All horses are the same colour; we can prove this by induction on the number of horses in a given set. Here's how: "If there's just one horse then it's the same colour as itself, so the basis is trivial. For the induction hypothesis, horses \(1\) through \(n - 1\) are the same colour, and similarly horses \(2\) through \(n\) are the same colour. But the middle horses, \(2\) through \(n - 1\), can't change colour when they're in different groups; these are horses, not chameleons. So horses \(1\) and \(n\) must be the same colour as well, by transitivity. Thus all \(n\) horses are the same colour; QED." What, if anything, is wrong with this reasoning?



Tim Ophelders solved this puzzle:

Denote by \(E(H)\) that all horses in a set \(H\) have the same color. Now consider the case of \(n = 2\) with horses \(\{a, b\}\), then the induction hypothesis must prove that \(E(\{a\})\) and \(E(\{b\})\) implies \(E(\{a, b\})\). Note that the middle set of horses is empty for \(n = 2\), such that \(E(\{\})\) holds no matter what color is attributed to the middle set of horses. The claim that the middle horses cannot change color is therefore misleading since they don't have a unique color.

Rino Raj solved this puzzle:

This apparent paradox is an old chestnut. The induction step needs horses \(2\) through \(n - 1\) to exist which is not possible for \(n = 2\).

Ed Murphy solved this puzzle:

The induction fails when going from one horse to two horses; for \(n = 2\), the set of horses \(1\) through \(n - 1\) and the set of horses \(2\) through \(n\) are disjoint sets, so the set of middle horses, \(2\) through \(n - 1\), is an empty set.

Wonderwice Margera solved this puzzle:

The argument assumes that the group of 'middle horses', i.e. those numbered from \(2\) to \(n - 1\), has at least one horse. This fails when \(n\) equals \(2\). That is why two horses (and therefore, even more than two horses) can have different colours.


This puzzle is taken from:

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