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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