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 βarithmeticβ, 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.
Theorem 6.3.2. Fundamental Theorem of Arithmetic.
The following are true:
Every integer
has a prime factorization.Every such factorization of a given
is the same if you put the prime factors in nondecreasing order (uniqueness).
More formally, we can say the following. Any positive integer
of primes, and further, if we can write a different such product
then
Proof.
We will prove this in Subsection 6.3.2.
Example 6.3.3.
For instance:
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
Example 6.3.4.
Usually we will implicitly assume the primes are in nondecreasing order, and write
Sometimes when the context is clear, one can even write
Using the same numbers as in the previous example:
Example 6.3.5.
Just to get this down, practice writing the following as a product of such prime powers.
See Exercise 6.6.14.
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 βlemmasβ 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.
Lemma 6.3.6.
If a prime
Proof.
Left to reader in Exercise 6.6.3; this is very closely related to Proposition 2.4.10.
Corollary 6.3.7.
If a prime
Proof.
By induction, left to reader in Exercise 6.6.4.
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):
Now let's look at another possible representation, possibly with different primes:
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.