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Explain the flaw in the following induction argument which shows all of Lucas’ toys are the same colour.

Proof: We will show by induction that: for every integer $n\ge1$, in any group of $n$ of Lucas’ toys, all the toys in this group are the same colour.

Basis: If Lucas had only one toy, then clearly all these toys are one colour, so the result holds for $n=1$.

Induction Hypothesis: Suppose for a fixed but arbitrary integer $k \ge 1$, in any group of $k$ of Lucas’ toys, all the toys in this group are the same colour.

Induction Step: Let $n = k + 1$. Take any fixed but arbitrary group of $k + 1$ of Lucas’ toys. Pick out an arbitrary toy (Toy A) from this group. Then all the remaining $k$ toys in the group are the same colour, by the induction hypothesis. Adding this toy back into the group and removing another toy (Toy B), we get a group of $k$ toys which contain Toy A. By the induction hypothesis, these $k$ toys are all the same colour. Thus all the toys in the group must be the same colour. Since the group chosen was arbitrary, in any group of $k + 1$ of Lucas’ toys, all the toys in this group are the same colour.

Conclusion: By the PMI, for all integers $n \ge 1$ any group of $n$ of Lucas’ toys are all the same colour. Therefore, since Lucas has a finite number of toys, all of Lucas’ toys are the same colour.

Jesse P Francis
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Pat
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  • The induction step fails for the case of k =1 because we have only two toys when k = 1. – Bob Nov 30 '15 at 01:39
  • This is a well-known puzzle, but usually stated as "proving" that all horses are the same color. – Paul Sinclair Nov 30 '15 at 04:13
  • I always heard it as all the marbles in any bag of marbles are the same color. I won a beer in my freshman year on it (which I never collected). I always answered it as "The inductive step assumes n - 1 > 0 which is never verified and is not true in the initial step n = 1". – fleablood Nov 30 '15 at 06:23

2 Answers2

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If you step back, one way to typify the issue is that the base case is faulty. We need to start with a base case such that the logic of the induction step holds when our base case does.

Suppose we try to apply the logic when we have $n=2$ toys, assuming that we have proven the base case for one toy. The inductive step implicitly relies on the fact that when we split it up into two groups of $k=1$ toys, there will be some mutual element shared within the group and hence, by transitivity, all toys would be of the same color.

However, when $n=2$, we don't have that mutual element, and hence the inductive step fails. To fix this proof, one would need to either change the inductive step to work for $k=1$ or prove the base case when $n=2$, of which both are clearl impossible since the result is absurd.

MT_
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The statement of your inductive hypothesis is incorrect. As written, it assumes what you want to prove. (I.e. that all subsets of the set of toys have the same color.) The correct statement of your inductive hypothesis is "There exists a set of $k$ toys all of the same color."

This inductive hypothesis is not strong enough to show that there is a set of $k+1$ toys all of the same color, which is the correct inadequacy for this argument to have.

A way to see that this is correct is that the base case only supports the existence of one element sets for each of which all of a set's elements are the same color. It makes no promise that any two sets have the same color, and it is correct not to do so.

Eric Towers
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    I agree with Soke that the real problem is that the base case is not the correct base case for this hypothesis, which should be the hypothesis for $k = 1$. Otherwise, there is nothing wrong with this hypothesis. If the correct base case for it were true, this would be a proper proof. – Paul Sinclair Nov 30 '15 at 04:22
  • ... and I disagree. The base case is true and valid. The error is believing that, for any $k>1$, the inductive hypothesis can ever follow from the base case. That is, the required inference that the base case implies any version of the inductive hypothesis is false. – Eric Towers Nov 30 '15 at 12:50
  • I don't see how the inductive hypothesis assumes what you want to prove. It assumes $P(k)$ and tries to show $P(k+1)$, as an induction proof does. I think you are misunderstanding the hypothesis. It doesn't say that every subset of $k$ toys has altogether the same color, but rather each subset of $k$ toys has its own unique color shared throughout that subset. This is indeed what the basis shows. – MT_ Nov 30 '15 at 17:15
  • @Soke: There are two universes. In the universe where all the toys are the same color, the given Inductive Hypothesis is true. In the universe where not all toys are the same color, the given inductive hypothesis is false. Consequently, the given inductive hypothesis is equivalent to the assumption that all toys have the same color. – Eric Towers Nov 30 '15 at 19:16