I was asked to recast an induction proof to a proof by well ordering princple. How are the 2 different? From my understanding the two are equivalent, so how will the proof be different? Thanks!
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possible duplicate of Well-Ordering and Mathematical Induction – Shobhit Sep 11 '13 at 04:43
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1Different statements can be logically equivalent. They are still different. – Zev Chonoles Sep 11 '13 at 04:44
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1Usually if you take a look at a proof by the well ordering principle it usually is used to prove that given a set that has positive integers as elements that it has a least element. Usually done so by contradiction. Take a look at a proof of the division alogrithm. While induction does imply the well ordering principle and vice versa. But if its going from an induction proof to a proof by the well ordering principle those are two different proof techniques. What was the proof you were trying to change? – user60887 Sep 11 '13 at 05:16
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@user60887 I'm trying to proof that "If I draw n straight lines on a piece of paper I cannot divide the piece of paper into more than n(n+1)/2 +1 regions" – whatdidthefoxsay Sep 11 '13 at 05:19
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1Because they are equivalent, you can transform one into the other. – Carsten S Dec 28 '13 at 22:13
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- For reference of user60887's comment, the proof of the division algorithm is done by proving existence using the well ordering principle and uniqueness using contradiction. See https://math.stackexchange.com/q/499789/1059606. 2. The Shobhit's reference link gives a proof of equivalence of these 2 theorems. This has a bit differences from proof "recast".
– An5Drama Apr 28 '24 at 01:50
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Since you already have an induction proof of some statement $\forall n \in \mathbb{N}: P(n)$, you presumable have some reasoning that concludes $P(0)$ and other reasoning that proves $P(n) \implies P(n+1)$ for arbitrary $n \in \mathbb{N}$.
This can be transformed into a proof using the well-ordering principle as follows:
Suppose that $\lnot \forall n \in \mathbb{N}: P(n)$. Then there is at least one $n \in \mathbb{N}$ for which $\lnot P(n)$; take the least such $n$. By [insert reasoning here that $P(0)$] $n \neq 0$. By [insert reasoning here that $P(n-1) \implies P(n)$, which we can do because $n \neq 0$] we now also have $\lnot P(n-1)$. This contradicts the fact that $n$ was minimal.
Therefore $\forall n \in \mathbb{N}: P(n)$.
(Note, this looks most natural if you original reasoning for $P(n) \implies P(n+1)$ actually works by assuming $\lnot P(n+1)$ and concluding $\lnot P(n)$.)
Magdiragdag
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More specifically if someone has questions, the last sentence means for the other direction, we need to assume $K:(\lnot \forall n \in \mathbb{N}: P(n)\implies P(n+1))$ instead of $\lnot \forall n \in \mathbb{N}: P(n)$. Then there is one subset $S\subseteq \mathbb{N}$ which has the minimal element satisfying $K$. Then follow the similar process "Well order implies induction" – An5Drama Apr 28 '24 at 02:29