"Rainer Deyke" <rainerd at eldwood.com> wrote in message news:kTEJb.724242$HS4.5376202 at attbi_s01... > Sean Ross wrote: > > a = bq + r and 0<=r<b [1] > > But 0 <= r < b is a contradiction when b < 0. > Right. But, the division algorithm states "Let a,b be integers with b>0" (which I mentioned in that post).