Talk:Method of Infinite Descent

As stated, the proof of this method requires the Axiom of Countable Choice, which is independent of ZF.

It could be refactored to instead use Well-Ordering Principle directly, keep the set of all smaller elements, or prove a long enough finite prefix of the sequence in the proof. CircuitCraft (talk) 12:44, 7 February 2023 (UTC)

I can't see why Axiom of Countable Choice could be needed. Given that $\map P {n_\alpha}$ holds for some arbitrary (finite) $n_\alpha$, there is only a finite number of numbers less than it. And choice from a finite set can be done in ZF, surely?

But in any case, we already do use Well-Ordering Principle . I don't understand any of this mathermetics stuff. --prime mover (talk) 18:55, 7 February 2023 (UTC)

The fact that there are only finitely many numbers less than $n$ isn't part of the proof. We are asserting the existence of an unending sequence, the construction of which requires ACC. The alternative proof using well-ordering directly is: consider the set of counterexamples. By WO, take the least one. By hypothesis, we can find a smaller one. Contradiction.

We could do it the original way if we take the least smaller counterexample by WO, but that's a little bit redundant.

Your suggestion is similar to my finite prefix example. We could construct $n + 1$ terms, which define an injection into at most $n$ elements. Contradiction. CircuitCraft (talk) 20:49, 7 February 2023 (UTC)

Right okay, I get where the problem is now. Can we craft the proof using Well-Ordering Principle and call it Proof 1, then one using ACC and call it Proof 2, and then the existing one can be kept as an "intuitive proof" which can be understood directly by a casual reader or a newbie mathematician? Some of the technical concepts (which can lead you down a rabbit-hole if you're not careful) can be daunting. So adding that "least smaller counterexample" bit to the original proof can also be worked through.

I will check back in the Bell work that this cites, it's such a long time since I wrote this page I can't remember if the proof came out of my head or from the book. In due course. --prime mover (talk) 21:01, 7 February 2023 (UTC)

It occurs to me, don't we also have to deduce the existence of $n_\gamma$ such that $0 < n_\gamma < n_\alpha$ for which $\neg \map P {n_\gamma}$? Sort of like "eventually you'll hit the bottom number, you can't avoid it", like in the case of the prime $5$ of the form $4 n + 1$. Hence it cannot be the case that there's always something smaller for which $P$ holds. If there is no $n_\gamma$ then it is possible for $P$ to hold for all $n$, if the case for the lowest $n$ of the descent construction is degenerate and admits no smaller $n \ge 0$. --prime mover (talk) 21:38, 7 February 2023 (UTC)

I don't think that proving a "bottom" is necessary. We're working with natural numbers, so zero is the bottom for everything. The idea behind the method is that descending sequences of natural numbers always terminate; whether you prove that termination because of a base case or by simply running out of numbers is inconsequential. Fermat probably just wanted to make it clear exactly where the process breaks down (speculation).

Nope, I don't understand a single thing about this. --prime mover (talk) 12:13, 8 February 2023 (UTC)