This question already has an answer here:
Upon reading of All horses have the same color "paradox", I began to wonder a couple of things.
First of all, to me the inductive step seems flawed. Just because I have $n$ white horses, does not mean that the $n+1$ will be white, too.
The only explanation to make the inductive step work seems to me that the actual statement we are assuming as base case it's not "there are at least $n$ horses of the same color", but rather "$P(n)$ = any grouping of $n$ horses is such all the horses in that grouping have the same color".
Now this makes the inductive step works, but it's completely useless, as $P(2)$ already implies that all the horses have the same color.
And of course the base case $P(2)$ is not true because it's equivalent to our problem. Setting to prove $P(2)$ is the same as proving that all horses have the same color.
So I can't understand why this has been considered a "false" proof by induction; either the inductive step does not work or it's completely useless.
The wikipedia page linked before explains the problem as
The problem in the argument is the assumption that because each of these two sets contains only one color of horses, the original set also contained only one color of horses. Because there are no common elements (horses) in the two sets, it is unknown whether the two horses share the same color.
Which does not make any sense to me.. It's like looking at the finger instead that at the moon.
Is there something wrong with my understanding of induction?
Admittedly, the base case is $P(1)$ and not $P(2)$. But the point holds;the only implication we need is $P(1) \implies P(2)$. All the subsequent implications are useless; a proof of this type is seldom called a proof by "induction". Otherwise all proofs are by induction! :)
What is even the point of establishing what $P(n)$ is? Just prove $P(2)$ and use as known fact $P(1)$ as you would do with any other known fact.
Since there has been some confusion about what I asked, I'll explain my thought process:
1) Begin reading the proof. Imagine that the statement is "P(n) = at least n horses have the same color", and if by induction on $n$ this hold for every $n$, then QED.
2) Realize the inductive step does not work
3) Understand that the proposition to be proved has been (silently) changed and now it's "P(n) = any grouping of $n$ horses is such all the horses in that grouping have the same color"."
4) Realize that $P(2)$ is what we want to prove.
5) Wonder why on earth we are trying to show $P(n)$ for $n > 2$. Induction proof usually works because the final statement to be proved come from the fact that it works for all $n$. Here $n=2$ suffice, so it is really odd.
6) Starting to think I've missed something here. But hey, it's a false proof, let's see what was wrong.
7) Turns out that the outline of the proof is fine but here's the catch: of all the implications $P(1) \implies P(2)$ is not valid hence the whole $P(n)$ cannot be valid.
8) Thinking (again) that this seems weird. Wonder if I've missed something .
By extended discussion it turns out that my understanding of induction was right. This lead me to believe that is is a bad example because it "seems" like a proof by induction but really some essential elements are not present. It's not that is wrong (apart from the failed implication), is that it gives a very bad idea of what a proof by induction is