This week's giveaway is in the EJB and other Java EE Technologies forum. We're giving away four copies of EJB 3 in Action and have Debu Panda, Reza Rahman, Ryan Cuprak, and Michael Remijan on-line! See this thread for details.

Theorem: All horses are of the same colour. Proof: We prove this by induction on the number of horses in a group. For n = 1, there is only one horse and hence nothing to prove. Assume that given any group of n horses, all are of the same colour. To prove the above statement for n + 1, consider a group of n + 1 horses. Remove one horse from the group. Then, by assumption, remaining n horses are of the same colour. Replace the horse and remove another one from the group. Now the remaining n are of the same colour. Hence all the n + 1 horses are of the same colour. [ October 15, 2003: Message edited by: Capablanca Kepler ]

This is easy. It fails when we have two horses. remove 1 horse. we have one left, and it must be the same color as itself. replace that one, and remove the other. it must be the same color as itself. BUT, there is nothing connecting the two sets togther. you have not proved that there is a RELATIONSHIP between the two groups. in other words, the color of one horse has no bearing on the color of another. [ October 16, 2003: Message edited by: fred rosenberger ]

There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors