Thursday, January 10, 2008

Everyone in Canada Is the Same Age

Just recently, I randomly came across a jewel of a fallacious proof by induction. Here it is, the beast:

In any group that consists of just one person, everybody in the group has the same age.

Therefore, statement S(1) is true.

The next stage in the induction argument is to prove that, whenever S(n) is true for one number (say n=k), it is also true for the next number (that is, n = k+1).

We can do this by (1) assuming that, in every group of k people, everyone has the same age; then (2) deducing from it that, in every group of k+1 people, everyone has the same age.

Let G be an arbitrary group of k+1 people; we just need to show that every member of G has the same age.

To do this, we just need to show that, if P and Q are any members of G, then they have the same age.

Consider everybody in G except P. These people form a group of k people, so they must all have the same age (since we are assuming that, in any group of k people, everyone has the same age).
Consider everybody in G except Q. Again, they form a group of k people, so they must all have the same age.

Let R be someone else in G other than P or Q.

Since Q and R each belong to the group considered in step 7, they are the same age.

Since P and R each belong to the group considered in step 8, they are the same age.

Since Q and R are the same age, and P and R are the same age, it follows that P and Q are the same age.

We have now seen that, if we consider any two people P and Q in G, they have the same age. It follows that everyone in G has the same age.

Q.E.D. We have shown that the statement is true for n=1, and we have shown that whenever it is true for n=k it is also true for n=k+1, so by induction it is true for all n.

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