
Re: Solving rational polynomials
Posted:
Aug 17, 2013 6:06 PM


On Saturday, August 17, 2013 1:26:18 AM UTC7, 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
A iff B means:
(if A then B) AND (if B then A)
But
If pq < 0 then p/q < 0 If p/q < 0 then pq < 0

