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Remember that *G* is a group of *k*+1 people. As long as *k *> 1,
*k*+1 > 2 and in this case there does exist a third person *R* in
*G*. And, the rest of the proof will work.

So what does this prove? It proves that *S*(*k*) implies *S*(*k*+1),
*as long as k>1*. In other words, it proves that

But nowhere have we shown that *S*(2) is true! The basis step of
this induction showed that *S*(1) was true, but our induction step
does not show that *S*(1) implies *S*(2).

Thinking about it in the following way may help. What this proof boils down to is this. It is showing that if every pair of people has the same age, then so does every group of three people, and so does every group of four people, and so on. But that's not anything startling; it should be fairly obvious if you think about it.

The problem is that it is not true that every pair of people has the same
age! It is true that, in any group consisting of just one person,
everyone in the group has the same age. However, the proof breaks
down when you try to use that fact to prove that every pair has
the same age. If {*P*,*Q*} is a pair of people, there's no third person
*R* in the pair to make this step of the argument work, and
therefore no way to
conclude that *P* and *Q* have the same age by comparing them each to *R*,
which is what the rest of the proof tries to do.

So: *S*(1) is true, and *S*(2) implies *S*(3) which implies *S*(4) etc.,
but *S*(1) does not imply *S*(2) and that is why the proof is fallacious.

To see the explanations for the other steps, finding out exactly why they're correct (or, in some cases, finding out which other steps have slight mistakes in them), go back to the list of steps in the proof. To try your hand at finding the fallacy in a different problem, go back to the Classic Fallacies index page.

This page last updated: May 26, 1998

Original Web Site Creator / Mathematical Content Developer: Philip Spencer

Current Network Coordinator and Contact Person: Joel Chan - mathnet@math.toronto.edu