[ahecht]

Originally Posted by Solace

(-1/1)^.5 = (1/-1)^.5

((-1)^.5) / (1^.5) = (1^.5) / ((-1)^.5)

Actually, if (-1/1)^.5 = (1/-1)^.5, then (+-)((-1)^.5) / (1^.5) = (+-)(1^.5) / ((-1)^.5)

Therefore, the final step of the proof will state (+-)1=(+-)1, which is true.


Here is another interesting one: a proof by induction that everyone who reads Chief Delphi is the same age!

First, a little refresher of proof by induction. In an inductive proof, we prove that statement S(n) is true when n=1, and then prove that if S(n) is true, then S(n + 1) is true. If we prove these two things, we have then proven that S(n) is true for all values of n.

Statement S(n): In any group of n Chief Delphi readers, everyone in that group has the same age.

First I prove that S(1) is true:

1. In any group that consists of just one Chief Delphi reader, everybody in the group has the same age, because after all there is only one person!
2. Therefore, S(1) is true.

Next, I prove that if S(n) is true, then S(n + 1) must also be true.

1. Let G be an arbitrary group of n+1 Chief Delphi readers; we just need to show that every member of G has the same age.
2. To do this, we just need to show that, if P and Q are any members of G, then they have the same age.
3. Consider everybody in G except P. These people form a group of n Chief Delphi readers, so they must all have the same age (since we are assuming that, in any group of n Chief Delphi readers, everyone has the same age).
4. Consider everybody in G except Q. Again, they form a group of n Chief Delphi readers, so they must all have the same age.
5. Let R be someone else in G other than P or Q.
6. Since Q and R each belong to the group considered in step 3, they are the same age.
7. Since P and R each belong to the group considered in step 4, they are the same age.
8. 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.
9. 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.

The proof is now complete: we have shown that the statement is true for n=1, and we have shown that whenever it is true for n it is also true for n+1, so by induction it is true for all n.