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This step is not the source of the fallacy.

This step is correctly stating what it means for everybody in G to have the same age: it means that every pair of members of G have the same age.

To prove that every pair of members of G have the same age, it is sufficient to let P and Q be arbitrary members of G and prove that they have the same age. As long as we do this in a way that doesn't depend on which two members we happened to pick (that's what it means to say that P and Q are "arbitrary"), then we have succeeded in proving that every pair of members of G have the same age.

Why don't you go back to the list of steps in the proof and see if you can identify which one is wrong, now that you know it isn't this one?
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

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