Theorem. All horses are the same color. Proof. We'll induct on the number of horses. Base case: 1 horse. Clearly with just 1 horse, all horses have the same color. Now, for the inductive step: we'll show that if it is true for any group of N horses, that all have the same color, then it is true for any group of N+1 horses Well, given any set of N+1 horses, if you exclude the last horse, you get a set o N horses. By the inductive step these N horses all have the same color. But by excluding the first horse in the pack of N+1 horses, you can conclude that the last N horses also have the same color. Therefore all N+1 horses have the same color. QED. Hmmn... clearly not all horses have the same color. So what's wrong with this proof by induction?

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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Theorem. All horses are the same color.
Proof. We'll induct on the number of horses. Base case: 1 horse. Clearly with
just 1 horse, all horses have the same color.
Now, for the inductive step: we'll show that if it is true for any group of N
horses, that all have the same color, then it is true for any group of N+1 horses.
Well, given any set of N+1 horses, if you exclude the last horse, you get a set of
N horses. By the inductive step these N horses all have the same color. But by
excluding the first horse in the pack of N+1 horses, you can conclude that the
last N horses also have the same color. Therefore all N+1 horses have the same
color. QED.
Hmmn... clearly not all horses have the same color. So what's wrong with this
proof by induction?
Transcribed Image Text:Theorem. All horses are the same color. Proof. We'll induct on the number of horses. Base case: 1 horse. Clearly with just 1 horse, all horses have the same color. Now, for the inductive step: we'll show that if it is true for any group of N horses, that all have the same color, then it is true for any group of N+1 horses. Well, given any set of N+1 horses, if you exclude the last horse, you get a set of N horses. By the inductive step these N horses all have the same color. But by excluding the first horse in the pack of N+1 horses, you can conclude that the last N horses also have the same color. Therefore all N+1 horses have the same color. QED. Hmmn... clearly not all horses have the same color. So what's wrong with this proof by induction?
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