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I'm going to prove that everyone's eyes are the same color. Ready?

If there is only one person, then it's obviously true; this person's eyes are the same color that this person's eyes.

Suppose it is established that $(n-1)$ persons must have the same eye color. Consider $n$ persons: the $(n-1)$ first have the same eye color, and the $(n-1)$ last have the same eye color. Since the two overlap, everyone has the same eye color.

My initialization is verified, and so is my induction. Since I have brown eyes, everyone has brown eyes. Wait a minute, what?

Jonathan H
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    This is the classical "every horse is the same color" problem. Variant thereof is covered here: http://math.stackexchange.com/questions/541852/basic-induction-probs – apnorton Apr 24 '14 at 01:39
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    Just when reading this question, the votes were - $$2\to1\to0\to1$$ –  Apr 24 '14 at 01:39
  • @anorton Sorry I didn't find this one, I just voted to close the post. Should I delete it instead? – Jonathan H Apr 24 '14 at 01:56
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    @Sh3ljohn Since it has an upvoted answer, you cannot delete it. It's probably better to close instead of delete, anyway; if it's closed as a dup., then there will be an obvious link between this question and the other one. Then, if someone googles "eye problem mathematical induction," they can easily find the answer here. – apnorton Apr 24 '14 at 01:57

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This works except when $n=2$.

That's why one can say that if any TWO people have eyes of the same color, then everyone's eyes have the same color.

Two commonplace forms of mathematical induction are these:

  • The case $n=1$ is trivial, and the hard part is the induction step;
  • The case $n=1$ is vacuously true; the induction step is trivial and relies on the case $n=2$ and on the induction hypothesis; and the hard part is the case $n=2$.

Your proof is of the second kind.

Michael Hardy
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