Mathematical induction using the first step as the base case is valid. The problem with the horses of one color problem is that you are using sloppy verbal reasoning that hides an unjustified assumption that n > 1. If you had tried to make a rigorous argument that the set of n+1 elements is the union of two of its subsets with n elements each, with those subsets having a non-empty intersection, this would be clear.
The problem with the horses of one color problem is that you are using sloppy verbal reasoning that hides an unjustified assumption that n > 1.
I’m not sure what you mean. I thought I stated it each time I was assuming n=1 and n=2.
In the induction step, we reason “The first horse is the same colour as the horses in the middle, and the horses in the middle have the same colour as the last horse. Therefore, all n+1 horses must be of the same colour”. This reasoning only works if n > 1, because if n = 1, then there are no “horses in the middle”, and so “the first horse is the same colour as the horses in the middle” is not true.
Mathematical induction using the first step as the base case is valid. The problem with the horses of one color problem is that you are using sloppy verbal reasoning that hides an unjustified assumption that n > 1. If you had tried to make a rigorous argument that the set of n+1 elements is the union of two of its subsets with n elements each, with those subsets having a non-empty intersection, this would be clear.
Induction based on n=1 works sometimes, but not always. That was my point.
I’m not sure what you mean. I thought I stated it each time I was assuming n=1 and n=2.
In the induction step, we reason “The first horse is the same colour as the horses in the middle, and the horses in the middle have the same colour as the last horse. Therefore, all n+1 horses must be of the same colour”. This reasoning only works if n > 1, because if n = 1, then there are no “horses in the middle”, and so “the first horse is the same colour as the horses in the middle” is not true.