Now, say you have a group of 3 horses, A, B, and C. You can use your knowledge about groups of 2 horses here: A and B must be the same color because they are a group of 2 horses. B and C must be the same color for the same reason. So all the horses are the same color.
You can now use the proof for groups of 3 horses to prove the same fact about groups of 4 horses, and so on.
The flawed inductive proof tries to generalize this argument to all groups of n and n+1 horses. That is, assume the property is true for groups of n horses, and show that it logically follows that it’s true for groups of n+1 horses. The structure of the argument is the same as my 2/3 horses example. However, you can’t use an argument like that for any value of n, as it isn’t true for n=1. That is, it’s not true that if every group of 1 horses is the same color (a true fact in nature) then it follows that every pair of horses is the same color.
The problem is that this fails when tried for N=2, because then each sub-group can have only one horse in it. We already know that each horse is the same color as itself, so we don't actually reveal any new information by doing this. Since it doesn't work for N=2, the base assumption used for the N=3 case is incorrect.
