If you know how to prove things by induction, then here is an amazing fact:

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?

Presentation Suggestions:

This delightful puzzle is an excellent test of student understanding of proofs by induction.

The Math Behind the Fact:

Hint: what could be wrong? You showed the base case. And you showed the inductive step, right?

Well actually, the argument in the inductive step breaks down in going from n=1 to n=2, because the first 1 horse and the last 1 horse have no horses in common, and therefore may not all have the same color.

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