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What is wrong with the following “proof” by mathematical induction that all cats are black?

Let P(n) denote the statement “In any group of n cats, if one cat is black, then they are all black.”

  • Step 1 The statement is clearly true for n = 1.

  • Step 2 Suppose that P(k) is true. We show that P(k+1) is true.

Suppose we have a group of k+1 cats, one of whom is black; call this cat “Tadpole.” Remove some other cat (call it “Sparky”) from the group. We are left with k cats, one of whom (Tadpole) is black, so by the induction hypothesis, all k of these are black. Now put Sparky back in the group and take out Tadpole. We again have a group of k cats, all of whom—except possibly Sparky—are black. Then by the induction hypothesis, Sparky must be black too. So all k+1 cats in the original group are black. Thus by induction p(n) is true for all n. Since everyone has seen at least one black cat, it follows that all cats are black.

Abdullah Alfaqir
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    Look at what is going on for $P(2)$ (that is assume $P(1)$ and try to prove $P(2)$). – Roland Jun 02 '17 at 15:42
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    What is wrong is that my neighbour has a golden brown cat. – Asinomás Jun 02 '17 at 15:44
  • It is $k=2$ that creates trouble. When we put Sparky back in "the group" and remove Tadpole, realize that the group consists of only Sparky, and hence if Sparky is not black, the induction hypothesis would not apply for the group consisting precisely of Sparky only. But very nice question, it will help your understanding of induction. – Sarvesh Ravichandran Iyer Jun 02 '17 at 15:45
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    This is a very old exercise. It was being bruited about during my undergraduate years, and that was in a prior geological era. – Lubin Jun 02 '17 at 15:50
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    This has it's own Wikipedia page :) – Blaza Jun 02 '17 at 16:22

3 Answers3

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Your proof breaks here:

"Now put Sparky back in the group and take out Tadpole. We again have a group of k cats, all of whom—except possibly Sparky—are black. Then by the induction hypothesis, Sparky must be black too."

You cannot use the induction hypothesis because k can be equal to 1 and you are not sure that Sparky is black.

gcc-6.0
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Your argument breaks down for $P(2)$. When you have a group of two cats, and you remove Sparky, only Tedpole is left. No other black cat can be proven to be in the group (because it has size one). So by adding Sparky back in and removing Tedpole you cannot be sure there is a black cat left.

Still a nice fake proof.

M. Winter
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Hint:

Using the inductive assumption, try to prove for $\;P(2)\;$ ...

DonAntonio
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