Re: Defining "<" for the rationals

Michael Stemper 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.

I could say "chose a member of [(a,b)] with a positive second element,"
or I could say "if b<0, change the signs of a and b." Neither of these
seem particularly elegant. Is there a cleaner way to define the less
than relation on rationals?


p < q  iff  q - p is positive.

[(a,b)]  is positive iff  a b  is.

--
Stephen J. Herschkorn                        sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey and Manhattan

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