NTIC The Fundamental Theorem of Arithmetic

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.

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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.

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Theorem 6.3.2. Fundamental Theorem of Arithmetic.

The following are true:

Every integer $n\gt 1$ has a prime factorization.
Every such factorization of a given $n$ is the same if you put the prime factors in nondecreasing order (uniqueness).

More formally, we can say the following. Any positive integer $N\gt 1$ may be written as a product

$\begin{equation*} N=\prod_{i=1}^n p_i \end{equation*}$

of primes, and further, if we can write a different such product

$\begin{equation*} N=\prod_{j=1}^m q_m \end{equation*}$

then $m=n$ and a reordering of the $q_j$ will make them the same as the $p_i\text{.}$

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Proof.

We will prove this in Subsection 6.3.2.

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Example 6.3.3.

For instance:

$30=2\cdot 3\cdot 5$
$24=2\cdot 3\cdot 2\cdot 2=2\cdot 2\cdot 2\cdot 3$

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.)

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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:

$30=2^1\cdot 3^1\cdot 5^1$
$24=2^3\cdot 3^1$

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Example 6.3.5.

Just to get this down, practice writing the following as a product of such prime powers.

$N=12100$
$N=1250$
$N=3072$

See Exercise 6.6.14.

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Subsection 6.3.2 Proof of the FTA

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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 number. (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.

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The last step requires the following lemma, which is Euclid's Book 7, Proposition 30.

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Lemma 6.3.6.

If a prime $p$ divides a product $ab\text{,}$ then $p$ divides at least one of $a$ or $b\text{.}$

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Proof.

Left to reader in Exercise 6.6.3; this is very closely related to Proposition 2.4.9.

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Corollary 6.3.7.

If a prime $p$ divides a finite product of primes, then $p$ divides at least one of them, i.e.

$\begin{equation*} p\Bigg\vert \prod_{k=1}^\ell a_k\text{ implies }p\mid a_k\text{ for at least one }k \end{equation*}$

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Proof.

By induction, left to reader in Exercise 6.6.4.

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Okay, now we need the details.

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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 induction we know $N+1$ 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 6.2.2.) In this case, $a$ 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 nondecreasing order:

\begin{equation*} N+1=\prod p_i \end{equation*}

Now suppose that (once again by induction) we have written all numbers up to $N$ uniquely as a product of primes. So let's look at another such representation,

\begin{equation*} N+1=\prod q_j \end{equation*}

At this point we need Corollary 6.3.7. By definition, $p_1$ divides $N+1\text{.}$ Hence $p_1$ divides at least one of the $q_j\text{.}$ But the only positive divisors of a prime are itself and 1, so $p_1=q_j\text{.}$

Cancel these from the product to get two different representations of (the integer) $\frac{N+1}{p_1}$ as a product of primes. By the induction hypothesis, these are unique up to reordering, so multiplying both by $p_1$ to get $N+1$ must be unique up to reordering.

By induction, we are done.

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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 artifact²In 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”.

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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.