2
$\begingroup$

Im trying to understand the solution about this, and all the other solutions are.. a bit weird for me and dont answer all my questions!

Reference : Questions on "All Horse are the Same Color" Proof by Complete Induction

Im using the answer there as reference, but my understanding is a bit different!

Find the error in the following proof that all horses are the same color. CLAIM: In any set of h horses, all horses are the same color.

PROOF: By induction on h.

Basis: For h = 1. In any set containing just one horse, all horses clearly are the same color. Induction step: For k > I assume that the claim is true for h = k and prove that it is true for h = k + 1. Take any set H of k + 1 horses. We show that all the horses in this set are the same color. Remove one horse from this set to obtain the set H1 with just k horses. By the induction hypothesis, all the horses in H, are the same color. Now replace the removed horse and remove a different one to obtain the set H2 . By the same argument, all the horses in H2 are the same color. Therefore all the horses in H must be the same color, and the proof is complete.

Here's what i have come up with :

Suppose a set of 1 horse. $H_a = { horse_{1} }$

Obviously, all the horses in set $H_a$ are the same, since theres just 1 horse.

So using that as the Basis of the induction we can proceed and try to prove its true for $N$ horses.

-- I dont fully understand the part below

As an example : set $H_0$ :

$H_0 = { horse_{1}, horse_{2}, horse_{3}... horse_{n} }$

In my understanding, its because, I can make a subset formed by 1 horse :

$K_1 = {horse_{1}}$ $K_2 = {horse_{2}}$ ...

I don't get it. I clearly know that this proof is a lie. And for me, its because you cant claim theres a relation on a set with size $>= 2$ , since :

$H_1 = { horse_{1}, horse_{2}}$

Theres nothing that says $horse_{1}$ equals $horse_{2}$. The basis was for one horse. But that alone is not a proof.

So what should I do ?

$\endgroup$
  • 1
    $\begingroup$ Hint: Can you prove, by induction, that the claim is true for $\;n=2\;$ ? $\endgroup$ – DonAntonio Apr 7 '17 at 15:08
  • $\begingroup$ No idea how i could. Theres no way to prove a horse1 is equal or different to horse2... $\endgroup$ – KenobiShan Apr 7 '17 at 15:18
  • $\begingroup$ Ah, that's precisely the gist of the problem with this "proof": you can't do that step! And good we can't: having all horses of the same color would be pretty boring. $\endgroup$ – DonAntonio Apr 7 '17 at 15:19
  • $\begingroup$ I have no idea how to prove mathmatically that " theres no way to prove a horse1 is equal or different to horse2..." $\endgroup$ – KenobiShan Apr 7 '17 at 15:25
  • $\begingroup$ @K It seems like you're having existencialist thoughts, yet the problem is way easier: take a general set with two horses: can you prove they have the same color? If you can then you can prove all horses have the same color. The gist is that you can not prove that, induction or not induction. That's all. $\endgroup$ – DonAntonio Apr 7 '17 at 16:05
3
$\begingroup$

The mistake is in the inductive step. The reasoning assumes that there is a $horse_1$ that you can remove, leaving you a set of $k$ horses that, by inductive hypothesis, all have the same color. OK, so far so good. But then it says we can put $horse_1$ back, and remove $horse_2$, and so this new set of $k$ horses have the same color as well. Also good!

OK, but why then doesn't this show that all $k+1$ horses have the same color?! Because if $k =1$, then $horse_1$ and $horse_2$ are the only two horses! And, yes, removing $horse_1$ leaves all remaining horses (which is just $horse_2$) the same color, and the same goes for removing $horse_2$: the only remaining horse is $horse_1$, and all those $k=1$ horses are the same color. But obviously we can not conclude that all $k+1 =2$ horses have the same color!

Of course, the reasoning would go through for $k>1$, for then you would have a $horse_3$ that, when $horse_1$ is removed must have the same color as $horse_2$, and that, when $horse_2$ is removed, must have the same color as $horse_1$, and thus $horse_1$, $horse_2$, $horse_3$, and any other horses would indeed have to have the same color.

But of course: if you assume that any $2$ horses always have the same color, then obviously any sized group of horses will all have the same color.

So, it is really just that 1 case (with 2 horses) that is the crucial case, and neither the base case nor the inductive step cover that case. Or, to be exact: the base doesn't cover it at all, and the reasoning in the inductive step fails for that particular case.

$\endgroup$
1
$\begingroup$

This is an infamous example of neglecting to determine the correct base case.

The inductive step is actually a perfectly valid argument. For $h \geq 3$. So the base case must include $h=2$ (and also $h=1$ and $h=0$).

So, the error is that the argument did not prove the proposition holds for the base case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.