my teacher gave the following as an example of a fallacious proof:

We'll prove that a group of $n$ people are either all male or all female.

For $n = 1$ The claim says that a group containing 1 person is a group containing either only males or only females, which is correct. Inductive Step: Let's assume P(n) is correct. Let $A = \{{a_1,a_2,...a_n,a_{n+1}}\}$ be a group of $n+1$ people. We'll define $A' = \{{a_1,a_2,...a_n}\}$ and $A'' = \{{a_2,...a_n,a_{n+1}}\}$. There are two options, either $a_2$ is male or female. Let's assume $a_2$ is a male. Since A' and A'' are n-sized groups, they contain all females or all males, and since $a_2$ is a member of both, they both contain all males. But $A = A'\cup A'' $, so A contains only males. A similiar approach is used if $a_2$ was a female.

What is the fallacy in this proof?


marked as duplicate by Asaf Karagila logic Feb 25 '15 at 11:07

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  • $\begingroup$ How you thought about it? Is the statement true? :) $\endgroup$ – Peter Franek Feb 25 '15 at 11:07
  • $\begingroup$ @PeterFranek he knows it isn't true! He wants to find where the proof is fallacious. $\endgroup$ – JP McCarthy Feb 25 '15 at 11:07
  • $\begingroup$ Hint:Does the inductive step work for all n values? Think of small ones... $\endgroup$ – Zamu Feb 25 '15 at 11:16
  • $\begingroup$ I'm sorry, I don't see why the proof fails at n=2. If n=2, then A' = {a1, a2}, A'' = {a2,a3}. Both sets have only one gender because their size is n, and both contain a2, and therefore are all male. $\endgroup$ – blz Feb 25 '15 at 14:57