To: WSFAlist at keithlynch.net Date: Fri, 21 Jun 2002 00:11:33 -0400 Subject: [WSFA] Goldbach's Conjecture From: ronkean at juno.com Reply-To: WSFA members <WSFAlist at keithlynch.net> On Wed, 19 Jun 2002 12:03:30 -0400 Steve Smith <sgs at aginc.net> writes: > ronkean at juno.com wrote: > > ... what is the difference between a human > > mathematician and a computer which explains why the human can do > > what the computer cannot? > > Because proofs are done by reasoning about the problem, not brute > calculation. There are specialized "therom proving" computer > programs, > but I don't know how effective they are in the general case. My > experience with similar programs is that it's almost as much work > to code the problem into a form that the computer can solve as it is > to solve it by hand. So it sounds like the difference may be that humans are more intelligent than computers. Usually, the ability to play chess competitively is considered a mark of intelligence. But an ordinary desktop computer, running a good chess-playing program, can probably consistently defeat 90% or more of all chess players, though most would concede that a desktop computer is less intelligent than a house cat. And we know that a few years ago, 'Big Blue', an IBM computer, defeated the world chess champion, though that computer had been programmed in consultation with chess experts who had helped program the machine to specifically address the playing style of that particular human champion. Nevertheless, it seems that technology by now has reached a level where a properly programmed computer can be expected to beat any human at chess. But in a broader sense, such a computer would not be considered to be very intelligent at all, inasmuch as it can't even conduct a conversation on a human level, so it may be that the idea that chess-playing ability is always related to intelligence, is false. Human chess players apply to the game principles derived from experience and study of past games, and estimations of the psychology of their opponents, methods which involve intelligence, whereas the chess playing skill of a non-intelligent machine is almost solely based on mere computational ability. But that may change with the development of artificial intelligence, and future chess playing machines may use both intelligence and computation to play the game. With artificial intelligence under development, there is the prospect that one day computers will be more intelligent than humans, in some meaningful sense, particularly that computers may become better than humans at devising mathematical proofs. Since logic figures heavily in mathematical proofs, it seems that an intelligent computer would have a better shot at doing mathematical proofs than it would at writing novels as skillfully as Ernest Hemingway (or was that Hemmingway?). When computers become much more intelligent than humans, perhaps a computer will devise a proof of Golbach's Conjecture. But ironically, it may happen that no human will be able to understand the proof. Many people are in that position today with mathematical proofs; they wait to see whether the math community has blessed the alleged proof, to decide whether to accept the proof as valid. Those who don't understand the proof don't really know that it is true in the same sense that those who do understand it may know - they are just taking others' word for it. If a computer devises a proof of Golbach's Conjecture which no human can understand, we would have to ask other computers if the proof is valid. If they all say yes, the validity of the proof would be confirmed, in a sense. This suggests that there is limit to how useful super-intelligent AI computers can be to humans, a limit imposed by the limited intelligence of humans themselves, and the fact that humans can't always understand what the computers are doing, or what they could do. Such computers might at best indulge humans as we do our pets. With your algebraic proof of the S = n(n+1)/2 formula, you provided an easily understndable example of how an algebraic proof can prove a theorem which applies to an infinite number of cases, without doing infinitely many calculations. I once had need to derive that formula for the sum of the integers 1 through n, and I did it geometrically. I drew a pattern like so: 1 * 2 ** 3 *** 4 **** 5 ***** 6 ****** and noticed that a triangle of dots was formed, and that the number of dots, where n is the number of numbers in the left column, is equal to n^2/2 + n/2, which can be re-written as n(n+1)/2, and that the number of dots is equal to the sum of the numbers in the left column. The reasoning is that if extra dots were filled in to make a square of dots, the number of dots would be n^2, like this: ****** ****** ****** ****** ****** ****** Then if we cut that square in half diagonally, cutting each of the dots in the central diagonal in half, and count up the dots, including counting each half dot as half a dot, we get n^/2 dots. n/2 dots needs to be added to make up for the missing n half dots in the central diagonal. Inspection of the diagram was proof enough for me that the formula is correct, but perhaps that was not a rigorous proof. Ron Kean . ________________________________________________________________