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The interplay of Euclid's lemma and the Fundamental Theorem of Arithmetic has me confused. In the Wikipedia article on the Fundamental Theorem of Arithmetic we have that every integer greater than 1 can be represented uniquely as a product of prime numbers, up to the order of the factors: $n=p_1p_2…p_r$ where any $p_k$ is prime, and order doesn't matter. Good. But then later it gives Euclid's lemma towards the proof:

If a prime $p$ divides the product $ab$, then $p$ divides either $a$ or $b$ or both.

The proofs I've seen say that by strong induction, any composite number in the basic form $ab$ can be seen as $a=p_1,p_2…p_j$ and $b=p_1,p_2…p_k$, i.e., they are each the products of primes. This seems circular, i.e., we're proving FToA with itself, albeit broken down (mysteriously for me) with strong induction. I found this post which states:

If $n$ is not prime, then we can choose some natural numbers $a$ and $b$, such that $n=ab$, where $a<n$ and $b<n$. Thus, by the inductive hypothesis, each of $a$ and $b$ is either prime or a product of primes, and since $n=ab$, it follows that $n$ is a product of primes.

seems to be getting to the heart of the matter, but how does it happen? My problem is how exactly is strong induction accomplishing this? It would seem to a programmer that all you're really doing is some drill-down recursion. So yes, how does strong induction bring about the necessary drill-down recursion behavior to break apart $a$ and $b$ over and over until all factors are prime? I've read other posts supposedly answering this, but no they don't. EL and FToA work together on this trick, so I'd like it specifically dealt with, not just canceled as duplicate.

147pm
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  • What do you understand "strong induction" to mean? – Robert Shore Dec 27 '24 at 17:36
  • SI as I've seen it wants us to prove a next case $k+1$ by having proved all previous cases, not just the previous case. This is confusing -- as is this application that seems to magically do a drill-down recursion. I'm a programmer and if I wrote a program to create prime factors, I'd run through the primes starting with 2, recursively applying 2 until it failed. That must be what strong induction is doing somehow. -- only attacking the composite first as just two factors, then investigating if each of the two is itself composite -- over and over if necessary. Basically, how is this happening? – 147pm Dec 27 '24 at 17:42
  • See the 2nd & 3rd dupes for detailed presentations of the inductive proof of the existence of prime factorizations (which does not use Euclid's Lemma). – Bill Dubuque Dec 27 '24 at 18:37
  • PLease don't duplicate your prior questions. $\ \ $ – Bill Dubuque Dec 27 '24 at 19:46
  • @BillDubuque: I saw your answer in duplicate #2, and yes, in the proof of your Lemma, you apply Peano's successor idea to make $S \subset \mathbb {N}$ actually $S = \mathbb{N}$. From there I can't follow how you're applying strong induction to get $n = jk\in S$ or why this does anything. I'm just not seeing the "inductively (recursively) generated structure by piggybacking on its inductive generation" aspect of all this. I'll study your third dup response next... – 147pm Dec 27 '24 at 21:18
  • @BillDubuque: Yes, again in the third duplicate, I see you touch my problem: If $N$ is not a prime, then $N=mn$, where $2 \leq m, n<N$. By induction both $m$ and $n$ are products of primes and thus so is $N$. How by induction? This is, as I keep saying, some sort of magic recursion that doesn't seem to be in the actual logic anywhere. – 147pm Dec 27 '24 at 21:44
  • Our strong induction hypothesis is that all smaller naturals $(> 1)$ have prime factorizations, hence both $,m,n,$ have prime factorizations, so appending them yields a prime factorization of $,N = mn.\ $ See also here for more on strong induction. $\ \ $ – Bill Dubuque Dec 27 '24 at 21:54
  • Yes, I saw that in Jay Cummings' book Proof..., i.e., you just state as your induction hypothesis that $m,n$ have prime factorizations since they're smaller than $N$. (Assume that all cases before or here smaller are true.) Then you concat $m$ and $n$ factorizations and, therefore, $N$ is also shown as prime factored. I suppose that's a deep as I can go with this. Thanks. – 147pm Dec 27 '24 at 22:18
  • Look at strong induction this way: Assume your statement is false for some $n$. Then because any set of positive integers has a smallest element, choose the smallest $n$ for which the statement is false. What can that smallest counterexample be? It can't be $1$ (or $2$) because we have proved the statement for the base case. But since we're assuming that $n$ is the smallest counterexample, our statement is true for all smaller positive integers. Our strong induction hypothesis tells us that if our statement is true for all positive integers smaller than $n$, it's also true for $n$. – Robert Shore Dec 27 '24 at 23:11
  • (Continued) This means that $n$ wasn't a counterexample after all, contradicting our assumption. But our assumption was that our statement had some counterexample. That assumption must therefore be false, so it has no counterexamples; i.e., our statement is true $\forall n \in \Bbb N$. – Robert Shore Dec 27 '24 at 23:12

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