Subject: lojban and group theory

From: Brook Conner <[email-masked]>
Date: Fri, 25 Jan 2002 00:00:47 -0500

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