prime factorization
†
of n!. But pi occurs a second time as a factor in each of
2
the multiples
of
p
are ≤ n, and these account for
i which†
†
Í ˙
an additional Í n2 ˙ occurrences of pi in the prime
Î pi ˚
†
factorization of n!. There are other occurrences of pi in
†
the factorization of n! due to small multiples of pi3, and
Ín˙ †
†
these account for Í 3 ˙ occurrences of pi in the prime
Î pi ˚
†
factorization of n!. And so on similarly for higher
†
Í ˙
k
powers of pi . Eventually, when pi > n, the terms Í nk ˙
Î pi ˚
†
†
equal 0, for this and all higher powers of pi . //
†
†
†
†