Negating "For All" Statement: Proving False

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To negate the statement effectively, one must change all "for all" quantifiers to "there exists" and adjust the inequalities accordingly, specifically changing "<" to "≤". The discussion emphasizes the importance of understanding the conditions within the statement to apply these transformations correctly. Clarification is sought on the reasoning behind these changes, indicating a need for deeper comprehension of the negation process. The conversation highlights the complexities involved in logical statements and their negations in mathematical contexts. Understanding these principles is crucial for accurately proving the original statement false.
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Homework Statement


Negate the following statement and thereby prove that it is false.
[PLAIN]http://img834.imageshack.us/img834/5020/74520479.png

Homework Equations


The Attempt at a Solution


I know the general rules but there are so many conditions in this statement that I don't know how to apply them.

My guess right now would be to change all the "for all" statements to "there existst" and to reverse the direction of the last inequality.
 
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keemosabi said:

Homework Statement


Negate the following statement and thereby prove that it is false.
[PLAIN]http://img834.imageshack.us/img834/5020/74520479.png


Homework Equations





The Attempt at a Solution


I know the general rules but there are so many conditions in this statement that I don't know how to apply them.

My guess right now would be to change all the "for all" statements to "there existst" and to reverse the direction of the last inequality.
Not exactly "reverse it". "<" should become "\le". Also that "there exist \delta&gt; 0" should become "for all \delta&gt; 0".
 
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HallsofIvy said:
Not exactly "reverse it". "<" should become "\le". Also that "there exist \delta&gt; 0" should become "for all \delta&gt; 0".
Thank you for the reply. I'm just wondering how you came up with that. Could you elaborate a little please?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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