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> it only works for n >= 2 Right, so it doesn't work -- either it's a correct proof of a non-sequitur ("true for n implies true for n+1, provided n meets some
by repsilat 6y ago
> it only works for n >= 2
Right, so it doesn't work -- either it's a correct proof of a non-sequitur ("true for n implies true for n+1, provided n meets some criteria"), or an incorrect proof of an inductive step ("true for n implies true for n+1").
Because it's claimed to be proof by induction, it's meant to be the latter -- the person doing the proving claimed to have proven the inductive step, and their proof of it was incorrect.
- pansa2 6y ago“True for n implies true for n+1, provided n >= 2” is a perfectly good inductive step. However, it must be coupled with a base case >= 2, which isn’t the case here. The only base case proven is 1. See also “Induction basis other than 0 or 1” [0]. [0] https://en.wikipedia.org/wiki/Mathematical_induction#Induction_basis_other_than_0_or_1 https://en.wikipedia.org/wiki/Mathematical_induction#Inducti...
- repsilat 6y agoSure, their mistaken proof of the n->n+1 implication could be taken for what it does prove, and such a theorem could be used in other circumstances to prove different things, but that doesn't really bear on any of the claims in the original context. Nobody was trying or claiming to be doing any other kind of induction. They said they had the base case and the inductive step, and they were clear about the base case and the inductive step. Their proof of the base case was correct, and their proof of the inductive step did not prove what they said it did. It doesn't matter what it did prove.
- brazzy 6y agoI think a better characterization of the fallacy is that it is indeed a non-sequitur, but it hides the "provided n meets some criteria" by bamboozling you with "arbitrarily chosen" elements from a set of n+1 so that you might fail to spot where you need 3 elements in a set that potentially has a cardinality of 2.