# The argument fails for two element sets!

There might not be anyone else in G other than P or Q!

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 if S(2) is true, then so is S(3), and if S(3) is true, so is S(4), and so on.

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.