* Theorem.* * (Fundamental
Theorem of Arithmetic)* Every integer greater than 1 can be
written in the form

In this product, and the 's are distinct primes. The factorization is unique, except possibly for the order of the factors.

For instance,

I need a couple of lemmas in order to prove the uniqueness part of the Fundamental Theorem. In fact, these lemmas are useful in their own right.

* Theorem.* If and , then .

* Proof.* Write

Then

Now and (since ), so .

* Theorem.* If p is prime and , then for some i.

For , the result says that if p is prime and
, then or . This is often called * Euclid's
lemma*.

* Proof.* The result is trivial if , since it says that if , then .

Do the case first. Suppose , and suppose . I must show .

, and p is prime, so or . If , then , which contradicts . Therefore, . By the preceding theorem, . This establishes the result for .

Assume , and assume the result is true when p divides a product of with less than n factors. Suppose that . Grouping the terms, I have

By the case , either or . If , I'm done. Otherwise, if , then p divides one of , , ..., , by induction. In either case, I've shown that p divides one of the 's, which completes the induction step and the proof.

Using these results, I'll prove the Fundamental Theorem of Arithmetic.

* Proof.* (* Fundamental Theorem
of Arithmetic*) First, I'll use induction to show that every
integer greater than 1 can be expressed as a product of primes.

is prime, so the result is true for .

Suppose , and assume every number less than n can be factored into a product of primes. If n is prime, I'm done. Otherwise, n is composite, so I can factor n as , where . By induction, a and b can be factored into primes. Then shows that n can, too.

Now I'll prove the uniqueness part of the Fundamental Theorem.

Suppose that

Here the p's are distinct primes, the q's are distinct primes, and all the exponents are greater than or equal to 1. I want to show that , and that each is for some b --- that is, and .

Consider . It divides the left side, so it divides the right side. By the last lemma, for some i. But is ( times), so again by the last lemma, . Since and are prime, .

To avoid a mess, renumber the q's so becomes and vice versa. Thus, , and the equation reads

If , cancel from both sides, leaving

This is impossible, since now divides the left side, but not the right.

For the same reason is impossible.

It follows that . So I can cancel the 's off both sides, leaving

Keep going. At each stage, I pair up a power of a p with a power of a q, and the preceding argument shows the powers are equal. I can't wind up with any primes left over at the end, or else I'd have a product of primes equal to 1. So everything must have paired up, and the original factorizations were the same (except possibly for the order of the factors).

Recall that the * least common multiple* of
nonzero integers a and b is the smallest positive integer divisible
by both a and b. The least common multiple of a and b is denoted .

For example,

Here's an interesting fact that is easy to derive from the Fundamental Theorem.

* Proposition.* If a and b are nonzero integers,
then .

* Proof.* (sketch) Factor a and b in products of
primes, but write out all the powers (e.g. write as ):

Here the q's are the primes a and b have in common, and the p's and r don't overlap. Picture:

From the picture,

Thus, .

You might see if you can prove this proposition without using the Fundamental Theorem (that is, without factoring a and b into products of primes.

Here's how this result looks for 36 and 90:

We have and ; notice that .

Copyright 2019 by Bruce Ikenaga