Rabbits Reproduce; Integers Don't

This is a classical conundrum, although the first time I heard it it was about horses of the same color, not rabbits as below. So, the idea is to use induction to prove that all rabbits are of the same color.

For the basis of induction pick any one rabbit. By default, the rabbit is of the same color with itself. For the inductive step, assume that any set of fewer than n + 1 rabbits have been proved to be monochromatic. Pick n + 1 rabbit. Temporarily remove one rabbit from the set, leaving a set, say A, of n rabbits. By the induction hypothesis,lemma,theorem,hypothesis,proposition, these n rabbits are of the same color. Put the "unused" rabbit back into the set and remove another one, leaving set B. As before, all rabbits in set B are of the same color. All the rabbits at hand (except for the special two) belong to both sets A and B, implying that the rabbits in both sets are of the same color. But the union A∪B contains all n + 1 rabbits which thus necessarily are all of the same color.

Edward Barbeau quotes a computer science major who gave the following analysis:

The faulty logic in the rabbit problem has to do with the behavior of rabbits and integers. For example, a set of rabbits is not isomorphic to a set of integers because their behavior is different. A set of integers is static while a set of rabbits is dynamic, i.e., integers don't reproduce themselves while rabbits do. Since the two sets are not isomorphic, we cannot expect the same laws to apply to both sets. Since we know hat the law of induction applies to the set of integers, we cannot expect the same law to apply to a set of rabbits that is not isomorphic. In the same manner, we cannot expect the laws of reproduction which apply to a set of rabbits to apply to a set of integers.

Discussion

References

  1. E. J. Barbeau, Mathematical Fallacies, Flaws, and Flimflam, MAA, 2000, p. 63

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What Went Wrong?

Proof by mathematical induction consists of two,one,two,three,foursteps: the base of induction wherein the statement to be proved is verified for one or more first admissible values of the parameter.

The induction is often compared to a chain of upstanding dominoes,dominoes,cubes,pyramids,Santa's reindeersthat fall down one after another once the first one was set in motion. The second (inductive) step need to be able to continue from the point verified at the first step. Since on the first step we only verified the statement for 1 rabbit, there is no way to proceed with the inductive step from n = 1 to n = 2. Indeed in this case, the intersection A∩B = Ø, i.e., empty. There are no rabbits that might have shared the colors,tastes,colors,habits,living quartersof the given two.

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Copyright © 1996-2018 Alexander Bogomolny

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