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I'm bugged by the following that's summarized on p. 109 of this PDF.

False theorem: All horses are the same color.

Proof by induction:

$\fbox{$P(n)$ is the statement: In every set of horses of size $n$, all $n$ horses are the same color.}$
$\fbox{Base Case or $P(1)$:}$ One horse is the same color as itself. This is true by inspection.
$\fbox{Induction Step:}$ Assume $P(k)$ for some $k \geq 1$.
$\fbox{Proof of $P(k + 1) :$}$
Since $\{H_1, H_2, ... , H_n\}$ is a set of $n$ horses, the induction hypothesis applies to this set. Thus, all the horses in this set are the same color.
Since $\{H_2, H_3, ... , H_{n+1}\}$ is also a set of $n$ horses, the induction step likewise holds for this set. Thus, all the horses in this set are the same color too.

Therefore, all $n +1$ horses in $\{H_1, H_2, H_3, ... , H_n , H_{n+1}\}$ are the same color. QED.

The issue the instructor was trying to point out is clearly valid. For the case $n = 1$, there is only ${H_1}$. So this case says nothing about possible overlapping elements of each set of $(n + 1)$, for instance $H_2$ in the above proof.

But it was proposed in the class discussion that this was the only problem. Had you been able to prove $P(2)$ true, then a proof of the above format would have been fine.

My interpretation is that yes, you could prove all horses are the same color, if you can prove that any set of two horses will be the same color. But this format would not work.

Why not? The problem I see is that the above proof is for the existence of at least one particular pair of sets of horses of sizes $n$ and $n + 1$, such that in each set, all horses are the same color. Particularly when the set of size $n$ is a subset of the set of size $n + 1$. In order to prove the induction step, don't you need to prove that sets of sizes $n$ and $n + 1$, do not necessarily contain the same, overlapping elements?

You could prove that any horse can be added to a set of 2 horses. Take the last two, and they must be the same color, and so on. Wwhatever color the first two happen to be, all other horses must thus be the same color.

Am I misinterpreting the example, or am I making a logical error? Thanks in advance.

NNOX Apps
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MVTC
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    P(1) is obviously _false_ by inspection. Many single horses are of more than one colour -- they're called piebalds or skewbalds. – Mike Scott May 15 '17 at 20:00

6 Answers6

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It’s clear from the question and from your discussion with @DonAntonio that you don’t actually understand the induction step of the argument. You seem to think that it involves comparing some previously chosen set of $n$ horses with some new set of $n+1$ horses, but it does not. Let me see if I can explain more clearly just how it does work.

We assume as our induction hypothesis that if $H$ is any set whatsoever of $n$ horses, then the horses in $H$ are all the same color. Now let $H=\{h_1,h_2,\dots,h_{n+1}\}$ be any set of $n+1$ horses; our goal is to show that all of the horses in $H$ are the same color. To do this, we look at the subsets

$$H_0=\{h_1,\dots,h_n\}\quad\text{and}\quad H_1=\{h_2,\dots,h_{n+1}\}\;.$$

Each of these subsets contains $n$ horses, so by hypothesis all of the horses in $H_0$ are the same color, and all of the horses in $H_1$ are the same color.

Note: The two $n$-horse sets that we’re looking at here were not given to us ahead of time: they are just two different $n$-horse subsets of $H$. And in fact any two different $n$-horse subsets of $H$ will work equally well for the argument.

Now if $n\ge 2$, the horse that I’ve called $h_2$ is in both $H_0$ and $H_1$. (In fact every horse except $h_1$ and $h_{n+1}$ is in both $H_0$ and $H_1$.) Since $h_2$ is in $H_0$, and we already know from the induction hypothesis that all of the horses in $H_0$ are the same color, it follows that every horse in $H_0$ is the same color as $h_2$. But $h_2$ is in $H_1$ as well, so by exactly the same reasoning we conclude that every horse in $H_1$ is the same color as $h_2$. But every horse in $H$ is in $H_0$ or $H_1$ (or both), so every horse in $H$ is the same color as $h_2$, and therefore the horses in $H$ are all the same color. This completes the induction step.


There is absolutely nothing wrong with that argument — provided that $n\ge 2$. In particular, it does not involve starting with some particular set of $n$ horses known to be the same color and trying to use that set to show that the horses in some arbitrary set of $n+1$ horses are all the same color. The argument starts with a set of $n+1$ horses and proceeds by looking at two $n$-horse subsets of the given set. It’s important here to realize that the induction hypothesis does not say that there is some particular set of $n$ horses that are all the same color: it says that in any set of of $n$ horses, all of the horses are the same color.

The argument fails only when $n=1$. In that case it fails because the sets $H_0=\{h_1\}$ and $H_1=\{h_2\}$ no longer overlap: there is no horse in both sets. Thus, while it’s true that all of the horses in $H_0$ are the same color and that all of the horses in $H_1$ are the same color, we can no longer infer that those two colors must be the same. But if we could somehow show that in every set of two horses, both horses were the same color, the induction argument as given would work just fine: we’d always be considering some $n\ge 2$, so the sets $H_0$ and $H_1$ would overlap.

Brian M. Scott
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I'm not sure I understand fully your doubt, but the actual problem with that, and many other examples of bogus inductive proofs out there, is that when you take the two sets

$$\{h_1,...,h_n\}\;,\;\;\{h_2,...,h_n,h_{n+1}\}$$

each of size $\,n\,$ and thus, by the Ind. Hypothesis, formed of horses of the same color, the leap to deduce all the horses altogether have the same color requires that the intersection of both sets above is not empty, something you can't prove (because it is false!) if $\,n=2\,$

Thus, if we had the claim "every two horse whatsoever are of the same color", then we could prove "all the horses are of the same color"

DonAntonio
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  • $n=1$, maybe, since otherwise they share $h_2$. – Arthur Jun 24 '13 at 09:16
  • No, for $\,n=1\,$ it works trivially. For $\,n=2\,$ we have the huge problem: each of the sets $\,\{h_1\}\;,\;\{h_2\}\;$ have all the horses of the same color, then (this is the bogus part) also $\,\{h_1,h_2\}\,$ has horses of the same color. From this impasse the whole thing crumbles down in this case. – DonAntonio Jun 24 '13 at 09:30
  • @DonAntonio, what is meant is that it works if you have P(2) as a base case. Then, you have the same setup where you prove that for an arb. n >= 2, p(n) -> p(n+1). Then for the first n, n = 2, P(n) gives {h1, h2} p(n+1) gives {h1, h2, h3}, and they share h2... My issue was that p(n+1) could also be {h4, h5, h6}, of which this proof doesn't address, but I guess after more thinking, it is obvious, that you add any element whatsoever to a set of 2, and it must be the same color, so all must be the same color. I just don't know how acceptable it would be without something more to it. – MVTC Jun 24 '13 at 10:10
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    But the above is assuming you could use and prove P(2) as a base case, which you can't. – MVTC Jun 24 '13 at 10:17
  • @DonAntonio I mean where you say "the intersection of both sets above is not empty, something you can't prove (because it is false!) if $n=2$." Here $n$ should be equal to $1$, otherwise the intersection contains $h_2$. That is what I meant. In the case $n=1$, you have two sets of uniform horses with empty intersection, thus you cannot go through with the induction step. – Arthur Jun 24 '13 at 10:24
  • I think there's a confussion, @Arthur: if $\,n=2\implies \{h_1,h_2\}\,$ ,then we form two sets with $\,n-1=1\,$ element each: $\,\{h_1\}\,,\,\{h_2\}\,$ . It's these two sets' intersection which is empty and thus the inductive hypothesis isn't appliable! Remember, in this proof's reasoning, $\,n\,$ is the number of horses, not of sets... – DonAntonio Jun 24 '13 at 11:00
  • Indeed so, @MVTC: if the case for $\,n=2\,$ horses was provab le then we would indeed easily prove *all* the horses are of the same color. That's what I remarked above that the real, tough (in fact, insoluble) problem is for $\,n=2\,$ – DonAntonio Jun 24 '13 at 11:03
  • @DonAntonio The confusion arises from the fact that you have written down the two sets $$ \{h_1,...,h_n\}\;,\;\;\{h_2,...,h_n,h_{n+1}\} $$ which, if $n=2$, have non-empty intersection. I might've misunderstood what part of the proof goes with that line of math, though – Arthur Jun 24 '13 at 15:09
  • @Arthur, what is $\,\{h_1\}\cap\{h_2\}\,$ ?! – DonAntonio Jun 24 '13 at 15:11
  • The fault that I perceive, besides the lack of intersection for n = 1, is that even for n = 2, The sets need not intersect. Set of size n = 2, {h1, h2}. Set of size (n + 1) = 3, {h1, h2, h3} are chosen, but you may also choose {x, y, z}. They must be arb. and not may or may not intersect. I see a flaw in the fact that necessarily overlapping sets were chosen. For example, if you say x is odd, and y i even, so x = 2k + 1, and y = 2k, this is a mistake, because you've k twice, x odd and y even does not mean x = y + 1. – MVTC Jun 24 '13 at 18:30
  • Of course the sets do not intersect for $\,n=2\,$, but they definitely do for $\,n\ge 3\,$ , since if we have $\;\{h_1,h_2,h_3,\ldots,h_n\}\;$ , then *by construction in the given proof*, we have the sets $\;A:=\{h_1,h_2,h_3,\ldots,h_{n-1}\}\;,\;B:=\{h_2,h_3,\ldots,h_n\}\;$ , and we always have $\;h_2\in A\cap B\neq\emptyset\;$ – DonAntonio Jun 24 '13 at 18:41
  • Your argument about even and odd integers is incomprehensible for me within the present discussion – DonAntonio Jun 24 '13 at 18:44
  • You cannot use k twice, just like you cannot use h2 twice was my point. – MVTC Jun 24 '13 at 18:50
  • Honest, @MVTC, I've no idea what you're talking about. As far as I can see, I haven't used the letter $\,k\,$ at all... – DonAntonio Jun 24 '13 at 18:51
  • If you let A be any set of size n, and B be any set of size n+1, then you have more than just two sets, you have maybe infinitely many of each size. You have to prove all of them intersect, not just the specific ones which happen to each include a few of the same horses. – MVTC Jun 24 '13 at 18:53
  • @MVTC, what? Do you know how induction works? I begin with an inductive hypothesis for *all* the sets with $\,n-1\,$ horses and from there I begin to work with a general set with $\,n\,$ horses and try to demonstrate this last set fulfills the same as any set with $\,n-1\,$ horses. The sets $\,A,B\,$ are just simplifying devices to denote two subsets with $\,n-1\,$ horses each that I construct from the original one with $\,n\,$ horses and etc. – DonAntonio Jun 24 '13 at 18:59
  • Your slightly confusing the problem by changing the format of the proof to p(n-1)->p(n) instead of p(n)->p(n+1), and thereby your statements are inconsistent with the work we are examining. But that's not really a problem. Anyways, if you read my question, your answer only repeats what I already stated, in a slightly confused way, while also misinterpreting my state of doubt, and missing the point of and nature of the question. – MVTC Jun 24 '13 at 19:16
  • @MVTC, I agree that for a beginner that can be confusing, but I see you were able to sort it out. I think you're missing, and big, the point and nature of the example of a bogus proof by induction given in that book, but I agree I could be misunderstanding your doubt, in particular after reading your comments above which seem to be far removed both from I tried to convey and from what the book's example was trying to teach. – DonAntonio Jun 24 '13 at 19:27
  • This is a very standard, old example of bad usage of the PMI, and perhaps you could try to re-write your post to clearly state what **exactly** your doubt is since, as you can see, not many have participated in this thread... – DonAntonio Jun 24 '13 at 19:30
  • @DonAntonio It's been a few comments since your last question to me, and my answer is that that intersection is empty. But if you look at it, you have chosen $n=1$ when you ask about $\{h_1\}\cap\{h_2\}$. – Arthur Jun 25 '13 at 01:09
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There is a more basic problem here : induction is irrelevant to the proof.

$P(2)$ is, by itself, a formalization of the statement that all horses are the same color. There is no reason to prove $P(2)$ to prove $P(3)$ to prove ... $P(n)$ and conclude from this chain of implications that all horses are the same color; one would try to prove $P(2)$ and in case of success, summarize that as "all horses have the same color".

If proving $P(n)$ were of interest for some reason, it can be derived in one step from $P(2)$, with no need for an induction.

The size of the set of horses is usually ignored in presentations of this argument and ones like it. But if there are $n$ horses in the universe, $P(k)$ is an empty statement for $k>n$, and is true regardless of the truth value $P(n)$. This shows that $P(n)$ is logically stronger than $P(n+1)$, and that the inductive step from $P(n) $ to $P(n+1)$ is vacuous for large $n$.

zyx
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Just want to add a quick summary.

The proof is "almost" perfect. It simply fails on n=1.

Yes obviously any set with 1 horse will have the same color. However, you can't extend that to n=2.

If you can somehow proof that any set of 2 horses have the same color then yes, you got something right, that any set with any number of horses will have the same color.

user4951
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As you pointed out, $P(1)$ is trivially true. And you are trying to prove $$P(n)\implies P(n+1)$$ to finish the proof. But It's clear that $P(1)$ does NOT implies $P(2)$. Thus the statement $$P(n)\implies P(n+1)$$ is not true. And this is where your proof falls.

Akira
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I just had a student come to me with this problem. I am not a mathematician by trade, so this was the first I'd heard of it. (It's a very good problem that forces the student to understand the method!)

My natural inclination was to negate that which was supposedly proven ("all horses are the same color"), and take the "contrapositive" of the entire inductive chain, and see if we do necessarily contradict the base case. (In other words, the inductive argument supposedly gives $P(1) \Rightarrow P(2) \Rightarrow \cdots\Rightarrow P$; I wanted to see if $\neg P \Rightarrow \cdots \Rightarrow \neg P(2) \Rightarrow \neg P(1) $.)

So, right off the bat, the negation of "all horses are the same color" is "there exists horse A and horse B such that A is a different color than B". From that, you immediately see that $P(1)$ is unsuitable base case because $\neg P(2)$ means "in a set of 2 horses, the two hoses are different colors". That does not contradict anything about $P(1)$ (which of course is logically unimpeachable).

What also comes out in the above line of reasoning confirms what @zyx said: induction is irrelevant to the statement being proven. At every step, the condition that the statement be false hinges on the existence of two different horses having different colors. So, if you had somehow shown $P(2)$ to be true, game over.