From: Stephen Turner (sret1_at_ntlworld.com)
Date: Tue Feb 05 2002 - 03:54:45 PST
We now have six players. Any more want to jump in? New Michael? Anyone who wants to do so has just over a day left. And James, you still have time to have another go if you're quick. -------------------------------------------------------------------------- ROUND 176 Round start: Wed 2002-01-30 15:32:50 Player Style Valid until ------ ----- ----------- Ed Murphy - 0.5 Mon 2002-02-11 22:23:30 Rich Holmes + 2.0 Fri 2002-02-08 02:42:41 Factitious + 2.0 Fri 2002-02-08 00:23:06 Jonathan Van Matre - 2.5 Wed 2002-02-06 18:03:06 Alan Riddell + 2.0 Wed 2002-02-06 16:50:44 Others 0 Wed 2002-02-06 15:32:50 James Willson - 2.0 Tue 2002-02-05 15:32:50 All times are in GMT and base ten. -------------------------------------------------------------------------- Rule By whom When Judgement Style ---- ------- ---- --------- ----- 176:1 Alan Riddell Wed 2002-01-30 16:50:44 VALID + 2.0 176:2 Jonathan Van Matre Wed 2002-01-30 18:03:06 VALID - 2.5 176:3 Ed Murphy Wed 2002-01-30 19:44:52 VALID - 1.5 176:4 Factitious Fri 2002-02-01 00:23:06 VALID + 2.0 176:5 Rich Holmes Fri 2002-02-01 02:42:41 VALID + 2.0 176:6 Ed Murphy Mon 2002-02-04 22:23:30 VALID + 1.0 177:7 James Willson Tue 2002-02-05 08:51:33 INVALID - 2.0 -------------------------------------------------------------------------- 176:1 VALID +2.0 Alan Riddell 2002-01-30 16:50:44 >>>>> Future rules shall write all numbers in base 11. e.g. 0,1,2,3,4,5,6,7,8,9,a,10,11,... and therefore for rest of this round, the round shall be known as "round 150". <<<<< Judgement: No problems. Style: A promising start, setting a direction and yet leaving future rules plenty of options. Nice and short too -- I like short rules. I give it +2. -------------------------------------------------------------------------- 176:2 VALID -2.5 Jonathan Van Matre 2002-01-30 18:03:06 >>>>> Define S as a set containing all base 11 numbers. Define U as a set containing the letter "R". Define US as the union of set S and set U. Then, All your base 11, "R", belong to US. QED. Future Fantasy Rules must use the term "QED", or its lengthier expression "quod erat demonstrandum". <<<<< Judgement: Note that eleven is spelled "10". "11" is twelve. However, that doesn't invalidate this rule. Style: Sorry, but I dislike this rule. "All your base are belong to us" is rather a tired joke by now. It can still be funny (you must have a look at http://www.randomdrivel.com/media/ayb3.swf some time) but this is just a feeble -- and ungrammatical -- contrived pun because the first rule happened to have the word "base" in it. In addition, the rule doesn't advance the round much: the restriction is easy to obey. "QED" is used wrongly (it should follow a proof, not a plain statement). And marks off for using "11" when you probably meant "10". I'm a judge who is not afraid to use the full range of style points. But I don't quite give this one the minimum because JVM is a newish player and because I realise that humour is subjective. -2.5. -------------------------------------------------------------------------- 176:3 VALID -1.5 Ed Murphy 2002-01-30 19:44:52 >>>>> A + a = 15 (A + a) - a = 15 - a A + (a - a) = 15 - a A = 15 - a A = 6 QED To avoid further alphanumeric confusion, future rules shall choose algebraic symbols from the ISO-6842 alphabet. This is the same as the standard English alphabet, except with these symbols omitted: O I Z E H S G L B Q A o i z e h s g l b q a <<<<< Judgement: A tough one to judge but I'm calling it invalid. The problem is the "ISO-6842". 6842 = 9000 base 10, and ISO 9000 is a well-known standard about business processes and quality control. Nothing to do with alphabets at all. I might have let you get away with it if this was one of the story telling rounds where we build a fantasy world, but here you should perhaps have chosen an unused number, or not tried to name a standard at all. Re-judgement: OK, I accept that I made a mistake here. I did consider this point -- that rules are only obliged to be consistent with previous rules, not with reality -- and concluded that to some extent at least, rules _do_ have to be consistent with reality. Otherwise whenever a rule was invalid, you could just say "ah, but the word means the opposite in my world". What I failed to take into account in this case, however, was that the word was _explicitly_ redefined in the fantasy world, and that has to be allowed. I reverse my previous judgement. Style: Again, we don't seem to be making much progress. In fact, the maths appears to be the same as in our real world -- why should the symbols +, - and = have their normal interpretations, for example? Also failing to obey your own restrictions is always unstylish. -------------------------------------------------------------------------- 176:4 VALID +2.0 Factitious 2002-02-01 00:23:06 >>>>> Fantasy math is equipped to deal with problems others consider insolvable. For example, dividing by 0 is often considered to be taboo. We can get around this by defining the constant j as the inverse of 0, so that 0 times j is equal to 1. This makes possible new solutions to many difficult equations, as can be shown by the following proof that there is a value x satisfying 0x - 4 = 9: 0x - 4 = 9 0x = 12 x = 12j QED Numbers containing j are not considered real, but are part of the set of fantasy numbers. Future rules must each contain at least 1 fantasy number. <<<<< Judgement: No problems. Style: Great, we've now got some new fantasy maths introduced -- we'll see what effect introducing infinities has in future. Also makes a restriction on future rules. And it uses base 11. -------------------------------------------------------------------------- 176:5 VALID +2.0 Rich Holmes 2002-02-01 02:42:41 >>>>> A classic "proof" that 0 = 1 goes as follows: 1j = 1j property of equality 1j-1j = 0 subtract 1j from both sides (1-1)j = 0 distributivity of subtraction 0j = 0 1-1 = 0 j = 0/0 = 0j divide both sides by 0 1 = 0 QED divide both sides by j The fallacy is of course that fantasy subtraction (and addition) are not distributive. Future rule writers would do well to bear this in mind. <<<<< Judgement: Valid. The usual phrase is "multiplication is distributive over addition" but the meaning is obvious here, so that's not enough to make it invalid. Style: A very nice reply to 176:4, working out the consequences of the new maths introduced there. You're absolutely right, of course, distributivity must fail when we allow infinities in our system. However, although it's very clever I do doubt whether the restriction is in fact restrictive at all. It may lead to some subtle traps but I suspect that it won't in fact cause any trouble. -------------------------------------------------------------------------- 176:6 VALID +1.0 Ed Murphy 2002-02-04 22:23:30 >>>>> Here is another "proof" that 1 = 0: j*0 = 1 (definition of j) (j*0)*0 = 1*0 (multiply both sides by 0) j*(0*0) = 0 (associativity of multiplication) j*0 = 0 (replace 0*0 with 0) 1 = 0 (definition of j) QED The fallacy is that multiplication is not associative. The next valid rule must either prove or disprove that fc + rc = (f+r)c unless c is a fantasy number and the next valid rule after that must either prove or disprove that (fr)c = f(rc) unless (f = 0 and c is a fantasy number) or (f is a fantasy number and c = 0) <<<<< Judgement: I can't see any problems. Style: It's a bit too similar to the previous rule, and there is no use of base 11. But it does make the next two rules do some specific maths, which is thematic, if maybe a bit too prescriptive. I would give it +1.5, but in fact I'm only going to give it +1 for a reason which I shall reveal later. -------------------------------------------------------------------------- 176:7 INVALID -2.0 James Willson 2002-02-05 08:51:33 >>>>> Perhaps the most famous fantasy maths problem is the proof that 1 = 0. It was commonly believed for centuries, but until the groundbreaking work of Set, all "proofs" that 1 = 0 were unsound. Set, of course, is the discoverer of k, a fantasy number. k is defined as 0^0. Set used this to construct the first sound proof that 1 = 0. k = k (all fantasy numbers equal themselves) 0^0 = 0^0 (definition of k) 1 = 0^0 (for all n, n^0 = 1) 1 = 0 (for all n, 0^n = 0) QED This has some interesting implications. Since 1 = k = 0, both 1 and 0 must be fantasy numbers. Set extended this result to show that all numbers are fantasy numbers. (This is called Set's Fantastic Completeness Theorem) We'll use this to prove the following trivially true: fc + rc = (f+r)c unless c is a fantasy number Since all numbers c are fantasy numbers, the proposition reduces to fc + rc = (f+r)c unless true and true -> b is true for arbitrary b. QED Set made many important contributions to fantasy maths. All future rules will describe some such contribution. <<<<< Judgement: Invalid. If true => b is true for all b then all statements are true. But there are false statements, e.g., multiplication is associative. Style: I was glad to find this blunder. The idea that 1 does in fact equal 0 is superficially amusing, but it implies that all numbers -- or at least all of our numbers -- are equal, which would cause the whole number system to collapse. (Of course the bigger leap in your proof is not the existence of 0^0, but the axioms that n^0 = 1 and 0^n = 0 for all n.) There are some positive aspects to this rule. First of all, it exploited the fact that fantasy numbers haven't yet been defined. (This was also the reason that I knocked off 0.5 points at 176:6 -- this loophole makes those theorems much easier to prove or disprove.) Secondly it introduces us to a mathematician. And finally the notion that all numbers are fantasy numbers is amusing, and doesn't seem to be problematic. Nevertheless, I think that this would have caused a lot of trouble, and I'm going to give it -2.0. -------------------------------------------------------------------------- -- Stephen Turner, Cambridge, UK http://homepage.ntlworld.com/adelie/stephen/ "This is Henman's 8th Wimbledon, and he's only lost 7 matches." BBC, 2/Jul/01 -- Rule Date: 2002-02-05 11:55:02 GMT
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