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Section 6.3 The Fundamental Theorem of Arithmetic

Subsection 6.3.1 Preliminaries and statement

Our biggest goal for this chapter, and the motive for introducing primes at this point, is the Fundamental Theorem of Arithmetic, or FTA. It should probably be called the Fundamental Theorem of Number Theory, but in older usage one said “arithmetic”, and the name has stuck.

Definition 6.3.1.

A factorization of an integer is a way of writing it as a product of integers. This nearly always refers to one of two things, which are mentioned explicitly if there is danger of ambiguity:
  • A product of prime numbers is called a prime factorization.
  • A product into positive powers of (distinct) primes is called a prime power factorization.


Example 6.3.3.

For instance:
  • \(\displaystyle 30=2\cdot 3\cdot 5\)
  • \(\displaystyle 24=2\cdot 3\cdot 2\cdot 2=2\cdot 2\cdot 2\cdot 3\)
Clearly (from normal experience) the only possible factorizations than these would just put the primes in a different order. Why doesn’t this work for \(N=1\text{?}\) (See Exercise 6.6.24.) As it happens, Euclid did not even consider one to be a number in the same sense as the others; see Joyce’s commentary

Example 6.3.4.

Usually we will implicitly assume the primes are in nondecreasing order, and write \(3^2\) instead of \(3\cdot 3\) (with the primes now necessarily in increasing order), so the following notation is common to express a prime power factorization:
\begin{equation*} N=\prod_{i=1}^n p_i^{e_i}\text{.} \end{equation*}
Sometimes when the context is clear, one can even write \(N = \prod p\) or \(N = \prod p^e\text{.}\)
Using the same numbers as in the previous example:
  • \(\displaystyle 30=2^1\cdot 3^1\cdot 5^1\)
  • \(\displaystyle 24=2^3\cdot 3^1\)

Example 6.3.5.

Just to get this down, practice writing the following as a product of such prime powers.
  • \(\displaystyle N=12100\)
  • \(\displaystyle N=1250\)
  • \(\displaystyle N=3072\)

Subsection 6.3.2 Proof of the FTA

This theorem is quite old, and of course Euclid has a nice proof of it
, along with various lemmata (the plural of lemma
, though I’ll also use “lemmas” in this text) that he needs to get there. The key ingredients are:
  • If a number is prime, that is the prime factorization.
  • If a number is composite, then it is by definition divisible by some smaller positive number other than one. (Euclid used the stronger fact that it is divisible by a prime in his proof of Infinitude of Primes.)
  • This process can be continued, but only finitely often.
  • Any other way in which you can write the same number as a product of primes is just a reordering of the one obtained in the previous step.
The last step requires the following lemma, which is Euclid’s Book 7, Proposition 30



Okay, now we need the details.

Proof of Theorem 6.3.2.

Let’s use induction on the size of \(N\text{.}\) So our base case is \(N=2\text{,}\) which is of course prime so it has (the) unique factorization \(2^1\text{.}\)
For the induction step, first suppose we have proved that all numbers up to \(N\) can be written as a product of primes (uniquely or not). Then we look at \(N+1\) to continue the induction.
  • If \(N+1\) is prime, that is its prime factorization, as with \(2\text{.}\)
  • If not, then by definition \(N+1\) is composite, so \(N+1=ab\text{,}\) where \(1\lt a,b\lt N+1\text{.}\) (Note why \(a,b\) are smaller! Recall the proof of the Sieve 6.2.3.) In this case, by the induction hypothesis, \(a\) and \(b\) have prime decompositions \(\prod p_i\) and \(\prod q_j\text{,}\) since they are less than \(N+1\) but not 1, and so \(N+1=\prod p_i \prod q_j\text{.}\)
By induction, this shows that a prime factorization exists for all numbers up to \(N+1\text{.}\) It remains to be shown that such a factorization is unique.
So first rewrite our factorization in a given order (such as nondecreasing):
\begin{equation*} N+1=\prod p_i,\; p_i\leq p_{i+1}\text{.} \end{equation*}
Now let’s look at another possible representation, possibly with different primes:
\begin{equation*} N+1=\prod q_j\text{.} \end{equation*}
At this point we need Corollary 6.3.7. By assumption, \(p_1\) divides \(N+1\text{.}\) Hence, by the corollary, \(p_1\) divides at least one of the \(q_j\text{.}\) But the only positive divisors of a prime are itself and 1, and \(p_1\) is prime (not one), so \(p_1=q_j\text{.}\)
Cancel these from both products to get two different representations of (the integer) \(\frac{N+1}{p_1}\) as a product of primes. By the induction hypothesis, since this number is less than \(N+1\text{,}\) these representations are unique up to reordering, so multiplying both by \(p_1\) to get \(N+1\) must also be unique up to reordering.
By induction, we are done.
Two comments about this proof are in order. First, some students may wonder about this induction proof, because in the induction hypothesis we do not simply assume a single \(N\) has a (unique) factorization (as in Example 1.2.4), but that all \(n\leq N\) do. But this is just an artifact
In my view, there is no pedagogical need for a separate notion of ‘strong induction’.
of the statement we are proving. The logical induction statement is not that “\(N\) has a factorization”, but that “all numbers less than \(N\) have a factorization”.
Second, if you are familiar with other algebraic structures, it is very important to note that other algebraic systems may not have unique factorization into primes, or even have a notion of prime elements! Even some structures very similar to the integers fail this; many interesting examples of this are just beyond the level of this course. For those who must know what this means now, try Exercise 6.6.30.