ie. If you assume S(2) is true, lets prove S(3), consider a set of 3 people, {a, b, c}.
apply S(2) to {a, b} are therefore the same age, apply S(2)_ to {b, c} are therefore the same age, this implies a.age == b.age == c.age, there for S(3) is true. The inductive step is done.
Thats what I thought made this a mind bender.
However, if it were possible to prove the case n = 2, we would have a valid inductive proof 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.
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#Inducti...
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.
I don't know if I'd call it a mind bender, but it does demonstrate that if one holds that all groupings of two individuals must be uniform with regard to an attribute A, then it follows that any grouping of individuals must be uniform with regard to A.
A simple proof by contradiction:
1) Assume some group of more than 2 individuals is not uniform with regard to A
2) There must be at least one pair of individuals for which attribute A differs
3) Create a subgrouping of one of those pairs
4) Now we have a grouping of 2 individuals which is non-uniform with regard to A
5) But we originally held that all groupings of 2 individuals must be uniform with regard to A
6) Contradiction! Hence a non-uniform grouping with regard to A is impossible, Q.E.D.