Apparently you didn't follow my analysis. I didn't start off my analysis with (q-p)3>0. I started my analysis of the remaining two cases with the case of q>p. Fundamental to that proof is that for any two real numbers p and q there are three mutually exclusive possibilities:
(1) p = q.
(2) q > p.
(3) q < p.
The (p-q)2 > 0 eliminates (1). (2) is the first case analyzed above. (3) is the second case analyzed above. Having analyzed all possible cases, the conclusion drawn was the result of that analysis.
eq0: (q-p)^3 > 0 => q <> 0, and q > p
eq1: eq1: (q-p)^2 => q <> 0
eq2: (q^3-p^3)/(pq^2-p^2q) > 3
OK, I didn’t follow your analysis after all. I do appreciate your help. My questions are not meant to be personally challenging. I did work with your response for quite some time before trimming things down to what I wrote. I don’t blame you if you become weary, correspondence like this takes much more effort than talking face to face and writing on a blackboard.
I will explain what I think the confusion was/is only because I believe it is and has been a significant educational opportunity beyond the original question asked.
The original question begin as an “If eq1 then show eq2 is a consequence” type of question. That this is not possible, i.e. using eq1 to prove e2, was shown by Subhotosh Khan by demonstrating that eq2 can be developed from eq0 not eq1. That this is the case is most clearly seen by working his case backwards to show that the conclusion is rooted in the revised premise.
Then I came along and accepted Subhotosh Khan’s new premise and analysis and wondered how the domains of p and q would need to be necessarily limited to accompany his development of eq2 from eq0. Unfortunately I missed the obvious factoring that (pq^2-p^2q) = pq(q-p), and was puzzled by how one would state the conditions on the domain of eq2.
The significant point here is that then I was no longer asking the original question, i.e not how to develop eq2 from the revised premise eq1, but rather given eq2, how to define its domain in light of it’s own form and that of the additional condition eq0.
At this point, you, Ishuda went back to the original equations eq1 and eq2 and demonstrated how one could prove the domain of eq1/eq2 was necessarily stated by p<> q and that p and q must both be positive or both negative.
The unfortunate thing for the less experienced reader (like me) was that you seemed to start off with the revised premise that the Subhotosh Khan had introduced, eq0. What was not obvious was that you were not borrowing his new premise but was recognizing that the author of the problem had almost certainly cooked up the problem by starting with the binomial expansion of (q-p)^3.
So this was the significant education point for me, an insight into seeing see one can cook up problems like this, to wit:
Take (p-q)^3 and use binomial expansion to write the identity:
eqA1: (p-q)^3 = q^3 – 3pq^2 + 3p^2q –p^2
Apply the same condition to each side of the equality, in this case:
eqA2: (p-q)^3 > 3
eqA3: q^3 – 3pq^2 + 3p^2q –p^2 > 3
disguise eqA3’s origin by twisting it around using the usual algebraic steps, to create, in this case:
eqA4: (q^3 – p^3)/((pq^2-qP^3) > 3
then use eqA2 as a premise to show that it implies eqA4 which it must since eqA4 is implied by eqA3, and eqA3 is implied by eqA2 since eqA1 shows that the LHS of each, eqA2 and eqA3, are identities of one another. Of course in the process of creating some eqA4 sort of disguise there is the likelihood of changing the domain.
Perhaps I still do not have everything straight, who knows, but I am sure that I have learned a good deal with wrestling with the details of this problem. Thanks again.
