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I understand that

FindInstance[{-((
k L p T β - 2 k p T^2 β - L p α γ + 
 2 p T α γ + c L α β γ - 
 k L T β μ + 
 L α γ μ)/(α (-L p + 2 p T + 
   c L β + L μ))) < 0, c > 0, k > 0, L > 0, 
p > 0, α > 0, γ > 0, μ > 0, T > 0}, {c, k, L, p, 
T, α, β, γ, μ}, 10000]

will give me 10000 instances that satisfy this set of inequalities. And I also understand that

Length[FindInstance[{-((
 k L p T β - 2 k p T^2 β - L p α γ + 
  2 p T α γ + c L α β γ - 
  k L T β μ + 
  L α γ μ)/(α (-L p + 2 p T + 
    c L β + L μ))) < 0, c > 0, k > 0, L > 0, 
p > 0, α > 0, γ > 0, μ > 0, T > 0}, {c, k, L, p, 
T, α, β, γ, μ}, 100]]

will give me an output of 100 because it counts the number of elements in the output list, and I asked for 100 elements.

How ever is there a way I can calculate the maximum number of instances there are for this inequality? I would like to do a comparison between the total number of instances for this inequality where the main function < 0, but divided by when the main function > 0 so if the ratio is < 1 then it proves that there are more instances when it is > 0 rather than < 0. I would like to do something like

Length[FindInstance[{System of Inequalities where main < 0},{values to solve for},MAX INSTANCES]]/Length[FindInstance[{System of Inequalities where main > 0},{values to solve for},MAX INSTANCES]]

Is there a way to count the total number of possible instances for this inequality?

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    $\begingroup$ Hi ! Please post code, not images :) $\endgroup$ – Sektor Dec 6 '14 at 22:09
  • $\begingroup$ Your variables are real numbers, so most probably the dimension of the result set is 1 or greater $\endgroup$ – Dr. belisarius Dec 6 '14 at 22:14
  • $\begingroup$ @Sektor I replaced the images with code, sorry :) @ belisarius So you're saying more than likely there are more instances where the inequality is < 0 than > 0? $\endgroup$ – Mr. F Dec 6 '14 at 22:15
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    $\begingroup$ Do you believe there are only a finite number of solutions? $\endgroup$ – bill s Dec 7 '14 at 4:16
  • $\begingroup$ That would be the main problem I run into, I should have addressed that I'm not sure if there are infinitely many solutions or not. Is there a way to check that with mathematica? $\endgroup$ – Mr. F Dec 7 '14 at 4:20
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Apply Reduce to your equation and you will see that there are many possible answers. Many of these answers have free variables: for instance, the first one I see is

0 < T < L/2

which means that for every possible value of L, you have an infinite number of possible T (assuming they are real-valued). Hence there are an infinite number of such answers.

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