From me@nellardo.com Thu Jan 24 21:00:49 2002 Return-Path: X-Sender: me@nellardo.com X-Apparently-To: lojban@yahoogroups.com Received: (EGP: mail-8_0_1_3); 25 Jan 2002 05:00:49 -0000 Received: (qmail 30855 invoked from network); 25 Jan 2002 05:00:48 -0000 Received: from unknown (216.115.97.172) by m10.grp.snv.yahoo.com with QMQP; 25 Jan 2002 05:00:48 -0000 Received: from unknown (HELO mail5.speakeasy.net) (216.254.0.205) by mta2.grp.snv.yahoo.com with SMTP; 25 Jan 2002 05:00:48 -0000 Received: (qmail 15055 invoked from network); 25 Jan 2002 05:00:48 -0000 Received: from unknown (HELO dsl027-135-047.nyc1.dsl.speakeasy.net) ([216.27.135.47]) (envelope-sender ) by mail5.speakeasy.net (qmail-ldap-1.03) with SMTP for ; 25 Jan 2002 05:00:48 -0000 Date: Fri, 25 Jan 2002 00:00:47 -0500 Subject: lojban and group theory Content-Type: text/plain; charset=US-ASCII; format=flowed Mime-Version: 1.0 (Apple Message framework v480) To: lojban@yahoogroups.com Content-Transfer-Encoding: 7bit In-Reply-To: <3C5081EB.3090405@reutershealth.com> Message-Id: <7EDC1C4F-1150-11D6-9015-003065B787D6@nellardo.com> X-Mailer: Apple Mail (2.480) From: Brook Conner X-Yahoo-Group-Post: member; u=66018878 X-Yahoo-Profile: nellardo For reference, I'm using the text *Topics in Algebra* by i. n. herstein, Wiley pub. as my reference for group, ring, and field theory. On Thursday, January 24, 2002, at 04:51 pm, John Cowan wrote: > Brook Conner wrote: >> (mathematically speaking, IEEE floating point numbers are none of >> these, >> are not even a group (I don't think NaN has an additive inverse, and >> Inf >> has very peculiar behavior)). > The additive inverse of NaN is NaN, I think, depending on your > definition of inverse: NaN + NaN is not 0, but 0 - NaN is definitely > NaN. A group is a set G and an operation (usually denoted +) such that: 1. For a,b in G, a + b is in G. IEEE FP does that. 2. For a, b, c in G, a + (b + c) = (a + b) + c). This is associativity. IEEE FP does this. 3. There exists an element 0 s. t. a + 0 = 0 + a = a. You can prove that 0 is unique, though it isn't part of the definition. If a is NaN, IEEE does that. 4. For any a in G, there exists one element b s. t. a + b = 0. NaN + NaN is NaN, and NaN - NaN is NaN. There is nothing you can add to NaN and get 0. Ergo, strictly speaking, IEEE FP numbers do not form a group in the algebraic sense. They have lots of other useful properties, like commutativity (a+b = b+a, something not necessarily true for a group). So when looking at the semantics of lojban in a programmatic sense, we need to be careful about exactly what we say those semantics are. And as all this IEEE bumpf shows, to some extent, the right formality/expressiveness/utility balance is domain-specific. Mathematicians would not be satisfied if mekso *had* to be IEEE floats or even doubles, though your average Perl programmer would probably be just fine with that. > As for not being a group, if they weren't a group over the defined > IEEE operations, that would mean that something not an IEEE-float > was being delivered, which is self-contradictory, since every > bit combination has an IEEE meaning. No, the fact that every bit combination means something just means the operation is closed. That's part 1 of the definition of a group. >> Another problem of semantics is choosing the subset of the vocabulary >> and making sure the user knows what it is. Are you going for an >> imperative model? Lots of "ko" running around. > > "ko" can be defaulted, though, if that's a convention established > between speaker and listener. Ah, missed that. Thanks. Brook