Concrete mathematics : a foundation for computer science

106 NUMBER THEORY
product of primes,
n = p,...pm = fiPk, Pl 6 .‘. 6 Pm. (4.10)
k=l
For example, 12=2.2.3; 11011 =7.11.11.13; 11111 =41.271. (Products
denoted by n are analogous to sums denoted by t, as explained in exercise
2.25. If m = 0, we consider this to be an empty product, whose value
is 1 by definition; that’s the way n = 1 gets represented by (4.10).) Such a
factorization is always possible because if n > 1 is not prime it has a divisor
nl such that 1 < nl < n; thus we can write n = nl .nz, and (by induction)
we know that nl and n2 can be written as products of primes.
Moreover, the expansion in (4.10) is unique: There’s only one way to
write n as a product of primes in nondecreasing order. This statement is
called the Fundamental Theorem of Arithmetic, and it seems so obvious that
we might wonder why it needs to be proved. How could there be two different
sets of primes with the same product? Well, there can’t, but the reason isn’t
simply “by definition of prime numbers!’ For example, if we consider the set
of all real numbers of the form m + nm when m and n are integers, the
product of any two such numbers is again of the same form, and we can call
such a number “prime” if it can’t be factored in a nontrivial way. The number
6 has two representations, 2.3 = (4 + &8 j(4 - fi 1; yet exercise 36 shows
that 2, 3, 4 + m, and 4 - m are all “prime” in this system.
Therefore we should prove rigorously that (4.10) is unique. There is
certainly only one possibility when n = 1, since the product must be empty
in that case; so let’s suppose that n > 1 and that all smaller numbers factor
uniquely. Suppose we have two factorizations
n = p, . . *Pm = ql...qk, Pl