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Re: Solving rational polynomials
Posted:
Aug 17, 2013 6:06 PM
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On Saturday, August 17, 2013 1:26:18 AM UTC-7, Virgil wrote:
> In article <0388f571-326f-437f-990f-b3000253e952@googlegroups.com>,
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> Kobu <kobu.selva@gmail.com> wrote:
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>
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> > Thm: Let p and q be functions. Then,
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> > p(x)/q(x) > 0 iff p(x)*q(x) > 0
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>
>
> Not quite, since when q(x) = 0,
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> p(x)/q(x) is not defined whereas
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> whereas p(x)*q(x) is defined and zero, so
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> "p(x)/q(x) > 0" is indeterminate but
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> "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
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