University B has a group of "perfectly logical" professors

This also belongs to a topic puzzles and paradoxes in math induction. We will discuss the puzzle and logic behind solving it but not mathematical induction.

I am pasting in the image of the question here:

Do not scroll past this if you do not want to see the answer.

I will not repeat the question so you better read it well. I am going to write answer below, if you do not want to look at the answer yet do not scroll.

Answer: All of the professors will resign.

Expanded answer:

1. Professors will meet for the last time in the last meeting of the year

2. They look at each other and realise the other person is not resigning

3. Deduce they must be the one who made the mistake and they all resign at the same time.

If you get it then you need not read the below.

Q. barbossa knows that jack isn't aware of his mistakes so how does he expect him to resign?

Let us consider that the two professors are Captain Jack Sparrow and Captain Hector Barbossa. They are sitting across the table

Let us run through jack's train of thoughts:

1. He sits at the table and looks at barbossa.

2. Professor X said one of you has committed a mistake

3. As jack knows barbossa has committed a mistake he says to himself "aha! barbossa be the black sheep" and sits back to drink his rum.

lets look at it the other(Opposite) way.

Jack thoughts:

1. If barbossa has NOT made a mistake then jack would NOT know about barbossa's mistake

2. He then says himself "One of us made the mistake, if he has done it then I'd known it. I haven't heard about his mistake. So I'm the one making the mistake".

3. That we know professor X's words "Which has been discovered by others in the department".

4. As soon as jack understands this he resigns.

The above did not happen and thus barbossa has made a mistake and jack knows it.

Barbossa then realizes this and the exact same seconds jack does too as barbossa too knows about his mistakes and then both of them resign!

All of this can be applied to as many sets of professors / pirates as you'd like. And the final result, every one resigns. Think about it you'll get it sooner or later!

Extra info:

When I was looking to add in images, funny enough I couldn't find one without copyright notice but I found a wikipedia article on pirates(the real ones), cool one. Have a look at it once you are done reading this. https://en.wikipedia.org/wiki/List_of_pirates

This puzzle is taken from http://www.math.cornell.edu/~mec/2008-2009/ABjorndahl/ppmi.pdf

I do NOT claim it as my own. All Copyrights to the respective owners.

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!