[diffeomorphism]

Disagree. The induction step works perfectly fine for n>=2. In particular, if you can give me the base case "any two horses have the same color", then this proof works and all horses have the same color.

The problem is that the base case is n=2, but the student checked n=0 and n=1.

[amitkgupta84]

The base case is not n=2, it’s clearly stated as n=1 in the proof.

The inductive step works fine if you introduce the premise n>=2. It is not valid to introduce that premise. The inductive step is therefore wrong.

The proof in the post is saying the base case is P(1) and the inductive case is that for all n, P(n) => P(n+1)

“P(1)” is true. “For all n, P(n) => P(n+1)” is false. The inductive step is wrong

A completely different proof not present in this post could be:

Base case: P(1) and P(2) Inductive case: for all n > 1, P(n) => P(n+1)

In that completely different proof, the inductive step would be correct and the base case would be wrong. That proof is not in the original post.

[diffeomorphism]

We don't even disagree about anything here.... We both say "P(n) => P(n+1)" does not work for n=1, but does work for n>1.

I would like to give the student partial credit for the "works for n>1" part and deduct points for the missing n=1 case. The two obvious ways to "fix" that are either giving a proof for "P(1)=>P(2)" (which is currently missing) or establishing P(2) another way and then using induction. Both are fixes of the same thing and not at all "completely different".

[Jensson]

The proof has errors, there is no fixing it, you can't prove that all horses have the same color given no data about horses. In this proof the base statement is 100% correct, in all sets of 1 horse all horses will have the same color. Hence the error has to be in the other part.

If a student was tasked with proving that all horses has the same color given no information about horses then the correct answer is "I can't prove this", the answer given in this article would give 0 points since the student obviously doesn't understand what they are doing. When you check that test, would you mark the base case as the source of the error of the proof, or would you mark the inductive step? The base case is correct, if you marked that part wrong the student would rightfully complain, their logic works there.

[shkkmo]

The base case is proven correctly, the inductive rules is proven OK except that the transitive set rule they invoke requires n>1

As such, the only data you would need yo prove that all sets of horses are the same color is that all pairs of horses are the same color.

I think zero points would be a very harsh grade given that an understanding of how to do an inductive proof was demonstrated and one minor error was made in an otherwise correct proof.

[Jensson]

> I think zero points would be a very harsh grade given that an understanding of how to do an inductive proof was demonstrated and one minor error was made in an otherwise correct proof.

Who says the question was to demonstrate an inductive proof? Being able to prove that all horses are the same color if all pairs of horses are the same color isn't particularly useful, as typically that would be the definition of all horses having the same color. All that humbug didn't add any extra information to the problem at all, it didn't make things clearer it just made them harder to understand.

[shkkmo]

To prove something like that you pretty much have to use a proof by induction or a proof by contradiction.

[diffeomorphism]

> The proof has errors, there is no fixing it,

You mean to tell me that horses are actually different colors? Really? /s

Anyways, as I said we agree already so no reason to waste further time on this.

[Jensson]

> You mean to tell me that horses are actually different colors? Really? /s

Right, the goal here is to find the logical error in an obviously wrong proof, not how to fix this proof as the proof is obviously wrong from the start. You trying to argue how to "fix" it is the one side-tracking the whole thing.

If you say "B wouldn't be wrong if you changed A, so A is wrong!" then that isn't how you find errors. If A is a correct statement and B is an incorrect statement then B is the wrong statement. If you can make B correct by changing A, then B is still an incorrect statement.

If this was a memorization question where you were supposed to find a particular A and B, then you could say that A is wrong even if the statement is correct, since you knew what A and B are supposed to be. But it isn't, we are just here to find the error in the statement to prove that the proof is wrong, we aren't here trying to find a correct version of this statement.

[wisty]

No, the induction step works for n>=3. The induction step fails for n=2.

For n=2, you can't prove h1 is the same colour as h2, because you don't have a h3 to compare it to.

[diffeomorphism]

tomayto tomahto

We are talking about the exact same case. The induction step is n-> n+1 and works perfectly well for n+1=3 horses.