
Re: Solving rational polynomials
Posted:
Aug 23, 2013 12:39 AM


On Wednesday, August 21, 2013 3:43:24 AM UTC7, Kobu wrote: > On Saturday, August 17, 2013 4:26:18 AM UTC4, Virgil wrote: > > > In article <0388f571326f437f990fb3000253e952@googlegroups.com>, > > > > > > Kobu <kobu.selva@gmail.com> wrote: > > > > > > > > > > > > > Thm: Let p and q be functions. Then, > > > > > > > p(x)/q(x) > 0 iff p(x)*q(x) > 0 > > > > > > > > > > > > Not quite, since when q(x) = 0, > > > > > > p(x)/q(x) is not defined whereas > > > > > > whereas p(x)*q(x) is defined and zero, so > > > > > > "p(x)/q(x) > 0" is indeterminate but > > > > > > "p(x)*q(x) > 0" is false
Look up the definition of the "iff' statement. > > > > Ok, but if someone has an equation like p(x)/q(x) > 0, then all the solutions of p(x)*q(x) > 0 will be solutions of the original one. > > > > So this is a good way to solve things where numerator and denominator are polynomials.
Or any other arbitrary functions

