Re: Defining "<" for the rationals

Bill Dubuque schrieb: 
Michael Stemper <mstemper@xxxxxxxxxxxxxxxx> wrote:

I've been looking at the rational numbers as an equivalence relation.
I've been able to show that the definitions of multiplication and addition
give the same answer, regardless of which member of an equivalance set
is chosen. I wanted to also define the "<" relation, but hit a stumbling
block. Most of the time, you can say (a,b)<(c,d) iff ad<bc. However, this
falls apart if b or d is negative.



You may use   a/b < c/d  <->  ad bd < bc bd


Oh well, I always like your fine comments.
My signum function solution was correct
but cumbersome. bd = |bd| * signum(bd)
and |bd| > 0 allow the simplification of
my suggestion 

[(a,b)] < [(c,d)] <-> signum(bd)*ad < signum(bd)*bc

by multiplying both sides of the inequality with
factor |bd|, yielding your elegant solution.

My best regards to you,
Rainer Rosenthal
r.rosenthal@xxxxxx
.

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