Question 1019960
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Assume that you have two positive numbers *[tex \Large x] and *[tex \Large b] and you calculate their product.  Then you have another pair of positive numbers *[tex \Large a] and *[tex \Large y] where *[tex \Large a\ <\ x] and *[tex \Large y\ <\ b].  It should be obvious to the casual observer that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ xb\ >\ ay]


for all positive real numbers *[tex \Large x], *[tex \Large y], *[tex \Large a], and *[tex \Large b] such that *[tex \Large a\ <\ x] and *[tex \Large y\ <\ b].


Since all the number are non-zero and non-negative, we can divide by any of them without creating an undefined number or changing the sense of the inequality.  Divide both sides by *[tex \Large b] and *[tex \Large y] yielding:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x}{y}\ >\ \frac{a}{b}]


for all positive real numbers *[tex \Large x], *[tex \Large y], *[tex \Large a], and *[tex \Large b] such that *[tex \Large a\ <\ x] and *[tex \Large y\ <\ b].


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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