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 . // † † † †