Theorem. (Fundamental Theorem of Arithmetic) Every integer greater than 1 can be written in the form
where and the 's are distinct primes. The factorization is unique, except possibly for the order of the factors.
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.
Lemma. If and , then .
Now and (since ), so .
Lemma. 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. Do the case first. Suppose , and suppose . I must show .
, and p is prime, so or . If , then , which contradicts . Therefore, . By the preceding lemma, . 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.
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.
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 .
Look at . 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).
Example. 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 .
Here's an interesting fact that is easy to derive from the Fundamental Theorem:
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,
Here's how this result looks for 36 and 90:
, , and .
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