how to point out errors in proof by induction I have searched for an answer to my question but no one seems to be talking about this particular matter..
I will use the all horses are the same color paradox as an example.
Everyone points out that the statement is false for n=2 and that if we want to prove the propositions we should use 2 as the base case for this proof.
But, (as I see it..), you have to use reason to figure that out. 
My question is, is there anything wrong with induction itself? 
except for the fact that we can use reason to understand why the proof is faulty.
This goes for all these problematic propositions..
Thank you all!


"But, (as I see it..), you have to use reason to figure that out. My question is, is there anything wrong with induction itself? except for the fact that we can use reason to understand why the proof is faulty."

I'm not entirely sure of what you mean with using 'reason', but I'm interpreting it as "is there way to spot the mistake in the argument without having the inspiration to check what happens for $n=2$?"
There is. Take Wikipedia's argument.

Inductive step
Assume that $n$ horses always are the same color. Let us consider a group consisting of $n+1$ horses.
First, exclude the last horse and look only at the first $n$ horses; all these are the same color since $n$ horses always are the same color. Likewise, exclude the first horse and look only at the last $n$ horses. These too, must also be of the same color. Therefore, the first horse in the group is of the same color as the horses in the middle, who in turn are of the same color as the last horse. Hence the first horse, middle horses, and last horse are all of the same color..

Consider a set of $n+1$ horses: $\{h_1, h_2,\ldots ,h_n, h_{n+1}\}$.
Wikipedia now tells you to consider $\{h_2, \ldots ,h_{n+1}\}$ and $\{h_1, \ldots h_n\}$ and apply the induction hypothesis on each of this sets, which then yields that $h_2, \ldots ,h_{n+1}$ are all of the same color and $h_1, \ldots ,h_{n}$ are all of the same color.
Up until this point there's nothing wrong. But now, to say that $h_1$ has the same color as $h_2$ and consequently as $h_3,\ldots ,h_{n+1}$ you need that $h_2\in \{h_1, \ldots ,h_n\}$, but this is only true if $n>2$, for $n+1=2$ one has $\{h_1 ,\ldots ,h_n\}=\{h_1\}$, so you can't make further progress.
