Thursday, February 22, 2018

real analysis - Properties of $>$ for rational numbers

This question comes from an introductory undergraduate course in Analysis. We have just started from defining the set of rational numbers and then we will construct the set of real numbers. We define the order relation "$> $" on the set of rationals as follows: $$(\frac{a}{b})>(\frac{c}{d})\Leftrightarrow (ad-bc)\in \mathbb{N}$$. I am trying to prove that the set of rationals with this order relation is an ordered set. I want to prove the transitive property, by doing the following: $$(\frac{a}{b})>(\frac{c}{d})\Leftrightarrow (ad-bc)\in \mathbb{N}$$



$$(\frac{c}{d})>(\frac{e}{f})\Leftrightarrow (cf-de)\in \mathbb{N}$$. I need to show that:
$$(\frac{a}{b})>(\frac{e}{f}), i.e\ (af-be)\in \mathbb{N}$$
I tried to prove the last statement, by saying that $(ad-bc)(cf-de)=(acdf-adde-bccf+bcde)\in \mathbb{N}$. Then, I didn't really what do next. Any help is highly appreciated?

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