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Let R be the set of real no.s and let f & g be functions from R into R .The negation of the statement
"For each s in R , there exists an r in R such that if f(r)>0 , then g(s)>0."
is which of the following
a)for each s in R ,there does not exist an r in R such that if f(r)>0 then g(s)>0
b)for each s in R there exists an r in R such that f(r)>0 & g(s) is less than or equal to 0
c)there exists an s in R such that for each r in R f(r)>0 & g(s) is less than or equal to zero

2007-03-22 16:39:23 · 1 answers · asked by Akash S 1 in Science & Mathematics Mathematics

1 answers

c

Original Statement:
∀s∈R [∃r∈R s.t. [f(r) > 0 → g(s) > 0]]

Negation:
~∀s∈R [∃r∈R s.t. [f(r) > 0 → g(s) > 0]]

Using DeMorgan's Law:
∃s∈R s.t. ~[∃r∈R s.t. [f(r) > 0 → g(s) > 0]]

Using it again:
∃s∈R s.t. [∀r∈R ~[f(r) > 0 → g(s) > 0]]

Using the property that an implication can only be false if its antecedant is true and its consequent is false:
∃s∈R s.t. [∀r∈R [f(r) > 0 and ~(g(s) > 0)]]

g(s) > 0 is false iff g(s) ≤ 0:
∃s∈R s.t. [∀r∈R [f(r) > 0 and g(s) ≤ 0]]

If you translate this back into English, it's the same as answer c.

2007-03-23 07:27:54 · answer #1 · answered by Phred 3 · 1 0

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