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I need to prove the existence part of unique factorization of integers (into primes) using the well-ordering principle.

Now, the couple of examples I've seen of proofs of this sort involve coming up with a non-empty set of integers, which is then declared to have a least element. Then we need to show that assuming the reverse of the desired property violates the assumption of the least element.

So I proceeded as follows:

Consider the set $S$ of all factors ($>1$) of the given number $n$. That is, $S = \{f_1, f_2, \ldots, f_n\}$. This set is non-empty because there is at least one element (the number itself). Also, let $f_1$ be the least element, by the well-ordering principle.

It is easy to show that $f_1$ is prime, for if it is not, then it can be decomposed into smaller factors, resulting in a new least-element. But I'm stuck here. What to do next? I can recursively argue that the remaining factors are either themselves prime or product of primes (using the theorem that every integer, if not prime, is divisible by a prime), but then I don't think I'm following the well-ordering principle anymore.

Please help.

ankush981
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    Here's a hint: suppose that not every integer has a prime factorization, i.e. that the set of positive integers lacking such a factorization is nonempty. – Kevin Carlson Dec 16 '13 at 00:27
  • To be clear, you prove that every integer has a prime factorization using the well-ordering principle, uniqueness is a different issue that you should settle afterward. – Dylan Yott Dec 16 '13 at 00:34
  • @DylanYott That is exactly what the question is about. Do you mean to say my approach is trying to tackle the uniqueness? – ankush981 Dec 16 '13 at 00:39
  • Oops, I should avoid reading things too quickly, you're exactly right. – Dylan Yott Dec 16 '13 at 00:41
  • @KevinCarlson Aha! Maybe this is the hook I was missing. Such a set has no least element, because its least element can always be decomposed into smaller factors. Is this right? (I feel it is). – ankush981 Dec 16 '13 at 00:41
  • @DylanYott No problem. Happens sometimes. :) – ankush981 Dec 16 '13 at 00:42
  • Yep, that sounds like the idea. – Kevin Carlson Dec 16 '13 at 01:01
  • @KevinCarlson Thank you! – ankush981 Dec 16 '13 at 01:24
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    @KevinCarlson Ah yes, UFT and WOP - surely this is to be done in one's sleep right? ;-) (Not sure if you remember me from Ross) – tc1729 Dec 16 '13 at 14:57
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    Hi Siddharth, certainly remember you! I was actually trying to figure out whether this could be Ross-related, as I don't know how common it is to prefer WOP over induction elsewhere. – Kevin Carlson Dec 17 '13 at 01:28

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Hint $\ $ Suppose for contradiction that some natural $> 1$ hs no prime factorization, and let $n$ be minimal such. Then $ n$ is not prime, so it has a proper factorization $\ n = ab.\ $ Since $a,b < n,\, $ what can you deduce about $\,a,b\,$ from the minimality hypothesis on $\,n\,$? $ $ Finally, what does the prior deduction imply about $\,n,\,$ given that is the product of $\,a\,$ and $\,b\,$?

Bill Dubuque
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