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It is stated in step 4 that we are trying to prove that, in every group of
*k*+1 people,
everyone has the same age.

In general, the way to prove that every "whatchamacallit" has a certain
property is to
consider a typical whatchamacallit *x*, and show that *x* has the
desired
property. As long as the argument works no matter which whatchamacallit is
being considered,
we will have proven that *all* whatchamacallits have the
property.

In our case, a "whatchamacallit" means a group of *k*+1 people.
So, we consider a typical watchamacallit, *G*, and we will show
that *G* has the desired property, namely that everyone in *G* is
the same age.

As long as our argument works no matter which group *G* of *k*+1
people
is being considered, we will have proven that, in every group of *k*+1
people,
everyone has 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