All horses are of same colour

All horses are of same colour

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We will establish a false statement that
Main clain

All horses are of same colour.

This is not correct but we will prove it true.
The above argument is not in the inductive form, we will convert it to look like one.
Inductive form (claim reformulation): “for every natural number n, every group consisting of n horses is monochromatic”
Base case every group consisting of 1 horse is monochromotic, because the set contains 1 horse and thus one colour only.
n case We assume Inductive hypothesis is true for n number of horses
n+1 case We take a set of n+1 horses, then we remove a horse call it rj.
There are n horses left. Let the newly formed group be called $G_{1}$. $G_1$ will have n horses and thus they all will be monochromatic as from assumption.
we may think rj is of a different colour may be, then we can remove some other horse call it cj and add in rj, then again it will contain n horses. Lets call them $G_{2}$. As $G_{2}$ also contains n horses and we know that if a group has n horses from above case are monochromatic and thus rj is also is of same colour and so are all horses of rj s colour as they all have been common through both the exchanges.
As cj’s color == rest of horses color == rj’s colour
We can now safely say that all horses are of same colour

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You may feel you are able to point some issues and thus prove that I was wrong. Like:

1. The claim reformulation you have done is wrong! No I have not, it is correct.
   If there are any finitely numbered horses, I can call them n number of horses and start working on the problem.
2. You can say “the concept of induction is false here because you can’t just assume n horses are of same colour and work on it. That this is a circular assumption”.
   You’d be correct but we knew right from the start that induction is a circular.

Now let us look at what went wrong, we considered the case of n = 1 horse and it being monochromatic. Now let us go with n = 1 being the n and n+1th case as n can be $\ge$ 1.
for n+1th case we assume there are 2 horses for n+1. The removing one of them at a time brings us to the base case scenario and thus we’ve proved that above claim is true under an argument with a hole in it.
The way we used induction is correct, but induction simply allow us to conceal the fact that we’re using n = 1, that’s it!