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Since both $a,b\in \mathbb{Q}^+$ and $a<b$, then of course $\frac{1}{a}$ is greater than $\frac{1}{b}$. However, I don't know how to prove that. I suppose I could do the greater than property in an ordered integral domain, such that $\frac{1}{a} - \frac{1}{b} >0$. But then, I don't know where to take that to.

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  • $\begingroup$ Multiply both sides by $\frac{1}{ab}$. $\endgroup$ – user194928 May 16 '16 at 16:59
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Divide both sides by $ab$. We get $\frac{1}{a}>\frac{1}{b}$.

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