What it means to negate the following statement?

[cris(c)]

Homework Statement

Suppose you have a collection of functions ${f_1,f_2,\ldots,f_m}$ and you make the following statement: $f_k ≤ 0$ for all k with strict inequality for at least some k.

Homework Equations

The negation of the above statement is $\exists k$ such that [itex]f_k>0 [\itex].<h2>The Attempt at a Solution</h2><br /> <br /> Does the negation of the statement above impose any restriction on the remaining functions (other than $f_k$? or I am free to assume anything about these other functions?[/itex]

 

[SammyS]

Homework Statement

Suppose you have a collection of functions ${f_1,f_2,\ldots,f_m}$ and you make the following statement: $f_k ≤ 0$ for all k with strict inequality for at least some k.

Homework Equations

The negation of the above statement is $\exists k$ such that $f_k>0$.

The Attempt at a Solution

Does the negation of the statement above impose any restriction on the remaining functions (other than $f_k$? or I am free to assume anything about these other functions?

That's not the negation .

What if $f_k=0$ for all k ?

 

[Curious3141]

Think of this as an AND statement. Two propositions A and B.

Proposition 1: $f_k \leq 0 \forall k$

AND

Proposition 2: $\exists k$ such that $f_k < 0$

Now apply De Morgan's Law: $\overline{AB} = \overline{A} + \overline{B}$

So the result is the negation of proposition 1 OR the negation or proposition 2.

NOT proposition 1: $\exists k$ such that $f_k > 0$

OR

NOT proposition 2: $f_k \geq 0 \forall k$

EITHER of these statements is an equally valid negation of the original statement. Either of those conditions, if met, will violate one of the propositions in the original statement, invalidating it.

Sorry for all the edits, but I think I finally have it right this time!

 

[cris(c)]

Think of this as an AND statement. Two propositions A and B.

Proposition 1: $f_k \leq 0 \forall k$

AND

Proposition 2: $\exists k$ such that $f_k < 0$

Now apply De Morgan's Law: $\overline{AB} = \overline{A} + \overline{B}$

So the result is the negation of proposition 1 OR the negation or proposition 2.

NOT proposition 1: $\exists k$ such that $f_k > 0$

OR

NOT proposition 2: $f_k \geq 0 \forall k$

EITHER of these statements is an equally valid negation of the original statement. Either of those conditions, if met, will violate one of the propositions in the original statement, invalidating it.

Sorry for all the edits, but I think I finally have it right this time!

Thanks a lot for your clarifying answer...this really helps a lot!